Question

if $\int$'(x)=(x-1)$(x-2{)}^{{}^2}$^, then for $\int$(x):

• A. x=2 is point of maxima
• B. x=2 is point of minima
• C. x=2 is either point of maxima or point of minima
• D. x= 2 is neither point of maxima nor point of minima

Answer 1

The correct option is B x=2 is point of minima

$f\left(x\right)=(x-1)(x-2{)}^2$

differentiating both sides w.r.t. 'x' we get

$f^{{}^{\prime }}\left(x\right)=2(x-1)(x-2)+(x-2{)}^2$

$f^{{}^{\prime }}\left(x\right)=(x-2)(2x-2+x-2)$

$f^{{}^{\prime }}\left(x\right)=(x-2)(3x-4)$

& $f^{\prime \prime }\left(x\right)=3(x-2)+(3x-4)$ [ again differentiating ]

for maxima or minima $f^{{}^{\prime }}\left(x\right)=0$

$\therefore (x-2)(3x-4)=0$

$\Rightarrow x=2\text{or}x=\frac{4}{3}$

Now according to the options putting x = 2 in $f^{\prime \prime }\left(x\right)$ we get, $f^{\prime \prime }\left(x\right)=2>0$

$\therefore x=2$ is the point of minima.

Option 3 is correct.