Let f:(0,∞)→R be a twice differentiable function. If for some a0, ∫_0^1 f(λ x) dλ = a f(x), f(1) =

Q: Why can't I just differentiate the integral directly with respect to $\lambda$?

That is a great question! The integral is defined with respect to $\lambda$, and the limits of integration are constants ($0$ and $1$). If you differentiate with respect to $\lambda$, you are not changing the integral's value with respect to $x$. To unlock the relationship between $f(x)$ and its derivative, you need to differentiate with respect to the variable $x$ that appears in the limits. That is why we first perform a substitution to make $x$ an upper limit of integration.

Q: How do I know to use the substitution $u = \lambda x$?

Whenever you see an integral of the form $\int_0^1 f(\lambda x) d\lambda$, the argument $\lambda x$ is a massive clue. By setting $u = \lambda x$, you transform the argument into a single variable $u$, and the limits of integration naturally become $0$ and $x$. This is a standard, powerful technique in JEE problems to convert integral equations into differential equations.

Q: Wait, why did we get $a=4$? Is it possible for $a$ to be something else?

The value of $a$ is not arbitrary; it is strictly constrained by the boundary conditions provided in the problem. Since $f(x)$ takes the form of a power function $x^n$, there is only one specific exponent $n$ (which depends on $a$) that allows the function to pass through both $(1, 1)$ and $(16, 1/8)$ simultaneously. Solving the resulting algebraic equation for $a$ yields $a=4$ as the unique solution.

Question

Let $f:(0,\infty)\rightarrow \mathbb{R}$ be a twice differentiable function. If for some $a>0$, $\int_{0}^{1}f(\lambda x)d\lambda=af(x)$, $f(1) = 1$ and $f(16)=\frac{1}{8}$, then $16-f^{\prime}(\frac{1}{16})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents an integral equation:
$$\int_{0}^{1} f(\lambda x) d\lambda = a f(x)$$
We are given that $f$ is twice differentiable, with boundary conditions $f(1) = 1$ and $f(16) = \frac{1}{8}$. Our objective is to determine the value of $16 - f^{\prime}(\frac{1}{16})$.

### Phase 1: The Transformation

To simplify the integral, we perform a substitution. Let $u = \lambda x$, which implies $du = x d\lambda$ or $d\lambda = \frac{du}{x}$.

Adjusting the limits of integration: when $\lambda = 0$, $u = 0$; when $\lambda = 1$, $u = x$. Substituting these into the original equation yields:
$$\frac{1}{x} \int_{0}^{x} f(u) du = a f(x)$$

Multiplying both sides by $x$, we obtain the simplified integral form:
$$\int_{0}^{x} f(u) du = ax f(x)$$

### Phase 2: The Calculus Bridge

We apply the **Newton-Leibniz theorem** by differentiating both sides with respect to $x$. The derivative of the left side is simply $f(x)$.

Applying the product rule to the right side, $ax f(x)$, we get:
$$f(x) = a [f(x) + x f^{\prime}(x)]$$

Rearranging the terms to isolate the derivative, we arrive at:
$$(1 - a) f(x) = ax f^{\prime}(x)$$

### Phase 3: Solving the Differential Equation

We now have a first-order separable differential equation:
$$\frac{f^{\prime}(x)}{f(x)} = \frac{1 - a}{ax}$$

Integrating both sides with respect to $x$ gives:
$$\ln f(x) = \frac{1 - a}{a} \ln x + C$$

Exponentiating both sides, we find the general form $f(x) = k x^{\frac{1-a}{a}}$. Using the condition $f(1) = 1$, we determine $k = 1$, resulting in:
$$f(x) = x^{\frac{1-a}{a}}$$

### Phase 4: Finding the Constants

We use the second condition, $f(16) = \frac{1}{8}$, to solve for $a$:
$$16^{\frac{1-a}{a}} = \frac{1}{8}$$

Expressing both sides as powers of $2$:
$$(2^4)^{\frac{1-a}{a}} = 2^{-3}$$

Equating the exponents, we have $4 \left( \frac{1-a}{a} \right) = -3$. Solving for $a$:
$$4 - 4a = -3a \Rightarrow a = 4$$

Substituting $a=4$ back into our function, we get:
$$f(x) = x^{\frac{1-4}{4}} = x^{-\frac{3}{4}}$$

### Phase 5: The Final Victory

We calculate the derivative $f^{\prime}(x)$:
$$f^{\prime}(x) = -\frac{3}{4} x^{-\frac{7}{4}}$$

Evaluating this at $x = \frac{1}{16}$:
$$f^{\prime}\left(\frac{1}{16}\right) = -\frac{3}{4} \left(\frac{1}{16}\right)^{-\frac{7}{4}} = -\frac{3}{4} (16)^{\frac{7}{4}} = -\frac{3}{4} (2^4)^{\frac{7}{4}} = -\frac{3}{4} (2^7) = -96$$

Finally, the requested value is:
$$16 - f^{\prime}\left(\frac{1}{16}\right) = 16 - (-96) = \mathbf{112}$$

Animated Solution for Mathematics - Definite Integration: Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a > 0$ , $\int_{0}^{1} f (λ x) d λ = a f (x)$ , $f (1) = 1$ and $f (16) = \frac{1}{8}$ , then $16 - f^{'} (\frac{1}{16})$ is equal to

Enter Numerical Value:

Visualized Solution

The Integral Equation

Substitution $u = λ x$

Transformed Equation

Differentiating both sides

Applying Product Rule

Separable Differential Equation

Integrating the Equation

Finding Constant $k$

Finding $a$

Solving for $a$ and $f (x)$

Finding the Derivative $f^{'} (x)$

Calculating $f^{'} (\frac{1}{16})$

Final Evaluation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Enter Numerical Value:

Visualized Solution

The Integral Equation

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=λx

Transformed Equation

Differentiating both sides

Applying Product Rule

Separable Differential Equation

Integrating the Equation

Finding Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(161​)