Given that for each a in (0,1),the limit g(a) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1.$ To find $g(a)$,we evaluate the integral $g(a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$.For $a = \frac{1}{2}$,$g\left(\

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(C) $1.$ To find $g(a)$,we evaluate the integral $g(a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$.For $a = \frac{1}{2}$,$g\left(\frac{1}{2}ight) = \int_0^1 t^{-1/2}(1-t)^{-1/2} dt = \int_0^1 \frac{dt}{\sqrt{t(1-t)}}$.Using the substitution $t = \sin^2 	heta$,$dt = 2 \sin 	heta \cos 	heta d	heta$,the integral becomes $\int_0^{\pi/2} \frac{2 \sin 	heta \cos 	heta}{\sin 	heta \cos 	heta} d	heta = \int_0^{\pi/2} 2 d	heta = \pi$.Thus,$g\left(\frac{1}{2}ight) = \pi$.$2.$ We observe that $g(1-a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{-(1-a)}(1-t)^{(1-a)-1} dt = \lim_{n ightarrow 0^{+}} \int_n^{1-n} t^{a-1}(1-t)^{-a} dt$.Using the property $\int_n^{1-n} f(t) dt = \int_n^{1-n} f(1-t) dt$,we get $g(1-a) = \lim_{n ightarrow 0^{+}} \int_n^{1-n} (1-t)^{a-1} t^{-a} dt = g(a)$.Since $g(1-a) = g(a)$,differentiating both sides with respect to $a$ using the chain rule gives $-g'(1-a) = g'(a)$.At $a = \frac{1}{2}$,$-g'\left(\frac{1}{2}ight) = g'\left(\frac{1}{2}ight)$,which implies $2g'\left(\frac{1}{2}ight) = 0$,so $g'\left(\frac{1}{2}ight) = 0$.Therefore,the correct pair is $(A, D)$.

Similar Questions

Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{1 + x}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$,then the value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to

If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

The number of continuous functions $f : [0, \frac{3}{2}] \rightarrow (0, \infty)$ satisfying the equation $4 \int_0^{3/2} f(x) dx + 125 \int_0^{3/2} \frac{dx}{\sqrt{f(x)+x^2}} = 108$ is

Which of the following inequalities is/are $TRUE$?
$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$
$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$
$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$
$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module