Application of Derivatives Class 12

Ex6.1, 1 teachoo.com
Find the rate of change of the area of a circle with respect to its
radius r when
(aj}r=3cm (b)r=4cm
Let Radius of circle =r
& Area of circle=A
We need to find rate of change of Area w. r. t Radius
i.e. we need to calculate “4
dr
We know that
Area of Circle = A= nr?

Ex 6.1, 2 teachoo.com
The volume of a cube is increasing at the rate of 8 cm?/s. How fast
is the surface area increasing when the length of an edge is 12 cm?
Let

x be length of side

V be Volume

t be time per second
We know that
Volume of cube = (Side)?

Vax
Given that
Volume of cube is increasing at rate of 8 cm?/sec.

Ex 6.1, 3 teachoo.com
The radius of a circle is increasing uniformly at the rate of 3 cm/s.
Find the rate at which the area of the circle is increasing when the
radius is 10cm.
Let r be the radius of circle .

& A be the Area of circle.
Given that
Radius of a circle is increasing at the rate of 3 cm/s

adr

Thus, a Z3em/sec ...(1)
We need to find rate of change of area of circle w. r.t time
when r=10cm

Ex6.1, 4 teachoo.com
An edge of a variable cube is increasing at the rate of 3 cm/s. How
fast is the volume of cube increasing when the edge is 10 cm long?
Let x be the edge of cube.
& V be the volume of cube.
Given that
Edge of cube is increasing at the rate of 3 cm/sec
. dx
“Gp = Bem/sec (1)
We need to calculate
how fast volume of cube increasing when edge is 10 cm
; _ av
i.e. we need to find az when x = 10 cm

Ex 6.1,5 teachoo.com

A stone is dropped into a quiet lake and waves move in circles at the

speed of 5 cm/s. At the instant when the radius of the circular wave

is 8 cm, how fast is the enclosed area increasing?
eee

Let r be the radius of circle

& A be the Area of circle

= ell

Given that

When stone is dropped into a lake waves move in a circle at

speed of 5 cm/sec

ie. Radius of circle increasing at a rate of 4 cm / sec.

ie. < =5 cm/sec (1)

We need find how fast area increasing when radius is 8 cm

Ex 6.1, 6 teachoo.com
The radius of a circle is increasing at the rate of 0.7 cm/s. What is
the rate of increase of its circumference?
Let r be the radius of circle

& C be the circumference of circle
Given that
Radius of a circle is increasing at the rate of 0.7 cm/sec

dr

ie —= (1)
Leo 0.7 cm/sec
We need find rate of change of circumference of circle w. r. t time
: dc
i.e. we need to calculate a

Ex 6.1, 7 teachoo.com
The length x of a rectangle is decreasing at the rate of 5 cm/minute
& the width y is increasing at the rate of 4 cm/minute. When x = 8
cm & y = 6cm, find the rates of change of
(a) the perimeter.
Let Length of rectangle = x
yom
& Width of rectangle = y
xcm
Given that
Length of rectangle is decreasing at the rate of 5 cm/min
dx . {As x is decreasing,
Me Sem/min (1) rate is negative)
And width of rectangle is increasing at the rate of 4 cm/min
ay _ w-(2)
Le. a 4cm/min

Ex 6.1, 8 teachoo.com
A balloon, which always remains spherical on inflation, is being
inflated by pumping in 900 cubic centimeters of gas per second. Find
the rate at which the radius of balloon increases when radius is 15 cm.
Since Balloon is spherical
Let r be the radius of spherical balloon SA
& V be the volume of spherical balloon. r
Given that
The balloon is inflated by pumping in in 900 cubic cm of gas per sec
i.e. Volume of balloon increasing at the rate of 900 cm?/sec
dv
] —s 3
eae 900 cm?/sec (1)

Ex 6.1, 9 teachoo.com
A balloon, which always remains spherical has a variable radius.
Find the rate at which its volume is increasing with the radius when
the later is 10 cm.
Since Balloon is spherical yr,
Let r be the radius of balloon .

& V be the volume of balloon. ¢
We need to find rate at which balloon volume is increasing when
radius is 10cm
i.e. We need to find change of volume w.r.t radius when r= 10
. _ ,av
i.e. we need to find a when r = 10cm

Ex 6.1, 10 teachoo.com
A ladder 5 m long is leaning against a wall. The bottom of the
ladder is pulled along the ground, away from the wall, at the rate
of 2 cm/s. How fast is its height on the wall decreasing when the
foot of the ladder is 4 m away from the wall ?
A
Let AB be the ladder
> x
& OB be the wall & OA be the ground. %
2cm/s
. B. _,O
Given y
Length of ladder is 5m
AB=5cm
LetOA= xcm& OB=ycm

Ex 6.1, 11 teachoo.com
A particle moves along the curve 6y = x° +2.
Find the points on the curve at which the y-coordinate is changing 8
times as fast as the x —coordinate.
Given that
A particular Moves along the curve
6y = x3 +2 ...(1}
We need to find points on the curve at which y coordinate is
changing 8 times as fast as the x — coordinate
ie. We need to find (x, y) for which
dy 4x
ae ae

Ex 6.1, 12 teachoo.com
The radius of an air bubble is increasing at the rate of 5 cm/s. At what
rate is the volume of the bubble increasing when the radius is 1 cm?
Since Air Bubble is spherical
Let r be the radius of bubble
& V be the volume of bubble
Given that
Radius of an air bubble is increasing at the rate of ; cm/s
. ar 4 /
i.e. 3) = om/sec

Ex 6.1, 13 teachoo.com
A balloon, which always remains spherical, has a variable diameter
; (2x +1). Find the rate of change of its volume with respect to x.
Let d be the diameter of the balloon
Given that
Diameter = d =5 (2x +1) SA
Let r be the radius of the balloon
d_3
r=554 (2x + 1)
The balloon is a spherical
4,3
Volume of the balloon = gar

Ex 6.1, 14 teachoo.com
Sand is pouring from a pipe at the rate of 12 cm?/s. The falling sand
forms a cone on the ground in such a way that the height of the
cone is always one-sixth of the radius of the base. How fast is the
height of the sand cone increasing when the height is 4 cm?
Given that sand is pouring from a pipe _
coe

& falling sand forms a cone LE
Let 7 be the radius ,

& h be height of the sand cone

& V be the volume of cone
Also, Sand is pouring from a pipe at the rate of 12cm?/sec
i.e. Rate of volume of acone w.rt time is 12cm /sec

Ex 6.1, 15 teachoo.com
The total cost C(x) in Rupees associated with the production of x
units of an item is given by
C(x) = 0.007x3 - 0.003x2 + 15x + 4000.
Find the marginal cost when 17 units are produced.

Since Marginal Cost is Rate of change in Total Cost w.r.t No of

units produced

Let MC be marginal cost

ac

So, MC = an

Given,

Total Cost = C(x) = 0.007x? — 0.003x? + 15x + 4000

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Ex 6.1, 16 "
The total revenue in Rupees received from the sale of x units of
a product is given by R(x) = 13x? + 26x + 15.
Find the marginal revenue when x = 7.
Marginal revenue is rate of change of total revenue w. r. t the
number of unit sold
Let MR be marginal revenue

dR
So, MR = x
Given,
Total revenue = R(x) = 13x? 4+ 26x+15 (1)

Ex 6.1, 17 teachoo.com
The rate of change of the area of a circle with respect to its
radius rat r=6cmis
(A) 10% (B) 127 (C) 8x (D) 117
Let r be the radius of circle
& A be the area of circle
We need to find
Rate of change of Area of circle w.rt toradius atr=6
dA
i.e. weneed tofind— atr=6
dr
We know that
Area of circle = A= 1?

Ex 6.1, 18 teachoo.com
The total revenue in Rupees received from the sale of x units of a
product is given by R(x) = 3x? + 36x + 5.
The marginal revenue, when x = 15 is
(A)116 (B)96 = (C}90.-—s (D) 126
Marginal revenue is rate of change of total revenue w. r. t the
number of unit sold
Let MR be marginal revenue
dR
So, MR = x
Given,
Total revenue = R(x) = 3x? + 36x+5

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Ex 6.2, 1 (Method 1)
Show that the function given by f (x) = 3x + 17 is strictly increasing
onR.

f(x) = 3x+17
Finding f(x)

F(x) =3
Since f’(x) > 0
Hence, f is strictly increasing on R

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Ex 6.2, 2
Show that the function given by f (x) = e?* is strictly increasing on R.
Let x, and x, be real numbers
Such that

X1< xX,
Multiplying both sides by 2

2x, < 2x,
Taking exponential both sides

e2%, < e2%,

Fx) <f(x,)
Hence, when x, < x, ,f(x,) < f(x)
Thus, f(x) is strictly increasing on R.

Ex 6.2, 3 teachoo.com
Find the intervals in which the function f given by f (x} = sin x is
(a) strictly increasing in (0 5)
f(x) = sin x
Rough
f(x) = cos x ou
cos0=1
cos = =
Since 4 ¥2
we
cos>=0
cos x>0forx € (0, 2) 2
Value of cos x > 0 for (0, 5)
* f’(x) <0 for x € (0,7)
. . . . . we
Thus, fis strictly increasing in (0, =)

Ex 6.2, 4 teachoo.com
Find the intervals in which the function f given by
f(x) = 2x? - 3xis
(a) strictly increasing (b) strictly decreasing
f (x) = 2x? -3x
Calculating f(x)
f' (x) =4x -3
Putting f’ (x) =0
4x -3=0
4x =3
3
x=-
4

Ex 6.2, 5 teachoo.com
Find the intervals in which the function f given by
f(x) = 2x3- 3x?- 36x+7is
(a) strictly increasing
(b) strictly decreasing
f(x) = 2x? -—3x? -36x+7
Calculating f(x)
f(x) = 6x? -6x -36+0
f(x) = 6 (x? -x -6)
f(x) =6(x* —3x +2x -6)
Fi'lx) = 6(x(x - 3) + 2 (x - 3))
F(x) = 6(x — 3) (x +2)

Ex 6.2, 6 teachoo.com
Find the intervals in which the following functions are strictly
increasing or decreasing:
(a) x? + 2x-5

f(x)=x24+2x -5
Calculating f’ (x)

f’(x}=2x +2

f(x) =2 (x +1)
Putting f’ (x) =0

2(x+1)=0

(x+1}=0

x=-1

Find the intervals in which the following functions are strictly increasing or decreasing: (b) 10 – 6𝑥 – 2𝑥2 Calculating f’(𝒙)
f(𝑥) = 10 – 6𝑥 – 2𝑥2
f’(𝑥) = –6 – 4𝑥
f’(𝑥) = −2(3 + 2𝑥)

Find the intervals in which the following functions are strictly increasing or decreasing: (c) –2𝑥3 – 9𝑥2 – 12𝑥 + 1 f(𝑥) = –2𝑥3 – 9𝑥2 – 12𝑥 + 1

Find the intervals in which the following functions are strictly increasing or decreasing: (d) 6 – 9𝑥 – 𝑥2 𝑓(𝑥) = 6 – 9𝑥 – 𝑥2

Find the intervals in which the following functions are strictly increasing or decreasing: (e) (𝑥 + 1)^3 (𝑥 – 3)^3 f(𝑥) = (𝑥+1)3 (𝑥−3)3

teachoo.co
Ex 6.2, 7 ma
Show that y = log(1 + x) -— ,x > —1, is anincreasing
function of x throughout its domain.
Given
2x
y=log (1+ x) sagt? -1
We need to show that y is strictly increasing function for x > -1
d
i.e. we need to show that 7 >Oforx> —-1
oo dy
Finding ax
2x
y=log (1+ x) Tax
dy a(log(1 +x)- ss)
dx dx

Ex 6.2, 8 teachoo.com
Find the values of x for which y = [x(x — 2)]? is an increasing function
y= [x(x - 2)/?
. | dy
Finding dx
y= [x(x — 2)/?
y=[x? — 2x}?
y= (x)* + (2x)? — 2(x?)(2x)
y=x" + 4x? — 4x3
Differentiating w.rt x
ay a(x* + 4x? — 4x7)
dx dx

