JEE(ADVANCED)-201

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: 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 $lim_{x \to 0} g (x) =$

Enter Numerical Value:

Visualized Solution

Analyze the Integrand

$g (x) = \int_{x}^{π /2} [f^{'} (t) csc t - cot t csc t f (t)] d t$
We need to evaluate: $lim_{x \to 0} g (x)$
Let's observe the complex expression inside the integral.

Recognize the Pattern

Notice the terms: $f^{'} (t)$ and $f (t)$
Also notice: $csc t$ and its derivative $- csc t cot t$
Recall the Product Rule: $\frac{d}{d t} [u (t) v (t)] = u^{'} (t) v (t) + u (t) v^{'} (t)$

Apply Product Rule

Let $u (t) = f (t)$ and $v (t) = csc t$
$\frac{d}{d t} [f (t) csc t] = f^{'} (t) csc t + f (t) (- csc t cot t)$
$\frac{d}{d t} [f (t) csc t] = f^{'} (t) csc t - f (t) csc t cot t$
The integrand is exactly the derivative of $f (t) csc t$ !

Integrate the Derivative

$g (x) = \int_{x}^{π /2} \frac{d}{d t} [f (t) csc t] d t$
By the Fundamental Theorem of Calculus:
$\int \frac{d}{d t} [F (t)] d t = F (t)$
$g (x) = [f (t) csc t]_{x}^{π /2}$

Evaluate Limits of Integration

$g (x) = f (\frac{π}{2}) csc (\frac{π}{2}) - f (x) csc x$
Since $csc (\frac{π}{2}) = 1$ and $csc x = \frac{1}{s i n x}$ :
$g (x) = f (\frac{π}{2}) - \frac{f ( x )}{s i n x}$

Set Up the Limit

We need to find $lim_{x \to 0} g (x)$
$lim_{x \to 0} g (x) = lim_{x \to 0} (f (\frac{π}{2}) - \frac{f ( x )}{s i n x})$
$f (\frac{π}{2})$ is a constant, so we focus on the second term.
Evaluate: $lim_{x \to 0} \frac{f ( x )}{s i n x}$

Check Indeterminate Form

As $x \to 0$ , $f (x) \to f (0) = 0$
As $x \to 0$ , $sin x \to sin 0 = 0$
$\frac{f ( x )}{s i n x}$ takes the indeterminate form $\frac{0}{0}$
We can apply L'Hopital's Rule.

Apply L'Hopital's Rule

$lim_{x \to 0} \frac{f ( x )}{s i n x} = lim_{x \to 0} \frac{f ^{'} ( x )}{c o s x}$
Substitute $x = 0$ : $\frac{f ^{'} ( 0 )}{c o s 0}$
Given $f^{'} (0) = 1$ and knowing $cos 0 = 1$ :
$\frac{1}{1} = 1$

Final Substitution

$lim_{x \to 0} g (x) = f (\frac{π}{2}) - 1$
The problem states $f^{'} (\frac{π}{2}) = 3$ , which is a known typo. It should be $f (\frac{π}{2}) = 3$ .
Substitute $f (\frac{π}{2}) = 3$ :
$lim_{x \to 0} g (x) = 3 - 1 = 2$

Conclusion

Final Answer: $2$
Key Takeaway: Always look for exact derivatives (Product/Quotient rule) in complex integrands.
Next Challenge: What if the integrand was $f^{'} (t) sec t + f (t) sec t tan t$ ?

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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The given integral is defined as:

g (x) = \int_{x}^{π /2} [f^{'} (t) csc t - cot t csc t f (t)] d t

At first glance, this expression appears complex. However, we must look for underlying relationships between the functions

f (t)

and the trigonometric terms.

The Detective Work

Recall the product rule for differentiation:

\frac{d}{d t} [u (t) v (t)] = u^{'} (t) v (t) + u (t) v^{'} (t)

If we set

u (t) = f (t)

and

v (t) = csc t

, we note that the derivative of

v (t)

v^{'} (t) = - csc t cot t

Substituting these into the product rule, we get:

\frac{d}{d t} [f (t) csc t] = f^{'} (t) csc t - f (t) csc t cot t

This matches the integrand perfectly. The entire expression is simply the derivative of the product

f (t) csc t

The Fundamental Theorem

By the Fundamental Theorem of Calculus, the integral of a derivative is the original function. We evaluate the expression at the given boundaries:

g (x) = [f (t) csc t]_{x}^{π /2}

g (x) = f (π /2) csc (π /2) - f (x) csc x

Since

csc (π /2) = 1

, the expression simplifies to:

g (x) = f (π /2) - \frac{f ( x )}{sin x}

The Final Limit

We now evaluate the limit as

x \to 0

x \to 0 lim g (x) = x \to 0 lim [f (π /2) - \frac{f ( x )}{sin x}]

The term

f (π /2)

is a constant. The term

\frac{f ( x )}{s i n x}

presents an indeterminate form of

\frac{0}{0}

because

f (0) = 0

and

sin 0 = 0

Applying L'Hopital's Rule, we differentiate the numerator and denominator:

x \to 0 lim \frac{f ^{'} ( x )}{cos x} = \frac{f ^{'} ( 0 )}{cos 0} = \frac{1}{1} = 1

Given the condition

f (π /2) = 3

, we perform the final subtraction:

g (0) = 3 - 1 = 2

The final result is 2.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or rather, why is the integrand a product rule?

Great question! It is all about pattern recognition. When you see

f^{'} (t)

paired with a function and

f (t)

paired with the derivative of that same function, it is almost always a product rule. Here,

csc t

is the function, and

- csc t cot t

is its derivative. The negative sign in the integrand matches perfectly with the derivative of

csc t

. It is a classic JEE setup designed to test if you can see the derivative hidden in plain sight.

Can we just use L'Hopital's Rule directly on the integral?

You could, but it is much harder! If you differentiate the integral using Leibniz's Rule, you get the integrand evaluated at the limits. However, evaluating the integral first is much cleaner. By integrating, you turn a complex calculus problem into a simple limit of a function, which is much easier to manage.

The problem says

f^{'} (π /2) = 3

, but that doesn't seem to be used. Is there a typo?

You have a sharp eye! Yes, that is a known typographical error in the original problem. The problem intended to state

f (π /2) = 3

. Without this value, the problem would be unsolvable. In competitive exams, if you encounter such a discrepancy, always look at the structure of the expression—it often reveals the intended value.

Enter Numerical Value:

Visualized Solution

Analyze the Integrand

Recognize the Pattern

Apply Product Rule

Integrate the Derivative

Evaluate Limits of Integration

Set Up the Limit

Check Indeterminate Form

Apply L'Hopital's Rule

Final Substitution

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Detective Work

The Fundamental Theorem

The Final Limit

Similar Questions

Comprehension Passage

The value of $g (\frac{1}{2})$ is

The value of $g^{'} (\frac{1}{2})$ is

Let $f : R \to R$ be a differentiable function such that its derivative $f^{'}$ is continuous and $f (π) = - 6$ . If $F : [0, π] \to R$ is defined by $F (x) = \int_{0}^{x} f (t) d t$ , and if $\int_{0}^{π} (f^{'} (x) + F (x)) cos x d x = 2$ , then the value of $f (0)$ is ____.

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π$ , then the value of $\int_{0}^{π /2} \frac{π ^{2} d t}{f ( t )}$ equals

Let $f : (0, 1) \to R$ be the functions defined as $f (x) = n$ if $x \in [\frac{1}{n + 1}, \frac{1}{n})$ where $n \in N$ . Let $g : (0, 1) \to R$ be a function such that $\int_{x^{2}}^{x} \frac{1 - t}{t} d t < g (x) < 2 x$ for all $x \in (0, 1)$ . Then $lim_{x \to 0} f (x) g (x)$

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

Let the function $f : [0, 2] \to R$ be defined as $f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l n x]}, x \in [0, 1) x \in [1, 2]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is

If $f (x) = {e^{c o s x} sin x, 2, for ∣ x ∣ \leq 2 otherwise$ then $\int_{- 2}^{3} f (x) d x =$

Enter Numerical Value:

Visualized Solution

Analyze the Integrand

Recognize the Pattern

Apply Product Rule

Integrate the Derivative

Evaluate Limits of Integration

Set Up the Limit

Check Indeterminate Form

Apply L'Hopital's Rule

Final Substitution

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Detective Work

The Fundamental Theorem

The Final Limit

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of g(21​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of g′(21​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0x​f(t)dt, where f is such that 21​≤f(t)≤1 for t∈[0,1] and 0≤f(t)≤21​ for t∈[1,2]. Then g(2) satisfies the inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2), f(0)=1 and f(2)=e2. Then the value of ∫02​f(x)dx is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(t)=∫0π​1−cos2tsin2x2xdx​,0<t<π, then the value of ∫0π/2​f(t)π2dt​ equals

.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 f(x)={ecosxsinx,2,​for ∣x∣≤2otherwise​ then ∫−23​f(x)dx=

The value of $g (\frac{1}{2})$ is

The value of $g^{'} (\frac{1}{2})$ is

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π$ , then the value of $\int_{0}^{π /2} \frac{π ^{2} d t}{f ( t )}$ equals

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 $f (x) = {e^{c o s x} sin x, 2, for ∣ x ∣ \leq 2 otherwise$ then $\int_{- 2}^{3} f (x) d x =$