Question

Match the definite integral in the Column-I with its value in Column-II

Column - I	Column - II
(A) ${lim}_{x\to \infty }\frac{{\left({\int }_0^x{e}^{x}dx\right)}^2}{{\int }_0^x{e}^{2x^2}dx}$ equals	(p) $\ln 6$
(B) The value of ${\int }_1^4\ln \left[x\right]dx$ is	(q) 2
(C) $\frac{4}{669}\left({\int }_{669}^{1338}\frac{f\left(x\right)}{f\left(x\right)+f(2007-x)}dx\right)$ is	(r) 0
(D) ${\int }_0^1\ln (\frac{1}{x}-1)dx$	(s) 1
	(t) A non negative integer

• A. A(r, t), B(p), C(q, t), D(r, t)
• B.

Answer 1

The correct option is A A(r, t), B(p), C(q, t), D(r, t)

(A) ${lim}_{x\to \infty }\frac{\left({\int }_0^x{e}^{x}dx\right)}{{\int }_0^x{e}^{2x^2}dx}$

$={lim}_{x\to \infty }\frac{(e^x-1{)}^2}{{\int }_0^x{e}^{2x^{2}dx}}={lim}_{x\to \infty }\frac{2(e^x-1)-e^x}{e^{2x^2}}$

$={lim}_{x\to \infty }\frac{2[2e^{2x}-e^x]}{e^{2x^2}.4x}=0$

(B) ${\int }_1^4\ln \left[x\right]dx$

$={\int }_1^2\log 1dx+{\int }_2^3\log 2dx+{\int }_0^4\log 3dx$

$=\log 2+\log 3=\ln 6$

(C) $\frac{4}{669}{\int }_{669}^{1338}\frac{f\left(x\right)}{f\left(x\right)+f(2007-x)}dx$

$=\frac{4}{669}{\int }_{669}^{1338}\frac{f(2007-x)}{f(2007-x)+f\left(x\right)}dx$

$=\frac{4}{669}\times \frac{669}{2}=2$

(D) ${\int }_0^1\ln (\frac{1}{x}-1)dx={\int }_0^1\ln \left(\frac{1-x}{x}\right)dx$

$={\int }_0^1\ln \left(\frac{x}{1-x}\right)dx=-{\int }_0^1\ln \left(\frac{1-x}{x}\right)dx$

$\Rightarrow {\int }_0^1\ln (\frac{1}{x}-1)dx=0$