Question

Assertion :Statement-1 : If $a,b$ in $R$ and $a<b$, then there is atleast one real number $c\in (a,b)$ such that $\frac{c}{a+b}=\frac{b^2+a^2}{4c^2}$.
Reason: Statement-2 : If $f\left(x\right)$ is continuous in $[a,b]$ and derivable in $(a,b)\&f^{\prime }\left(c\right)=0$ for atleast one $c\in (a,b)$, then it necessarily implies that $f\left(a\right)=f\left(b\right)$.

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
• C. Statement-1 is True, Statement-2 is False.
• D. Statement-1 is False, Statement-2 is True.

Answer 1

The correct option is C Statement-1 is True, Statement-2 is False.

Statement-1 : Consider $f\left(x\right)=x^4$
which is continuous and differentiable everywhere.
So, it is continuous on $[a,b]$ and differentiable on $(a,b)$
Applying LMVT in $[a,b]$, we get $c\in (a,b)$ such that
$4c^3=\frac{b^4-a^4}{b-a}$
$\Rightarrow 4c^3=\frac{(b^2-a^2)(b^2+a^2)}{b-a}$
$\frac{c}{a+b}=\frac{b^2+a^2}{4c^2}$
Hence, statement-1 is true.
Statement-2 is converse of Rolle's theorem which is not true.
We can show by an example .
Let $f\left(x\right)=x^3$ on $[-1,1]$
Clearly, f(x) is continuous on [-1,1] and differentiable in $(-1,1)$
$f^{\prime }\left(x\right)=3x^2$
$f^{\prime }\left(0\right)=0$
But, there are no two distinct points $d,e$ in [-1,1] such that $f\left(d\right)=f\left(e\right)$
Hence, statement 2 is not correct.