JEE Main 2019 (10 January)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Q: Why can't I just integrate $f(t)$ to solve this?

That is a classic trap! We are solving for the function $f(t)$ itself. If we try to integrate it, we would need to know what $f(t)$ looks like beforehand. We don't! The integral is a container holding the function. Differentiation is the tool that 'breaks' the container, allowing us to see the function inside. Always look for the derivative when you see an integral equation with variable limits.

Q: Can I just plug in $x=1/2$ at the very beginning?

No, absolutely not! If you plug in $x=1/2$ before differentiating, you are just evaluating the integral at a specific point, which gives you a constant value, not the function $f(x)$. You must differentiate first to find the expression for $f(x)$, and only then can you find $f'(x)$. Think of it as peeling an onion—you have to remove the integral layer before you can get to the function layer.

Select Answer:

Progressive Hints

Hint 1

Look at the structure of the equation:

\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t

. You have an unknown function

f (t)

trapped inside an integral. Do not attempt to integrate

f (t)

directly; you don't know its form yet. Instead, treat the integral as a function of its limits. What operator can 'undo' an integral?

Hint 2

Recall the Newton-Leibniz rule. It is the bridge between integration and differentiation. The formula is

\frac{d}{d x} \int_{g (x)}^{h (x)} F (t) d t = F (h (x)) h^{'} (x) - F (g (x)) g^{'} (x)

. Apply this to both sides of your equation. This is the key to unlocking the function

f (x)

Hint 3

When you differentiate the right-hand side, be extremely careful with the integral

\int_{x}^{1} t^{2} f (t) d t

. Since

x

is the lower limit, the derivative will introduce a negative sign. Your starting equation after differentiation should look like

f (x) = 2 x - x^{2} f (x)

. From here, it is just simple algebra to isolate

f (x)

Visualized Solution

Analyze the Integral Equation

Given equation: $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$
Goal: Find $f^{'} (\frac{1}{2})$
Strategy: Differentiate both sides with respect to $x$ using the Newton-Leibniz Rule.

The Newton-Leibniz Rule

Formula: $\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$
This rule allows us to differentiate an integral with variable limits.

Differentiate the LHS

Applying Leibniz Rule to LHS: $\frac{d}{d x} \int_{0}^{x} f (t) d t$
LHS Derivative: $f (x) \cdot \frac{d}{d x} (x) - f (0) \cdot \frac{d}{d x} (0)$
Result: $f (x) \cdot 1 - 0 = f (x)$

Differentiate the RHS

Applying Leibniz Rule to RHS: $\frac{d}{d x} (x^{2} + \int_{x}^{1} t^{2} f (t) d t)$
Derivative of $x^{2}$ : $2 x$
Derivative of $\int_{x}^{1} t^{2} f (t) d t$ : $1^{2} f (1) \cdot (0) - x^{2} f (x) \cdot (1)$
RHS Derivative: $2 x - x^{2} f (x)$

Form the Algebraic Equation

Equating the derivatives of both sides:
$f (x) = 2 x - x^{2} f (x)$

Isolate $f (x)$

Rearrange to group $f (x)$ terms:
$f (x) + x^{2} f (x) = 2 x$
Factor out $f (x)$ :
$f (x) (1 + x^{2}) = 2 x$
Solve for $f (x)$ :
$f (x) = \frac{2 x}{1 + x ^{2}}$

Apply the Quotient Rule

To find $f^{'} (x)$ , use the Quotient Rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$
Let $u = 2 x ⟹ u^{'} = 2$
Let $v = 1 + x^{2} ⟹ v^{'} = 2 x$

Differentiate $f (x)$

Substitute into the formula:
$f^{'} (x) = \frac{2 ( 1 + x ^{2} ) - ( 2 x ) ( 2 x )}{( 1 + x ^{2} ) ^{2}}$

Simplify $f^{'} (x)$

Expand the numerator:
$f^{'} (x) = \frac{2 + 2 x ^{2} - 4 x ^{2}}{( 1 + x ^{2} ) ^{2}}$
Combine like terms:
$f^{'} (x) = \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}}$

Substitute $x = \frac{1}{2}$

Substitute $x = \frac{1}{2}$ into $f^{'} (x)$ :
$f^{'} (\frac{1}{2}) = \frac{2 - 2 ( \frac{1}{2} ) ^{2}}{( 1 + ( \frac{1}{2} ) ^{2} ) ^{2}}$

Evaluate Numerator and Denominator

Numerator: $2 - 2 (\frac{1}{4}) = 2 - \frac{1}{2} = \frac{3}{2}$
Denominator: $(1 + \frac{1}{4})^{2} = (\frac{5}{4})^{2} = \frac{25}{16}$

Final Computation

Combine the results:
$f^{'} (\frac{1}{2}) = \frac{3/2}{25/16}$
Multiply by the reciprocal:
$f^{'} (\frac{1}{2}) = \frac{3}{2} \cdot \frac{16}{25} = \frac{24}{25}$
Final Answer: Option (2) is correct.

00:00 / 00:00

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

We are tasked with solving for

f (t)

in the following integral equation:

\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t

This equation presents a function trapped within integral boundaries. To liberate

f (t)

, we must employ the Newton-Leibniz rule.

The Magic Wand

The Newton-Leibniz Rule

The Newton-Leibniz rule provides the bridge between accumulation and change. It is defined as:

\frac{d}{d x} \int_{g (x)}^{h (x)} F (t) d t = F (h (x)) h^{'} (x) - F (g (x)) g^{'} (x)

Applying this to the left side,

\int_{0}^{x} f (t) d t

, the derivative with respect to

x

simplifies elegantly to

f (x)

The Surgery

Differentiating the Right Side

Now, we differentiate the right side,

x^{2} + \int_{x}^{1} t^{2} f (t) d t

. The derivative of

x^{2}

2 x

For the integral term

\int_{x}^{1} t^{2} f (t) d t

, the upper limit is a constant (

1

), so its derivative is

0

. The lower limit is

x

, so we subtract the integrand evaluated at

x

\frac{d}{d x} \int_{x}^{1} t^{2} f (t) d t = 0 - (x^{2} f (x)) (1) = - x^{2} f (x)

Combining these, our equation transforms into a simple algebraic form:

f (x) = 2 x - x^{2} f (x)

The Algebraic Cleanup

To isolate

f (x)

, we move all terms containing the function to one side:

f (x) + x^{2} f (x) = 2 x

Factoring out

f (x)

, we obtain:

f (x) (1 + x^{2}) = 2 x

Dividing by

(1 + x^{2})

, we find the explicit form of our function:

f (x) = \frac{2 x}{1 + x ^{2}}

The Final Sprint

The Quotient Rule

To find

f^{'} (1/2)

, we apply the quotient rule,

\frac{d}{d x} (\frac{u}{v}) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}

, where

u = 2 x

and

v = 1 + x^{2}

. Given

u^{'} = 2

and

v^{'} = 2 x

, we have:

f^{'} (x) = \frac{2 ( 1 + x ^{2} ) - ( 2 x ) ( 2 x )}{( 1 + x ^{2} ) ^{2}}

Simplifying the numerator:

f^{'} (x) = \frac{2 + 2 x ^{2} - 4 x ^{2}}{( 1 + x ^{2} ) ^{2}} = \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}}

The Victory Lap

Finally, we substitute

x = 1/2

into the derivative:

f^{'} (1/2) = \frac{2 - 2 ( 1/4 )}{( 1 + 1/4 ) ^{2}} = \frac{2 - 1/2}{( 5/4 ) ^{2}}

This simplifies to:

f^{'} (1/2) = \frac{3/2}{25/16} = \frac{3}{2} \cdot \frac{16}{25} = \frac{24}{25}

The final result is 24/25.

