Question

Let $f( x )=( x -1)( x -2)( x -3)$ $( x - n ), n \in N$ and $\int \frac{f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2}{f^2(x)} d x=g(x)+C$, where $C$ is arbitrary constant.

Column I		Column II
A	If $f^{\prime}( n )=5040$, then $n$ is divisible by	P	4
B	If $g(x)$ is discontinuous at 9 points, $\forall x \in R$ then $n$ is greater than	Q	6
C	If $g(x)=5$ has 8 solutions, then $n$ may be equal to	R	8
D	If the number of roots of equation $f ^{\prime}( x )=0$, be $( n -5)^2( n -1),( n >1)$	S	9 then possible values of $n$ is/are

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

Answer 1

Answer: (D)

(A) $\ln ( f ( x ))=\ln ( x -1)+\ln ( x -2)+\ldots \ldots+\ln ( x - n )$
$f ^{\prime}( x )= f ( x )\left[\frac{1}{ x -1}+\frac{1}{ x -2}+\ldots \ldots+\frac{1}{ x - n }\right] $
$f ^{\prime}( x )=( x -2)( x -3) \ldots .( x - n )+( x -1)( x -3) \ldots .( x - n )+\ldots . .+( x -1)( x -2) \ldots . .( x -( n -1))$
$f ^{\prime}( n )=( n -1)( n -2)( n -3) \cdot 3 \cdot 2 \cdot 1 (\text { all other factors except the last vanishes when } x = n ) $
$5040=( n -1) ! $
$7 !=( n -1) ! \Rightarrow n =8$
(B)$g ( x )= \frac{ f ^{\prime}( x )}{ f ( x )}=\frac{1}{ x -1}+\frac{1}{ x -2}+\ldots \ldots \ldots \frac{1}{ x - n }+ C ; $
$\text { hence } g ^{\prime}( x )<0 \text { (think!) }$
$\Rightarrow g ( x ) \text { is discontinuous at } n \text { points } n =9$
(C) $g ( x )=5$ has either $n -1$ or $n$ roots
$g(x)=\frac{f^{\prime}(x)}{f(x)} \text { or } g(x)=\frac{f^{\prime}(x)}{f(x)}+C$
$\therefore n =8 \text { or } 9$
(D) No. of roots $f ^{\prime}( x )=0$ is $n -1$ (use L.M.V.T.)
$\Rightarrow (n-1)(n-5)^2=n-1 \Rightarrow(n-1)(n-6)(n-4)=0$