Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& \text{The equation x log x =}3-x\text{has atleast one root in}& \text{(p)}& (0,1)\\ \text{(b)}& \text{If 27 a+ 9b+ 3c+ d= 0, then the equation}4a{x}^3+3b{x}^2+2cx+d=0\text{has atleast one}& \text{(q)}& (1,3)\\ & \text{root in}& & \\ \text{(c)}& \text{If c}=\sqrt{3}\text{and}f\left(x\right)=x+\frac{1}{x},\text{then interval of x in which LMVT is applicable for}& \text{(r)}& (0,3)\\ & f\left(x\right)\text{is}& & \\ \text{(d)}& \text{If c}=\frac{1}{2}\text{and}f\left(x\right)=2x-x^2,\text{then interval of x in which LMVT is applicable for}& \text{(s)}& (-1,1)\\ & f\left(x\right)\text{is}& & \end{array}$

Which of the following is correct?

• A. A-Q,B-R,C-S,D-Q
• B. A-P,B-R,C-P,D-Q
• C. A-Q,B-R,C-S,D-P
• D. A-Q,B-R,C-Q,D-P

Answer 1

The correct option is D

A-Q,B-R,C-Q,D-P

(a) Let f'(x) $=logx-\frac{3}{x}+1\Rightarrow f\left(x\right)=(x-3)logx+c\therefore f\left(1\right)=f\left(3\right)$

From LMVT,

$f^{\prime }\left(c\right)=0,wherec\in (1,3)$

(b) Let f'(x) $=4a{x}^3+3b{x}^2+2cx+d\therefore f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e\therefore f\left(0\right)=f\left(3\right)=e\because 27a+9b+3c+d=0$

From LMVT,

$f^{\prime }\left(c\right)=0,wherec\in (0,3)$

(c) $\frac{f\left(b\right)-f\left(a\right)}{b-a}=f^{\prime }\left(\sqrt{3}\right)=\frac{2}{3}\Rightarrow \frac{ab-1}{ab}=\frac{2}{3}$

$\text{Only}a=1;b=3\text{satisfies from given options}$

(d) $\frac{f\left(b\right)-f\left(a\right)}{b-a}=f^{\prime }\left(\frac{1}{2}\right)\Rightarrow a+b=1$

$\text{Only}a=0;b=1\text{satisfies from given options}$