The area of the region bounded by the curve y = |x|, x|x - 2| [cite: 699], the x-axis and the lines x = -2 and x = 4 is…

More Integrals Previous-Year Questions — Page 7

Q68 jee_main_2025_04_april_evening Properties of Definite Integrals

Let

f(x) + 2f((1)/(x)) = x² + 5

and

2 g (x) - 3 g ((1)/(2)) = x, x > 0

. If

α = \int_ 1 ^ 2 f (x) d x

, and

β = \int_ 1 ^ 2 g (x) d x

, then the value of

9α + β

is:

A. $1$ $1$
B. $0$ $0$
C. $10$ $10$
D. $11$ $11$

Solution

Core Logic

We have two functional equations to solve before integrating.

Equation 1: $f(x) + 2f((1)/(x)) = x² + 5$ Replace $x$ $x$ with $(1)/(x)$ $\frac{1}{x}$:

f((1)/(x)) + 2f(x) = (1)/(x²) + 5

Multiplying this new equation by $2$ $2$ and subtracting the original Equation 1 eliminates the $f((1)/(x))$ $f\left(\frac{1}{x}\right)$ term:

4f(x) + 2f((1)/(x)) - (f(x) + 2f((1)/(x))) = 2((1)/(x²) + 5) - (x² + 5)

3f(x) = (2)/(x²) - x² + 5 f(x) = (2)/(3x²) - (x²)/(3) + (5)/(3)

Step 1: Finding alpha

Integrate $f(x)$ $f(x)$ from $1$ $1$ to $2$ $2$:

α = \int₁² ( (2)/(3x²) - (x²)/(3) + (5)/(3) ) dx = [ -(2)/(3x) - (x³)/(9) + (5x)/(3) ]₁²

α = ( -(1)/(3) - (8)/(9) + (10)/(3) ) - ( -(2)/(3) - (1)/(9) + (5)/(3) ) = (19)/(9) - (8)/(9) = (11)/(9)

Thus, $9α = 11$ $9\alpha = 11$.

Step 2: Solving for g(x) and finding beta

We are given $2g(x) - 3g((1)/(2)) = x$ . Substitute $x = (1)/(2)$ $x = \frac{1}{2}$:

2g((1)/(2)) - 3g((1)/(2)) = (1)/(2) -g((1)/(2)) = (1)/(2) g((1)/(2)) = -(1)/(2)

Substitute this constant value back into the original equation:

2g(x) - 3(-(1)/(2)) = x 2g(x) + (3)/(2) = x g(x) = (x)/(2) - (3)/(4)

Now find $β$ $\beta$:

β = \int₁² ( (x)/(2) - (3)/(4) ) dx = [ (x²)/(4) - (3x)/(4) ]₁² = ( 1 - (3)/(2) ) - ( (1)/(4) - (3)/(4) ) = -(1)/(2) - (-(1)/(2)) = 0

Step 3: Calculating 9alpha + beta

Combining our values:

9α + β = 11 + 0 = 11

Pattern Recognition

Functional equations involving $x \to (1)/(x)$ $x \to \frac{1}{x}$ are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations

Q72 jee_main_2025_04_april_evening Integration by Substitution

\int (\sqrt(1 + x²) + x)¹⁰(\sqrt(1 + x²) - x)⁹dx = (1)/(m) (( 1 + x ^ 2 + x) ^ n (n 1 + x ^ 2 - x)) + C

where

C

is the constant of integration and

m,nin N

, then

m + n

is equal to

Numerical Answer. Answer: 379 to 379

Solution

Core Logic

Let's simplify the integrand by rationalizing the denominator term block. Notice that:

(\sqrt(1+x²) - x)(\sqrt(1+x²) + x) = (1+x²) - x² = 1

1\sqrt(1+x²) - x = \sqrt(1+x²) + x

Substituting this back into the denominator expression column:

I = \int (\sqrt(1+x²) + x)¹⁰ \cdot (\sqrt(1+x²) + x)⁹ dx = \int (\sqrt(1+x²) + x)¹⁹ dx