Ex 6.2, 9 teachoo.com
4sin @ Tw
Prove that y = ————— - 6 is an increasing function of 8 in|, | .
(2 + cos 8) 2
4sin@
yO 4 cos 8) mad
a
We need to prove that function is increasing function of 6 in (0 , =)
. dy via
ie. we need to prove that oS >dforee]0, 2
Differentiating w.rt 6
dy _d ( 4 sin @ )
d@ d@\2+cos@
dy 5 (sree ) dé
dod \2 +cos 0 d0

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Ex 6.2, 10 CAEHOOLOM
Prove that the logarithmic function is strictly increasing on (0, 00).
F(x) = log (x)
We need to prove f(x) in increasing on x € (0 , 00)
i.e. we need to showf’(x) > 0 for x € (0, 00)
Now,
f(x) = log x
"xy at
FO)=;
When x > 0
+, 0
x

teachoo.co
Ex 6.2, 11 ma
Prove that the function f given by f(x) = x*- x + 1isneither
strictly increasing nor strictly decreasing on (-1, 1).
Given
f(x) ax?-x+1
Finding f’(x)
f(x) =2x-1
Putting f’(x) =0
2x -1=0
2x=1
1
xeT
2

Ex 6.2, 12 teachoo.com
Which of the following functions are strictly decreasing on (0 5p
(A) cos x
f(x)=cosx
F(x) =-sinx
Since,
sinx >Oforx € (0,2)
So,
-sinx<Ofor x€ (0,2)
« f'(x)<Oforx € (0 7)
. . . . we
So, fis strictly decreasing in (0 2)

Which of the following functions are strictly decreasing on (0,𝜋/2) ? (B) cos 2𝑥 Let f(𝑥) = cos 2𝑥

Which of the following functions are strictly decreasing on (0,𝜋/2) ? (C) cos 3𝑥
Let f(𝑥) = cos 3𝑥

Which of the following functions are strictly decreasing on (0,𝜋/2) ? (D) tan 𝑥f(𝑥) = tan 𝑥
f’(𝒙) = sec2 𝒙

Ex 6.2, 13 teachoo.com
On which of the following intervals is the function f given by
f(x) = x1 + sinx -1 strictly decreasing ?
(A) (0, 1) (B) G, n) (C) (0, 4) (D) None of these
fix) = x'© + sinx-1
f'(x) = 100x% + cos x
Let’s check sign of f’(x) in different intervals
(A) (0,1)
For O<x<1
Checking sign of 100x°* and cos x

Ex 6.2, 14 teachoo.com
Find the least value of a such that the function f given by f (x) =
x? + ax + Lis strictly increasing on (1, 2).
We have

f(x) =x? +ax4+1
And,

f(x)=2x+a
Given f is strictly increasing on (1 , 2)

« f(x) >0o0n (1,2)

Putting value of f’(x)

2x+a>0 on(1,2)

Ex 6.2, 15 teachoo.com
Let I be any interval disjoint from [-1, 1]. Prove that the function f
given by f(x) = x + =i strictly increasing on I.
Lis any interval disjoint from [—1, 1]
Let I= (—o, -1) U (1,0) [1,1]
—_ —_]Y+—
—0o -1 1 fee]
(-~,—-1) (1, %)
Given
1
f(x) =x+ 5
We need to show f(x) is strictly increasing on I
ie. we need to show f’(x) > 0 for x € (—«, —1) U (1,0)
Finding f’(x)

Ex 6.2, 16 teachoo.com
Prove that the function f given by f (x) = log sin x is strictly increasing
on (0,2) and strictly decreasing on G.n)

f(x) = log sin x
We need to show that f(x) is strictly increasing on (0 5) & strictly

. we
decreasing on G ,t)
i.e. We need to showf’(x) > 0 for x € (0 5)
y z
& f(x) <0 forx € ¢ 10)

Finding f’(x)

f(x) = Clog. sinxy

Ex 6.2, 17 teachoo.com
Prove that the function f given by f (x) = log cos x is strictly
decreasing on (0,2) and strictly increasing on(5, 7)

f(x) = log cos x
We need to show that f(x) is strictly decreasing on (0 5) &
Strictly increasing on G ,t)
i.e. We need to showf’(x) <O for x € (0 5)

& f(x) > 0 forx € G 10)

Finding f’(x)

f(x) = log. cos x)’

Ex 6.2, 18 teachoo.com
Prove that the function given by f(x) = x°- 3x? + 3x-100
is increasing in R.
We need to show f(x) is strictly increasing on R
ie. we need to show f’(x) > 0
Finding f’ (x)
f(x) =3x? -6x+3 -0
= 3(x? -2x+1)
= 3((x)? + (1)? -— 2@)(4))
=3(x-1)?

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Ex 6.2, 19
The interval in which y = x? e~ is increasing is
(A) (— 99, 00) (B) (2, 0) (C) (2, 29) (D) (0, 2)
Let f(x) =x*e™*
Finding f’(x)

f(x) = @?e*)
Using product rule

£@) = @?)'e* + (0 *Y @)

f'(@) = (2x) e* + (e™) (?)

f(x) = 2x e* — e* x?

f(@) =xe*(2—x)

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Ex 6.3, 1 (Method 1)
Find the maximum and minimum values, if any, of the following
functions given by
()fG) = @x- 1)? +3
f(x) = (2x-1)? +3
Square of number cant be negative
it can be O or greater than 0
Hence,
Minimum value of (2x — 1)? =0
Minimum value of (2x — 17) + 3=0+3=3

Find the maximum and minimum values, if any, of the following functions given by (ii) f (𝑥) = 9𝑥2+12𝑥+2Finding f’(x)
f (𝑥)=9𝑥2+12𝑥+2
Diff. w.r.t 𝑥
f’(𝑥)=18𝑥+12
f’(𝑥)=6(3𝑥+2)

Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = –(𝑥 – 1)^2+10 f(𝑥)=−(𝑥−1)^2+10

Find the maximum and minimum values, if any, of the following functions given by (iv) f(𝑥) = 𝑥3 + 1f(𝑥)=𝑥^3+1

Ex 6.3, 2 teachoo.com

Find the maximum and minimum values, if any, of the following

functions given by

(i} f &) = |x + 2|-1

f@)=|x + 2|-1

Minimum value of |x + 2| =0

Minimum value of f(x) = minimum value of |x + 2|-1
=0-1
=-1

Hence minimum value of f(x) = —1

And there is no maximum value of f(x}

Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(𝑥)= – | 𝑥 +1|+3f(𝑥)= – | 𝑥 +1|+3

Find the maximum and minimum values, if any, of the following functions given by (iii) ℎ(𝑥)= sin ⁡(2𝑥)+ 5ℎ(𝑥)= sin ⁡(2𝑥)+ 5

Find the maximum and minimum values, if any, of the following functions given by (iv) 𝑓 (𝑥)=|sin⁡4𝑥+3|𝑓 (𝑥)=| sin⁡4𝑥+3|

Find the maximum and minimum values, if any, of the following functions given by (v) ℎ(𝑥)=𝑥 + 1 , 𝑥 ∈ (−1 , 1) Drawing graph of f(𝑥)=𝑥+1

EX 6.3, 3 teachoo.com
Find the local maxima and local minima, if any, of the following
functions. Find also the local maximum & the local minimum values,
as the case may be:
(i) f (x) = x?

f(x) = x?
Finding f’(x)

f’(x) = 2x
Putting f(x) = 0

2x=0

x=0

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (ii) 𝑔(𝑥)=𝑥3 –3𝑥𝑔(𝑥)=𝑥3 –3𝑥

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iii) ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥<𝜋/2 ℎ(𝑥)=sin⁡𝑥+cos⁡𝑥, 0<𝑥<𝜋/2

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iv) f (𝑥)=sin⁡𝑥 –cos⁡𝑥, 0<𝑥<2 𝜋f (𝑥)=sin⁡𝑥 –cos⁡𝑥, 0<𝑥<2 𝜋

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (v) 𝑓 (𝑥)=𝑥3 –6𝑥2+9𝑥+15Putting f’(𝒙)=𝟎
3(𝑥^2−4𝑥+3)=0
𝑥^2−4𝑥+3=0
𝑥^2−3𝑥−𝑥+3=0
𝑥(𝑥−3)−1(𝑥−3)=0
(𝑥−1)(𝑥−3)=0
So, x = 1 & x = 3

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (vi) g (𝑥) = 𝑥/2 + 2/𝑥 , 𝑥 > 0g (𝑥) = 𝑥/2 + 2/𝑥, 𝑥 > 0

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (vii) g (𝑥) = 1/(𝑥^2 + 2)Finding g’(𝒙)
g’(𝑥)=𝑑/𝑑𝑥 (1/(𝑥^2 + 2))
g’(𝑥)=(𝑑(𝑥^2 + 2)^(−1))/𝑑𝑥
g’(𝑥)=−1(𝑥^2+2)^(−1−1) × (2𝑥+0)
g’(𝑥)=−2𝑥(𝑥^2+2)^(−2)
g′(𝑥)=( −2𝑥 )/(𝑥^2 + 2)^2

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (viii) f(𝑥) = 𝑥√(1−𝑥), 𝑥 > 0= √(1−𝑥) + 1/(2√(1 − 𝑥)) (0 −1) . 𝑥
= √(1−𝑥) – 𝑥/(2√(1 − 𝑥))
= (2(√(1 − 𝑥) )^2− 𝑥)/(2√(1 − 𝑥))
= (2(1 − 𝑥) − 𝑥)/(2√(1 − 𝑥))
= (2 − 2𝑥 − 𝑥)/(2√(1 − 𝑥))
= (2 − 3𝑥)/(2√(1 − 𝑥))

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Ex 6.3, 4
Prove that the following functions do not have maxima or minima:
(i) f (x) = e*
Given f (x) = e*
Finding maxima or minima
f(x) = e
Putting f* (x) =0

ex=0
This is not possible for any value of x.
« f (x) does not have a maxima or minima.

Prove that the following functions do not have maxima or minima: (ii) g(x) = log xGiven g (x) = log x

Prove that the following functions do not have maxima or minima: (iii) ℎ(𝑥)= 𝑥^3+𝑥^2+𝑥+1Here 𝑎 = 3, b = 2, & c = 1
x = (− 2 ± √(4 − 4(3)(1)))/6
x = (− 2 ± √(4 − 12))/6
x = (−2 ± √(− 8))/6
x = (− 2 ± 2√(− 2))/6
x = (−𝟏 ± √(− 𝟐))/𝟑

Ex 6.3, 5 teachoo.com
Find the absolute maximum value and the absolute minimum
value of the following functions in the given intervals:
(J f(x) = x3,x € [- 2,2]
Finding f’(x)

rey — 229)

fa)= 7

f(x) = 3x?
Putting f’(x) = 0

3x7 =0

x? =0

x=0

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (𝑥) = sin⁡𝑥 + cos⁡𝑥 , 𝑥 ∈ [0, 𝜋 ] Finding f’(𝒙)
f’(𝑥)=𝑑(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)/𝑑𝑥
f’(𝑥)=cos⁡〖𝑥 −sin⁡𝑥 〗

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(𝑥) = 4𝑥 – 1/2 𝑥2 , 𝑥 ∈ [−2, 9/2] Putting f’(𝒙)=𝟎
4 – 𝑥=0
𝑥=4
∴ 𝑥=4 is only critical point

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f (𝑥) = (𝑥 −1)2 + 3, 𝑥 ∈ [−3,1] f (𝑥) = (𝑥 −1)2 + 3

Ex 6.3, 6 teachoo
Find the maximum profit that a company can make, if the profit
function is given by p(x) = 41- 72x - 18x?
The profit function is given by
p(x) = 41 - 72x - 18x?
Finding p’(x)
p’(x) = -72 - 36x
Putting p‘ (x) = 0
-72 - 36x =0
-36x = 72
-72
x=
36
x=-2

Ex 6.3, 7 teachoo.com
Find both the maximum value and the minimum value of
3x4 - 8x? + 12x? - 48x +25 on the interval (0, 3].
Let f(x) = 3x4 - 8x3 + 12x?-48x+25, where x € [0, 3]
Finding f’(x)
, _ d(3x* - 8x3 + 12x? - 48x +25)
f(x) = ax
f(x) = 3 x 4x3 —8 x 3x24+12 x 2x—484+0
f(x) = 12x3 — 24x? + 24x — 48
f(x) = 12(x3 — 2x? + 2x—4)

