Question

Let $\alpha ={\int }_0^{\frac{\pi }{4}}\frac{\cos x}{{\sin }^{3}x+{\cos }^{3}x}dx;\beta ={lim}_{n\to \infty }{\left(\frac{{\prod }_{r=1}^n(n^3+r^3)}{n^{3}n}\right)}^{\frac{1}{n}}\text{and}\mu ={\int }_0^1\frac{dx}{1+x^3}$

Column - I	Column - II
(P) If $\mu -k\alpha =0,$ then k is	(1) 0
(Q) If $\ln \beta =\ln a-3+b\mu ,$ then the value of $(a+b)$ is	(2) 1
(R) If $\mu =\frac{1}{3}(\ln a+\frac{\pi }{\sqrt{b}}),$ then the value of $\frac{2b}{a}$ is	(3) 2
(S) The value of $\left[\ln \left(\frac{4}{\beta }\right)\right]$ is (where [.] denotes the greatest integer function)	(4) 3
	(5) 4
	(6) 5

Then the correct option is

• A. P(4); Q(6); R(2); S(1)
• B. P(2); Q(6); R(4); S(2)
• C. P(2); Q(5); R(4); S(2)
• D. P(2); Q(5); R(2); S(3)

Answer 1

The correct option is B P(2); Q(6); R(4); S(2)

(P) $\alpha {\int }_0^{\frac{\pi }{4}}\frac{\cos x}{{\sin }^{3}x+{\cos }^{3}x}dx={\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}x}{1+{\tan }^{3}x}dx$

let $\tan x=t,{\sec }^{2}xdx=dt$

$\alpha ={\int }_0^1\frac{dt}{1+t^3}=\mu \Rightarrow \mu -\alpha =0$

(Q) $\ln \beta ={lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^n\ln (1+{\left(\frac{r}{n}\right)}^3)$

$\Rightarrow \ln \beta ={\int }_0^1\ln (1+x^3)dx$

${\begin{array}{c}=x\ln (1+x^3)\end{array}|}_0^1-{\int }_0^1\frac{3x^2}{1-x^3}xdx$

$\Rightarrow \ln \beta =\ln 2-3{\int }_0^1\frac{x^3}{1+x^3}dx$

$\Rightarrow \ln \beta =\ln 2-3[1-{\int }_0^1\frac{1}{1+x^3}dx]$

$\Rightarrow \ln \beta =\ln 2-3+3\mu$

(R) $\mu ={\int }_0^1\frac{dx}{(1+x)(x^2-x+1)}$

$={\int }_0^1[\frac{1}{3}.\frac{1}{x+1}-\frac{1}{6}\frac{(2x-1)-3}{x^2-x+1}]dx$

$\Rightarrow \mu ={\left[\frac{1}{3}\ln \right(x+1)-\frac{1}{6}\ln (x^2-x+1)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)]}_0^1$

$\Rightarrow \mu =\frac{1}{3}(\ln 2+\frac{\pi }{\sqrt{3}})$

(S) $\because \ln \beta =\ln 2-3+3\mu$

$=\ln 2-3+(\ln 2+\frac{\pi }{\sqrt{3}})$

$\ln \left(\frac{4}{\beta }\right)=3-\frac{\pi }{\sqrt{3}}$

$\left[\ln \left(\frac{4}{\beta }\right)\right]=1$