Let f ( x ) = lim n → ∞ n n ( x + n ) x + n 2 … x + n n n ! x 2 + n 2 x 2 + n 2 4 ⋯ x 2 + n 2 n 2 x n , for all x > 0 . Then

f ' ( 3 ) f ( 3 ) ≥ f ' ( 2 ) f ( 2 )

For each positive integer n , let y n = 1 n n + 1 n + 2 n + 3 . . . . . n + n 1 n For x ∈ R , let · be the greatest integer less than or equal to x . If lim n → ∞ y n = L , then the value of L is

We have, y n = 1 n n + 1 n + 2 n + 3 . . . . . n + n 1 n ⇒ y n = n + 1 n n + 2 n n + 3 n . . . . . n + n n 1 n ⇒ y n = 1 + 1 n 1 + 2 n 1 + 3 n . . . . . 1 + n n 1 n ⇒ y n = ∑ r = 1 n 1 + r n 1 n ⇒ lim n → ∞ y n = lim n → ∞ ∑ r = 1 n 1 + r n 1 n ⇒ L = lim n → ∞ ∑ r = 1 n 1 + r n 1 n ⇒ log L = 1 n lim n → ∞ ∑ r = 1 n log 1 + r n ⇒ log L = ∫ 0 1 log 1 + x d x ⇒ log L = log 1 + x x 0 1 - ∫ 0 1 x 1 + x d x ⇒ log L = log 2 - ∫ 0 1 x + 1 - 1 1 + x d x ⇒ log L = log 2 - x - log 1 + x 0 1 ⇒ log L = log 2 - 1 - log 2 ⇒ log L = 2 log 2 - 1 ⇒ log L = log 4 - 1 ⇒ L = 4 e ⇒ L = 1

The value of the integral ∫ 0 1 2 1 + 3 ( x + 1 ) 2 ( 1 - x ) 6 1 4 d x is

Let I = ∫ 0 1 / 2 1 + 3 ( x + 1 ) 2 ( 1 - x ) 6 1 / 4 d x ⇒ I = ∫ 0 1 / 2 1 + 3 1 - x 2 1 + x 1 - x 2 1 / 4 d x Put 1 + x 1 - x = v ⇒ d d x 1 + x 1 - x = d v d x ⇒ 1 · 1 - x - - 1 1 + x 1 - x 2 = d v d x ⇒ d x 1 - x 2 = d v 2 When x = 0 , v = 1 and x = 1 2 , v = 3 ∴ I = ∫ 1 3 ( 1 + 3 ) ( v ) 2 / 4 d v 2 ⇒ I = ( 1 + 3 ) 2 ∫ 1 3 v - 1 2 d v ⇒ I = ( 1 + 3 ) 2 2 v 1 3 ⇒ I = ( 1 + 3 ) ( 3 - 1 ) ⇒ I = 3 - 1 ⇒ I = 2

Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t . Then

f 2 x is an odd function of x on R .

f x is monotonically decreasing on 0 , 1 .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 25

HARD

JEE Main/Advance

IMPORTANT

Earn 100

The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t^{2}}{1 + t^{4}} d t = 2 x - 1$ is ___.

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Important Questions on Definite Integration

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 26

The value of

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x

is equal to

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 30 - Definite Integration>Exercise-3>Q 27

$Let f (x) = \lim_{n \to \infty} {[\frac{n^{n} (x + n) (x + \frac{n}{2}) \dots (x + \frac{n}{n})}{n! (x^{2} + n^{2}) (x^{2} + \frac{n^{2}}{4}) \dots (x^{2} + \frac{n^{2}}{n^{2}})}]}^{\frac{x}{n}}$ , for all $x > 0 .$ Then