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JEE Advanced Mathematics Limits 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Mathematics Question (2024) — Solution

Question

Let f(x) be a continuously differentiable function on the interval (0, ) such that f(1)=2 and t x t^ 10 f(x)-x^ 10 f(t) t^9-x^9 =1 for each x>0. Then, for all x>0, f(x) is equal to

Options

  1. A. 31 11 x - 9 11 x^ 10
  2. B. 9 11 x + 13 11 x^ 10
  3. C. -9 11 x + 31 11 x^ 10
  4. D. 13 11 x + 9 11 x^ 10

Answer

B. 9 11 x + 13 11 x^ 10

Step-by-step solution

aligned & _ t x t^ 10 f(x)-x^ 10 f(t) t^9-x^9 =1 \\ & _ t x 10 t^9 f(x)-x^ 10 f^ (t) 9 t^8 =1 aligned aligned & 10 x f(x)-x^2 f^ (x)=9 \\ & x^2 f^ (x)=10 x f(x)-9 \\ & f^ (x)= 10 f(x) x - 9 x^2 \\ & d y d x - 10 x y=- 9 x^2 \\ & y 1 x^ 10 = - 9 x^2 1 x^ 10 d x \\ & y x^ 10 = 9 11 x^ 11 +c...(1) aligned aligned & f (1)=2 2 1 = 9 11 + c c = 13 11 \\ & f ( x )= 9 11 x + 13 11 x ^ 10 aligned Option (2) is correct.

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