Step 1: Implementing the Substitution Path

Let $t = \sqrt(1+x²) + x$ $t = \sqrt{1+x^2} + x$. Then:

dt = ( x\sqrt(1+x²) + 1 ) dx = ( x + \sqrt(1+x²)\sqrt(1+x²) ) dx = t\sqrt(1+x²) dx

dx = \sqrt(1+x²)t dt

Since $\sqrt(1+x²) + x = t$ $\sqrt{1+x^2} + x = t$ and $\sqrt(1+x²) - x = (1)/(t)$ $\sqrt{1+x^2} - x = \frac{1}{t}$, adding both gives:

2\sqrt(1+x²) = t + (1)/(t) \sqrt(1+x²) = (1)/(2)(t + (1)/(t))

Thus, $dx = (1)/(2t)(t + (1)/(t)) dt = (1)/(2)(1 + (1)/(t²)) dt$ .

Step 2: Integrating with respect to t

Substitute these back into the integral:

I = \int t¹⁹ \cdot (1)/(2)(1 + (1)/(t²)) dt = (1)/(2) \int (t¹⁹ + t¹⁷) dt

I = (1)/(2) ( t²⁰20 + t¹⁸18 ) + C = t¹⁸4 ( (t²)/(10) + (1)/(9) ) + C = t¹⁸360 (9t² + 10) + C

Step 3: Matching Form and Finding m + n

To match the template format, let's pull out a factor of $t$ $t$:

I = t¹⁹360 ( 9t + (10)/(t) ) + C = t¹⁹360 ( 9(\sqrt(1+x²)+x) + 10(\sqrt(1+x²)-x) ) + C

I = (\sqrt(1+x²)+x)¹⁹360 ( 19\sqrt(1+x²) - x ) + C

Comparing this directly with the given answer format, we identify:

$m = 360$ $m = 360$
$n = 19$ $n = 19$

Computing $m + n$ $m + n$:

m + n = 360 + 19 = 379

Pattern Recognition

Expressions containing conjugate factors like $\sqrt(1+x²) \pm x$ $\sqrt{1+x^2} \pm x$ frequently simplify under rationalization because their product equals 1. This dynamic quickly reduces fractional components into single power blocks.

Chapter Mix

Class 12 Mathematics: Indefinite Integrals

Q66 jee_main_2025_04_april_morning Properties of Definite Integrals

The value of

\int₋₁¹ (1 + \sqrt(|x| - x))e^x + (\sqrt(|x| - x))e-xe^x + e-x dx

is equal to

A. $3 - 2\sqrt(2)3$ $3 - \frac{2\sqrt{2}}{3}$
B. $2 + 2\sqrt(2)3$ $2 + \frac{2\sqrt{2}}{3}$
C. $1 - 2\sqrt(2)3$ $1 - \frac{2\sqrt{2}}{3}$
D. $1 + 2\sqrt(2)3$ $1 + \frac{2\sqrt{2}}{3}$

Solution

Related Formula

King's property of definite integrals:

\intₐb f(x)dx = \intₐb f(a+b-x)dx

Core Logic

Let the given integral be $I$ $I$. Apply King's property by substituting $x \to -x$ $x \to -x$:

I = \int₋₁¹ (1 + \sqrt(|x| + x))e-x + (\sqrt(|x| + x))exe-x + ex dx

Add both integral expressions $2I = I + I$ $2I = I + I$:

2I = \int₋₁¹ (e^x + e-x) + (\sqrt(|x| - x) + \sqrt(|x| + x))(e^x + e-x)e^x + e-x dx

2I = \int₋₁¹ (1 + \sqrt(|x| - x) + \sqrt(|x| + x)) dx

Step 1: Apply Symmetry Properties

The integrand is completely even. Hence, convert intervals:

2I = 2\int₀¹ (1 + \sqrt(|x| - x) + \sqrt(|x| + x)) dx

For $x in [0,1]$ $x \in [0,1]$, $|x| = x \sqrt(|x| - x) = 0$ $|x| = x \implies \sqrt{|x| - x} = 0$ and $\sqrt(|x| + x) = \sqrt(2x)$ $\sqrt{|x| + x} = \sqrt{2x}$:

