JEE Advanced 2015

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f^{'} (x) = \frac{192 x ^{3}}{2 + s i n ^{4} π x}$ for all $x \in R$ with $f (\frac{1}{2}) = 0$ . If $m \leq \int_{1/2}^{1} f (x) d x \leq M$ , then the possible values of $m$ and $M$ are

Q: Why can't I just integrate $f'(x)$ directly to find $f(x)$?

That is a great question! If you try to integrate $\frac{192x^3}{2 + \sin^4(\pi x)}$, you will quickly realize that there is no elementary antiderivative for this expression. JEE Advanced often includes problems that look like standard integration tasks but are actually designed to test your ability to use bounding techniques and inequalities instead of brute-force calculus.

Q: My calculated interval is $[2.6, 3.9]$. How do I choose the correct option?

The problem asks for $m$ and $M$ such that $m \le \int_{1/2}^1 f(x) dx \le M$. This means the interval $[2.6, 3.9]$ must be a subset of the interval $[m, M]$. You are looking for the option where $m \le 2.6$ and $M \ge 3.9$. Option 4, which is $[1, 12]$, is the only one that contains our calculated range.

Select Answer:

Visualized Solution

Understanding the Problem Setup

We are given the derivative: $f^{'} (x) = \frac{192 x ^{3}}{2 + s i n ^{4} π x}$
We also have an initial condition: $f (\frac{1}{2}) = 0$
Our goal is to find the bounds $m$ and $M$ such that: $m \leq \int_{1/2}^{1} f (x) d x \leq M$

Bounding the Trigonometric Term

Direct integration of $f^{'} (x)$ is extremely difficult due to the $sin^{4} π x$ term.
Instead, we can bound the denominator using the standard range of the sine function.
For any real $x$ , we know: $0 \leq sin^{4} π x \leq 1$

Adding the Constant Term

We have: $0 \leq sin^{4} π x \leq 1$
Add $2$ to all parts of the inequality:
$2 + 0 \leq 2 + sin^{4} π x \leq 2 + 1$
This simplifies to: $2 \leq 2 + sin^{4} π x \leq 3$

Inverting the Inequality

We have: $2 \leq 2 + sin^{4} π x \leq 3$
Taking the reciprocal reverses the inequality signs:
$\frac{1}{3} \leq \frac{1}{2 + s i n ^{4} π x} \leq \frac{1}{2}$

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

Multiply the entire inequality by $192 x^{3}$ :
$\frac{192 x ^{3}}{3} \leq \frac{192 x ^{3}}{2 + s i n ^{4} π x} \leq \frac{192 x ^{3}}{2}$
Since $x \in [\frac{1}{2}, 1]$ , $192 x^{3}$ is strictly positive, so the inequality direction remains unchanged.
Simplifying the constants: $64 x^{3} \leq f^{'} (x) \leq 96 x^{3}$

Integrating the Lower Bound

To find $f (x)$ , we integrate the lower bound $64 t^{3}$ from $\frac{1}{2}$ to $x$ :
$\int_{1/2}^{x} 64 t^{3} d t \leq \int_{1/2}^{x} f^{'} (t) d t$
Using the fundamental theorem of calculus:
$[16 t^{4}]_{1/2}^{x} \leq f (x) - f (\frac{1}{2})$
Since $f (\frac{1}{2}) = 0$ , we get: $16 x^{4} - 16 (\frac{1}{16}) \leq f (x) ⟹ 16 x^{4} - 1 \leq f (x)$

Integrating the Upper Bound

Similarly, integrate the upper bound $96 t^{3}$ from $\frac{1}{2}$ to $x$ :
$\int_{1/2}^{x} f^{'} (t) d t \leq \int_{1/2}^{x} 96 t^{3} d t$
$f (x) - f (\frac{1}{2}) \leq [24 t^{4}]_{1/2}^{x}$
Using $f (\frac{1}{2}) = 0$ : $f (x) \leq 24 x^{4} - 24 (\frac{1}{16}) ⟹ f (x) \leq 24 x^{4} - 1.5$

Bounded Function Curves

We have established the function bounds:
$16 x^{4} - 1 \leq f (x) \leq 24 x^{4} - 1.5$
Both curves start at $(0.5, 0)$ and diverge as $x$ increases towards $1$ .

Integrating the Lower Bound from $\frac{1}{2}$ to $1$

Lower bound integral: $I_{l o w er} = \int_{1/2}^{1} (16 x^{4} - 1) d x$
$I_{l o w er} = [\frac{16}{5} x^{5} - x]_{1/2}^{1}$
At $x = 1$ : $\frac{16}{5} (1) - 1 = 3.2 - 1 = 2.2$
At $x = \frac{1}{2}$ : $\frac{16}{5} (\frac{1}{32}) - \frac{1}{2} = 0.1 - 0.5 = - 0.4$
$I_{l o w er} = 2.2 - (- 0.4) = 2.6$

