Let f: (0,) R be a twice differentiable function. If for some a 0, 0^a f(x) dx = f(a), f(1) = 1, f(16) = 18, then 16 - f^-1 ( 116) is equal to:

Given: 0^a f(x) dx = f(a), f(1)=1, f(16)= 18 Find: 16 - f^-1 ( 116) From the solution, the conclusion is that 16 - f^-1 ( 116) = 4 Therefore, the required value is 4.

Let f: (0,) R be a twice differentiable function. If for some a 0… — Solution | JEE Mathematics

NVAMediumJEE 2025Linear Differential Equations

Let $f : (0, \infty) \to \mathbb{R}$ be a twice differentiable function. If for some $a \neq 0$ ,

\int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then} \quad 16 - f^{-1}\left( \frac{1}{16} \right)

is equal to:

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given:

\int_0^a f(x) \, dx = f(a), \quad f(1)=1, \quad f(16)=\frac{1}{8}

Find:

16 - f^{-1}\left(\frac{1}{16}\right)

From the solution, the conclusion is that

16 - f^{-1}\left(\frac{1}{16}\right) = 4

Therefore, the required value is $4$ .

Common mistakes

Differentiating the condition incorrectly with respect to $a$ . By the fundamental theorem of calculus, $\frac{d}{da}\int_0^a f(x)\,dx = f(a)$ , and differentiating the right side gives $f'(a)$ , so the relation becomes $f(a)=f'(a)$ . Do not treat the integral as a constant.
Confusing $f^{-1}\left(\frac{1}{16}\right)$ with $\frac{1}{f(16)}$ . The inverse function asks for the input at which the function value is $\frac{1}{16}$ , whereas $\frac{1}{f(16)} = 8$ is unrelated.

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Answer

Step-by-step solution

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