CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
Question 2
Question 3 Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13
Question 14
Question 15 Important
Question 16
Question 17
Question 18 Important
Question 19 Important
Question 20 Important
Question 21 (A)
Question 21 (B) Important
Question 22 (A) Important
Question 22 (B) Important
Question 23
Question 24 Important
Question 25
Question 26 (A) Important
Question 26 (B) Important
Question 27
Question 28 Important You are here
Question 29 Important
Question 30 (A)
Question 30 (B) Important
Question 31
Question 32 (A) Important
Question 32 (B) Important
Question 33 - Part 1
Question 33 - Part 2 Important
Question 34 Important
Question 35 (A) Important
Question 35 (B)
Question 36 (i) - Case Based
Question 36 (ii) Important
Question 36 (iii) (A)
Question 36 (iii) (B) Important
Question 37 (i) - Case Based
Question 37 (ii)
Question 37 (iii) (A) Important
Question 37 (iii) (B) Important
Question 38 (i) - Case Based Important
Question 38 (ii)
Question 38 (iii) (A)
Question 38 (iii) (B) Important
CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
Last updated at Dec. 13, 2024 by Teachoo
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Question 28 If πΌ and π½ are zeroes of a polynomial 6π₯2 β 5x + 1 then form a quadratic polynomial whose zeroes are πΌ2 and π½2 .Let p(x) = 6π₯2 β 5x + 1 Roots of p(x) are πΌ and π½ So, Sum of roots = πΌ + π½ = (β(β5))/6 = π/π And, Product of roots = πΌπ½ = π/π We need to find quadratic polynomial whose zeroes are πΌ2 and π½2 . So, Required polynomial = q(x) = x2 β (Sum of Zeroes)x + Product of Zeroes = x2 β (πΌ2 + π½2)x + πΌ2 Γ π½2 Using (a2 + b2) = (a + b)2 β 2an = x2 β [(πΌ + π½)2 β 2πΌπ½] x + (πΌπ½)2 Putting πΌ + π½ = 5/6 & πΌπ½ = 1/6 = π₯^2β[(5/6)^2β2 Γ1/6]π₯+(1/6)^2 = π₯^2β[25/36β1/3]π₯+1/36 = π₯^2β[(25 β 12)/36]π₯+1/36 = π^πβ[ππ/ππ]π+π/ππ We can multiply by 36 to make our equation cleaner = 36(π₯^2β[13/36]π₯+1/36) = πππ^πβπππ+π