COMPASSLearning

Talk to us Register Login

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Definite Integrals

Let

f (x) + 2 f (\frac{1}{x}) = x^{2} + 5

and

2 g (x) - 3 g (\frac{1}{2}) = x, x > 0

. If

α = \int_{1}^{2} f (x) d x

, and

β = \int_{1}^{2} g (x) dx

, then the value of

9 α + β

is:

(A)1
(B)0
(C)10
(D)11✓ Correct

Answer: (D)

Solution

Replacing

x

by

\frac{1}{x}

in

f (x) + 2 f (\frac{1}{x}) = x^{2} + 5

and eliminating

f (\frac{1}{x})

gives

f (x) = \frac{2}{3 x ^{2}} - \frac{x ^{2}}{3} + \frac{5}{3}

, so

α = \int_{1}^{2} f d x = \frac{11}{9}

. Putting

x = \frac{1}{2}

in

2 g (x) - 3 g (\frac{1}{2}) = x

gives

g (\frac{1}{2}) = - \frac{1}{2}

, hence

g (x) = \frac{x}{2} - \frac{3}{4}

and

β = \int_{1}^{2} g d x = 0

. Therefore

9 α + β = 11

.

JEE Main · 2025 · 04 Apr 2025 · 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 →