Let f be a differentiable function such that for x geq 0. Then f(2) is equal to _______

Q12 jee_main_2026_21_jan_morning Properties of Definite Integrals with Modulus

The value of

\int-π/6π/6( π+4x¹¹1- (|x|+π/6))dx

is equal to

A. $2π$ $2\pi$
B. $4π$ $4\pi$
C. $8π$ $8\pi$
D. $6π$ $6\pi$

Solution

Related Formula

\int₋ₐa f(x) dx = \int₀a [f(x) + f(-x)] dx

Core Logic

Let $I = \int-π/6π/6 π+4x¹¹1- (|x|+π/6)dx$ . The denominator $1 - (|x| + π/6)$ $1 - \sin(|x| + \pi/6)$ is an even function. The numerator can be split into an even part ( $π$ $\pi$) and an odd part ( $4x¹¹$ $4x^{11}$).

\int₋ₐa 4x¹¹1- (|x|+π/6) dx = 0 (Since integrand is odd)

Step 1: Simplify to Even Integral

We are left with the even part:

I = \int-π/6π/6 (π)/(1 - (|x| + π/6)) dx

Using even function property $\int₋ₐa f(x) dx = 2 \int₀a f(x) dx$ :

I = 2π \int₀π/6 (1)/(1 - (x + π/6)) dx

Step 2: Substitution

Let $t = x + (π)/(6) \Rightarrow dt = dx$ . Limits: when $x = 0 \Rightarrow t = π/6$ $x = 0 \Rightarrow t = \pi/6$, when $x = π/6 \Rightarrow t = π/3$ $x = \pi/6 \Rightarrow t = \pi/3$.

I = 2π \intπ/6π/3 (dt)/(1 - t)

Step 3: Solve the Integral

Rationalize the denominator:

I = 2π \intπ/6π/3 (1 + t)/((1 - t)(1 + t)) dt

I = 2π \intπ/6π/3 (1 + t)/( ² t) dt

I = 2π \intπ/6π/3 ( ² t + t t) dt

Integrate directly:

I = 2π [ t + t ]π/6π/3

Evaluate limits: Upper limit ( $π/3$ $\pi/3$): $(π/3) + (π/3) = \sqrt(3) + 2$ Lower limit ( $π/6$ $\pi/6$): $(π/6) + (π/6) = 1\sqrt(3) + 2\sqrt(3) = 3\sqrt(3) = \sqrt(3)$

I = 2π [(\sqrt(3) + 2) - \sqrt(3)] = 2π (2) = 4π

Pattern Recognition

Symmetric limits $[-a, a]$ $[-a, a]$ instantly demand testing for odd/even parity. Any mixed polynomial like $c + k xodd$ $c + k x^{\text{odd}}$ over an even denominator guarantees the odd power term strictly vanishes, halving calculation time.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Q25 jee_main_2026_21_jan_morning Absolute Value Integrals

6\int₀π|( 3x+ 2x+ x)|dx

is equal to.....

Numerical Answer. Answer: 17 to 17

Solution

Related Formula

A + B = 2 ((A+B)/(2)) ((A-B)/(2))

2x = 2 x x

2x = 2 ² x - 1

Core Logic

Let $I = 6\int₀π| 3x + x + 2x| dx$ . Apply sum-to-product on $3x + x$ $\sin 3x + \sin x$: $3x + x = 2 (2x) (x)$ $\sin 3x + \sin x = 2 \sin(2x) \cos(x)$

So the expression becomes: $|2 (2x) x + 2x| = | 2x (2 x + 1)| = |2 x x (2 x + 1)|$ Since $x in [0, π]$ $x \in [0, \pi]$, $x \geq 0$ $\sin x \geq 0$. We can pull it out of the modulus. $I = 12 \int₀π x |2 ² x + x| dx$

Step 1: Coordinate Substitution

Substitute $t = x$ $t = \cos x$, then $dt = - x dx$ $dt = -\sin x dx$. Limits: when $x = 0$ $x = 0$, $t = 1$ $t = 1$. When $x = π$ $x = \pi$, $t = -1$ $t = -1$.

I = 12 \int₋₁¹ |2t² + t| dt

Step 2: Resolve Modulus Intervals

The roots of $2t² + t = 0$ $2t^2 + t = 0$ are $t = 0$ $t = 0$ and $t = -1/2$ $t = -1/2$. The quadratic $2t² + t$ $2t^2 + t$ is negative in the interval $(-1/2, 0)$ $(-1/2, 0)$ and positive elsewhere. Split the integral:

I = 12 [ \int₋₁-1/2 (2t² + t) dt - \int-1/2⁰ (2t² + t) dt + \int₀¹ (2t² + t) dt ]

Integration of modulus polynomial function diagram for Q25 - JEE Main 2026 Morning

Step 3: Evaluate Integrals

Anti-derivative: $F(t) = (2t³)/(3) + (t²)/(2)$ $F(t) = \frac{2t^3}{3} + \frac{t^2}{2}$. $F(1) = 2/3 + 1/2 = 7/6$ $F(1) = 2/3 + 1/2 = 7/6$ $F(0) = 0$ $F(0) = 0$ $F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$ $F(-1) = -2/3 + 1/2 = -1/6$ $F(-1) = -2/3 + 1/2 = -1/6$

Evaluate each segment:

$\int₋₁-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$
$-\int-1/2⁰ = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$
$\int₀¹ = F(1) - F(0) = 7/6 - 0 = 28/24$

Sum of parts inside bracket: $(5)/(24) + (1)/(24) + (28)/(24) = (34)/(24) = (17)/(12)$

Step 4: Final Output

I = 12 \times (17)/(12) = 17

Pattern Recognition

Whenever an integral features a cascading sum of sine frequencies like $(kx)$ $\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under $t = x$ $t = \cos x$.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Q22 jee_main_2026_21_jan_evening Properties of Definite Integrals

If

\int₀¹4 ⁻¹(1-2x+4x²)dx=a ⁻¹(2)-b ₑ(5)

, where

a, b in N

, then

(2a+b)

is equal to ____.

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

⁻¹(y) = ⁻¹((1)/(y))

⁻¹((x - y)/(1 + xy)) = ⁻¹x - ⁻¹y

King's Property: \int₀^a f(x)dx = \int₀^a f(a-x)dx

Core Logic

Let $I = \int₀¹ ⁻¹(1-2x+4x²) dx$ . Convert $⁻¹$ $\cot^{-1}$ to $⁻¹$ $\tan^{-1}$:

⁻¹(1 + 2x(2x-1)) = ⁻¹( (1)/(1 + 2x(2x-1)) )

Notice that $2x - (2x-1) = 1$ $2x - (2x-1) = 1$. Thus, the integrand is $⁻¹( (2x - (2x-1))/(1 + 2x(2x-1)) )$ .

I = \int₀¹ ( ⁻¹(2x) - ⁻¹(2x-1) ) dx

Step 1: Apply Definite Integral Properties

Applying King's property to the second term $\int₀¹ ⁻¹(2x-1) dx$ $\int_0^1 \tan^{-1}(2x-1) dx$:

x \to 1-x \int₀¹ ⁻¹(2(1-x)-1) dx = \int₀¹ ⁻¹(1-2x) dx = -\int₀¹ ⁻¹(2x-1) dx

Wait, this implies $\int₀¹ ⁻¹(2x-1) dx = 0$ $\int_0^1 \tan^{-1}(2x-1) dx = 0$! Thus, $I = \int₀¹ ⁻¹(2x) dx$ $I = \int_0^1 \tan^{-1}(2x) dx$.

Step 2: Integration by Parts

Solve $\int₀¹ ⁻¹(2x) \cdot 1 dx$ $\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$:

I = [ x ⁻¹(2x) ]₀¹ - \int₀¹ x (2)/(1+4x²) dx

I = ⁻¹(2) - (1)/(4) \int₀¹ (8x)/(1+4x²) dx

Let $1+4x² = t 8x dx = dt$ $1+4x^2 = t \implies 8x dx = dt$. At $x=0, t=1$ $x=0, t=1$; at $x=1, t=5$ $x=1, t=5$.

I = ⁻¹(2) - (1)/(4) \int₁⁵ (dt)/(t) = ⁻¹(2) - (1)/(4) ln(5)

Step 3: Compare and Calculate Result

The original integral has a factor of $4$ $4$:

4I = 4 ⁻¹(2) - ln(5)

Compare with $a ⁻¹(2) - b ₑ(5)$ $a\tan^{-1}(2) - b\log_e(5)$:

a = 4, b = 1

Therefore, $2a + b = 2(4) + 1 = 9$ $2a + b = 2(4) + 1 = 9$.

Pattern Recognition

Always convert $⁻¹$ $\cot^{-1}$ quadratic inputs into $⁻¹(x-y)/(1+xy)$ $\tan^{-1}\frac{x-y}{1+xy}$ forms. Apply King's property on symmetric limits; often one piece vanishes entirely.

Chapter Mix

Class 12 Maths: Definite Integrals Class 12 Maths: Inverse Trigonometric Functions

Q15 jee_main_2026_22_january_morning Properties of Definite Integrals

The value of

\int-(π)/(2)(π)/(2)((1)/([x]+4))dx

, where

[ ]

denotes the greatest integer function, is

A. $(1)/(60)(21π-1)$ $\frac{1}{60}(21\pi-1)$
B. $(1)/(60)(π-7)$ $\frac{1}{60}(\pi-7)$
C. $(7)/(60)(3π-1)$ $\frac{7}{60}(3\pi-1)$
D. $(7)/(60)(π-3)$ $\frac{7}{60}(\pi-3)$

Solution

Related Formula

The greatest integer function [x] is piecewise constant on intervals [n, n+1).

\intₐb f(x) dx is broken into sub-intervals where f(x) is constant.

Core Logic

The integral bounds are from $-π/2 \approx -1.57$ $-\pi/2 \approx -1.57$ to $π/2 \approx 1.57$ $\pi/2 \approx 1.57$. We must split the integral at every integer point between these bounds.

Intervals:

$[-π/2, -1) [x] = -2$ $[-\pi/2, -1) \implies [x] = -2$
$[-1, 0) [x] = -1$ $[-1, 0) \implies [x] = -1$
$[0, 1) [x] = 0$ $[0, 1) \implies [x] = 0$
$[1, π/2) [x] = 1$ $[1, \pi/2) \implies [x] = 1$

Step 1: Splitting the Integral

I = \int-π/2π/2(1)/([x]+4)dx

I = \int-π/2⁻¹ (1)/(-2 + 4) dx + \int₋₁⁰ (1)/(-1 + 4) dx + \int₀¹ (1)/(0 + 4) dx + \int₁π/2 (1)/(1 + 4) dx

Step 2: Evaluating Sub-integrals

I = \int-π/2⁻¹ (1)/(2) dx + \int₋₁⁰ (1)/(3) dx + \int₀¹ (1)/(4) dx + \int₁π/2 (1)/(5) dx

Evaluate the limits for each constant integral:

I = (1)/(2) ( -1 - (-(π)/(2)) ) + (1)/(3) (0 - (-1)) + (1)/(4) (1 - 0) + (1)/(5) ( (π)/(2) - 1 )

I = (1)/(2) ( (π)/(2) - 1 ) + (1)/(3) + (1)/(4) + (1)/(5) ( (π)/(2) - 1 )

Step 3: Simplifying the Expression

Group the $( (π)/(2) - 1 )$ $\left( \frac{\pi}{2} - 1 \right)$ terms:

I = ((1)/(2) + (1)/(5)) ( (π)/(2) - 1 ) + (1)/(3) + (1)/(4)

I = (7)/(10) ( (π)/(2) - 1 ) + (7)/(12)

I = (7π)/(20) - (7)/(10) + (7)/(12)

Find a common denominator for the constants (LCD is 60):

-(7)/(10) + (7)/(12) = -(42)/(60) + (35)/(60) = -(7)/(60)

Rewrite $(7π)/(20)$ $\frac{7\pi}{20}$ with denominator 60:

(7π)/(20) = (21π)/(60)

So,

I = (21π)/(60) - (7)/(60) = (7(3π - 1))/(60) = (7)/(60)(3π - 1)

Pattern Recognition

Integration of step functions always transforms into a simple sum of rectangle areas $(cᵢ \times Δ xᵢ)$ $(c_i \times \Delta x_i)$. Immediately break the bounds at integers, turning a calculus problem into elementary arithmetic.

Chapter Mix

Class 12 Maths: Definite Integration

Q23 jee_main_2026_22_january_morning Integration by Substitution

If

\int( x)(-11)/(2)( x)(-5)/(2)dx= - p₁q₁( x)(9)/(2)- p₂q₂( x)(5)/(2)- p₃q₃( x)(1)/(2)+ p₄q₄( x)(-3)/(2)+C

, where

pᵢ

and

qᵢ

are positive integers with

(pᵢ, qᵢ)=1

for i=1, 2, 3, 4 and C is the constant of integration, then

15p₁p₂p₃p₄q₁q₂q₃q₄

is equal to ____.

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

When powers of sine and cosine add up to a negative even integer, extract ² x and substitute x = t

Core Logic

Integral: $I = \int -11/2 x -5/2 x dx$

The sum of powers is $-(11)/(2) - (5)/(2) = -8$ $-\frac{11}{2} - \frac{5}{2} = -8$. Convert the integrand entirely into terms of $x$ $\tan x$ and $x$ $\sec x$ by dividing and multiplying by $-11/2 x$ $\cos^{-11/2} x$.

I = \int (( x)/( x))-11/2 -11/2 x -5/2 x dx

I = \int ( x)-11/2 ( x)⁻⁸ dx

I = \int ( x)-11/2 ⁸ x dx

Step 1: Integration by Substitution

Rewrite $⁸ x = ( ² x)³ ² x = (1 + ² x)³ ² x$ .

I = \int ( x)-11/2 (1 + ² x)³ ² x dx

Substitute $t = x dt = ² x dx$ .

I = \int t-11/2 (1 + t²)³ dt

Expand $(1 + t²)³ = 1 + 3t² + 3t⁴ + t⁶$ $(1 + t^2)^3 = 1 + 3t^2 + 3t^4 + t^6$.

I = \int t-11/2 (1 + 3t² + 3t⁴ + t⁶) dt

I = \int (t-11/2 + 3t-7/2 + 3t-3/2 + t1/2) dt

Step 2: Evaluating the Anti-derivatives

Integrate term by term:

I = t-9/2-9/2 + 3 t-5/2-5/2 + 3 t-1/2-1/2 + t3/23/2 + C

I = -(2)/(9)t-9/2 - (6)/(5)t-5/2 - 6t-1/2 + (2)/(3)t3/2 + C

Substitute back $t = x = (1)/( x)$ $t = \tan x = \frac{1}{\cot x}$, which implies $t-a = ( x)^a$ $t^{-a} = (\cot x)^a$.

I = -(2)/(9)( x)9/2 - (6)/(5)( x)5/2 - (6)/(1)( x)1/2 + (2)/(3)( x)-3/2 + C

Step 3: Variable Assignment

Comparing with $-(p₁)/(q₁)( x)9/2 - (p₂)/(q₂)( x)5/2 - (p₃)/(q₃)( x)1/2 + (p₄)/(q₄)( x)-3/2$ :

$p₁ = 2, q₁ = 9$ $p_1 = 2, q_1 = 9$ $p₂ = 6, q₂ = 5$ $p_2 = 6, q_2 = 5$ $p₃ = 6, q₃ = 1$ $p_3 = 6, q_3 = 1$ $p₄ = 2, q₄ = 3$ $p_4 = 2, q_4 = 3$

Calculate $(15 p₁ p₂ p₃ p₄)/(q₁ q₂ q₃ q₄)$ :

= (15(2)(6)(6)(2))/((9)(5)(1)(3)) = (15 \times 144)/(135) = (2160)/(135) = 16

Pattern Recognition

If $\int ^m x ⁿ x dx$ $\int \sin^m x \cos^n x \,dx$ has $m+n$ $m+n$ as a negative even integer, unconditionally extract $|m+n|x$ $\sec^{|m+n|}x$ and set $x = t$ $\tan x = t$. The expansion expands gracefully into polynomial power rules without any trig substitution hassle.

Chapter Mix

Class 12 Maths: Indefinite Integration

Let f$f$ be a differentiable function such that 2(x+2)²f(x) - 3(x+2)² = 10∫₀x(t+2)f(t)dt$$2(x+2)^{2}f(x) - 3(x+2)^{2} = 10\int_{0}^{x}(t+2)f(t)dt$$ for x ≥ 0$x \geq 0$. Then f(2)$f(2)$ is equal to ________.

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve the First-Order Differential Equation

Step 2: Apply the Boundary Condition

Step 3: Evaluate at target point x = 2

Pattern Recognition

Chapter Mix

Reference Study Guides

More Definite Integrals Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Simplify to Even Integral

Step 2: Substitution

Step 3: Solve the Integral

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Coordinate Substitution

Step 2: Resolve Modulus Intervals

Step 3: Evaluate Integrals

Step 4: Final Output

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Definite Integral Properties

Step 2: Integration by Parts

Step 3: Compare and Calculate Result

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Splitting the Integral

Step 2: Evaluating Sub-integrals

Step 3: Simplifying the Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration by Substitution

Step 2: Evaluating the Anti-derivatives

Step 3: Variable Assignment

Pattern Recognition

Chapter Mix

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Let $f$ $f$ be a differentiable function such that $2(x+2)²f(x) - 3(x+2)² = 10\int₀x(t+2)f(t)dt$ for $x \geq 0$ $x \geq 0$. Then $f(2)$ $f(2)$ is equal to ________.