Common Traps & FAQs

Why can't I just integrate

f (t)

to solve this?

That is a classic trap! We are solving for the function

f (t)

itself. If we try to integrate it, we would need to know what

f (t)

looks like beforehand. We don't! The integral is a container holding the function. Differentiation is the tool that 'breaks' the container, allowing us to see the function inside. Always look for the derivative when you see an integral equation with variable limits.

Why is there a negative sign in the derivative of the integral

\int_{x}^{1} t^{2} f (t) d t

Great observation! The Newton-Leibniz rule states that the derivative of

\int_{g (x)}^{h (x)} F (t) d t

F (h (x)) h^{'} (x) - F (g (x)) g^{'} (x)

. Here, our lower limit is

g (x) = x

, so

g^{'} (x) = 1

. The term

F (g (x)) g^{'} (x)

becomes

F (x) \cdot 1

. Because it is the lower limit, it is subtracted. That negative sign is crucial—forgetting it is the most common mistake in JEE Advanced problems!

Can I just plug in

x = 1/2

at the very beginning?

No, absolutely not! If you plug in

x = 1/2

before differentiating, you are just evaluating the integral at a specific point, which gives you a constant value, not the function

f (x)

. You must differentiate first to find the expression for

f (x)

, and only then can you find

f^{'} (x)

. Think of it as peeling an onion—you have to remove the integral layer before you can get to the function layer.

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Select Answer:

Visualized Solution

Analyze the Integral Equation

The Newton-Leibniz Rule

Differentiate the LHS

Differentiate the RHS

Form the Algebraic Equation

Isolate $f (x)$

Apply the Quotient Rule

Differentiate $f (x)$

Simplify $f^{'} (x)$

Substitute $x = \frac{1}{2}$

Evaluate Numerator and Denominator

Final Computation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Magic Wand

The Surgery

The Algebraic Cleanup

The Final Sprint

The Victory Lap

Similar Questions

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

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

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

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-

Let $f$ be a differentiable function in $(0, \frac{π}{2})$ . If $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$ , then $\frac{1}{3} f^{'} (\frac{1}{3})$ 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 $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

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

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0x​f(t)dt=x2+∫x1​t2f(t)dt, then f′(1/2) is :

Select Answer:

Visualized Solution

Analyze the Integral Equation

The Newton-Leibniz Rule

Differentiate the LHS

Differentiate the RHS

Form the Algebraic Equation

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

Apply the Quotient Rule

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

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

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=21​

Evaluate Numerator and Denominator

Final Computation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Magic Wand

The Surgery

The Algebraic Cleanup

The Final Sprint

The Victory Lap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0x​f(t)dt=x+∫x1​tf(t)dt, then the value of f(1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→R and F(x)=∫0x​f(t)dt. If F(x2)=x2(1+x), then f(4) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function and f(1)=4. Then the value of limx→1​∫4f(x)​x−12t​dt is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R is a differentiable function and f(2)=6, then limx→2​x−2∫6f(x)​2tdt​ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a differentiable function in (0,2π​). If ∫cosx1​t2f(t)dt=sin3x+cosx, then 3​1​f′(3​1​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a continuous function satisfying ∫0t2​(f(x)+x2)dx=34​t3,∀t>0. Then f(4π2​) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a non-negative function defined on the interval [0, 1]. If ∫0x​1−(f′(t))2​dt=∫0x​f(t)dt, 0≤x≤1, and f(0)=0, then

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Isolate $f (x)$

Differentiate $f (x)$

Simplify $f^{'} (x)$

Substitute $x = \frac{1}{2}$

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

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

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

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-

Let $f$ be a differentiable function in $(0, \frac{π}{2})$ . If $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$ , then $\frac{1}{3} f^{'} (\frac{1}{3})$ is equal to :

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 $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

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