Question

Let $f\left(x\right)=(x-1{)}^4(x-2{)}^n,n\in N$. Then $f\left(x\right)$ has

• A. A maximum at $x=1$ if $n$ is odd.
• B. A maximum at $x=1$ if $n$ is even.
• C. A minimum at $x=1$ if $n$ is even.
• D. A minimum at $x=2$ if $n$ is even.

Answer 1

The correct option is D A minimum at $x=2$ if $n$ is even.

Given : $f\left(x\right)={(x-1)}^4{(x-2)}^n,n\in N$

$f\left(0\right)=(-2{)}^n,f(0.5)=(0.5{)}^4(-1.5{)}^n$

$f\left(1.5\right)=\left(0.5{)}^4\right(-0.5{)}^n,f\left(1.75\right)=\left(0.75{)}^4\right(-0.25{)}^n$

0.5>0

$f\left(0.5\right)>f\left(0\right)$ if n is odd $\Rightarrow f^{\prime }\left(x\right)$ is positive

$f\left(0.5\right)<f\left(0\right)$ if n is even $\Rightarrow f^{\prime }\left(x\right)$ is negative

1.5<1.75

$f\left(1.5\right)>f\left(1.75\right)$ if n is odd $\Rightarrow f^{\prime }\left(x\right)$ is negative

$f\left(1.5\right)<f\left(1.75\right)$ if n is even $\Rightarrow f^{\prime }\left(x\right)$ is positive

$\therefore$ at $x=1f\left(x\right)$ has minimum if n is even

at $x=1f\left(x\right)$ has maximum if n is odd

($\because$ positive slope to negative slope means maximum

negative slope to positive means minimum)

$f\left(3\right)=2^4$

$f\left(4\right)=3^4\ast 2^n$

$f\left(4\right)>f\left(3\right)$ irrespective of n $\Rightarrow f^{\prime }\left(x\right)$ is always positive

f(x) has minimum at $x=2$ if n is even.