Question

If ${(x-a)}^{2m}{(x-b)}^{2n+1}$, where $m$ and $n$ are positive integers and $a>b$, is the derivative of a function $f$, then-

• A. $x=a$ gives neither a maximum nor a minimum
• B. $x=a$ gives a maximum
• C. $x=b$ gives neither a maximum nor a minimum
• D. None of these

Answer 1

The correct option is A $x=a$ gives neither a maximum nor a minimum

$f^{\prime }\left(x\right)=2m(x-a{)}^{2m-1}(x-b{)}^{2n+1}+(2n+1)(x-a{)}^{2m}.(x-b{)}^{2n}$
$=(x-a{)}^{2m-1}(x-b{)}^{2n}\left[2m\right(x-a)+(2n+1\left)\right(x-a\left)\right]$
$=(x-a{)}^{2m-1}(x-b{)}^{2n}\left[2x\right(m+n)-a(2m+2n+1\left)\right]$
$f^{\prime \prime }\left(x\right)$
$=(2m-1)(x-a{)}^{2m-2}(x-b{)}^{2n}\left[\right[2x(m+n)-a(2m+2n+1)]+2n(x-a{)}^{2m-1}(x-b{)}^{2n-1}[2x(m+n)-a(2m+2n+1)]+(x-a{)}^{2m-1}(x-b{)}^{2n}2(m+n)$
Now
$f^{\prime \prime }(x{)}_{x=a}=0$
Hence neither a maximum nor a minimum at $x=a$.