Ex 6.3, 8 teachoo.com
At what points in the interval (0, 27 ], does the function sin 2x
attain its maximum value?
Let f(x) =sin2x, x €[0,27]
Finding f’(x)
, _ d(sin 2x)
f(x) = a
f(x) = 2 cos 2x
Putting f’(x) = 0
2cos 2x = 0
cos 2x =0
cos 2x = cos>

teachoo.com
Ex 6.3, 9
What is the maximum value of the function sinx + cos x?
Let f(x) = sinx + cosx
Consider the interval x € [0 , 27]
Finding f’(x)
, _ d(sin x + cos x)
f'(v) = ”
f'(x) = cosx — sinx
Putting f’(x) = 0
cosx —sinx =0
cosx = sinx

teachoo.com
Ex 6.3, 10
Find the maximum value of 2x° - 24x + 107 in the interval
[1, 3]. Find the maximum value of the same function in [-3, —-1].
Let f(x) = 2x3 — 24x + 107
Finding f’(x)
d(2x3 — 24x + 107
f(x) = (2x x )
dx
=2x3x?—24
= 6x? — 24
=6(x?-4)

Ex 6.3, 11 teachoo.com
It is given that at x = 1, the function x*- 62x? + ax + 9 attains
its maximum value, on the interval [0, 2]. Find the value of a.
We have
f(x) = x*- 62x + ax+ 9
Finding f’(x)
, d(x*- 62x? +.ax+9)
f(x) = a
=4x3 62 x2x+a
=4x3 -124x+4
Given that at x = 1,
f(x) = x4 — 62x? + ax + 9 attain its Maximum Value

Ex 6.3, 12 teachoo.com
Find the maximum and minimum values of x + sin 2x on [0, 2n ]
Let f(x) = x +sin 2x
Finding f’ (x)
, _ d(x + sin 2x)
fi) = Sets
=1+4+2cos2x
Putting f(x) = 0
1+2cos2x=0
2cos2x =—-1
cos 2x = —
2

Ex 6.3, 13 teachoo.com
Find two numbers whose sum is 24 and whose product is as large
as possible.
Let first number be x
Now, given that
First number + Second number = 24
x + second number = 24
Second number = 24 -—x

Product = (first number ) x (second number)

= x (24-—x)
Let P(x) =x (24— x)

teachoo.com

Ex 6.3, 14 (Method 1)
Find two positive numbers x and ysuch that x + y= 60 and xy? is
maximum.
Given two number x and y, such that

x+y =60

y=60—-x ..(1)
Let P = xy?
We need to maximize P
Now,

P= xy?
Putting value of y from (1)

P= x(60—x)3

Ex 6.3, 15 (Method 1) teachoo.com
Find two positive numbers x and y such that their sum is 35
and the product x? y® is a maximum.
Given two number are x & y
Such that
x + y=35
y=35 -x ...(1)
Let P=x? y°
We need to maximize P
Finding P’(x)
P(x) = x*y®
P(x) = x2(35 —x)°

Ex 6.3, 16 teachoo.com
Find two positive numbers whose sum is 16 and the sum of whose
cubes is minimum.
Let first number be x
Now,
First number + second number =16

x + second number = 16

second number = 16 -x

Now,
Sum of Cubes = (first number )° + (second number )*
Let S(x) = x? + (16 —x)?
We Need to Find Minimum Value of s(x)

Ex 6.3, 17 teachoo.com
A square piece of tin of side 18 cm is to be made into a box without}
top, by cutting a square from each corner and folding up the flaps
to form the box. What should be the side of the square to be cut
off so that the volume of the box is the maximum possible.
4 | |x
I N
|| >
1 0
x| | 18-2 | |x Le
Let x cm be the length of a side of the removed square
Hence,
Length after removing = 18 -x - x=18-2x
Breadth after removing = 18 - x - x=18-2x
Height of the box = x

Ex 6.3, 18 teachoo.com
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box
without top, by cutting off square from each corner and folding up
the flaps. What should be the side of the square to be cut off so
that the volume of the box is maximum ?

x{_| ------------}*

—
x{_| 45 - 2x [|x Box a
Let x be the length of a side of the removed square
Thus,

Length after removing = 45 —x -x = 45 -2x
Breadth after removing = 24 -x -x =24-2x
Height of the box = x

Ex 6.3, 19 teachoo.com
Show that of all the rectangles inscribed in a given fixed circle, the
square has the maximum area.
D Cc
Let radius be r of the circle
& let x be the length
& y be the breadth of the rectangle A 5
Now,
A ABC is right angle triangle
(AB)? + (BC)? = (AC)?

x+y? = (2r)? (As AC is diameter of circle}

x? +y? = 4r2

y= 4r?- x? (1)

Ex 6.3, 20 teachoo.com
Show that the right circular cylinder of given surface and maximum
volume is such that its height is equal to the diameter of the base.
Let r,h be the Radius & Height of Cylinder respectively
& V,Sbe the Volume & Surface area of Cylinder respectively
r
—_ +
Given
Surface Area of Cylinder = 2nr?+ 2nrh h
S = 2nr*+ 2nrh
_———
S-2nr* = 2arh
S —2nr? =h
2nr
h= S—2nr? | (1)
2nr

Ex 6.3, 21 teachoo.com
Of all the closed cylindrical cans (right circular), of a given volume of
100 cubic centimeters, find the dimensions of the can which has the
minimum surface area?
Let r,&hbe the radius & height of cylinder respectively
& V&Sbe the volume & surface area of cylinder respectively
Given volume = 100 cm? r
We know that
h

Volume of a Cylinder = mr7h

V= arth

100 = ar7h

Ex 6.3, 22 teachoo.com
A wire of length 28 m is to be cut into two pieces. One of the pieces
is to be made into a square and the other into a circle. What should
be the length of the two pieces so that the combined area of the
square and the circle is minimum?
Awire of length 28m cut into two parts x 28-x

Let x be the length of first part / \

Length of 2"¢ part = (28 — x)m || C )
Given that ~

4

one part is converted into a square
Let length of x m be converted into a square

Ex 6.3, 23 teachoo.com
Prove that the volume of the largest cone that can be inscribed in a
sphere of radius R is - of the volume of the sphere.
Cone of largest volume inscribed in the sphere of radius R
A
Let OC = x
Radius of cone = BC
Height of cone=h=OC+0A
B

Finding OC
A BOC is a right angled triangle
Using Pythagoras theorem in ABOC

OB? = BC? + 0C?

R2= BC2 + x?

Ex 6.3, 24 teachoo.com
Show that the right circular cone of least curved surface and given
volume has an altitude equal to V2 time the radius of the base
Let r& h be the radius & height of a cone respectively
And V &S be the volume & curved surface area of cone respectively}
Given volume of cone is constant
Volume of cone = 5m(radius)” (height)
1

v= 5n(r)*h

3V r

—=r*h

Tw

n=(“)3

“Ags r2

Ex 6.3, 25 teachoo.com
Show that the semi-vertical angle of the cone of the maximum
volume and of given slant height is tan -t V2
Let J be the slant height & 8 be the semi vertical angle of the cone.
Now, I
Height of cone = h = 1 cos 8
Radius of cone=r=Isin@
T

We need to maximize volume of cone

Velarzh

3
Ve 5rl?sin?@ lcos@
V= 5 l3sin?@ cos @

Ex 6.3, 26 teachoo.com
Show that semi-vertical angle of right circular cone of given surface
area and maximum volume is sin + 6)
Letr,h &/be the radius, height & slant height
of a cone respectively
I
And Let V&S be the volume & surface area
& 8 be a semi vertical angle of a cone
r
Given surface Area of a cone is constant
Surface Area of a cone = nr? + nrl
S= nr? +arl
S-nr? = nrl

teachoo.com
Ex 6.3, 27 (Method 1)
The point on the curve x? = 2y which is nearest to the point (0, 5) is
(A)(2V2,4) — (B)(2V2,0) — (C)(0, 0) (D) (2, 2)
Let (hk) be the point on the curve x? = 2y = 2y
Where is nearest to the point (0,5)
h, k)
(0, 5)
Since (h, k) lie onthe curve x? = 2y
> (A k) will satisfy the equation of curve x? = 2y
=> Puttingx =h & y=kin equation
h?=2k _ ...(1)
We need to minimize the distance of a point (A, kK) from(0, 5)

Ex 6.3, 28 teachoo.com
1-xt+x?
For all real values of x, the minimum value of is
1+x+x
(ayo (BL (CJB_ (DY
Let f(x) = 172+#
Ltxt+x
Finding f'(x)
f(x) =1o2t#
44x42?
Using quotient Rule
Uu ' ulv—vlu
As () 7 py
pie = Goet) (te etx?) (ant ean xt)!
~ (+x+x7)2

Ex 6.3, 29 teachoo.com
The maximum value of [x(x — 1) + 1] 0 < xs lis
(y(t) @F (a (D)0
Let f(x) = [x(x —1) +1
Finding f’(x)
f(x) = bebe - 1] + 1
fe) =? —x +17
fio) _ a(x? == +1)8
fe) =3@? x4 pat deat) a2+4)

Example 1 teachoo.com
Find the rate of change of the area of a circle per second with
respect to its radius rwhen r=5 cm.
We have to find rate of change of area of circle with respect to
radius

. . d(Area of circle) _ da

i.e. we need to find d (radius of circle)» dr
We know that

Area of circle = nr?
A=nr
dA

Finding —

inding =

Example 2 teachoo.com
The volume of a cube is increasing at a rate of 9 cubic
centimeters per second. How fast is the surface area increasing
when the length of an edge is 10 centimeters ?
Let
x be length of side
V be Volume
t be time per second
We know that
Volume of cube = (side)?
Vax (1)

Example 3 teachoo.com

A stone is dropped into a quiet lake and waves move in circles at a

speed of 4.cm per second. At the instant, when the radius of the

circular wave is 10 cm, how fast is the enclosed area increasing?
————————s

Let r be the radius of circle

& A be the Area of circle

ee |

Given that

When stone dropped into a lake wavers movie in circle at speed to 4

per sec.

i.e. Radius of circle increasing at a rate of 4 cm / sec.

ie. = =4cm/sec_...(1)

We need to calculate how fast area increasing when waves is 10 cm

Example 4 teachoo.com
The length x of a rectangle is decreasing at the rate of 3 cm/minute
and the width y is increasing at the rate of 2 cm/minute. When x =
10 cm and y = 6 cm, find the rates of change of (a) the perimeter
and (b) the area of the rectangle.
Let Length of rectangle = x cm yom
& Width of rectangle = ycm
xcm
Given length x is decreasing at the rate of 3 cm/minute
dx . (As x is decreasing,
a3 cm/min (1) rate is negative)
and width y is increasing at the rate of 2 cm/min
dy _ i 2
= 2cm/min (2)

Example 5 teachoo.com
The total cost C(x) in Rupees, associated with the production of x
units of an item is given by C(x) = 0.005 x? — 0.02 x? + 30x + 5000
Find the marginal cost when 3 units are produced, where by
marginal cost we mean the instantaneous rate of change of total
cost at any level of output.
Marginal cost is the rate of change of total cost w.r.t output
(unit produced)
Let MC be marginal cost

dc
« MC = x
It is given
C(x) = 0.005x? — 0.02x? + 30x + 5000
We need to find marginal cost when 3 unit produced

Example 6 teachoo.com
The total revenue in Rupees received from the sale of x units of a
product is given by R(x) = 3x? + 36x + 5. Find the marginal revenue,
when x = 5, where by marginal revenue we mean the rate of change
total revenue with respect to the number of items sold at an instant.
Marginal revenue is rate of change of total revenue w. r. t the
number of unit sold
Let MR be marginal revenue

aR
So, MR = -
Given,
Total revenue= R(x) = 3x? + 36x+5 (1)

Example 7 (Method 1) teachoo.com
Show that the function given by f(x) = 7x —3 is strictly increasing
onR.

f(x)=7x -3
Finding f(x)

f’(x) = (7x - 3)

f'(x)=7
Since f’(x) >0
Hence, f is strictly increasing on R

Example 8 teachoo.com
Show that the function f given byf (x) = x3- 3x? + 4x,xER
is strictly increasing on R.

f(x) = x2 -—3x? +4x
Finding f(x)

f(x) = 3x?-3.2x4+4

f(x) =3x?-6x+4

f(x) = 3x*-6x +341

f(x) =3 (x?-2x4+1)4+1

f(x) =3 (x-1)? +1
As square is a positive number,
The value of f’(x) will be always positive for every real number

teachoo.com
Example 9
Prove that the function given by f (x) = cos x is
(a) strictly decreasing in (0, 7)
f(x) =cosx
f(x) =-sinx
Since,
sin x > 0 for x € (0, n)
So,
-sinx <Ofor x € (0, nm)
« f’(x) <0 for x € (0, 1m)
So, fis strictly decreasing in (0, ™)

teachoo.

Example 10 CACHOO.LOM
Find the intervals in which the function f given by f (x) = x? — 4x + 6 is
(a) strictly increasing
(b) strictly decreasing
f (x) =x?-4x+6
Calculate f’(x)
f' @&) = 2x-4
Calculate f’(x) = 0

2x-4=0

2x=4

xat=2

2

Example 11 teachoo.com
Find the intervals in which the function f given by
f (x) = 423 — 6x?-72x + 30 is
(a) strictly increasing
(b) strictly decreasing.
f (x) = 423 — 6x?-72x + 30
Calculating f’(x)
f (x) = 423 — 6x?- 72x + 30
f'(x) = 12x? — 12x - 72x
f'(x) = 12(?-x- 6)
f'(x) = 12(x? — 3x + 2x -6)
f' (x) = 12(x(x — 3) + 2(x -3))
f'@O) =12(¢42)(-3)

Example 12 teachoo.com
Find intervals in which the function given by f (x) = sin 3x,
wT).
x, € [0,5] is
(a) increasing (b) decreasing.
f(x) =sin 3x where x € [o 5
Finding f’(x)
, _ d(sin 3x)
fa)= a
f’(x) = cos 3x x3
f(x) =3.cos 3x

Example 13 teachoo.com
Find the intervals in which the function f given by
f(x) =sinx + cosx ,0 <x < 27 is strictly increasing or strictly
decreasing.
f(x) =sinx+cosx
Finding f’ (x)

Hy) = “gi

f(x) = 7 (sin x + cos x)

f(x) - d(sin x) + d(cos x)

dx dx
f(x) = cos x + (—sinx)
f’(x) =cosx - sinx

Example 14 teachoo.com
Find the maximum and the minimum values, if any, of the function
fgiven byf(x) = x?,x ER
Y

f(x) = x?

First we plot the Graph of x”
xX
ie)

Atx =0

f(0) =0
And, we note that

f(x) > 0 for all x except 0
Thus, Minimum value of f(x) = 0 at x=0

Example 15 teachoo.com
Find the maximum and minimum values of f, if any, of the function
given by f(x) = |x], x ER. Y
—x, x <0
Fo) = lal =f So
x
ie)
Atx =0
f(0) =0
And, we note that
f(x) > 0 for all x except 0
Thus, Minimum value of f(x) = 0 at x =0

Example 16 teachoo.com
Find the maximum and the minimum values, if any, of the
function given by f (x) = x, x € (0, 1).
Y
We first draw graph of f(x) = x
f(x) =x, xE(0,D
Here, points 0 and 1 are not included x
ie) 1
Thus,
f will have
Minimum value of point closest to 0
& Maximum value of point closest to 1
but its not possible to locate such points

Example 17 (Method 1) teachoo.com
Find all points of local maxima and local minima of the function
f given by f (x) = x8- 3x + 3.

f@) =x- 3x+3
Finding f’ (x)

f'() =3x*-34+0

fe) =3@? -1)
Putting f’(x) =0

3(x?-1) =0

x?-1=0

(-D@+1)=0

Thus x =1, -1 are only critical points

teachoo.com
Example 18 (Method 1)
Find all the points of local maxima and local minima of the function
f given byf (x) = 2x3 - 6x2 + 6x +5.
f (x) = 2x3 -6x? + 6x45
Finding f’ (x)
' _ d(2x3 - 6x? + 6x +5)
f (x) ~ dx
f'(x) = 6x? -12x+6+4+0
f'(~) = 6? —2x4+1)
Putting f’(x) =0
6(x* —2x +1) =0
x*-—2x+1=0

Example 19 (Method 1) teachoo.com

Find local minimum value of the function f given by f (x) = 3+ |x]

XER.

F(x)=3+ |x|

Since Value of |x| > 0

So, Minimum value of |x| = 0

Now,

Minimum value of f (x) = 3 + Minimum value of [|x|

=3+0
=3

Hence minimum value of f (x) = 3

Example 20 teachoo.com
Find local maximum and local minimum values of the function f
given by f (x) = 3x + 4x3- 12x? +12
f(x) = 3x4 + 4x3 - 12x7+12
Finding f’ (x)
, _ d(3x* + 4x? - 12x? + 12)

f () ~ dx

fF (x) = 12x93 +12x? - 24x+0

f (x) = 12(3 +: x? — 2x)

f (x) = 12x(x? +x — 2)

f (x) = 12x? + 2x -—x—-2)

f (x) = 12x (x(x + 2) - 1 + 2))

f () = 12x (x- 1) +2)

Example 21 (Method 1) teachoo.com
Find all the points of local maxima and local minima of the
function f given by f(x) = 2x3 -6x2 + 6x + 5.
f(®) = 223 -6x? + 6x45
Finding f’(x)
, _ d(2x? — 6x? + 6x +5)

£@) =,

f(x) = 6x? —12x+64+0

f'@c) = 6(x? — 2x +1)

f'@) = 6(@) + G)? - 2@)))

f(x) = 6(«— 1

Example 22 teachoo.com
Find two positive numbers whose sum is 15 and the sum of whose
squares is minimum.
Let first number be x
Since

Sum of two positive numbers is 15

x + 2°¢ number = 15
2™4 number = 15 - x

Let S(x)} be the sum of the squares of the numbers

S(x) = (1t number)? + (2° number)?

S(x) = x7 + (15 — x)”
We need to minimize S(x)

Example 23 teachoo.com
Find the shortest distance of the point (0, c}) from the parabola
y =x*,whereO<cs 5. 3
X=y
(0, c}
Let (A , k) be any point on parabola y = x? (h, k)
Let D be required Distance between (h, k} & (0, c)
Distance between two (x1, ¥1) & (x2, ¥2) point is
d= V & — ¥)? + - y1)?
D=/(0—h}* + (c —ky
D=/(-h}? + (c —k)*
D=/h? +(c —k}* (1)

Example 24 teachoo.com
Let AP and BQ be two vertical poles at points A and B, respectively.
If AP = 16 m, BQ = 22 mand AB = 20 m, then find the distance of a
point R on AB from the point A such that RP? + RQ? is minimum
Given AP & BQ be two poles Q
where AP = 16m, BQ = 22m
P
& AB=20m
22m
16m
Let R be a point on AB
A R B
Let AR =x m. 20m
RB =AB —AR
RB=20 -x (Given AB = 20m)
Here, 2 A=90° & 2 B = 90° as they are vertical poles

Example 25 teachoo.com
If length of three sides of a trapezium other than base are equal to
10 cm, then find the area of the trapezium when it is maximum.
D 10cm C

Let ABCD be given trapezium

€ 2)

S, %
Given length of sides other
than base is 10
A xX ?p Q * B
-- AD = DC=CB = 10cm.
Draw a perpendicular DP & CQ on AB.
Let AP = x cm
By symmetry
QB=xcm

Example 26 teachoo.com
Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that cone.
A
Let OC =r be the radius of cone
& OA=h be height of cone 7 h
B < Q
& Z OAQ= a be the semi-vertical angle ina)
ee
And, ee
Let OE = x be the radius of cylinder
Height of cylinder = 00’
Since cone is given, radius (r) and height (h) of cone are constant
And, radius (x) and height (00’) of cylinder is variable

Example 27 teachoo.com
Find the absolute maximum and minimum values of a function f
given by f (x) = 2x3- 15x? + 36x +1 0n the interval [1, 5].
f(x) = 2x? - 15x? + 36x +1
Finding f’(x)
f(x) = 6x? — 30x + 36
Putting f’(x) = 0
6x? - 30x +36=0
6(x? —5x+6)=0
(x? —5x+6)=0

Example 28 teachoo.com
Find absolute maximum and minimum values of a function f
4 1
given by f(x) =12x3- 6x3,x €[-1,1]
4 i
f(x) =12 x3- 6x3
Finding f’(x)
fC ) _ aliox3 — 6x3)
x)= ax
4 1
= 12 x tx3-t~-6 x tx37t
3 3
4-3 1-3
=4x4x 3 —2x 3
i =
= 16x3 —2x3

Example 29 teachoo.com
An Apache helicopter of enemy is flying along the curve given by
y= x? + 7. Asoldier, placed at (3, 7), wants to shoot down the
helicopter when it is nearest to him. Find the nearest distance.
Given curve
yax+7
yax?+7
(x] y)
(3, 7)
Let (x, y) be any point on parabola y = x? + 7
Let D be required Distance between (x, y) & (3,7)
D=/(3—x)?+(7 —y)?
=J/9+x2 —6x+49+ y2—-14y
=x? +y?-—6x—-—14y +58

Example 30 teachoo.com
A car starts from a point P at time t = 0 seconds and stops at point Q.
The distance x, in metres, covered by it, in t seconds is given by
—+2f9_£
x=t (2 *).
Find the time taken by it to reach Q and also find distance b/w P & Q
Given Distance ip
P x Q
t
— 72 -_-
x=t (2 )
At points P and Q, the Velocity of the car is 0
Let v be the velocity of the car
v = Change in Distance w.rt ttime
_&
Ua

Example 31 teachoo.com
A water tank has the shape of an inverted right circular cone with
its axis vertical and vertex lowermost. Its semi-vertical angle is
tan“ (0.5). Water is poured into it at a constant rate of 5 cubic
meter per hour. Find the rate at which the level of the water is
rising at the instant when the depth of water in the tank is 4 m.
Water tank is in shape of cone Cy 5
Let r be the radius of cone, \aL/

h be the height of cone, at
& a be the semi-vertical angle \V

O

Given vos
Semi-vertical angle is tan~1(0.5)

Example 32 teachoo.com
A man of height 2 meters walks at a uniform speed of 5 km/h away
from a lamp post which is 6 meters high. Find the rate at which the
length of his shadow increases.
B

Let AB be the lamp post ras

& MN be the man of height 2m. Lt

& AM = x meter 6

& MS is the shadow of the man

A x _

Let length of shadow MS = s meter woe “Show
Given man walks at speed of 5 km/h
2 S=5km/h

Example 33 teachoo.com
Find intervals in which the function given by
fx) <2 xt —43- 3924 2 edd
10 5 5
is (a) strictly increasing (b) strictly decreasing
= 3 yd 4 43_ 3424 36
Fx) = 5% gx 3xe+ xed
Finding f’(x)
Hy) = 3 3_4 2_ 36
fR=F x Ax - X 3x* —3 x 2x + 5 +0
"(x)= 3 By 2 36
Pa) =5x exe 6x +
ity) = 843 125.2 36
F(x) = 5x Bt 6x + =
"(x)= 6(% 2 4 8
FO)= 6( 5 5 xt =)

Example 34 teachoo.com
Show that the function f given by f (x) = tan“(sin x + cos x), x > Ois
always an strictly increasing function in (0,5)
f(x) = tan“1(sin x + cos x)
Finding f’(x)
, d(tan~1(sin x +cos x))
x) =
r@ -
_ 1 x d(sin x + cos x)
~ 4+ (sin x + cos x)? dx
: ( inx)
= > OO * (cS & — sin x
1+ (sin?x + cos? x +2 sin x cos x) *
1 ( inx) 2 2
=—_———-. x (cos x — sinx i =
1+(14+2sinx cos x) (sin* x + cos? x = 1)
1 .
=———_—_——— _ x (cos x — sin x)
2+2sinx cosx

Example 35 teachoo.com
A circular disc of radius 3 cm is being heated. Due to expansion, its
radius increases at the rate of 0.05 cm/s. Find the rate at which its
area is increasing when radius is 3.2 cm.
Let r be the radius of circle . i
Pie
& A be the Area of circle. ‘
Given that
Radius increases at the rate of 0.05 cm/s
dr
Thus, a 0.05 cm /sec
We need to find rate of change of area of circle w. r. t time when
r=3.2cm

Example 36 teachoo.com
An open topped box is to be constructed by removing equal
squares from each corner of a 3 meter by 8 meter rectangular
sheet of aluminum and folding up the sides. Find the volume of
the largest such box.
a
1 R
IN
tu) 4
------- Vv
xL | 8-2 | |x * Sa
8-2x
Let x m be the length of a side of the removed square
Hence,
Length after removing = 8 -x - x=8-2x
Breadth after removing = 3 - x - x =3-2x
Height of the box = x

Example 37 teachoo.com
Manufacturer can sell x items at a price of rupees (s - =)
each. The cost price of x items is Rs G + 500) Find the number
of items he should sell to earn maximum profit.
Let S(x) be the Selling Price of x items.
& C(x) be the cost Price of x item.

Given

», . x
Manufacture sell x items at a price of rupees (s - =) each

x
S(x) =X x (s -=)
Pra
=5x Too

Example 14 teachoo.com
Find the slope of the tangent to the curve y = x3- xatx = 2.
We know that
=
Slope of tangent = ae
— 2G8- x)
~ dx
= 3x7-1
Puttingx = 2
dy = 2
el = 3(2)?-1
= 3(4)-1=12-1=11
Thus, slope of the tangent is 11

teachoo.com
Example 15
Find the point at which the tangent to the curve y=v4x —3-1
. 2
has its slope 3:
Given, Slope of the tangent to the curve is ;
We know that
=
Slope of tangent = ax
20 dy
3. dx
dy _2
dx 3
a(vV4x—3-1) 2
dx 3

Example 16 teachoo.com
Find the equation of all lines having slope 2 and being tangent
2
tothe curve y+ —— =0
x-3
Equation of curve is
2
Yt

Differentiating both sides w.rt x

dy “=(S) =

dx + dx\x-3) — 0

dy ad (>)

dx dx \x —3

dy _ _ a _ -1

dx 2 dx (x 3)

Example 17 teachoo.com
. . x2 yy? .
Find points on the curve 7th 1 at which the tangents are
(i) parallel to x-axis (ii) parallel to y-axis.
Given curve
x? y? _
41257 1
Slope of the tangent is —
ope of the tangent is 7
ge ay
Finding ax
2x 2y dy_
at 2s * axe 0
x 2y dy _
27 dsax~ 0

Example 18 teachoo.com
Find the equation of the tangent to the curve y = Gone at the
point where it cuts the x-axis.
Slope of the tangent to the curve is

dy (x —7)'[@-2) @-3)]- @-7)[@- 3) @-2)]'

dx (x — 2)? (x — 3)2

dy 1x @—2)(«-3)- @—7)[@- 3)! @-2)+@- 3) @—2)']

dx (x — 2)? (« — 3)2

dy) @-2)@-3)- @- DE x @-2)+@-3)x1]

dx (x — 2)2 ( — 3)?

dy_ (x — 2) («-3)-(«#-7)(2x-5)

dx (x — 2)2 (x — 3)?

dy _ (x - 2) -3) _ (x — 7)(2x -—5)

dx” (x-2)32 (@-3)% (x-2)2 @-3)2

Example 19 teachoo.com
2 2
Find the equations of the tangent and normal to the curve xs + y3
=2 at (1,1).
Given curve
2 2
xa tys=2
Differentiating both sides w.rt x
2 1- 2 2 1 2 dy _
rn? Ar ra 0
234 22 ay
2.24 dy _-2 =
ay ° ax a te

Example 20 teachoo.com
Find the equation of tangent to the curve given by x= asin*t,y
=bcos*t ata point where t = =
The curve is given as
X=asin*t, y=bcos*t
dy
Slope of the tangent = —
dx
Here,
dy
dy _ at
dx ax
dt

Example 21 teachoo.com
Use differential to approximate ¥36.6
Let y= x
where x = 36 & Ax=0.6
Since y = yx
dy _ a(x) 1
dx dx 2x
Now,
dW
Ay= dx Ax
1

Example 22 teachoo.com
1
Use differential to approximate (25)3
i
lety = x3
Where x = 27 and Ax = —2
a
Since y = x3
a
dy _ axa)
dx” dx
1 2. 1
=- X3
3
1 2
=-X3
3

Example 23 teachoo.com
Find the approximate value of f (3.02), where f (x) = 3x? + 5x + 3.
Given f(x) = 3x? + 5x +3
where x = 3 and Ax = 0.02
Finding f’(x)
f’(x) = 6x +5
Now,
Ay = f'(x) Ax
= (6x + 5) 0.02
Putting x =3
=(6x 345) 0.02
= 23 x 0.02

Example 24 teachoo.com
Find the approximate change in the volume V of a cube of side x
meters caused by increasing the side by 2%.
Given

Side of the cube = x metres
Also,

Increase in side = 2% of side

= 0.02x

Hence, Ax = 0.02 x
And,

Volume of the cube = V = x3 m?

teachoo.co

Example 25 ma
If the radius of a sphere is measured as 9 cm with an error of 0.03
cm, then find the approximate error in calculating its volume.
Given

Radius of sphere =r =9 cm

Error in radius = Ar = 0.03 cm
We need to find Error in calculating Volume

Let Volume of sphere = V= jar

«. We need to find AV

Example 45 teachoo.com
Find the equation of the normal to the curve x? = 4y which passes
through the point (1, 2).
Given Curve
x? = dy
x? = ay
(1, 2)
Differentiating w.rt. x
Ad
x=
dx
ay _x Same as
dx 2 .
* Misc 4
-1 -1 -2
oo Slope of normal ay =a = x
* G)

Example 46 teachoo.com
Find the equation of tangents to the curve y = cos (x + y),
—2n <x S 2m that are parallel to the line x + 2y = 0.
Given curve is
y=cos(x+y) (1)
We need to find equation of tangent which is parallel to the line
x+2y =0
_ dy
We know that slope of tangent is an
y =cos(x+y)
Diff wart. x
dy | d(cos(x + y))
dx ax

Misc 1 teachoo.com
Show that the function given by f(x) = “e* is maximum at x = e.
r
Let f(x) = 24
x
Finding f’(x)
, d flogx
f(x) = dx ( x )
a(log x) a(x) log x Using quotient rule
f(x) = ar (“) _ ulv-vlu
as (7) =a
1
, Pia log x
*@) ="
, 1-logx
f(x) =a

Misc 2 teachoo.com
The two equal sides of an isosceles triangle with fixed base b are
decreasing at the rate of 3 cm per second. How fast is the area
decreasing when the two equal sides are equal to the base ?
A
Let x be the equal sides of isosceles triangle
ie. AB=AC=x x x
And, Base = BC=b
B
—
b
Given that equal side of Triangle decreasing at 3 cm per second
. dx
ie. — = -3 cm/sec.
dt
We need to find how fast is the area decreasing when the two
equal sides are equal to the base

Misc 3 teachoo.com
Find the intervals in which the function f given by
f(x) = 4sinx —2x —xcosx
2+cosx
is (i) increasing (ii} decreasing.
Asinx —2x —xcosx
fx) ~ 2+cosx
Let’s consider the interval [0 , 277]
Finding f’(x)
Asinx —2x —xcosx
f(x) ~ 2+ cosx
4sin x — x(2 + cos x)
fx) ~ 2+ cosx

Misc 4 teachoo.com
Find the intervals in which the function f given by f (x) =x? + yak
# Ois (i) increasing (ii) decreasing.
= 34
f@)= +5
Finding f’(x)
f= 8 +x)
dx
= 3x? +(—3)-3-+
= 3x2 —-3x74
3
= 3x? _ a
-afy2 —-1
~ a(x x)

Misc 5 teachoo.com
Find the maximum area of an isosceles triangle inscribed in the
. x2 y? ae . .
ellipse = + = 1 with its vertex at one end of the major axis.
@ Db

Given equation of ellipse is /\™

x? y? _ 7

@ tpt A A

(a, 0)

where Major axis of ellipse is along x-axis
Here,

Coordinate of A= (a, 0)

Coordinate of A’ = (-a, 0)

Misc 6 teachoo.com
A tank with rectangular base and rectangular sides, open at the top
is to be constructed so that its depth is 2m and volume is 8 m*. If
building of tank costs Rs 70 per sq. meters for the base and Rs 45 pe
square meter for sides. What is the cost of least expensive tank?
Given
Depth of tank=h=2m
h=2 hi,
& Volume of tank = V=8 m3 = bey
=x
Let Length of Tank = xm y, =i
=-<A~AFvr: h DEY
Breadth of Tank = ym l=x
We know that
Volume of tank = Length x Breadth x Height

Misc 7 teachoo.com
The sum of the perimeter of a circle and square is k, where k is
some constant. Prove that the sum of their areas is least when the
side of square is double the radius of the circle.
Let x be radius of circle
& y be side of the square
y
Given
Sum of perimeter of circle & square is k
Circumference of circle + Perimeter of square = k
2n (Radius) + 4 (Side) =k
anx + 4y =k
4y =k —2nx

Misc 8 teachoo.com
A window is in the form of a rectangle surmounted by a semicircular
opening. The total perimeter of the window is 10 m. Find the
dimensions of the window to admit maximum light through the
whole opening.
Let Length of rectangle be x (
x
& Breadth of rectangle be y y
Here x
Diameter of semicircle = x
-. Radius of semicircle = ;

Misc 9 teachoo.com
A point on the hypotenuse of a triangle is at distance a and b from
the sides of the triangle. Show that the minimum length of the
2 2 3
hypotenuse is (a3 + bs) 2
Let P be the point on the hypotenuse AC
Given point P is at distance a & b from sides of triangle
A
Let’s construct PL 1 AB & PM 1 BC
«PL=a & PM=b
P
Lb as
Thus, Z APL=@ (Corresponding angles) M ~
M

Misc 10 teachoo.com
Find the points at which the function f given by
fd) = («—-2)* 4+ 1)? has
{i) local maxima
(ii) local minima
(iii) point of inflexion
Fa) = (—2)*@+ D3
Finding f’(x)
, a (@— 2)* + 1)?)

f')= ae
Using product rule
as (uv) =u'vt+v'u

= ((@ —2)*)'@ + 3 + (@ +13) @—-2)4

Misc 11 teachoo.com
Find the absolute maximum and minimum values of the function f
given by f(x) = cos?x + sinx,x € [0,7]
f(x) = cos*x+sinx , x€[0,7]
Finding f’(x)
1 _ d(cos? x + sin x)
£Q) = ——
= 2c0s x. atcos x) +cosx
dx
= 2cos x(—sin x) + cosx
= cos x (—2sin x + 1)

Misc 12 teachoo.com
Show that the altitude of the right circular cone of maximum
volume that can be inscribed in a sphere of radius r is = .
Given,
Radius of sphere =r
Let R be the radius of the cone and H be its height.
A
Let 2 BOC = 6
Now, h
AC=AO+0C
H=r+rcos@ Nes
H=r(1+cos 6)

Misc 13 teachoo.com
Let f be a function defined on [a, b] such that f’ (x) > 0, for all x
€ (a, b). Then prove that f is an increasing function on (a, b).
We have to prove that function is always increasing
ie. f(xy) < f(xz) for x, < x2
where x, , x, € [a,b]

Proof
Let x, ,X2 be two numbers in the interval [a, b]
ie. x, ,X_ € [a,b]
And,

Xy <X,

Misc 14 teachoo.com
Show that the height of the cylinder of maximum volume that
can be inscribed in a sphere of radius R is ae Also find the
maximum volume.
Given €
Radius of sphere=R
h
Let h be the height
& x be the diameter of cylinder A B
—_S—
Ind ABC
Using Pythagoras theorem
(CB)? + (AB)? = (AC)?
h2 +x? =(R+R)2

Misc 15 teachoo.com
Show that height of the cylinder of greatest volume which can be
inscribed in a right circular cone of height 4 and semi vertical angle a
is one-third that of the cone and the greatest volume of cylinder is
* wh tan’ a
27
P
Given A
Height of cone=h /*
A B
Semi-vertical angle of cone = a h
Let Radius of Cylinder = x
_——
Sc
Now, Q B R
x
Height of cylinder = OO’
=PO -PO’

Misc 16 teachoo.com
A cylindrical tank of radius 10 m is being filled with wheat at the
rate of 314 cubic meter per hour. Then the depth of the wheat is
increasing at the rate of
(A}1m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h
Let r be the radius of cylindrical tank
& V be the volume of cylindrical tank h
& h be the depth of the cylindrical tank
Given that cylindrical tank of radius 10 m being filled with
wheat at the rate of 314 cubic meter per hour
i.e. Change if volume of tank is 314 m3/hr. when r = 10
ie. “ = 314 m?/hr. when r= 10

teachoo.
Misc 1 ‘eachoo.com
Using differentials, find the approximate value of each of the
following:
i
17\4
(=)
1 1 1
(2) _(17)4 (ane
aoe
81 (g1)4 3
i
Lety = x4
Where x = l6andAx=1
4
Since y = x4
1
dy _ alxa)
dx” dx

Misc 1 Using differentials, find the approximate value of each of the following: (b) 〖(33)〗^(− 1/5)〖(33)〗^(− 1/5) = 1/(33)^(1/5)

Misc 4 teachoo.com
Find the equation of the normal to the curve x? = 4y which passes
through the point (1, 2).
Given Curve
x? = dy
x? = ay
(1, 2)
Differentiating w.rt. x
Ad
2x = =
dx
Same as
dy _%
dx 2 Example 45
-1 -1 -2
oo Slope of normal ay =a = x
* G)

teachoo.com
Misc 5
Show that the normal at any point 6 to the curve
x=acos@+a@sin6,y=asind-aécosé
is at a constant distance from the origin.
Given curve
x=acosd+aé@sin0,y=asin@-aé@cosé
We need to show distance of a normal from (0, 0) is constant
First , calculating Equation of Normal
We know that
dy

Slope of tangent is —

P' 8 dx

Misc 20 teachoo.com
The slope of the tangent to the curve x = t? + 3t-8, y= 2t?-2t
-5 at the point (2,- 1) is
22 6 7 -6
(A> (BYE (Cle (D)
We need to find slope of tangent at (2, -1)
._ dy
We know that slope of tangent is an
dy
dy _ ae
dx &
dt

Misc 21 teachoo.com
The line y = mx + 1 is a tangent to the curve y* = 4x if the value
of mis
1
(A) 1 (B) 2 (C) 3 {D)>
Let (h ,k) be the point at which tangent is to be taken
& Given Equation of tangent y = mx +1 Ny
x
2th
& Curve is y? = 4x 2
(h, k)
We know that
y? = 4x
_ ay
Slope of tangent to the Curve is ax
y? = 4x

Misc 22 teachoo.com
The normal at the point (1,1) on the curve 2y + x? = 3 is
(A)x + y = 0 (B)x-y = 0
(C)x+y+1=0 (D)x-y =1
Since (1, 1) lies on normal, it will satisfy its equation
Optioni:x+y=0
Putting x = 1 & y = 1 in equation
1+1=0
2=0
This is false
So, it is not equation of Normal

Misc 23 teachoo.com
The normal to the curve x* = 4 passing (1, 2) is
(Ax +y =3 (B)x-y = 3
()x+y=1 (D)x-y=1
Since (1, 2) lies on normal, it will satisfy its equation
Option 1: x+y =3
Putting x = 1 & y = 2 in equation

14+2=3

3=3
This is true
So, it is the equation of Normal

Misc 24 teachoo.com
The points on the curve 9y? = x, where the normal to the curve
makes equal intercepts with the axes are
8 -8 3 3
ay(4t5) (4) ((44t5) (i(+42)
Since Normal makes equal intercepts with the axes
It’s equation will be
24% e a
atp | (h, k)
Putting b=a
~4Y a4 Oy? = x2
a a
xty=a Equation of line is
y=-x+a XyYuy
a b
where a is x-intercept
«. Slope of Normal = -1 & bis y—intercept

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed by the following equation
y = 4x – 1/2 x
2
where x is the number of days exposed to sunlight.
Question 1
The rate of growth of the plant with respect to sunlight is ______ .
(a) 4x – 1/2 x
2
(b) 4 – x
(c) x – 4
(d) x – 1/2 x
2
Question 2
What is the number of days it will take for the plant to grow to the maximum height
(a) 4
(b) 6
(c) 7
(d) 10
Question 3
What is the maximum height of the plant?
(a) 12 cm
(b) 10 cm
(c) 8 cm
(d) 6 cm
Question 4
What will be the height of the plant after 2 days?
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
Question 5
If the height of the plant is 7/2 cm, the number of days it has been exposed to the sunlight is ______ .
(a) 2
(b) 3
(c) 4
(d) 1

P(x) = -5x
2
+125x + 37500 is the total profit function of a
company, where x is the production of the company.
Question 1
What will be the production when the profit is maximum?
(a) 37500
(b) 12.5
(c) –12.5
(d) –37500
Question 2
What will be the maximum profit?
(a) Rs 38,28,125
(b) Rs 38281.25
(c) Rs 39,000
(d) None
Question 3
Check in which interval the profit is strictly increasing .
(a) (12.5, ∞)
(b) for all real numbers
(c) for all positive real numbers
(d) (0, 12.5)
Question 4
When the production is 2 units, what will be the profit of the company?
(a) 37,500
(b) 37,730
(c) 37,770
(d) None
Question 5
What will be production of the company when the profit is Rs 38,250?
(a) 15
(b) 30
(c) 2
(d) data is not sufficient to find

A potter made a mud vessel, where the shape of the pot is based on f(x)=|x-3|+|x-2|, where f(x) represents the height of the pot.
Question 1
When x > 4, What will be the height in terms of x ?
(a) x – 2
(b) x – 3
(c) 2x – 5
(d) 5 – 2x
Question 2
Will the slope vary with x value?
(a) Yes
(b) No
Question 3
What is dy/dx at x = 3
(a) 2
(b) –2
(c) Function is not differentiable
(d) 1
Question 4
When the x value lies between (2, 3) then the function is
(a) 2x – 5
(b) 5 – 2x
(c) 1
(d) 5
Question 5
If the potter is trying to make a pot using the function f(x) = [x], will he get a pot or not? Why?
(a) Yes, because it is a continuous function
(b) Yes, because it is not continuous
(c) No, because it is a continuous function
(d) No, because it is not continuous

An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides.
Question 1
Find the volume of that open box?
(a) 4x
3
– 96x
2
+ 576x
(b) 4x
3
+ 96x
2
– 576x
(c) 2x
3
– 48x
2
+ 288x
(d) 2x
3
+ 48x
2
+ 288x
Question 2
Find the value of dV/dx
(a) 12(x
2
+ 16x – 48)
(b) 12(x
2
– 16x + 48)
(c) 6(x
2
+ 8x – 24)
(d) 6(x
2
– 8x + 24)
Question 3
Find the value of (d
2
V)/(dx
2
)
(a) 24(x + 8)
(b) 12(x – 4)
(c) 24(x – 8)
(d) 12(x + 4)
Question 4
Find the value of x other than 12?
(a) 3
(b) 9
(c) 1
(d) 4
Question 5
Volume is maximum at what height of that open box?
(a) 3 cm
(b) 9 cm
(c) 1 cm
(d) 4 cm

A right circular cylinder is inscribed in a cone.
S = Curved Surface Area of Cylinder.
Based on the above information answer the following questions:
Question 1
(r )/r
1
= ?
(A) (h - h
1
)/h
1
(B) (h
1
- h)/h
1
(C) (h - h
1
)/h
(D) (h + h
1
)/h
1
Question 2
Find the value of ‘S’?
(A) 2πr/h (h
1
- h)h
(B) 2πr/h
1
(h
1
- h)h
(C) (2πr
1
)/h
1
(h
1
- h)h
(D) (2πr
1
)/h
1
(h
1
+ h)h
Question 3
Find the value of dS/dh ?
(A) (2πr
1
)/h (h
1
- 2h)
(B) (2πr
1
)/h
1
(h - 2h
1
)
(C) 2πr/h (h
1
- 2h)
(D) 2πr1/h
1
(h
1
- 2h)
Question 4
Find the value of (d^2 S)/(dh^2 )
(A) − (4πr
1
)/h
1
(B) − 4πr/h
(C) − (4πr
1
)/h
(D) (4πr
1
)/h
Question 5
What is the relation between r
1
and r?
(A) r
1
= r
/
2
(B) 2r
1
= 3r
(C) r
1
= 2r
(D) r
1
/2 = r/3

There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm.
Based on the above information answer the following:
This question is inspired from
Example 37 - Chapter 6 Class 12 (AOD)
- Maths
Question 1
What is the value of DP?
(A)
√(
100
-
x
2
)
(B)
√(
x
2
-
100
)
(C) 100
-
x
2
(D) x
2
− 100
Question 2
What is the area of trapezium A(x)?
(A) (x
-
10
)√(
100
-
x
2
)
(B)
(
x
+
10)
√(
100 - x
2
)
(C)
(
x
-
10
)
(100 - x
2
)
(D)
(
x
+
10)(100 - x
2
)
Question 3
If A'(x) = 0, then what are the values of x?
(A) 5,-10
(B)
-
5, 10
(C)
-
5,-10
(D) 5,10
Question 4
What is the value of maximum Area?
(A) 75 √2 cm
2
(B) 75 √3 cm
2
(C) 75 √5 cm
2
(D) 75 √7 cm
2

The shape of a toy is given as f(x)=6(2x
4
- x
2
). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.
Question 1
Which value from the following may be abscissa of critical point?
(a) ± 1/4
(b) ± 1/2
(c) ± 1
(d) None
Question 2
Find the slope of the normal based on the position of the stick.
(a) 360
(b) –360
(c) 1/360
(d) (-1)/360
Question 3
What will be the equation of the tangent at the critical point if it passes through (2,3)?
(a) x + 360y = 1082
(b) y = 360x – 717
(c) x = 717y + 360
(d) none
Question 4
Find the second order derivative of the function at x = 5.
(a) 598
(b) 1176
(c) 3588
(d) 3312
Question 5
At which of the following intervals will f(x) be increasing?
(a) (-∞", " (-1)/2) ∪ (1/2 ", " ∞)
(b) (1/2,0) ∪ (1/2 ", " ∞)
(c) (0", " 1/2) ∪ (1/2 ", " ∞)
(d) (-∞", " (-1)/2) ∪ (0", " 1/2)

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:
10 cm2/s (B) 3 cm2/s
(C) 10√𝟑 cm2/s (D) 10/3 cm2/s

A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is:
(A) 1/10 radian/sec (B) 1/20 radian/sec
(C) 20 radian/sec (D) 10 radian/sec
Let AB be the ladder
& OA be the wall & OB be the ground.

The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of
(A) 𝜋/4 (B) 𝜋/3
(C) 𝜋/2 (D) 𝜋/6
Same as Exactly same Question 2 – NCERT Exemplar MCQ
So, the correct answer is (C)

The interval on which the function f(𝑥) = 2𝑥3 + 9𝑥2 + 12𝑥 - 1
is decreasing is:
(A) [−1,∞) (B) [–2, –1]
(C) (−∞,−2] (D) [–1, 1]

Let the f : R → R be defined by f (x) = 2x + cos x, then f :
(A) has a minimum at x = 𝜋 (B) has a maximum, at x = 0
(C) is a decreasing function (D) is an increasing function
Given
f (𝒙) = 2𝒙 + cos 𝒙

y = x (x – 3)2 decreases for the values of x given by :
1 < x < 3 (B) x < 0
(C) x > 0 (D) 0 < x < 3/2
𝑓(𝑥)=𝑥〖(𝑥−3)〗^2

The function f (x) = 4 sin3 x – 6 sin2 x + 12 sin x + 100 is strictly
Increasing in (𝜋, 3𝜋/2) (B) decreasing in (𝜋/2,𝜋)
(C) decreasing in ((−𝜋)/2 ", " 𝜋/2) ` (D) decreasing in ("0, " 𝜋/2)
f (𝑥) = 4 sin^3⁡〖𝑥−6 sin^2⁡〖𝑥+12 sin⁡〖𝑥+100〗 〗 〗

Which of the following functions is decreasing on (0,𝜋/2)
(A) sin 2x (B) tan x
(C) cos x (D) cos 3x
To check decreasing, we check if 𝒇^′ (𝒙)<𝟎 in (0,𝜋/2)

The function f (x) = tan x – x
always increases
(B) always decreases
(C) never increases
(D) sometimes increases and sometimes decreases.
Given 𝑓(𝑥) = tan 𝑥 − 𝑥

If x is real, the minimum value of x2 – 8x + 17 is
–1 (B) 0
(C) 1 (D) 2
𝑓(𝑥) =𝑥^2−8𝑥+17

The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is
126 (B) 0
(C) 135 (D) 160
𝒇(𝑥)=𝑥^3−18𝑥^2+96𝑥

The function f (x) = 2x3 – 3x2 – 12x + 4, has
two points of local maximum
(B) two points of local minimum
(C) one maxima and one minima
(D) no maxima or minima
f (𝑥) = 2𝑥3 – 3𝑥2 – 12𝑥 + 4

The maximum value of sin x . cos x is
(A) 1/4 (B) 1/2
(C) √2 (D) 2 √2

At x = 5𝜋/6, f (x) = 2 sin 3x + 3 cos 3x is :
maximum
(B) minimum
(C) zero
(D) neither maximum or minimum
Since, we have to check maximum and minimum value at x = 5π/6
So, we will find f ” (x)

Maximum slope of the curve y = –x3 + 3x2 + 9𝑥 – 27 is:
0 (B) 12
(C) 16 (D) 32
Given
y = − 𝑥^3 + 3𝑥^2 + 9𝑥− 27

f (x) = xx has a stationary point at
(A) x = e (B) x = 1/𝑒
(C) x = 1 (D) x = √𝑒
A stationary point of a function is a point where
𝒇′(𝒙) = 0

The maximum value of (1/𝑥)^𝑥 is:
(A) e (B) ee
(C) 𝑒^(1/𝑒) (D) 〖1/𝑒〗^(1/𝑒)
Let f (𝑥) = (1/𝑥)^𝑥
To find maximum value, we need to differentiate f(x)

The abscissa of the point on the curve 3y = 6x – 5x3, the normal at which passes through origin is:
1 (B) 1/3
(C) 2 (D) 1/2
Given curve
3𝑦=6𝑥−5𝑥^3

The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 = 2
(A) touch each other (B) cut at right angle
(C) cut at an angle 𝜋/3 (D) cut at an angle 𝜋/4

The tangent to the curve given by x = et . cost, y = et . sint at t =
𝜋/4 makes with x-axis an angle:
0 (B) 𝜋/4 (C) 𝜋/3 (D) 𝜋/2
Given curve
𝒙=𝒆^𝒕.𝒄𝒐𝒔𝒕 and 𝒚=𝒆^𝒕.𝒔𝒊𝒏𝒕

The equation of the normal to the curve y = sin x at (0, 0) is:
x = 0 (B) y = 0
(C) x + y = 0 (D) x – y = 0
𝑦=sin⁡𝑥
Since Slope of normal =(−1)/(𝑑𝑦/𝑑𝑥)

The point on the curve y2 = x, where the tangent makes an angle of 𝜋/4 with x-axis is
(A) (1/2, 1/4) (B) (1/4, 1/2)
(C) (4, 2) (D) (1, 1)
𝑦^2=𝑥
Slope of the tangent is 𝒅𝒚/𝒅𝒙

The curve y = 𝑥^(1/5) has at (0, 0)
(A) a vertical tangent (parallel to y-axis)
(B) a horizontal tangent (parallel to x-axis)
(C) an oblique tangent
(D) no tangent
Since options are about finding tangent
So, finding slope of tangent i.e. 𝒅𝒚/𝒅𝒙 at (0, 0)

The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is
(A) 3x – y = 8 (B) 3x + y + 8 = 0
(C) x + 3y ± 8 = 0 (D) x + 3y = 0
Since, the normal to the curve is parallel to the line 𝑥+3𝑦=8
∴ Slope of normal = Slope of line

If the curve ay + x2 = 7 and x3 = y, cut orthogonally at (1, 1), then the value of a is:
1 (B) 0 (C) –6 (D) 6
Two curves cut each other orthogonally means the cut each other at right angles

If y = x4 – 10 and if x changes from 2 to 1.99, what is the change in y
0.32 (B) 0.032 (C) 5.68 (D) 5.968

The equation of tangent to the curve y (1 + x2) = 2 – x, where it crosses x-axis is:
(A) x + 5y = 2 (B) x – 5y = 2
(C) 5x – y = 2 (D) 5x + y = 2
First, let us find the point at which curve cuts 𝑥−axis
Since 𝒚=𝟎 on 𝑥−axis
∴ Point =(𝒙,𝟎)

The points at which the tangents to the curve y = x3 – 12x + 18 are parallel to x-axis are:
(A) (2, –2), (–2, –34) (B) (2, 34), (–2, 0)
(C) (0, 34), (–2, 0) (D) (2, 2), (–2, 34)
Given Curve
"y = x3 – 12x + 18"

The tangent to the curve 〖𝑦=𝑒〗^2𝑥 at the point (0, 1) meets x-axis at:
(0, 1) (B) − 1/2, 0
(C) (2, 0) (D) (0, 2)
Given
〖𝑦=𝑒〗^2𝑥

The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is:
(A) 22/7 (B) 6/7
(C) − 6/7 (D) −6

Ex 6.3, 1 teackoo
Find the slope of the tangent to the curve y = 3x*- 4x atx = 4
Given
y = 3xt- Ax

We need to find Slope of tangent at x = 4
We know that

— oy
Slope of tangent = ax

=3 x 4x3 -4

=12x3 —4

Puttingx =4

teachoo.com
Ex 6.3, 2
Find the slope of the tangent to the curve
yas, x#2 atx =10
x-1
y= yo2 x#2
. ay
We know that Slope of tangent is os
x-1
dy _ (3)
dx dx
Using Quotient Rule
Uu ' uly - vu
As (®) 7 ps
dy _ (= 1)'(e-2)- @-2)'@-1)
dx (x - 2)?

Ex 6.3, 3 teachoo.com
Find the slope of the tangent to curve y = ext
at the point whose x — coordinate is 2.
y=xe-x+1
We know that
. dy

slope of tangent is

dy _ a(x3-x+1)

ax ax

d

& = 3x?-1+0

dx
We need to find “ at the point whose x — coordinate is 2

teachoo.com
Ex 6.3, 4
Find the slope of the tangent to the curve y = x? — 3x42
at the point whose x — coordinate is 3
y=xi-3x42
We know that
_ ay

slope of tangent = x

dy _ 2_
Since x — coordinate is 3
Putting x = 3 in (1)

Ex 6.3,5 teachoo.com
Find the slope of the normal to the curve x = acos? 6,
y =asin’ @ at g=*
Given Similarly
x = acos*@ y=asin3@
Differentiating w.rt. 6 Differentiating w.rt. 0
ax _ afacos* 0) dy _ a(asin’ 0)
wo a0 do do
dy d(sin® 6)
ax d(cos? 0) == gq.
qe * a do a0
dy . 2
da . v=
& = a.3cos* 9. (—sind) 70 a.3sin* 6.(cos 8)
dx 2 ® = [email protected]
3a =~ 3asin @ cos 6 do .

Ex 6.3, 6 teachoo.com
Find the slope of the normal to the curve x = 1—asiné,
y =bcos*@ atO= 5
_ ay
Slope of tangent is x
dy
dy _ a6
dx ax
a6
y = bcos" 6
x=1-asin@ Differentiating w.rt. 6
Differentiating w.rt. 8 dy _ a(b cos? 9)
dx d(1—-—asin 0) do do
ao a0
* _b 2cos4 Aicos #)
dx do do
707 0-—acos@
® ~ 2bcosé. (— sind)
dx dé
— = -—acos@
a0 dy ;
— =-—2bsin@ cosé
do

Ex 6.3, 7 teackoo.com
Find points at which the tangent to the curve
y =x? — 3x? — 9x + 7is parallel to the x-axis
Equation of Curve is
y=x?-3x2-9x+7
Differentiating w.rt. x

dy _ a(x?—3x?-9x%+7)

dx ax

d

* = 3x? -6x-9+0

dx

d

“Y = 3x2 — 6x —9

dx

GY _ 27,2 _ ay —

mn 3(x* — 2x — 3)

Ex 6.3, 8 teachoo.com
Find a point on the curve y = (x — 2)? at which the tangent is
parallel to the chord joining the points (2, 0) and (4, 4).
Given Curves is
y=(x-2y
Let AB be the chord joining the Point (2,0) & (4,4)
& CD be the tangent to the Curve y = (x — 2)?
Given that tangent is Parallel to the chord
i.e. CD || AB
If two lines are parallel,
«. Slope of CD = Slope of AB if p
then their slopes are equal

teachoo.com

Ex 6.3, 9
Find the point on the curve y = x? — 11x + 5 at which the
tangentis y = x — 11.
Equation of Curve is

y=x?-11x+5
We know that
Slope of tangent is =

ope of tangent is

dy _ a(x?-11x+5)

dx ax

d.

Pa 3x27 -11 0 1)

dx

Ex 6.3, 10 teachoo.com
Find the equation of all lines having slope —1 that are tangents to
the curve y = —— ,xX#1.
x-1
Equation of Curve is
_ 1
y ~~ e-4
SI ft ti oy
ope of tangentis —
1
dy _ (523)
ax ax
dy _ a _ -1
dx dx (x 1)
dy _ _4)-1-1
_ 1x («-1)

teachoo.com
Ex 6.3, 11
Find the equation of all lines having slope 2 which are tangents to
1
the curve y= yaar + 3.
The Equation of Given Curve is :
_ 1
vr y3
We know that
_ dy
Slope of tangent is x
1
ay _ (523)
dx dx

Ex 6.3, 12 teachoo.com
Find the equations of all lines having slope 0 which are tangent to
th = —1
@ curve y=
Given C isy= 1
iven Curve is y = 3
Slope of tangent is
ope of tangentis
1
dy = a(a- 2x + 3)
dx dx
dy _ d(@x’-2x+3)7
dx dx
dy _ 2 2 d(x?-2x+3)
en 1¢c? — 2x + 3) a
d _
a= —(x? — 2x + 3)72(2x — 2)

Ex 6.3, 13 teachoo.com
. . x2 yy? .
Find points on the curve athe 1 at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.
2 2
aan a
9 16
2 x2
a al
16 9
Differentiating w.r.t. x
2 2
ye a
a(t) _ 40-5)
ax ax
2
140%) _ aay _ (5)
16 dx ax ax

Ex 6.3, 14 teachoo.com
Find the equations of the tangent and normal to the given
curves at the indicated points:
(i) y = x4 — 6x3 + 13x? —10x+5 at (0,5)
y =x* — 6x94 13x? — 10x45
Differentiating w.rt.x
& = 4x3 18x2426x-10 (1)
dx
Now Point Given is (0 , 5)
Hencex=0, y=5
Putting x = 0 in (1}

Find the equations of the tangent and normal to the given curves at the indicated points: (ii) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (1, 3) Given Curve is
𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5

Find the equations of the tangent and normal to the given curves at the indicated points: (iii) 𝑦=𝑥3 𝑎𝑡 (1, 1)Given Curve is
𝑦=𝑥3

Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=𝑥2 𝑎𝑡 (0, 0) Given Curve is
𝑦=𝑥^2

Find the equations of the tangent and normal to the given curves at the indicated points: (v) 𝑥=cos⁡𝑡, 𝑦=sin⁡𝑡 𝑎𝑡 𝑡= 𝜋/4At 𝒕= 𝝅/𝟒
x = cos 𝜋/4 = 1/√2
y = sin 𝜋/4 = 1/√2
∴ At 𝑡=𝜋/4 , the point is (1/√2 " ," 1/√2)

Ex 6.3, 15 teachoo.com
Find the equation of the tangent line to the curve
y =x? —2x+7 whichis:
(a) parallel to the line 2x -y+9=0
We know that Slope of tangent is °
y=x? —2x+7
Differentiating w.rt.x
dy _ _
amex 2 (1)
Finding Slope of line 2x —-y +9 =0
2x-yt+9=0
y=2xt+9

Ex 6.3, 16 teachoo.com
Show that the tangents to the curve y = 7x? + 11 at the points
where x = 2 andx = —2 are parallel.
We know that 2 lines are parallel y
Slope of 1% line = Slope of 2"? line
m =m,
We know that Slope of tangent is °
Given Curve is
y=7x3 +11
Differentiating w.rt.x
dy _ d(7x? + 11)
dx dx

Ex 6.3, 17 teachoo.com
Find the points on the curve y = x? at which the slope of the
tangent is equal to the y-coordinate of the point
Let the Point be (h, k) on the Curve

yar
Where Slope of tangentat (hk ,k) = y — coordinate of (h,k)
. dy _
Le. x | (h,k) = k
Given y = x3
Differentiating w.rt.x

dy _ 4,2

7 3x

Ex 6.3, 18 teachoo.com
For the curve y = 4x3 — 22°, find all the points at which the tangent
passes through the origin.
Let (h ,k) be the Required Point on the Curve at which tangent
is to be taken
Given Curve is
y = 4x3 — 2x%
Since Point (h ,k) is on the Curve
=> (h,k) will satisfy the Equation of Curve
Puttingx =h ,y=k
k= 4h? —-2h° (1)

Ex 6.3, 19 teachoo.com
Find the points on the curve x2 + y?_ — 2x —3 =0 at which
the tangents are parallel to the x — axis
Given that
Tangent is parallel to the x — axis
«. Slope of tangent = Slope of x — axis
We know that
Slope of tangent is =
lope of tangent is =
4 dy
Finding ax
x? +y* —2x -3=0

teachoo.co
Ex 6.3, 20 ”
Find the equation of the normal at the point (am?, am?*) for
the curve ay? = x?
We know that
. dy
Slope of tangentis a
Given
ay? =x3
Differentiating w.rt.x
a(ay?) _ a(x?)
dx dx
2?) _ a?)
dx dx

Ex 6.3, 21 teachoo.com
Find the equation of the normal to the curve y = x? + 2x +6
which are parallel to the line x + 14y + 4 = 0.
Let (hk, k) be the point on the Curve at which Normal is to be
taken
Given Curve is

y=xi+2x+6
Since point (h , k) is on the Curve
« (h, k) will satisfies the Equation of Curve
Puttingx =h, y=k

k=h3+2h+6 ~~ ~C(1)

Ex 6.3, 22 teachoo.com
Find the equations of the tangent and normal to the parabola
y? = 4ax at the point (at?, Zat).
Given Curve is y* = 4ax
We need to find equation of tangent & Normal at (at?, 2at)
We know that
Slope of tangent is
ope of tangent is
y? = 4ax
Differentiating w.rt.x
d{y?) _ d(4ax)
dx —s dx
d(y? d d
dx dy dx

teachoo.co

Ex 6.3, 23
Prove that the curves x = y? & xy = k cut at right angles if 8k?=1
We need to show that the curves cut at right angles
Two Curve intersect at right angle if
the tangents to the curves at the point of intersection are
perpendicular to each other
First we Calculate the point of intersection of Curve (1) & (2)

x=y? ..(1)
Putting x = y? in (2)

xy=k

Ex 6.3, 24 teachoo.com
Find the equations of the tangent and normal to the hyperbola
x2 2
are = Lat the point (x9, Yo)
We know that
Slope of tangentis —
ope of tangentis
aa dy
Finding ax
x2 y?
ep!
y? _ 1 2
“p22
2 2
yu _
be a2 1

teachoo.com

Ex 6.3, 25
Find the equation of the tangent to the curve ¥3x — 2 which is
parallel to the line 4x - 2y+5=0.
Let (h , k) be the point on Curve from tangent to be taken
We know that Equation of tangent is °

y=v3x -2
Differentiating w.rt.x

i

ay _ a(3x —2)2

dx dx

dy 38

dx 2y3x—2

Ex 6.3, 26 teachoo.com
The slope of the normal to the curve y = 2x? + 3 sin x at x = O is
1 1
(a)3 (8) 5 (c) -3 (0)-5
Slope of tangent is =
ope of tangent is
y = 2x* +3sinx
Differentiating w.rt. x
dy _ d(2x? +3 sinx)
dx dx
d
& = 4x + 3cosx
dx
We know that
Slope of tangent x Slope of Normal = —1

teachoo.com
Ex 6.3, 27
The line y = x + 1is a tangent to the curve y? = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, - 2} (D} (- 1, 2)
Given Curve is
y? = 4x
Differentiating w.rt. x
d{y?) _ d(4x)
dx dx
d 2
ay?) y 4
dy dx
dy _
2y x 7 4

Ex 6.4, 1 teackoo
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
(i) ¥25.3
Let y= x
Thus,
V25.3=ytAy
Here,
2
Ay= ax Ax
where x = 25 & Ax =0.3

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
(ii) V49.5
Lety=/x
where x = 49 & Ax=0.5
Since y = /x
dy _ Ax) 1
dx dx 2x
Now,
mas
Ay= ax Ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
(iii) V0.6
Let y=Jx
where x = 0.64 & Ax = —0.04
Since y = fx
dy ax) 1
dx dx 2x
Now,
-
Ay = ax Ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
{iv) (0.009)3
i
Let y = (x)3
where x = 0.008 & Ax = 0.001
Differentiating w.rt.x
1
ay _ dlx3) _ 1 1
ax 3 rs
dx dx 3 3x3
ng Ay = 22
Using Ay = x Ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(v) (0.999)10
a
Let y = x 10
where x = 1, Ax = —0.001
Now,
1
y= x 10
Differentiating w.rt.x
1
dy ale i0) 1 2 1
= Foo Se? = TT
dx dx 10 10 x10

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
i
(vi) (15)4
i
Lety=x4
where x = 16,Ax =—-1
Now,
i
yous
Differentiating w.rt.x
1
dy divs) 1 22# 4 23
= —_ = -X 4 =-K 4
dx dx 4 4

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
i
(vii) (26)3
i
Let y = (x)3
where x = 27 & Ax =—1
Now,
i
y=xsa
Differentiating w.rt.x
1
dy alx 3) 1 iy 1 72 1
= = — =-X3 =-X 3 =z
dx ax 3 3 2
3x3

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(viii) (255)4
i
Let y = (x)}4
where x = 256 & Ax =—1
Now,
i
y= (a
Differentiating w.rt.x
1
dy a(x) 1 4-4 175% 1
—_— = — =-X4 =-xK 4=—7F
ax ax 4 4 axé

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
i
(ix} (82)4
aa
Let y = (x)+
where x = 81 Ax =1
Now,
i
y=)
Differentiating w.rt.x
1
dy alx i) 1 4-4 123 1
— = — = -X4 =-X 4=>-— Wy
dx dx 4 4 3
4x4

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(x) (401)2
i
Let y = x2
where x = 400 & Ax=1
Now,
dy 1
dx 2Vx
. _ ay
Using Ay = ix Ax
_ ay
Ay = aE Ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(xi) (0.0037)2
Let y=/x
Let x = 0.0036 & A x= 0.0001
Since y = fx
ay ax)
dx dx 2x
Now,
_W
Ay = x Ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(xii} (26.57)3
i
lety=x3
where x = 27 & Ax = —0.43
Now,
i
yHxes
Differentiating w.rt.x
1
dy alx 3) 1 2-4 172 1
— = m= -X3 =r-xX 3=>——745
dx dx 3 3 2
3x3

Ex 6.4,1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(xiii} (81.5)4
i
Let y = (x)#
where x = 81 & Ax =0.5
Now,
i
y=xue
Differentiating w.rt.x
1
dy alx 3) 1 1-4 1 = 1
—=—_ =- x4 =-x 4=—;Ax
ax dx 4 4 ax

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
3
(xiv) (3.968)2
3
Let y = x2
where x = 4 & Ax = —0.032
Now,
3
y=.
Differentiating w.rt.x
3
dy _ alx 2)
dx ax
1
dy 345
dx 2

Ex 6.4, 1 teachoo.com
Using differentials, find the approximate value of each of the
following up to 3 places of decimal.
1
(xv) (32.15)5
2
Let y = (x)5
where x = 32 & Ax =0.15
Now,
2
y= (x)s
Differentiating w.rt.x
1
dy alx 5) 1 t_-14 1 == 1
— = =H -X5 =-x 5 =F
dx dx 5 5 =
5x5

Ex 6.4, 2 teachoo.com
Find the approximate value of f (2.01), where f (x) = 4x? + 5x + 2.
Let x= 2 and Ax=0.01
Given f(x) = 4x2 + 5x +2
f(x) = 8x +5
Now,
Ay = f’(x) Ax = (8x +5) 0.01
Also,
Ay = f (x + Ax) - f(x)
f(x+ Ax) =f (x) + Ay

teachoo.co

Ex 6.4, 3 eae «om
Find the approximate value of f (5.001), where f (x) = x? — 7x? +15.
Letx=5 and Ax=0.001
Given f (x} =x? - 7x? +15

f'(x) = 3x? - 14x
Now,

Ay = f’(x) Ax = (3x? - 14x) 0.001
Also,

Ay = f (x + Ax) - f(x)

f(x + Ax) = f(x) + Ay

Ex 6.4, 4 teachoo.com
Find the approximate change in the volume V of a cube of side x
meters caused by increasing the side by 1%.
Let side of the cube = x metres
Increase in side = 1% = 0.01 x
Hence, Ax = 0.01 x
Volume of the cube = V = x? m?
We need to find approximate change in volume v of the cube
ie. AV
Now,
dv
AV= x Ax

Ex 6.4, 5 teachoo.com
Find the approximate change in the surface area of a cube of side x
meters caused by decreasing the side by 1%.
Let side of the cube = x meters.
Given
Decrease in side = 1% = - 0.01 x
Hence, Ax =-0.01 x
Surface area of the cube = S = 6x? m?
We need to find the approximate change in the surface area of
the cube
ie. AS

Ex 6.4, 6 teachoo.com
If the radius of a sphere is measured as 7 m with an error of 0.02
m, then find the approximate error in calculating its volume.
Let r be the radius of the sphere
Given r=7m
Error in measurement of radius = Ar
Ar=0.02 m
Volume of the sphere = V = sur?
We need to find error in calculating the volume that is Av
AV= @xar
dr

Ex 6.4, 7 teachoo.com
If the radius of a sphere is measured as 9 m with an error of 0.03 m,
then find the approximate error in calculating its surface area.
Let r be the radius of the sphere

Givenr=9m
Error in measurement of radius = Ar

Ar = 0.03 m
Surface area of the sphere = S = 4ar?
We need to find the error in calculating the surface area AS

As=& x Ar

dr
2
- d(Anr dy Ar
dr

Ex 6.4, 8 teachoo.com
If f(x) = 3x? + 15x +5, then the approximate value of f (3.02) is
(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66
Let x = 3 and Ax = 0.02
f’(x) = 6x +15
Now,
Ay = f’(x) Ax = (6x + 15) 0.02
Also,
Ay =f (x + Ax) - f(x)
f (x + Ax) = f (x) + Ay
f (3.02) = 3x? + 15x +5 + (6x + 15) (0.02)

Ex 6.4, 9 teachoo.com
The approximate change in the volume of a cube of side x
meters caused by increasing the side by 3% is
(A) 0.06 x? m? (B) 0.6 x? m? (C) 0.09 x? m? (D) 0.9 x? m3
Let side of the cube = x meters
Increase in side = 3% = 0.03 x
Hence, Ax = 0.03 x
Volume of the cube = V = x? m3?
We need to find approximate change in volume V of the cube
ie. AV

Chapter 6 Class 12 - Application of Derivatives
- MCQ Worksheet 1
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- Mix Questions Worksheet 1
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Chapter 6 Class 12 - Application of Derivatives
- Assertion and Reasoning
Worksheet 1
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Chapter 6 Class 12 - Application of Derivatives
- Case Based Question
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