GATE Electrical 2021 | Question: 3

Question

Let $f\left ( x \right )$ be a real-valued function such that ${f}'\left ( x_{0} \right )=0$ for some $x _{0} \in\left ( 0,1 \right ),$ and ${f}''\left ( x \right )> 0$ for all $x \in \left ( 0,1 \right )$. Then $f\left ( x \right )$ has

• A. no local minimum in $(0,1)$
• B. one local maximum in $(0,1)$
• C. exactly one local minimum in $(0,1)$
• D. two distinct local minima in $(0,1)$

Comments

Correct me if I am wrong

f(x)=(x-0.5)^2

here f'(x)=0 at x=0.5 and f''(x)>0 for every x in the interval (0,1) as f''(x)=2

So, a global minimum occurs according to the curve of f(x) which also implies a local minimum in the interval (0,1) and since it is the only one, we can go for option C

Answer 1

Ans :- option C 

Concept : If f′′(x)>0 everywhere on an interval, f is strictly convex on that interval. A critical point (where derivative zero) of a strictly convex function is a unique global (hence local) minimum.