Let f:R R be defined f(x) = ae2x + bex + cx . If f(0) = -1 , f'( 2) = 21 and 0 4 (f(x)

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Questions	18	26	18	62

Q53 jee_main_2025_29_jan_evening Area Enclosed by Two Curves

Let the area enclosed between the curves

|y| = 1 - x²

and

x² + y² = 1

be

α

. If

9α = β π + γ

,

β, γ

are integers, then the value of

|β - γ|

equals

A. $27$ $27$
B. $18$ $18$
C. $15$ $15$
D. $33$ $33$

Solution

Related Formula

Area enclosed between two symmetric functions across four quadrants:

α = 4 \times ( Area of circle in 1st quadrant - \int₀¹ yparabola dx )

Core Logic

The curves are symmetric across the axes. The region is enclosed between the unit circle $x² + y² = 1$ $x^2 + y^2 = 1$ and the parabolas $y = \pm(1 - x²)$ $y = \pm(1 - x^2)$.

Area Enclosed by Two Curves diagram for Q53 - JEE Main 2025 Evening

By symmetry, the area $α$ $\alpha$ in the first quadrant is the area under the circle minus the area under the parabola from $x=0$ $x=0$ to $x=1$ $x=1$.

Step 1: Calculate the Area Components

Area of a circle quadrant with $r=1$ $r=1$ is $(π)/(4)$ $\frac{\pi}{4}$.

Area under the parabola:

\int₀¹ (1 - x²) dx = [ x - (x³)/(3) ]₀¹ = 1 - (1)/(3) = (2)/(3)

Step 2: Evaluate the Total Area

α = 4 [ (π)/(4) - (2)/(3) ] = π - (8)/(3)

Multiplying both sides by 9:

9α = 9π - 24

Comparing with $9α = βπ + γ$ $9\alpha = \beta\pi + \gamma$, we get:

β = 9, γ = -24

Hence,

|β - γ| = |9 - (-24)| = 33

Pattern Recognition

Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem.

Chapter Mix

Class 12 Mathematics: Integral Calculus

Q71 jee_main_2025_29_jan_evening Integration of Modulus and Greatest Integer Functions

If

24\int₀(pi)/(4)( |4x - (pi)/(12)| + [2 x])dx = 2π +α

, where

[\cdot ]

denotes the greatest integer function, then

α

is equal to

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

Additivity property of definite integrals over split interval boundary points:

\intₐc g(x) dx = \intₐb g(x) dx + \intbc g(x) dx

Core Logic

Separate the integration tracking flow into two component operations:

I₁ = \int₀(pi)/(4) |4x - (pi)/(12)| dx

I₂ = \int₀(pi)/(4) [2 x] dx

Step 1: Solve Modulus Integral Term

The argument changes sign inside absolute value bounds when $4x - (pi)/(12) = 0 x = (pi)/(48)$ :

I₁ = \int₀(pi)/(48) - (4x - (pi)/(12)) dx + \int(pi)/(48)(pi)/(4) (4x - (pi)/(12)) dx

= (1)/(4) [ (4x - (pi)/(12)) ]₀(pi)/(48) - (1)/(4) [ (4x - (pi)/(12)) ](pi)/(48)(pi)/(4)

Evaluating across numerical boundaries gives:

24 \cdot I₁ = 24((1)/(2)) = 12

Step 2: Solve Greatest Integer Component and Combine

Analyze step limits inside greatest integer block $[2 x]$ $[2\sin x]$: For $x in [0, (pi)/(6))$ $x \in \left[0, \frac{pi}{6}\right)$, $0 \leq 2 x < 1 [2 x] = 0$ . For $x in [(pi)/(6), (pi)/(4)]$ , $1 \leq 2 x < \sqrt(2) [2 x] = 1$ .

I₂ = \int₀(pi)/(6) 0 dx + \int(pi)/(6)(pi)/(4) 1 dx = (pi)/(4) - (pi)/(6) = (pi)/(12)

Combine both sections aggregated by multiplier 24:

Total = 12 + 24((pi)/(12)) = 2π + 12

Comparing directly with expression statement parameters $2π + α$ $2\pi + \alpha$ isolates response value: $α = 12$ $\alpha = 12$

Pattern Recognition

Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q jee_main_2025_28_jan_morning Properties of Definite Integrals (King's Property)

If

\int-(π)/(2)(π)/(2)(96x² ²x)/((1 + e^x)) dx = π (α π² +β),α ,β in Z,

then

(α + β)²

equals:

A. 144
B. 196
C. 100
D. 64

Solution

Related Formula

King's property for definite integration:

\intₐ^b f(x) dx = \intₐ^b f(a+b-x) dx

Core Logic

Apply the identity $x \to -x$ $x \to -x$ to the integral:

I = \int-(π)/(2)(π)/(2) (96x² ² x)/(1 + e^x) dx = \int-(π)/(2)(π)/(2) 96x² ² x1 + e-x dx

Step 1: Adding both integral variations

Adding the equations eliminates the exponential denominator term ( $1+e^x$ $1+e^x$):

2I = \int-(π)/(2)(π)/(2) 96x² ² x \cdot [(1)/(1+e^x) + (e^x)/(1+e^x)] dx

I = 48 \int₀(π)/(2) x² (1 + 2x) dx

Step 2: Evaluating the integrated components

Integrating by parts gives:

I = π (2π² - 12)

Matching coefficients with the template: $α = 2$ $\alpha = 2$ and \beta = -12.

(α + β)² = (2 - 12)² = (-10)² = 100

Pattern Recognition

Exponential denominators like $1+e^x$ $1+e^x$ in symmetric integral intervals are prime candidates for simplification using King's property.

Chapter Mix

Class 12 Maths: Definite Integrals

Q65 jee_main_2025_03_april_morning Methods of Integration by Substitution

Let

f(x) = \int x³\sqrt(3 - x²) dx

. If

5f(\sqrt(2)) = -4

, then

f(1)

is equal to:

A. $- 2\sqrt(2)5$ $-\frac{2\sqrt{2}}{5}$
B. $- 8\sqrt(2)5$ $-\frac{8\sqrt{2}}{5}$
C. $- 4\sqrt(2)5$ $-\frac{4\sqrt{2}}{5}$
D. $- 6\sqrt(2)5$ $-\frac{6\sqrt{2}}{5}$

Solution

Related Formula

Method of algebraic parameter substitution: Set $3-x² = t² -2xdx = 2tdt xdx = -tdt$

Core Logic

Perform the specified variable parameter replacement steps:

3 - x² = t² x dx = -t dt

Rewrite the internal integral block components:

f(x) = \int x² \cdot \sqrt(3-x²) \cdot (x dx) = \int (3-t²) \cdot t \cdot (-t dt)

= \int (t⁴ - 3t²) dt = (t⁵)/(5) - t³ + C

Return to original reference variable $x$ $x$:

f(x) = (3-x²)5/25 - (3-x²)3/2 + C

Step 1: Constant integration resolving

Evaluate function boundary conditions at $x = \sqrt(2)$ $x = \sqrt{2}$:

f(\sqrt(2)) = (3-2)5/25 - (3-2)3/2 + C = (1)/(5) - 1 + C = -(4)/(5) + C

Given $5f(\sqrt(2)) = -4 f(\sqrt(2)) = -(4)/(5)$ .

-(4)/(5) + C = -(4)/(5) C = 0

Step 2: Numeric tracking value

Evaluate final targeted definition state value at $x=1$ $x=1$:

f(1) = (3-1)5/25 - (3-1)3/2 = 25/25 - 23/2

= 23/2((2)/(5) - 1) = 2\sqrt(2)(-(3)/(5)) = - 6\sqrt(2)5

Pattern Recognition

Splitting powers of $x$ $x$ to create a direct match with internal derivative differential flags speeds up the integration transformation sequence.

Chapter Mix

Class 12 Mathematics: Integrals

Q67 jee_main_2025_03_april_morning Definite Integral of Greatest Integer Function

Let the domain of the function

f(x) = ₂ ₄ ₆(3 + 4x - x²)

be

(a, b)

. If

\int₀b-a[x²]dx = p - \sqrt(q) - \sqrt(r)

, where

p, q, r in N

and

(p, q, r) = 1

, and

[\cdot]

represents the greatest integer function, then

p + q + r

is equal to:

A. 10
B. 8
C. 11
D. 9

Solution

Related Formula

Domain of log chain iterations: For $₂ ₄(M) > 0$ $\log_2\log_4(M) > 0$, we require $₄(M) > 1 M > 4$ $\log_4(M) > 1 \implies M > 4$.

Core Logic

Trace internal arguments outward sequentially:

₄ ₆(3 + 4x - x²) > 0 ₆(3 + 4x - x²) > 1

3 + 4x - x² > 6¹ x² - 4x + 3 < 0

(x-1)(x-3) < 0 x in (1, 3)

Thus, determine limits:

a = 1, b = 3 b - a = 2

Step 1: Setting up the greatest integer function integration

We need to evaluate $\int₀² [x²] dx$ $\int_0^2 [x^2] \, \mathrm{d}x$. Identify step boundary switch locations inside range $[0, 2]$ $[0, 2]$:

For $x in [0, 1): [x²] = 0$ $x \in [0, 1): [x^2] = 0$
For $x in [1, \sqrt(2)): [x²] = 1$ $x \in [1, \sqrt{2}): [x^2] = 1$
For $x in [\sqrt(2), \sqrt(3)): [x²] = 2$ $x \in [\sqrt{2}, \sqrt{3}): [x^2] = 2$
For $x in [\sqrt(3), 2): [x²] = 3$ $x \in [\sqrt{3}, 2): [x^2] = 3$

Set up separate boundary component integrations:

\int₀² [x²] dx = \int₀¹ 0 dx + \int₁\sqrt(2) 1 dx + \int\sqrt(2)\sqrt(3) 2 dx + \int\sqrt(3)² 3 dx

= 0 + (\sqrt(2) - 1) + 2(\sqrt(3) - \sqrt(2)) + 3(2 - \sqrt(3))

= \sqrt(2) - 1 + 2\sqrt(3) - 2\sqrt(2) + 6 - 3\sqrt(3) = 5 - \sqrt(2) - \sqrt(3)

Step 2: Matching coefficients

Compare values with requested answer template shape:

5 - \sqrt(2) - \sqrt(3) = p - \sqrt(q) - \sqrt(r)

p = 5, q = 2, r = 3

Final Sum = p + q + r = 5 + 2 + 3 = 10

Pattern Recognition

Integrals over greatest integer configurations change value exactly where the inner expression tracks through integer milestones. Mapping boundaries accurately resolves calculations smoothly.

Chapter Mix

Class 12 Mathematics: Integrals

Solution & Explanation

Related Formula

Core Logic

Step 1: Applying the Integral Condition

Step 2: Solving the System of Equations

Step 3: Finding Final Absolute Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Integral Calculus Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate the Area Components

Step 2: Evaluate the Total Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve Modulus Integral Term

Step 2: Solve Greatest Integer Component and Combine

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Adding both integral variations

Step 2: Evaluating the integrated components

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Constant integration resolving

Step 2: Numeric tracking value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the greatest integer function integration

Step 2: Matching coefficients

Pattern Recognition

Chapter Mix

More Integral Calculus Questions — jee_main_2024_30_january_evening

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