Integrating the Upper Bound from $\frac{1}{2}$ to $1$

Upper bound integral: $I_{u pp er} = \int_{1/2}^{1} (24 x^{4} - 1.5) d x$
$I_{u pp er} = [\frac{24}{5} x^{5} - 1.5 x]_{1/2}^{1}$
At $x = 1$ : $\frac{24}{5} (1) - 1.5 = 4.8 - 1.5 = 3.3$
At $x = \frac{1}{2}$ : $\frac{24}{5} (\frac{1}{32}) - 1.5 (\frac{1}{2}) = 0.15 - 0.75 = - 0.6$
$I_{u pp er} = 3.3 - (- 0.6) = 3.9$

Combining the Integral Bounds

We have found: $2.6 \leq \int_{1/2}^{1} f (x) d x \leq 3.9$
We are given: $m \leq \int_{1/2}^{1} f (x) d x \leq M$
Therefore, the interval $[2.6, 3.9]$ must be a subset of $[m, M]$ .

Finding the Correct Option

Let's check the options for $[m, M]$ :
Option 1: $[13, 24]$ (Does not contain $[2.6, 3.9]$ )
Option 2: $[\frac{1}{4}, \frac{1}{2}] = [0.25, 0.5]$ (Does not contain $[2.6, 3.9]$ )
Option 3: $[- 11, 0]$ (Does not contain $[2.6, 3.9]$ )
Option 4: $[1, 12]$ (Contains $[2.6, 3.9]$ since $1 \leq 2.6$ and $3.9 \leq 12$ )
Correct Option: $m = 1, M = 12$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

We are given the derivative

f^{'} (x) = \frac{192 x ^{3}}{2 + s i n ^{4} ( π x )}

. The goal is to determine the bounds for the integral of

f (x)

over the interval

[1/2, 1]

, given

f (1/2) = 0

Many students panic when encountering such a complex denominator. However, in JEE Advanced, an "impossible" integral is often a signal to use inequality bounding rather than direct integration.

The Art of Bounding

Consider the denominator

2 + sin^{4} (π x)

. We know that for any real

x

, the range of

sin (π x)

[- 1, 1]

. Consequently, the range of

sin^{4} (π x)

[0, 1]

By adding

2

to this range, we trap the denominator:

2 \leq 2 + sin^{4} (π x) \leq 3

Taking the reciprocal of these positive values reverses the inequality signs:

\frac{1}{3} \leq \frac{1}{2 + sin ^{4} ( π x )} \leq \frac{1}{2}

The Calculus Bridge

Now, we multiply this inequality by the numerator

192 x^{3}

. Since we are working in the interval

[1/2, 1]

x

is positive, ensuring the inequality signs remain unchanged:

64 x^{3} \leq f^{'} (x) \leq 96 x^{3}

To find

f (x)

, we integrate these bounds from

1/2

x

. Given the initial condition

f (1/2) = 0

, we perform the integration:

\int_{1/2}^{x} 64 t^{3} d t \leq f (x) \leq \int_{1/2}^{x} 96 t^{3} d t

This yields the following polynomial bounds for

f (x)

16 x^{4} - 1 \leq f (x) \leq 24 x^{4} - 1.5

The Final Integration

We now integrate these bounds over the interval

[1/2, 1]

to find the range of the integral of

f (x)

For the lower bound:

\int_{1/2}^{1} (16 x^{4} - 1) d x = [\frac{16}{5} x^{5} - x]_{1/2}^{1} = (3.2 - 1) - (0.1 - 0.5) = 2.2 + 0.4 = 2.6

For the upper bound:

\int_{1/2}^{1} (24 x^{4} - 1.5) d x = [\frac{24}{5} x^{5} - 1.5 x]_{1/2}^{1} = (4.8 - 1.5) - (0.15 - 0.75) = 3.3 + 0.6 = 3.9

The value of the integral is trapped in the interval

[2.6, 3.9]

. Any option

[m, M]

that contains this range is valid. Option 4, $[1, 12]$ , is the correct choice as it satisfies

1 \leq 2.6

and

3.9 \leq 12

Common Traps & FAQs

Why can't I just integrate

f^{'} (x)

directly to find

f (x)

That is a great question! If you try to integrate

\frac{192 x ^{3}}{2 + s i n ^{4} ( π x )}

, you will quickly realize that there is no elementary antiderivative for this expression. JEE Advanced often includes problems that look like standard integration tasks but are actually designed to test your ability to use bounding techniques and inequalities instead of brute-force calculus.

Why does the inequality sign flip when I take the reciprocal?

Think of it as a seesaw. If you have two positive numbers, say

2

and

3

, we know

2 < 3

. However, when you look at their reciprocals,

\frac{1}{2}

0.5

and

\frac{1}{3}

is approximately

0.33

. Clearly,

0.5 > 0.33

. Whenever you take the reciprocal of positive values, the order of the inequality must reverse. It is a classic trap, so stay vigilant!

My calculated interval is

[2.6, 3.9]

. How do I choose the correct option?

The problem asks for

m

and

M

such that

m \leq \int_{1/2}^{1} f (x) d x \leq M

. This means the interval

[2.6, 3.9]

must be a subset of the interval

[m, M]

. You are looking for the option where

m \leq 2.6

and

M \geq 3.9

. Option 4, which is

[1, 12]

, is the only one that contains our calculated range.

