Let Max _ 0 ⩽ x ⩽ 2 \ 9 - x ^ 2 5 - x \ = α and Min _ 0 ⩽ x ⩽ 2 \ 9 - x ^ 2 5 - x \ = β . If ∫ _ β - (8)/(3) ^ 2α - 1 Max \ 9 - x ^ 2 5 - x , x\ dx = α _ 1 + α _ 2 _ e ((8)/(15)) , then α _ 1 + α _ 2 is equal to ______

Correct answer: 0 Here y = 9 - x ^ 2 5 - x = 5 + x + (16)/(x - 5) (dy)/(dx) = 1 - 16 (x - 5) ^ 2 When (dy)/(dx) = 0 ⇒ 1 - 16 (x - 5) ^ 2 = 0 So critical point is x = 1 in [0, 2] Now we get y(0) = (9)/(5), y(1) = 2, y(2) = (5)/(3) So α = 2 and β = (5)/(3) I = ∫ _ - 1 ^ 3 max( 9 - x ^ 2 5 - x , x) I = ∫ _ - 1 ^ (9)/(5) (5 + x + (16)/(x - 5))dx + ∫ _ (9)/(5) ^ 3 xdx = [5x + x ^ 2 2 + 16 (x - 5)] _ - 1 ^ (9)/(5) + [ x ^ 2 2 ] _ (9)/(5) ^ 3 I = 18 + 16 ((8)/(15)) α _ 1 = 18 and α _ 2 = 16 ⇒ α _ 1 + α _ 2 = 34

JEE Main 202224 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 24 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2022 · 24 June · Morning Shift · Mathematics · Definite Integration

Let Max _ 0 ⩽ x ⩽ 2 \ 9 - x ^ 2 5 - x \ = α and Min _ 0 ⩽ x ⩽ 2 \ 9...

Let

\mathop{Max}_{0 ⩽ x ⩽ 2}\left\{\dfrac{9 - {x}^{2}}{5 - x}\right\} = \alpha

and

\mathop{Min}_{0 ⩽ x ⩽ 2}\left\{\dfrac{9 - {x}^{2}}{5 - x}\right\} = \beta

. If

{∫}_{\beta - \dfrac{8}{3}}^{2\alpha - 1} Max \left\{\dfrac{9 - {x}^{2}}{5 - x}, x\right\}dx = {\alpha}_{1} + {\alpha}_{2}\log_{e}\left(\dfrac{8}{15}\right)

, then

{\alpha}_{1} + {\alpha}_{2}

is equal to ______

Correct answer

Step-by-step solution

Here $y = \dfrac{9 - {x}^{2}}{5 - x} = 5 + x + \dfrac{16}{x - 5}$

$\dfrac{dy}{dx} = 1 - \dfrac{16}{{\left(x - 5\right)}^{2}}$

When $\dfrac{dy}{dx} = 0 ⇒ 1 - \dfrac{16}{{\left(x - 5\right)}^{2}} = 0$

So critical point is x = 1 in \left[0, 2\right] $Now we get$ y\left(0\right) = \dfrac{9}{5}, y\left(1\right) = 2, y\left(2\right) = \dfrac{5}{3} $So$ \alpha = 2 $and$ \beta = \dfrac{5}{3} $$ I = {∫}_{- 1}^{3}max\left(\dfrac{9 - {x}^{2}}{5 - x}, x\right) $$ I = {∫}_{- 1}^{\dfrac{9}{5}}\left(5 + x + \dfrac{16}{x - 5}\right)dx + {∫}_{\dfrac{9}{5}}^{3}xdx $$ = {\left[5x + \dfrac{{x}^{2}}{2} + 16\ln \left(x - 5\right)\right]}_{- 1}^{\dfrac{9}{5}} + {\left[\dfrac{{x}^{2}}{2}\right]}_{\dfrac{9}{5}}^{3} $$ I = 18 + 16\ln \left(\dfrac{8}{15}\right) $$ {\alpha}_{1} = 18 $and$ {\alpha}_{2} = 16 $$ ⇒ {\alpha}_{1} + {\alpha}_{2} = 34

Practice Definite Integration on Quantrex Academy →

More from Definite Integration

The value of the definite integral ∫_ 0 ^ 2 1 3^ (x+3)/(2) \, dx is JEE Advanced 2026 The value of the integral ∫_ 0 ^ 2 x(x^2+x+1) ( x+1 )( x^4+x^2+1 ) \, dx is equal to: JEE Main 2026 If ∫_ π/6 ^ π/4 ( (x-(π)/(3)) (x+(π)/(3))+1)dx = α _e( 3 -1) , then 9α^2 is equal to… JEE Main 2026 Let A = bmatrix 1 & 3 & -1 2 & 1 & α 0 & 1 & -1 bmatrix be a singular matrix. Let f(x) =… JEE Main 2026 Let f: ℝ arrow ℝ be such that f(xy) = f(x)f(y) , for all x, y ∈ ℝ and f(0) ≠ 0 . Let g… JEE Main 2026 The value of the integral ∫ _ -1 ^ 1 ((x^3 + |x| + 1)/(x^2 + 2|x| + 1)) dx is equal to : JEE Main 2026 The value of the integral ∫_ -π/4 ^ π/4 ( 32 ^4 x 1 + e^ x )dx is: JEE Main 2026 Let (2^ 1-a + 2^ 1+a ) , f(a) , (3^a + 3^ -a ) be in A.P. and α be the minimum value of… JEE Main 2026 Let f: [1, ∞) arrow R be a differentiable function defined as f(x) = ∫_1^x f(t)\,dt +… JEE Main 2026 The value of the integral ∫_0^∞ ( _e(x))/(x^2 + 4)\,dx is: JEE Main 2026 The value of the integral ∫_ π/6 ^ π/3 ((4 - ^2 x)/( ^4 x)) dx is: JEE Main 2026 The integral ∫_ 0 ^ 1 ^ -1 (1+x+x^2)dx is equal to: JEE Main 2026 Let f be a twice differentiable function such that f(x)=∫_ 0 ^ x (t-x)dt-∫_ 0 ^ x f(t)… JEE Main 2026 Let ∫_ -2 ^ 2 (| x| + [x x])\,dx = 2(3 - 2) + β , where [·] is the greatest integer… JEE Main 2026 The value of ∫_0^ 20π ( ^4 x + ^4 x) \, dx is equal to: JEE Main 2026 Let [·] denote the greatest integer function. Then the value of ∫_0^3 ( e^x + e^ -x [x]!… JEE Main 2026 Full Definite Integration list Definite Integration JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs