JEE Main 2022 (27 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: 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 :

Q: Can we just divide by $\sin x$ when simplifying the equation?

Yes, we can! We are given that $x \in (0, \frac{\pi}{2})$. In this interval, $\sin x$ is strictly positive and never zero. Since we are not dividing by zero, the operation is perfectly valid. Always check the domain before performing such steps in JEE problems.

Q: Why did we choose $t = \cos x$ as the substitution?

We chose $t = \cos x$ because our equation was in terms of $f(\cos x)$. By setting $t = \cos x$, we transform the function into $f(t)$, which is exactly what we need to differentiate. It bridges the gap between the trigonometric world of $x$ and the algebraic world of $t$.

Select Answer:

Progressive Hints

Hint 1

The Spark: You are looking at an integral equation where the function

f (t)

is trapped inside the integral. To 'free' it, you need to differentiate both sides with respect to

x

. Remember the Newton-Leibniz Rule:

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

Hint 2

The Arsenal: Once you differentiate, you will have an equation involving

f (cos x)

. To find

f (t)

, you need to perform a change of variable. Let

t = cos x

. You will need to express

tan x

and

sec^{2} x

in terms of

t

using trigonometric identities like

tan^{2} x = sec^{2} x - 1

Hint 3

The Execution: After substituting

t = cos x

, you will arrive at an explicit algebraic expression for

f (t)

. Differentiate this expression with respect to

t

using the quotient rule, and finally, substitute

t = \frac{1}{3}

into your derivative

f^{'} (t)

Visualized Solution

The Problem Setup & Newton-Leibniz Rule

Given equation: $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$
We need to find $\frac{1}{3} f^{'} (\frac{1}{3})$ .
To extract $f (t)$ from inside the integral, we differentiate both sides with respect to $x$ .
Newton-Leibniz Rule: $\frac{d}{d x} (\int_{g (x)}^{h (x)} w (t) d t) = w (h (x)) \cdot h^{'} (x) - w (g (x)) \cdot g^{'} (x)$

Differentiating the Left Hand Side (LHS)

LHS: $\frac{d}{d x} \int_{c o s x}^{1} t^{2} f (t) d t$
Upper limit is a constant ( $1$ ), so its derivative is $0$ .
Lower limit is $cos x$ , its derivative is $- sin x$ .
Applying the rule: $= (1)^{2} f (1) \cdot (0) - (cos x)^{2} f (cos x) \cdot (- sin x)$
Simplifies to: $sin x cos^{2} x f (cos x)$

Differentiating the Right Hand Side (RHS)

RHS: $\frac{d}{d x} (sin^{3} x + cos x)$
Use chain rule for $sin^{3} x$ : $3 sin^{2} x \cdot cos x$
Derivative of $cos x$ is $- sin x$
Result: $3 sin^{2} x cos x - sin x$

Equating and Simplifying

Equating LHS and RHS: $sin x cos^{2} x f (cos x) = 3 sin^{2} x cos x - sin x$
Factor out $sin x$ on RHS: $sin x (3 sin x cos x - 1)$
Since $x \in (0, \frac{π}{2})$ , $sin x \neq = 0$ . We can safely divide both sides by $sin x$ .
$cos^{2} x f (cos x) = 3 sin x cos x - 1$

Isolating $f (cos x)$

Divide by $cos^{2} x$ to isolate $f (cos x)$ :
$f (cos x) = \frac{3 s i n x c o s x}{c o s ^{2} x} - \frac{1}{c o s ^{2} x}$
Simplify the terms:
$f (cos x) = 3 tan x - sec^{2} x$

Variable Substitution ( $t = cos x$ )

Let $t = cos x$ . We need to express $tan x$ and $sec^{2} x$ in terms of $t$ .
$sec x = \frac{1}{t} ⟹ sec^{2} x = \frac{1}{t ^{2}}$
Using identity $tan^{2} x = sec^{2} x - 1$ :
$tan x = \frac{1}{t ^{2}} - 1 = \frac{1 - t ^{2}}{t}$ (Positive since $x \in (0, \frac{π}{2})$ )

Expressing $f (t)$ Explicitly

Substitute these back into our function:
$f (t) = 3 (\frac{1 - t ^{2}}{t}) - \frac{1}{t ^{2}}$
Now we have a clear, explicit function for $f (t)$ .

