Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f'(0) = 4$ and $f(x) + \int_{0}^{x} (x-t) f'(t) dt = (e^{2x} + e^{-2x}) \cos 2x + \frac{2}{a} x$, then $(2a+1)^5 a^2$ is equal to

### Analyzing the Setup Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a problem that might look like a tangled web of exponentials, trigonometric functions, and integrals. We are given the equation: $$f(x) + \int_{0}^{x} (x-t) f'(t) dt = (e^{2x} + e^{-2x}) \cos 2x + \frac{2}{a} x$$ In the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to strip away that mask and find the constant $a$. ### The Surgical Precision of Leibniz The core challenge here is the integral term $\int_{0}^{x} (x-t) f'(t) dt$. We have the variable $x$ trapped both in the limit and inside the integrand. To break this open, we reach for the **Leibniz Rule** for differentiation under the integral sign. The formula is: $$\frac{d}{dx} \int_{u(x)}^{v(x)} g(x, t) dt = \int_{u(x)}^{v(x)} \frac{\partial}{\partial x} g(x, t) dt + g(x, v(x)) \frac{dv}{dx} - g(x, u(x)) \frac{du}{dx}$$ ### Taming the Left-Hand Side Let us apply this to our left-hand side (LHS). We differentiate $f(x) + \int_{0}^{x} (x-t) f'(t) dt$ with respect to $x$. The derivative of $f(x)$ is $f'(x)$. Applying Leibniz to the integral, the partial derivative of $(x-t)f'(t)$ with respect to $x$ is $f'(t)$. The upper limit term becomes $(x-x)f'(x) = 0$, and the lower limit term is $0$. Thus, the derivative of the integral simplifies to $\int_{0}^{x} f'(t) dt$. By the Fundamental Theorem of Calculus, this is $f(x) - f(0)$. Our LHS derivative is: $$f'(x) + f(x) - f(0)$$ ### The Right-Hand Side and the Substitution Now, we differentiate the right-hand side (RHS): $(e^{2x} + e^{-2x}) \cos 2x + \frac{2}{a} x$. Using the product and chain rules, we obtain: $$\frac{d}{dx} RHS = 2(e^{2x} - e^{-2x}) \cos 2x - 2(e^{2x} + e^{-2x}) \sin 2x + \frac{2}{a}$$ Equating the two sides, we have: $$f'(x) + f(x) - f(0) = 2(e^{2x} - e^{-2x}) \cos 2x - 2(e^{2x} + e^{-2x}) \sin 2x + \frac{2}{a}$$ ### The Final Revelation We are given $f'(0) = 4$. If we substitute $x=0$ into our differentiated equation, the terms $f(x)$ and $f(0)$ cancel out, leaving: $$f'(0) = \frac{2}{a}$$ Since $f'(0) = 4$, we find $4 = \frac{2}{a}$, which implies **$a = \frac{1}{2}$**. Finally, we calculate $(2a+1)^5 a^2$: $$(2(\frac{1}{2}) + 1)^5 (\frac{1}{2})^2 = (1+1)^5 (\frac{1}{4}) = 32 \cdot \frac{1}{4} = 8$$ We have arrived at the answer: **8**.

JEE Main 2022 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

Q: Wait, why can't we just integrate the function $f(x)$ directly?

That is a great question! The problem states that $f$ is a twice differentiable function, but it does not give us the explicit form of $f(x)$. We are dealing with an integral equation, which is a functional equation. Our goal is not to find $f(x)$ itself, but to use the properties of the equation to extract the value of the constant $a$. Differentiating is the most direct path to 'unlock' the information hidden inside the integral.

Enter Numerical Value:

Visualized Solution

Analyzing the Integral Equation

Given equation: $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$
Given condition: $f^{'} (0) = 4$
Objective: Find the value of $(2 a + 1)^{5} a^{2}$ by determining the constant $a$ .

