Question

Column I		Column II
A	The least value of 'a' for which the equation, $\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$ has atleast one solution on the interval $(0,\pi /2)$ is	P	20
B	A closed vessel tapers to a point both at its top $E$ and its bottom $F$ and is fixed with EF vertical when the depth of the liquid in it is $xcm$, the volume of the liquid in it is, $x^2(15-x)cu$. cm. The length $EF$ is	Q	13
C	If Rolle's theorem is applicable to the function $f(x)=\frac{\ln x}{x}(x>0)$ over the interval $[a,b]$ where $a,b\in I$, then the value of $\left(a^2+b^2\right)$ is equal to	R	10
		S	9

• A. (A) S ; (B) R ; (C) P
• B. (A) S ; (B) R ; (C) Q
• C. (A) Q ; (B) R ; (C) P
• D. (A) S ; (B) P; (C) S

Answer 1

Answer: (C)

(A) $\frac{dy}{dx}=-\frac{4\cos x}{{\sin }^{2}x}+\frac{\cos x}{(1-\sin x{)}^2}=0$ gives $\sin x=\frac{2}{3}$
note that $f(x)\to \infty$ as $x\to 0^{+}$or $x\to \frac{{\pi }^{-}}{2}$ and between two maxima we have a minima.
(B) ${\frac{dv}{dx}=3x(10-x)=0\Rightarrow x=0;x=10;\frac{d^{2}v}{d{x}^2}]}_{x=10}<0\Rightarrow v$ is $\max$ at $x=10$ $\Rightarrow EF=10cm$. Ans.
(C) Clearly, for $f(a)=f(b)$ where $a,b\in I$
So, a must be 2 .

$\Rightarrow \frac{\ln b}{b}=\frac{\ln 2}{2}$
$\Rightarrow b=4$
$\text{ So, }(a+b)=6$