I = \int₀¹ (1 + \sqrt(2x)) dx

Step 2: Final Integration Execution

I = [ x + \sqrt(2) \cdot x3/23/2 ]₀¹ = [ x + 2\sqrt(2)3x3/2 ]₀¹

I = 1 + 2\sqrt(2)3

Pattern Recognition

When functions involve combinations of exponential components ( $e^x, e-x$ $e^x, e^{-x}$) over symmetric boundaries, adding the variable reflection eliminates exponential fractions instantly.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q57 jee_main_2025_07_april_evening Area Under Curves

If the area of the region

(x,y):1 + x²\leq y\leq x + 7,11 - 3x

is A, then 3A is equal to

A. $50$ $50$
B. $49$ $49$
C. $46$ $46$
D. $47$ $47$

Solution

Related Formula

The area enclosed between upper bounding function $yupper$ $y_{\text{upper}}$ and lower function $ylower$ $y_{\text{lower}}$ is:

Area = \intₐb (yupper - ylower) dx

Core Logic

We need to find the intersection points of the curves to understand the $x+7, 11-3x$ $\min\{x+7, 11-3x\}$ boundary transition:

$x+7 = 11-3x 4x = 4 x = 1$ .

Hence, the line switches behavior at $x=1$ $x=1$.

Intersecting $1+x²$ $1+x^2$ with $x+7$ $x+7$:

x² - x - 6 = 0 (x-3)(x+2) = 0 x = -2 or x = 3

Intersecting $1+x²$ $1+x^2$ with $11-3x$ $11-3x$:

x² + 3x - 10 = 0 (x+5)(x-2) = 0 x = 2 or x = -5

Area Under Curves diagram for Q57 - JEE Main 2025 Evening

Step 1: Set up Integrals

The transition points show that from $x = -2$ $x = -2$ to $1$ $1$, the upper line is $x+7$ $x+7$, and from $x = 1$ $x = 1$ to $2$ $2$, the upper line is $11-3x$ $11-3x$.

A = \int₋₂¹ ((x + 7) - (1 + x²)) dx + \int₁² ((11 - 3x) - (1 + x²)) dx

A = \int₋₂¹ (x + 6 - x²) dx + \int₁² (10 - 3x - x²) dx

Step 2: Integration Evaluation

Evaluating the first integral:

[ (x²)/(2) + 6x - (x³)/(3) ]₋₂¹ = ((1)/(2) + 6 - (1)/(3)) - (2 - 12 + (8)/(3)) = (37)/(6) - (-(22)/(3)) = (27)/(2)

Evaluating the second integral:

[ 10x - (3x²)/(2) - (x³)/(3) ]₁² = (20 - 6 - (8)/(3)) - (10 - (3)/(2) - (1)/(3)) = (34)/(3) - (49)/(6) = (19)/(6)

Total Area $A$ $A$:

A = (27)/(2) + (19)/(6) = (81 + 19)/(6) = (100)/(6) = (50)/(3)

Step 3: Calculate 3A

Multiplying the total area by 3:

3A = 3 \cdot ((50)/(3)) = 50

Pattern Recognition

When a boundary contains a $\min\{\}$ or $\max\{\}$ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals.

Chapter Mix

Class 12 Mathematics: Integral Calculus

The area of the region bounded by the curve y = |x|, x|x - 2|$y = \max\left\{|x|, x|x - 2|\right\}$ [cite: 699], the x-axis and the lines x = -2$x = -2$ and x = 4$x = 4$ is equal to[cite: 699, 702]:

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up separate area integral blocks

Pattern Recognition

Chapter Mix

Reference Study Guides

More Integrals Previous-Year Questions — Page 7

Solution

Core Logic

Step 1: Finding alpha

Step 2: Solving for g(x) and finding beta

Step 3: Calculating 9alpha + beta

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Implementing the Substitution Path

Step 2: Integrating with respect to t

Step 3: Matching Form and Finding m + n

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Symmetry Properties

Step 2: Final Integration Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up Integrals

Step 2: Integration Evaluation

Step 3: Calculate 3A

Pattern Recognition

Chapter Mix

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The area of the region bounded by the curve $y = |x|, x|x - 2|$ $y = \max\left\{|x|, x|x - 2|\right\}$ [cite: 699], the x-axis and the lines $x = -2$ $x = -2$ and $x = 4$ $x = 4$ is equal to[cite: 699, 702]: