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

Q28 jee_main_2024_01_february_morning Properties of Definite Integrals

If

\int-π/2π/2 8\sqrt(2) x dx(1+ex)(1+ ⁴x)=απ+β ₑ(3+2\sqrt(2)),

where

α, β

are integers, then

α²+β²

equals

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

King's Property for symmetric integration limits:

\int₋ₐa f(x) dx = \int₋ₐa f(-x) dx

Core Logic

Let the given definite integral be $I$ $I$:

I = \int-π/2π/2 8\sqrt(2) x dx(1+ex)(1+ ⁴x) (1)

Apply King's property by replacing $x$ $x$ with $-x$ $-x$ (since $-π/2 + π/2 = 0$ $-\pi/2 + \pi/2 = 0$):

I = \int-π/2π/2 8\sqrt(2) (-x) dx(1+e(-x))(1+ ⁴(-x))

I = \int-π/2π/2 8\sqrt(2) x dx(1+e- x)(1+ ⁴x) = \int-π/2π/2 8\sqrt(2) x \cdot ex dx(ex+1)(1+ ⁴x) (2)

Step 1: Simplify by Adding Expressions

Adding equations (1) and (2):

2I = \int-π/2π/2 8\sqrt(2) x(1 + ex) dx(1+ex)(1+ ⁴x)

2I = \int-π/2π/2 8\sqrt(2) x dx1+ ⁴x

Since the integrand is even, we can change the limits from $0$ $0$ to $π/2$ $\pi/2$:

2I = 2 \int₀π/2 8\sqrt(2) x dx1+ ⁴x I = \int₀π/2 8\sqrt(2) x dx1+ ⁴x

Step 2: Substitution and Algebraic Deconstruction

Let $x = t x dx = dt$ $\sin x = t \implies \cos x \, dx = dt$. Limits change from $0$ $0$ to $1$ $1$:

I = \int₀¹ 8\sqrt(2) dt1+t⁴ = 4\sqrt(2) \int₀¹ (2 dt)/(1+t⁴)

Dividing the numerator and denominator by $t²$ $t^2$, we write it as two distinct expressions:

I = 4\sqrt(2) [ \int₀¹ (1+(1)/(t²))/(t²+(1)/(t²)) dt - \int₀¹ (1-(1)/(t²))/(t²+(1)/(t²)) dt ]

I = 4\sqrt(2) [ \int-\infty⁰ (dz)/(z²+2) - \int\infty² (dk)/(k²-2) ]

where $z = t - (1)/(t)$ $z = t - \frac{1}{t}$ and $k = t + (1)/(t)$ $k = t + \frac{1}{t}$.

Step 3: Integrate and Evaluate Parameters

Evaluating standard anti-derivatives:

I = 4\sqrt(2) [ 1\sqrt(2) ⁻¹( z\sqrt(2)) ]-\infty⁰ - 4\sqrt(2) [ 12\sqrt(2)ln| k-\sqrt(2)k+\sqrt(2)| ]\infty²

I = 4\sqrt(2)( π2\sqrt(2)) - 2 ln| 2-\sqrt(2)2+\sqrt(2)| = 2π - 2ln(\sqrt(2)-1)²

I = 2π + 2ln(3+2\sqrt(2))

Matching parameters yields $α = 2$ $\alpha = 2$ and $β = 2$ $\beta = 2$. Thus:

α² + β² = 2² + 2² = 8

Pattern Recognition

Sees: Exponential variables causing asymmetry in symmetric integration bounds. Shortcut: Using King's rule completely eliminates the confusing $ex$ $e^{\sin x}$ factor, leaving behind a straightforward rational trigonometric configuration.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q10 jee_main_2024_29_january_evening Indefinite Integration

If

\int (3)/(2) x + (3)/(2) x\sqrt( ³ x ³ x (x - θ)) d x = A \sqrt( θ x - θ) + B \sqrt( θ - θ x) + C

, where

C

is the integration constant, then

AB

is equal to

A. $4sec(2θ)$ $4\mathrm{sec}(2\theta)$
B. $4 θ$ $4\sec \theta$
C. $2 θ$ $2\sec \theta$
D. $8secec(2θ)$ $8\mathrm{sec}\mathrm{ec}(2\theta)$

Solution

Related Formula

(x - θ) = x θ - x θ

Core Logic

Let us partition the given integral $I$ $I$ into two clean parts $I₁$ $I_1$ and $I₂$ $I_2$:

I = \int 3/2 x\sqrt( ³ x ³ x ( x θ - x θ)) dx + \int 3/2 x\sqrt( ³ x ³ x ( x θ - x θ)) dx

Factoring out appropriate powers of $x$ $\sin x$ and $x$ $\cos x$ from inside the roots:

I₁ = \int ² x\sqrt( x θ - θ) dx

I₂ = \int ² x\sqrt( θ - x θ) dx

Step 1: Evaluating the Integrals

For $I₁$ $I_1$, substitute $x θ - θ = t² ² x dx = (2t dt)/( θ)$ :

I₁ = \int (2t dt)/(t θ) = (2t)/( θ) = 2 θ \sqrt( x θ - θ)

For $I₂$ $I_2$, substitute $θ - x θ = z² ² x dx = (2z dz)/( θ)$ :

I₂ = \int (2z dz)/(z θ) = (2z)/( θ) = 2 θ \sqrt( θ - x θ)

Step 2: Combining Coefficients

Comparing with the given expression, we find the coefficients:

A = 2 θ, B = 2 θ

Multiplying them together:

AB = 4 θ θ = (4)/( θ θ) = (8)/(2 θ θ) = 8 (2θ)

Pattern Recognition

When integrating roots of trigonometric functions involving $(x - θ)$ $(x - \theta)$, try dividing/multiplying fields by $ⁿ x$ $\cos^n x$ or $ⁿ x$ $\sin^n x$ to force $x / ² x$ $\tan x / \sec^2 x$ templates.

Chapter Mix

Class 12 Mathematics: Integrals

Q24 jee_main_2024_29_january_evening Definite Integration

If

\int(π)/(6)(π)/(3)\sqrt(1 - 2x) dx = α +β \sqrt(2) +γ \sqrt(3)

where

α ,β

and

γ

are rational numbers, then

3α + 4β - γ

is equal to

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

\sqrt(1 - 2x) = \sqrt(( x - x)²) = | x - x|

Core Logic

The absolute function $| x - x|$ $|\cos x - \sin x|$ flips sign at $x = (π)/(4)$ $x = \frac{\pi}{4}$ within the boundary range:

For $x in [(π)/(6), (π)/(4)]$ : $x \geq x | x - x| = x - x$
For $x in [(π)/(4), (π)/(3)]$ : $x \geq x | x - x| = x - x$

Step 1: Boundary Splitting Integration

I = \int(π)/(6)(π)/(4) ( x - x) dx + \int(π)/(4)(π)/(3) ( x - x) dx

I = [ x + x ](π)/(6)(π)/(4) + [ - x - x ](π)/(4)(π)/(3)

I = ( 1\sqrt(2) + 1\sqrt(2) - ((1)/(2) + \sqrt(3)2) ) + ( -(1)/(2) - \sqrt(3)2 - (- 1\sqrt(2) - 1\sqrt(2)) )

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

Step 2: Coefficient Matrix Matching

Comparing with the given baseline layout:

α = -1, β = 2, γ = -1

Evaluating the required target metrics:

3α + 4β - γ = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6

Pattern Recognition

Never forget that $\sqrt(f(x)²) = |f(x)|$ $\sqrt{f(x)^2} = |f(x)|$. Skipping modulus checks inside definite root boundaries leads to incorrect answers.

Chapter Mix

Class 12 Mathematics: Integrals

Q jee_main_2024_27_jan_morning Properties of Definite Integrals

If

(a, b)

be the orthocentre of the triangle whose vertices are

(1, 2)

,

(2, 3)

and

(3, 1)

, and

I₁=\intₐbx~sin(4x-x²)dx

,

I₂=\intₐbsin(4x-x²)dx

, then

36 I₁I₂

is equal to:

A. $72$ $72$
B. $88$ $88$
C. $80$ $80$
D. $66$ $66$

Solution

Related Formula

\intₐ^b f(x) dx = \intₐ^b f(a+b-x) dx (King's Rule)

Core Logic

First, find the orthocentre $(a,b)$ $(a,b)$ of $Δ ABC$ $\Delta ABC$ with vertices $A(1, 2)$ $A(1, 2)$, $B(2, 3)$ $B(2, 3)$, and $C(3, 1)$ $C(3, 1)$. Slope of $AB = (3-2)/(2-1) = 1$ $AB = \frac{3-2}{2-1} = 1$. The altitude from $C$ $C$ onto $AB$ $AB$ must be perpendicular to $AB$ $AB$, so its slope is $-1$ $-1$. Equation of altitude from $C(3,1)$ $C(3,1)$:

y - 1 = -1(x - 3) \Rightarrow x + y = 4

The orthocentre $(a, b)$ $(a, b)$ lies on all altitudes, including this one. Thus, it satisfies $a + b = 4$ $a + b = 4$.

Step 1: Applying Definite Integral Properties

Given $I₁ = \intₐ^b x (4x-x²) dx$ $I_1 = \int_a^b x \sin(4x-x^2) dx$, let's rewrite the argument of sine:

4x - x² = x(4-x)

Apply King's Rule replacing $x$ $x$ with $(a+b-x)$ $(a+b-x)$. Since we proved $a+b = 4$ $a+b = 4$, substitute $x$ $x$ with $(4-x)$ $(4-x)$:

I₁ = \intₐ^b (4-x) ((4-x)(4 - (4-x))) dx

I₁ = \intₐ^b (4-x) ((4-x)x) dx

I₁ = \intₐ^b (4-x) (4x-x²) dx

Step 2: Evaluating the Integral Ratio

Expand the newly formed integral:

I₁ = 4 \intₐ^b (4x-x²) dx - \intₐ^b x (4x-x²) dx

Notice that the second term is $I₁$ $I_1$ and the first integral is $I₂$ $I_2$:

I₁ = 4I₂ - I₁ \Rightarrow 2I₁ = 4I₂ \Rightarrow (I₁)/(I₂) = 2

Step 3: Final Output Evaluation

We need the value of $36 (I₁)/(I₂)$ $36 \frac{I_1}{I_2}$:

36 \times 2 = 72

Pattern Recognition

Whenever you see $\intₐ^b x \cdot f(x(a+b-x)) dx$ $\int_a^b x \cdot f(x(a+b-x)) dx$, immediately apply King's Rule to factor out $x$ $x$. You rarely need the individual values of the integration bounds, only their sum.

Chapter Mix

Class 11 Maths: Straight Lines Class 12 Maths: Definite Integration

Q9 jee_main_2024_27_jan_morning Integration of Irrational Functions

If

\int₀¹ 1\sqrt(3+x)+\sqrt(1+x)dx=a+b\sqrt(2)+c\sqrt(3)

, where

a

,

b

,

c

are rational numbers, then

2a+3b-4c

is equal to:

A. $4$ $4$
B. $10$ $10$
C. $7$ $7$
D. $8$ $8$

Solution

Related Formula

\int xⁿ dx = xⁿ⁺¹n+1

Core Logic

To evaluate integrals with sum of square roots in the denominator, multiply and divide by the conjugate to rationalize it.

I = \int₀¹ \sqrt(3+x)-\sqrt(1+x)(\sqrt(3+x)+\sqrt(1+x))(\sqrt(3+x)-\sqrt(1+x))dx

I = \int₀¹ \sqrt(3+x)-\sqrt(1+x)(3+x) - (1+x)dx

I = (1)/(2) \int₀¹ (\sqrt(3+x) - \sqrt(1+x)) dx

Step 1: Integration and Bounds Setup

Integrate the resulting expression:

I = (1)/(2) [ (3+x)3/23/2 - (1+x)3/23/2 ]₀¹

I = (1)/(2) \cdot (2)/(3) [ (3+x)3/2 - (1+x)3/2 ]₀¹

I = (1)/(3) [ ((4)3/2 - (2)3/2) - ((3)3/2 - (1)3/2) ]

Step 2: Term Simplification

Evaluate the boundary powers: $43/2 = 8$ $4^{3/2} = 8$ $23/2 = 2\sqrt(2)$ $2^{3/2} = 2\sqrt{2}$ $33/2 = 3\sqrt(3)$ $3^{3/2} = 3\sqrt{3}$ $13/2 = 1$ $1^{3/2} = 1$

Substitute back into the expression:

I = (1)/(3) [ 8 - 2\sqrt(2) - 3\sqrt(3) + 1 ] = (1)/(3) [ 9 - 2\sqrt(2) - 3\sqrt(3) ]

I = 3 - (2)/(3)\sqrt(2) - \sqrt(3)

Step 3: Finding Co-efficients

Comparing with $a+b\sqrt(2)+c\sqrt(3)$ $a+b\sqrt{2}+c\sqrt{3}$ yields: $a = 3$ $a = 3$, $b = -(2)/(3)$ $b = -\frac{2}{3}$, $c = -1$ $c = -1$

Compute $2a+3b-4c$ $2a+3b-4c$:

2(3) + 3(-(2)/(3)) - 4(-1)

$6 - 2 + 4 = 8$ $6 - 2 + 4 = 8$

Pattern Recognition

Whenever you see a sum of square roots in the denominator of an integrand, the immediate algorithmic next step is rationalization.

Chapter Mix

Class 12 Maths: Definite 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 — Page 10

Solution

Related Formula

Core Logic

Step 1: Simplify by Adding Expressions

Step 2: Substitution and Algebraic Deconstruction

Step 3: Integrate and Evaluate Parameters

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating the Integrals

Step 2: Combining Coefficients

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Boundary Splitting Integration

Step 2: Coefficient Matrix Matching

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Definite Integral Properties

Step 2: Evaluating the Integral Ratio

Step 3: Final Output Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration and Bounds Setup

Step 2: Term Simplification

Step 3: Finding Co-efficients

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 ________.