Differentiating $f (t)$ - First Term

We need $f^{'} (t)$ . Let's differentiate $f (t) = 3 \frac{1 - t ^{2}}{t} - t^{- 2}$
First term derivative using quotient rule:
$\frac{d}{d t} (\frac{1 - t ^{2}}{t}) = \frac{t \cdot \frac{- 2 t}{2 1 - t ^{2}} - 1 - t ^{2} \cdot 1}{t ^{2}}$
$= \frac{\frac{- t ^{2}}{1 - t ^{2}} - 1 - t ^{2}}{t ^{2}}$

Simplifying the First Term Derivative

Simplify the numerator by taking common denominator:
$= \frac{- t ^{2} - ( 1 - t ^{2} )}{t ^{2} 1 - t ^{2}} = \frac{- 1}{t ^{2} 1 - t ^{2}}$
Multiply by the constant $3$ :
First part of $f^{'} (t)$ is $\frac{- 3}{t ^{2} 1 - t ^{2}}$

Differentiating Second Term and Combining

Second term derivative: $\frac{d}{d t} (- t^{- 2}) = - (- 2) t^{- 3} = \frac{2}{t ^{3}}$
Combine both parts to get the full derivative $f^{'} (t)$ :
$f^{'} (t) = \frac{- 3}{t ^{2} 1 - t ^{2}} + \frac{2}{t ^{3}}$

Substituting $t = \frac{1}{3}$

We need to evaluate at $t = \frac{1}{3}$ .
Pre-calculate components:
$t^{2} = \frac{1}{3}$
$t^{3} = \frac{1}{3 3}$
$1 - t^{2} = 1 - \frac{1}{3} = \frac{2}{3}$

Computing $f^{'} (\frac{1}{3})$

Substitute the components into $f^{'} (t)$ :
$f^{'} (\frac{1}{3}) = \frac{- 3}{( \frac{1}{3} ) \frac{2}{3}} + \frac{2}{\frac{1}{3 3}}$
Simplify the fractions:
$= \frac{- 9}{\frac{2}{3}} + 63 = - 9 \frac{3}{2} + 63$

Final Calculation

The question asks for $\frac{1}{3} f^{'} (\frac{1}{3})$ .
Multiply our result by $\frac{1}{3}$ :
$= \frac{1}{3} (- 9 \frac{3}{2} + 63)$
Distribute $\frac{1}{3}$ :
$= - \frac{9}{2} + 6 = 6 - \frac{9}{2}$

Conclusion

Final Answer: $6 - \frac{9}{2}$
Key Takeaways:
1. Use Newton-Leibniz Rule to differentiate integrals with variable limits.
2. Always check the domain (here $x \in (0, \frac{π}{2})$ ) before canceling terms or taking square roots.
3. Careful algebraic substitution simplifies complex functional equations.

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

We are tasked with finding the derivative of a function

f (t)

defined implicitly by the integral equation:

\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x

To extract

f (t)

, we employ the Leibniz Integral Rule. This rule allows us to differentiate an integral with respect to its variable limits.

The Master Equation

Differentiating both sides of the equation with respect to

x

yields:

- (cos x)^{2} f (cos x) \cdot \frac{d}{d x} (cos x) = 3 sin^{2} x cos x - sin x

Since

\frac{d}{d x} (cos x) = - sin x

, the equation simplifies to:

sin x cos^{2} x f (cos x) = 3 sin^{2} x cos x - sin x

Given

x \in (0, \frac{π}{2})

, we know $\sin x eq 0$. Dividing both sides by

sin x

, we obtain:

cos^{2} x f (cos x) = 3 sin x cos x - 1

Solving for the Function

To isolate

f (cos x)

, we divide by

cos^{2} x

f (cos x) = 3 \frac{sin x}{cos x} - \frac{1}{cos ^{2} x} = 3 tan x - sec^{2} x

Substituting

t = cos x

, we note that

tan x = \frac{1 - t ^{2}}{t}

and

sec^{2} x = \frac{1}{t ^{2}}

. Thus, the function is:

f (t) = 3 \frac{1 - t ^{2}}{t} - \frac{1}{t ^{2}}

Final Calculation

To find

f^{'} (t)

, we differentiate with respect to

t

f^{'} (t) = 3 [\frac{t \cdot \frac{- t}{1 - t ^{2}} - 1 - t ^{2}}{t ^{2}}] + \frac{2}{t ^{3}}

Simplifying the expression, we get:

f^{'} (t) = \frac{- 3}{t ^{2} 1 - t ^{2}} + \frac{2}{t ^{3}}

Evaluating at

t = \frac{1}{3}

, where

t^{2} = \frac{1}{3}

and

1 - t^{2} = \frac{2}{3}

f^{'} (\frac{1}{3}) = \frac{- 3}{\frac{1}{3} \cdot \frac{2}{3}} + \frac{2}{( \frac{1}{3} ) ^{3}} = - 9 \frac{3}{2} + 63

The final result is:

$63 - \frac{9 3}{2}$

Common Traps & FAQs

Wait, why does the upper limit term in the Newton-Leibniz rule disappear?

That is a great observation! The Newton-Leibniz rule states that we multiply by the derivative of the limit. Since our upper limit is the constant

1

, its derivative is

0

. Therefore, the entire term

w (1) \cdot 0

vanishes. It is a beautiful simplification that saves us a lot of work!

Can we just divide by

sin x

when simplifying the equation?

Yes, we can! We are given that

x \in (0, \frac{π}{2})

. In this interval,

sin x

is strictly positive and never zero. Since we are not dividing by zero, the operation is perfectly valid. Always check the domain before performing such steps in JEE problems.

Why did we choose

t = cos x

as the substitution?

We chose

t = cos x

because our equation was in terms of

f (cos x)

. By setting

t = cos x

, we transform the function into

f (t)

, which is exactly what we need to differentiate. It bridges the gap between the trigonometric world of

x

and the algebraic world of

t

Animated Solution for Mathematics - Definite Integration: 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 :

Select Answer:

Visualized Solution

The Problem Setup & Newton-Leibniz Rule

Differentiating the Left Hand Side (LHS)

Differentiating the Right Hand Side (RHS)

Equating and Simplifying

Isolating $f (cos x)$

Variable Substitution ( $t = cos x$ )

Expressing $f (t)$ Explicitly

Differentiating $f (t)$ - First Term

Simplifying the First Term Derivative

Differentiating Second Term and Combining

Substituting $t = \frac{1}{3}$

Computing $f^{'} (\frac{1}{3})$

Final Calculation

Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Master Equation

Solving for the Function

Final Calculation

Similar Questions

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

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

$lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ is equal to

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If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Let $f : R \to R$ be defined as $f (x) = e^{- x} sin x$ . If $F : [0, 1] \to R$ is a differentiable function such that $F (x) = \int_{0}^{x} f (t) d t$ , then the value of $\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x$ lies in the interval

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

$lim_{x \to 0} \frac{\int _{0}^{x^{2}} s i n t d t}{x ^{3}}$ is equal to:

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

Select Answer:

Visualized Solution

The Problem Setup & Newton-Leibniz Rule

Differentiating the Left Hand Side (LHS)

Differentiating the Right Hand Side (RHS)

Equating and Simplifying

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

Variable Substitution ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=cosx)

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

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

Simplifying the First Term Derivative

Differentiating Second Term and Combining

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=3​1​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(3​1​)

Final Calculation

Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Master Equation

Solving for the Function

Final Calculation

Similar Questions

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Isolating $f (cos x)$

Variable Substitution ( $t = cos x$ )

Expressing $f (t)$ Explicitly

Differentiating $f (t)$ - First Term

Substituting $t = \frac{1}{3}$

Computing $f^{'} (\frac{1}{3})$

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

$lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ 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 $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

If $f (x) = \int_{0}^{x} t (sin x - sin t) d t$ then :

$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

$lim_{x \to 0} \frac{\int _{0}^{x^{2}} s i n t d t}{x ^{3}}$ is equal to:

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