Find the critical point of the function.


CBSE Class 12 Sample Paper for 2023 Boards
CBSE Class 12 Sample Paper for 2023 Boards
Last updated at Dec. 13, 2024 by Teachoo
Transcript
Question 37 (ii) Find the critical point of the function.Now A =(" " 𝟒𝒃)/𝒂 " " 𝒙√((𝒂^𝟐 − 𝒙^𝟐 ) ) We need to maximise A, but A has a square root Which will be difficult to differentiate Let Z = A2 Z = ((" " 𝟒𝒃)/𝒂 𝒙√((𝒂^𝟐 − 𝒙^𝟐 ) ))^2 Z = (16𝑏^2)/𝑎^2 × 𝑥^2 (𝑎^2−𝑥^2 ) Z = (𝟏𝟔𝒃^𝟐)/𝒂^𝟐 × (𝒂^𝟐 𝒙^𝟐−𝒙^𝟒) Since A is positive, A is maximum if A2 is maximum So, we maximise Z = A2 Differentiating w.r.t x 𝑑𝑍/𝑑𝑥= 𝑑((𝟏𝟔𝒃^𝟐)/𝒂^𝟐 (𝑎^2 𝑥^2−𝑥^4))/𝑑𝑥 𝑑𝑍/𝑑𝑥=(16𝑏^2)/𝑎^2 × 𝑑(𝑎^2 𝑥^2−𝑥^4 )/𝑑𝑥 𝑑𝑍/𝑑𝑥=(16𝑏^2)/𝑎^2 × (𝑎^2 × 2𝑥−4𝑥^3) 𝑑𝑍/𝑑𝑥=(16𝑏^2)/𝑎^2 × (〖2𝑎〗^2 𝑥−4𝑥^3) 𝒅𝒁/𝒅𝒙=(𝟑𝟐𝒃^𝟐)/𝒂^𝟐 × (𝒂^𝟐 𝒙−𝟐𝒙^𝟑) Putting 𝒅𝒁/𝒅𝒙= 0 (𝟑𝟐𝒃^𝟐)/𝒂^𝟐 × (𝒂^𝟐 𝒙−𝟐𝒙^𝟑 )=𝟎 𝒂^𝟐 𝒙−𝟐𝒙^𝟑=0 2𝑥^𝟑=𝒂^𝟐 𝒙 𝑥^𝟑/𝒙=𝒂^𝟐/𝟐 𝑥^2=𝒂^𝟐/𝟐 𝒙=±𝒂/√𝟐 Since x is length, it will be positive ∴ Critical point is 𝒙=𝒂/√𝟐