Animated Solution for Mathematics - Definite Integration: Let $f^{'} (x) = \frac{192 x ^{3}}{2 + s i n ^{4} π x}$ for all $x \in R$ with $f (\frac{1}{2}) = 0$ . If $m \leq \int_{1/2}^{1} f (x) d x \leq M$ , then the possible values of $m$ and $M$ are

Select Answer:

Visualized Solution

Understanding the Problem Setup

Bounding the Trigonometric Term

Adding the Constant Term

Inverting the Inequality

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

Integrating the Lower Bound

Integrating the Upper Bound

Bounded Function Curves

Integrating the Lower Bound from $\frac{1}{2}$ to $1$

Integrating the Upper Bound from $\frac{1}{2}$ to $1$

Combining the Integral Bounds

Finding the Correct Option

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Bounding

The Calculus Bridge

The Final Integration

Similar Questions

If $f (x) = A sin (\frac{π x}{2}) + B$ , $f^{'} (\frac{1}{2}) = 2$ and $\int_{0}^{1} f (x) d x = \frac{2 A}{π}$ , then constants $A$ and $B$ are

Let $f : R \to R$ be a differentiable function such that $f (0) = 0$ , $f^{'} (\frac{π}{2}) = 3$ and $f^{'} (0) = 1$ . If $g (x) = \int_{x}^{π /2} [f^{'} (t) cosec t - cot t cosec t f (t)] d t$ for $x \in (0, \frac{π}{2})$ , then $lim_{x \to 0} g (x) =$

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

If $\int_{0}^{π} x f (sin x) d x = A \int_{0}^{π /2} f (sin x) d x$ , then $A$ is

Comprehension Passage

The correct statement(s) is(are)

If $\int_{1}^{3} x^{2} F^{'} (x) d x = - 12$ and $\int_{1}^{3} x^{3} F^{''} (x) d x = 40$ , then the correct expression(s) is (are)

If $I = \int_{0}^{\frac{π}{2}} \frac{s i n ^{\frac{3}{2}} x}{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x} d x$ , then $\int_{0}^{\frac{π}{2}} \frac{x s i n x c o s x}{s i n ^{4} x + c o s ^{4} x} d x$ equals :

If $24 \int_{0}^{\frac{π}{4}} (sin ∣4 x - \frac{π}{12} ∣ + [2 sin x]) d x = 2 π + α,$ where [ ] denotes the greatest integer function, then a is equal to

$\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ is

Select Answer:

Visualized Solution

Understanding the Problem Setup

Bounding the Trigonometric Term

Adding the Constant Term

Inverting the Inequality

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

Integrating the Lower Bound

Integrating the Upper Bound

Bounded Function Curves

Integrating the Lower Bound from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

Integrating the Upper Bound from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

Combining the Integral Bounds

Finding the Correct Option

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Bounding

The Calculus Bridge

The Final Integration

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=Asin(2πx​)+B, f′(21​)=2​ and ∫01​f(x)dx=π2A​, then constants A and B are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying f(x)+f(π−x)=π2,∀x∈R. Then ∫0π​f(x)sinxdx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0π​xf(sinx)dx=A∫0π/2​f(sinx)dx, then A is

Comprehension Passage

The correct statement(s) is(are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫13​x2F′(x)dx=−12 and ∫13​x3F′′(x)dx=40, then the correct expression(s) is (are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If I=∫02π​​sin23​x+cos23​xsin23​x​dx, then ∫02π​​sin4x+cos4xx sin x cos x​dx equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 24∫04π​​(sin∣4x−12π​∣+[2 sin x])dx=2π+α, where [ ] denotes the greatest integer function, then a is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫−ππ​1+cos2x2x(1+sinx)​dx is

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

Integrating the Lower Bound from $\frac{1}{2}$ to $1$

Integrating the Upper Bound from $\frac{1}{2}$ to $1$

If $f (x) = A sin (\frac{π x}{2}) + B$ , $f^{'} (\frac{1}{2}) = 2$ and $\int_{0}^{1} f (x) d x = \frac{2 A}{π}$ , then constants $A$ and $B$ are

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

If $\int_{0}^{π} x f (sin x) d x = A \int_{0}^{π /2} f (sin x) d x$ , then $A$ is

If $\int_{1}^{3} x^{2} F^{'} (x) d x = - 12$ and $\int_{1}^{3} x^{3} F^{''} (x) d x = 40$ , then the correct expression(s) is (are)

If $I = \int_{0}^{\frac{π}{2}} \frac{s i n ^{\frac{3}{2}} x}{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x} d x$ , then $\int_{0}^{\frac{π}{2}} \frac{x s i n x c o s x}{s i n ^{4} x + c o s ^{4} x} d x$ equals :

If $24 \int_{0}^{\frac{π}{4}} (sin ∣4 x - \frac{π}{12} ∣ + [2 sin x]) d x = 2 π + α,$ where [ ] denotes the greatest integer function, then a is equal to

$\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ is