Let a differentiable function f satisfy f(x)+∫3ˣ f(t)/t d t=√(x+1), x ≥ 3. Then 12 f(8) is equal to :

Solution

Maths sir · 5 steps

Differentiate to kill the integral, then the integrating factor x turns it into a product rule.

1. Differentiate. By the Leibniz rule, the derivative of ∫₃ˣ f(t)/t dt is f(x)/x. So f'(x) + f(x)/x = 1/2√(x+1).
2. Starting value. Put x = 3 in the given equation. The integral from 3 to 3 is zero, so f(3) = √4 = 2.

   d/dx(x f(x)) = x f'(x) + f(x) = x/2√(x+1)

3. Integrate. Multiplying the equation by x gives the line above. Put u = √(x+1), so x = u² - 1 and dx = 2u du. The right side becomes (u² - 1) du, so x f(x) = u³/3 - u + C.
4. Find C. At x = 3, u = 2: 3 × 2 = 8/3 - 2 + C, so C = 16/3.
5. So. At x = 8, u = 3: 8 f(8) = 9 - 3 + 16/3 = 34/3, so f(8) = 17/12 and 12 f(8) = 17, option (c).

The idea to keep. When f appears both outside and inside an integral with a variable limit, differentiate first. Then read the starting value by putting the lower limit in for x.

Answer from the NTA final answer key. Solution by PrepWiser.