COMPASSLearning

Talk to us Register Login

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Definite Integrals

Let

α > 0

. If

\int_{0}^{α} \frac{x}{x + α - x} dx = \frac{16 + 20 2}{15}

, then

α

is equal to :

(A)2✓ Correct
(B)4
(C) $2$
(D) $22$

Answer: (A)

Solution

Rationalising,

\frac{x}{x + α - x} = \frac{x ( x + α + x )}{α} .

Thus the integral is

\frac{1}{α} (\int_{0}^{α} x x + α d x + \int_{0}^{α} x^{3/2} d x) .

The substitution

u = x + α

gives

\int_{0}^{α} x x + α d x = \frac{4 2 + 4}{15} α^{5/2}

, and

\int_{0}^{α} x^{3/2} d x = \frac{2}{5} α^{5/2} .

Adding and dividing by

α

gives

\frac{10 + 4 2}{15} α^{3/2} .

Equating to

\frac{16 + 20 2}{15}

,

α^{3/2} = \frac{16 + 20 2}{10 + 4 2} = 22 = 2^{3/2}

, so

α = 2.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Definite Integrals →

More JEE Main questions on Definite Integrals

The value of $\int_{- π /6}^{π /6} (\frac{π + 4 x ^{11}}{1 - s i n ( ∣ x ∣ + π /6 )}) d x$ is equal to
JEE Main 2026View solution →
$6 \int_{0}^{π} ∣ (sin 3 x + sin 2 x + sin x) ∣ d x$ is equal to.......
JEE Main 2026View solution →
Let $[\cdot]$ denote the greatest integer function and $f (x) = lim_{n \to \infty} \frac{1}{n ^{3}} \sum_{k = 1}^{n} [\frac{k ^{2}}{3 ^{x}}]$ . Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal …
JEE Main 2026View solution →
If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) dx = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to ____.
JEE Main 2026View solution →
Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is
JEE Main 2026View solution →
The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[ x ] + 4}) dx$ , where [•] denotes the greatest integer function, is
JEE Main 2026View solution →