Let y = f(x)$y = f(x)$ be a thrice differentiable function in (-5, 5)$(-5, 5)$ . Let the tangents to the curve y = f(x)$y = f(x)$ at (1, f(1))$(1, f(1))$ and (3, f(3))$(3, f(3))$ make angles (π)/(6)$\frac{\pi}{6}$ and (π)/(4)$\frac{\pi}{4}$ , respectively with positive x-axis. If 27∫₁³((f'(t))² + 1)f''(t)dt = α + β √(3) where α, β are integers, then the value of α +β equals$$27\int_{1}^{3}\left(\left(f'(t)\right)^{2} + 1\right)f''(t)dt = \alpha + \beta \sqrt{3} \quad \text{where } \alpha, \beta \text{ are integers, then the value of } \alpha +\beta \text{ equals}$$

Related Formula

Integration by parts formula: ∫ u dv = uv - ∫ v du$\int u \, dv = uv - \int v \, du$.

Core Logic

Let I(x) = ∫₀x t² (x - t) dt$I(x) = \int_{0}^{x} t^2 \sin(x - t) dt$. Use integration by parts. Let u = t²$u = t^2$ and dv = (x - t) dt v = (x - t)$dv = \sin(x - t) dt \implies v = \cos(x - t)$.

Step 1: Second Integration by Parts

Now integrate ∫₀x 2t (x - t) dt$\int_{0}^{x} 2t \cos(x - t) dt$ by parts again: Let u = 2t$u = 2t$ and dv = (x - t) dt v = - (x - t)$dv = \cos(x - t) dt \implies v = -\sin(x - t)$.

Step 2: Equating and Solving

Substitute this back into the first equation:

We are given I(x) = x²$I(x) = x^2$. Therefore:

The solutions for x = 1$\cos x = 1$ are x = 2nπ$x = 2n\pi$, where n$n$ is an integer. We need the roots in the interval [0, 100]$[0, 100]$. 0 ≤ 2nπ ≤ 100 0 ≤ n ≤ (100)/(2π) ≈ (100)/(6.28) ≈ 15.92$0 \leq 2n\pi \leq 100 \implies 0 \leq n \leq \frac{100}{2\pi} \approx \frac{100}{6.28} \approx 15.92$.

Since n$n$ must be an integer, n$n$ takes values 0, 1, 2, , 15$0, 1, 2, \dots, 15$. The total number of solutions is 16$16$.

Pattern Recognition

Convoluted-looking definite integrals containing a shift (x-t)$(x-t)$ often rapidly unpack through successive integration by parts where the polynomial variable diminishes until exhaustion.

Chapter Mix

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

Related Formula

Core Logic

Let ₑ t = x ⇒ t = e^x ⇒ dt = e^x dx$\log_e t = x \Rightarrow t = e^x \Rightarrow dt = e^x dx$.

Step 1: Integration

Let g(x) = - (x/2)$g(x) = -\cot(x/2)$. Then g'(x) = (1)/(2)cosec²(x/2)$g'(x) = \frac{1}{2}\operatorname{cosec}^2(x/2)$. The integral fits the standard form ∫ e^x (g(x) + g'(x)) dx$\int e^x (g(x) + g'(x)) dx$.

Step 2: Finding Constant of Integration

Given f(eπ/2) = -eπ/2$f(e^{\pi/2}) = -e^{\pi/2}$:

Step 3: Calculating Alpha

Find f(eπ/4)$f(e^{\pi/4})$:

Recall (π/8) = √(2) + 1$\cot(\pi/8) = \sqrt{2} + 1$.

Thus, α = -(1 + √(2)) = -1 - √(2)$\alpha = -(1 + \sqrt{2}) = -1 - \sqrt{2}$.

Pattern Recognition

Logarithmic inputs in integrals are the classic cue to substitute t = e^x$t = e^x$ which inherently sets up the e^x(f(x) + f'(x))$e^x(f(x) + f'(x))$ format.

Chapter Mix

Class 12 Maths: Integrals Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Expand the integral by distributing f(y)$f(y)$:

Since the integrals evaluate to constants, let:

Thus, the function becomes:

Step 1: Setting up equation for A

Substitute f(y) = e^y + A + B y e^y$f(y) = e^y + A + B y e^y$ back into the integral for A$A$:

Integrate by parts: ∫₀¹ y e^y dy = [y e^y - e^y]₀¹ = (e - e) - (0 - 1) = 1$\int_0^1 y e^y dy = [y e^y - e^y]_0^1 = (e - e) - (0 - 1) = 1$ ∫₀¹ y² e^y dy = [y² e^y - 2y e^y + 2e^y]₀¹ = (e - 2e + 2e) - 2 = e - 2$\int_0^1 y^2 e^y dy = [y^2 e^y - 2y e^y + 2e^y]_0^1 = (e - 2e + 2e) - 2 = e - 2$ ∫₀¹ A y dy = (A)/(2)$\int_0^1 A y dy = \frac{A}{2}$

So,

Step 2: Setting up equation for B

Substitute f(y)$f(y)$ back into the integral for B$B$:

Integrate terms: ∫₀¹ e^y dy = e - 1$\int_0^1 e^y dy = e - 1$ ∫₀¹ A dy = A$\int_0^1 A dy = A$ ∫₀¹ B y e^y dy = B(1) = B$\int_0^1 B y e^y dy = B(1) = B$

So,

0 = e - 1 + A A = 1 - e$0 = e - 1 + A \implies A = 1 - e$

Step 3: Calculating f(0)

Substitute$

Step 3: Calculating f(0)

Substitute $A = 1 - einto the expression for$ into the expression for $f(0): From the general equation,$: From the general equation, $f(x) = e^x + A + B x e^x.

Core Logic

Split the integral into two terms:

Step 1: Integration by Parts

Take the first integral I₁ = ∫ (1)/( ⁵ x) · ² x dx$I_1 = \int \frac{1}{\sin^5 x} \cdot \sec^2 x dx$. Apply Integration by Parts taking u = (1)/( ⁵ x)$u = \frac{1}{\sin^5 x}$ and dv = ² x dx$dv = \sec^2 x dx$:

Step 2: Re-substituting into Total Integral

The integral terms exactly cancel out:

Thus, f(x) = ( x)/( ⁵ x)$f(x) = \frac{\tan x}{\sin^5 x}$.

Step 3: Function Evaluation

Calculate f((π)/(6))$f\left(\frac{\pi}{6}\right)$:

Calculate f((π)/(4))$f\left(\frac{\pi}{4}\right)$:

Difference:

Pattern Recognition

Integrals of the form ∫ (complicated term 1 - complicated term 2) dx$\int (\text{complicated term 1} - \text{complicated term 2}) dx$ often cancel perfectly when Integration by Parts is applied to just one of the terms. Look for derivative relationships between the numerators.

Chapter Mix

Class 12 Mathematics: Integrals

Core Logic

We need to find the explicit form of f(t)$f(t)$. Let x² + 1 = t$x^2 + 1 = t$. This implies x² = t - 1$x^2 = t - 1$. Substitute this back into the given equation f(x²+1) = x⁴ + 5x² + 2$f(x^2+1) = x^4 + 5x^2 + 2$:

Step 1: Evaluate the Integral

Now, we compute the definite integral:

Substitute upper limit 3$3$:

Chapter Mix

Class 12 Mathematics: Integrals Class 11 Mathematics: Functions