Question

If $f^{\prime }(x)=(x-a{)}^{2n}(x-b{)}^{2p+1}$ where n and p are positive integers, then :

• A. x = a is a point of minimum
• B. x = a is a point of maximum
• C. x is not a point of maximum or minimum
• D. none of these

Answer 1

Answer: (C)

Given, $f^{\prime }(x)=(x-a{)}^{2n}(x-b{)}^{2p+1}$
$\therefore$ For maxima or minima
f '(x) = 0
$\Rightarrow {\left(x-a\right)}^{2n}=0$ or ${\left(x-b\right)}^{2p+1}=0$
$\Rightarrow x=a$ and $x=b$
Now, $f^{\prime }\left(a-h\right)=\left[{\left(a-h-a\right)}^{2n}.{\left(a-h-b\right)}^{2p+1}\right]$
$=h^{2n}{\left(a-b-h\right)}^{2p+1}$
and $f^{\prime }\left(a+h\right)=\left[\left(a+h-a\right)2n.{\left(a+h-b\right)}^{2p+1}\right]$
$=h^{2n}{\left(a-b+h\right)}^{2p+1}$
Thus, f '(x) does not change sign as x passes through 'a', so x is not a point of maximum or minimum.