Let f(x)= x/sinx, x∈(0,1) 1, x=0 , Consider the integral I_n=√n∫01/nf(x)e^(-nx)dx Then limn→∞I_n

Correct answer: (B) exists and is 0. f(x) is an increasing function. so, f(x)∈1,1/sin1 ∀x∈[0,1) Now, √n∫01/ne^(-nx)dx≤√n∫01/nf(x)e^(-nx)dx≤√nsin1∫01/ne^(-nx)dx ⇒limn→∞(1-1e√n≤limn→∞I_n≤1-1e)/((sin1)√n) ⇒0≤limn→∞I_n≤0 ⇒limn→∞I_n=0

Q39 / 91 · KVPY 2020 (SB/SX) · Mathematics

Let $f (x) = \{\begin{matrix} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{matrix},$ Consider the integral $I_{n} = \sqrt{n} \int_{0}^{1 / n} f (x) e^{- n x} d x$
Then $\lim_{n \to \infty} I_{n}$

A
does not exist
Correct answer
B
exists and is $0$
C
exists and is $1$
D
exists and is $1 - e^{- 1}$

Solution

$f (x)$ is an increasing function.
so, $f (x) \in [1, \frac{1}{\sin 1}) \forall x \in [0, 1)$
Now,

$\sqrt{n} \int_{0}^{1 / n} e^{- n x} d x \leq \sqrt{n} \int_{0}^{1 / n} f (x) e^{- n x} d x \leq \frac{\sqrt{n}}{\sin 1} \int_{0}^{1 / n} e^{- n x} d x$
$\Rightarrow \lim_{n \to \infty} \frac{1 - \frac{1}{e}}{\sqrt{n}} \leq \lim_{n \to \infty} I_{n} \leq \frac{1 - \frac{1}{e}}{(\sin 1) \sqrt{n}}$
$\Rightarrow 0 \leq \lim_{n \to \infty} I_{n} \leq 0$
$\Rightarrow \lim_{n \to \infty} I_{n} = 0$

Hence, the correct answer is option: 'B'.

Let f(x)=xsinx,x∈(0,1)1,x=0, Consider the integral In=n∫01/nf(x)e-nxdx Then limn→∞In

Solution

More Definite Integration Questions

Let $f (x) = \{\begin{matrix} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{matrix},$ Consider the integral $I_{n} = \sqrt{n} \int_{0}^{1 / n} f (x) e^{- n x} d x$
Then $\lim_{n \to \infty} I_{n}$