The Leibniz Rule Tool

Leibniz Rule Formula:
$\frac{d}{d x} \int_{u (x)}^{v (x)} g (x, t) d t = \int_{u (x)}^{v (x)} \frac{\partial}{\partial x} g (x, t) d t + g (x, v (x)) \frac{d v}{d x} - g (x, u (x)) \frac{d u}{d x}$

Differentiating the LHS

Differentiating LHS: $\frac{d}{d x} [f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t]$
$= f^{'} (x) + \frac{d}{d x} \int_{0}^{x} (x - t) f^{'} (t) d t$

Simplifying the Integral Derivative

Partial derivative part: $\int_{0}^{x} \frac{\partial}{\partial x} [(x - t) f^{'} (t)] d t = \int_{0}^{x} f^{'} (t) d t$
Upper limit term: $[(x - x) f^{'} (x)] \cdot 1 = 0$
Lower limit term: $[(x - 0) f^{'} (0)] \cdot 0 = 0$
Result: $\frac{d}{d x} \int_{0}^{x} (x - t) f^{'} (t) d t = \int_{0}^{x} f^{'} (t) d t$

Fundamental Theorem of Calculus

Using FTC: $\int_{0}^{x} f^{'} (t) d t = [f (t)]_{0}^{x} = f (x) - f (0)$
Final LHS derivative: $f^{'} (x) + f (x) - f (0)$

Differentiating the RHS

RHS: $(e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$
Derivative: $\frac{d}{d x} [(e^{2 x} + e^{- 2 x}) cos 2 x] + \frac{d}{d x} (\frac{2}{a} x)$
$= 2 (e^{2 x} - e^{- 2 x}) cos 2 x - 2 (e^{2 x} + e^{- 2 x}) sin 2 x + \frac{2}{a}$

The Complete Differentiated Equation

Equating LHS and RHS derivatives:
$f^{'} (x) + f (x) - f (0) = 2 (e^{2 x} - e^{- 2 x}) cos 2 x - 2 (e^{2 x} + e^{- 2 x}) sin 2 x + \frac{2}{a}$

Substituting $x = 0$

Substitute $x = 0$ into the equation:
LHS: $f^{'} (0) + f (0) - f (0) = f^{'} (0)$
RHS: $2 (e^{0} - e^{0}) cos 0 - 2 (e^{0} + e^{0}) sin 0 + \frac{2}{a}$

Solving for $a$

Simplifying RHS at $x = 0$ :
$2 (1 - 1) (1) - 2 (1 + 1) (0) + \frac{2}{a} = 0 - 0 + \frac{2}{a} = \frac{2}{a}$
Equating LHS and RHS: $f^{'} (0) = \frac{2}{a}$
Given $f^{'} (0) = 4 ⟹ 4 = \frac{2}{a} ⟹ a = \frac{1}{2}$

Final Expression Setup

Expression to evaluate: $(2 a + 1)^{5} a^{2}$
Substitute $a = \frac{1}{2}$ :
$(2 (\frac{1}{2}) + 1)^{5} (\frac{1}{2})^{2}$

The Final Answer

Simplify the bracket: $2 (\frac{1}{2}) + 1 = 1 + 1 = 2$
Evaluate powers: $2^{5} \cdot (\frac{1}{2})^{2} = 32 \cdot \frac{1}{4}$
Final calculation: $\frac{32}{4} = 8$
Final Answer: 8

00:00 / 00:00

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a problem that might look like a tangled web of exponentials, trigonometric functions, and integrals.

We are given the equation:

f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x

In the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to strip away that mask and find the constant

a

The Surgical Precision of Leibniz

The core challenge here is the integral term

\int_{0}^{x} (x - t) f^{'} (t) d t

. We have the variable

x

trapped both in the limit and inside the integrand.

To break this open, we reach for the Leibniz Rule for differentiation under the integral sign. The formula is:

\frac{d}{d x} \int_{u (x)}^{v (x)} g (x, t) d t = \int_{u (x)}^{v (x)} \frac{\partial}{\partial x} g (x, t) d t + g (x, v (x)) \frac{d v}{d x} - g (x, u (x)) \frac{d u}{d x}

Taming the Left-Hand Side

Let us apply this to our left-hand side (LHS). We differentiate

f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t

with respect to

x

The derivative of

f (x)

f^{'} (x)

. Applying Leibniz to the integral, the partial derivative of

(x - t) f^{'} (t)

with respect to

x

f^{'} (t)

. The upper limit term becomes

(x - x) f^{'} (x) = 0

, and the lower limit term is

0

Thus, the derivative of the integral simplifies to

\int_{0}^{x} f^{'} (t) d t

. By the Fundamental Theorem of Calculus, this is

f (x) - f (0)

. Our LHS derivative is:

f^{'} (x) + f (x) - f (0)

The Right-Hand Side and the Substitution

Now, we differentiate the right-hand side (RHS):

(e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x

. Using the product and chain rules, we obtain:

\frac{d}{d x} R H S = 2 (e^{2 x} - e^{- 2 x}) cos 2 x - 2 (e^{2 x} + e^{- 2 x}) sin 2 x + \frac{2}{a}

Equating the two sides, we have:

f^{'} (x) + f (x) - f (0) = 2 (e^{2 x} - e^{- 2 x}) cos 2 x - 2 (e^{2 x} + e^{- 2 x}) sin 2 x + \frac{2}{a}

The Final Revelation

We are given

f^{'} (0) = 4

. If we substitute

x = 0

into our differentiated equation, the terms

f (x)

and

f (0)

cancel out, leaving:

f^{'} (0) = \frac{2}{a}

Since

f^{'} (0) = 4

, we find

4 = \frac{2}{a}

, which implies $a = \frac{1}{2}$ .

Finally, we calculate

(2 a + 1)^{5} a^{2}

(2 (\frac{1}{2}) + 1)^{5} (\frac{1}{2})^{2} = (1 + 1)^{5} (\frac{1}{4}) = 32 \cdot \frac{1}{4} = 8

We have arrived at the answer: 8.

Common Traps & FAQs

Wait, why can't we just integrate the function

f (x)

directly?

That is a great question! The problem states that

f

is a twice differentiable function, but it does not give us the explicit form of

f (x)

. We are dealing with an integral equation, which is a functional equation. Our goal is not to find

f (x)

itself, but to use the properties of the equation to extract the value of the constant

a

. Differentiating is the most direct path to 'unlock' the information hidden inside the integral.

How do I know when to use the Leibniz Rule?

Think of the Leibniz Rule as your surgical tool for calculus. Whenever you see an integral where the variable of integration (

t

) and the variable of differentiation (

x

) are mixed together inside the integrand (like in our term

(x - t) f^{'} (t)

), the Leibniz Rule is the standard, most efficient way to handle it. It allows you to differentiate 'through' the integral sign.

Why does the integral term vanish when we differentiate?

It doesn't vanish entirely, but it simplifies beautifully! When you differentiate

\int_{0}^{x} (x - t) f^{'} (t) d t

, the Leibniz Rule gives you two parts: the integral of the partial derivative of the integrand, and the evaluation of the integrand at the limits. Because the integrand

(x - t) f^{'} (t)

becomes zero when

t = x

, the upper limit term vanishes. The lower limit is a constant, so its derivative is zero. This leaves you with a much simpler integral, which you can then solve using the Fundamental Theorem of Calculus.

Animated Solution for Mathematics - Definite Integration: Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Integral Equation

The Leibniz Rule Tool

Differentiating the LHS

Simplifying the Integral Derivative

Fundamental Theorem of Calculus

Differentiating the RHS

The Complete Differentiated Equation

Substituting $x = 0$

Solving for $a$

Final Expression Setup

The Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Surgical Precision of Leibniz

Taming the Left-Hand Side

The Right-Hand Side and the Substitution

The Final Revelation

Similar Questions

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

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t (d t)$ for all $x \in R$ and $f : [0, 1/2] \to [0, \infty)$ be a continuous function. For $α \in [0, 1/2]$ , if $F^{'} (α) + 2$ is the area of the region bounded by $x = 0, y = 0, y = f (x)$ and $x = α$ , then $f (0)$ is

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

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Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

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Enter Numerical Value:

Visualized Solution

Analyzing the Integral Equation

The Leibniz Rule Tool

Differentiating the LHS

Simplifying the Integral Derivative

Fundamental Theorem of Calculus

Differentiating the RHS

The Complete Differentiated Equation

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

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

Final Expression Setup

The Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Surgical Precision of Leibniz

Taming the Left-Hand Side

The Right-Hand Side and the Substitution

The Final Revelation

Similar Questions

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Substituting $x = 0$

Solving for $a$

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

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Let $f : R \to R$ be a differentiable function and $f (1) = 4$ . Then the value of $lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

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Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

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Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then