GATE DA 2025 | Question: 41

Question

​​​​Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice-differentiable function and suppose its second derivative satisfies $f^{\prime \prime}(x)>0$ for all $x \in \mathbb{R}$. Which of the following statements is/are ALWAYS correct?

• A. $f$ has a local minima
• B. There does not exist $x$ and $y, x \neq y$, such that $f^{\prime}(x)=f^{\prime}(y)=0$
• C. $f$ has at most one global minimum
• D. $f$ has at most one local minimum

Comments

we know that it is a convex(concave up) function so option C,D is always true
on the other hand
option B: we know that f''(x) is always greater than 0, hence whenever f'(x) is so it will be a local minima in this case it would be the global minima too as it is convex, hence there can be ATMOST ONE value that f'(x) can be zero? so option B is also correct

correct me if i am wrong

Answer 1

Suppose $f: \mathbb{R}\mapsto \mathbb{R}$ and $f''(x) >0$ for every $x \in \mathbb{R}$. The immediate conclusion is $f(x)$ is concave in $(\infty, \infty)$.

A. Consider $f(x) = e^x$, then  $f''(x) = e^x > 0, \forall x \in \mathbb{R}$. But $f(x)$ doen't have any local minima in $(-\infty, \infty)$. So option A is false.

 

B. For a contradiction let's assume there is  $x, y$ such that $x\neq y$ and $f'(x) = f'(y)=0$, without loss of generality let  $x < y$.

Lagrange mean value theorem:- If $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then $\exists c\in(a,b)$ such that $g'(c) = \dfrac{g(b)-g(a)}{b-a}$.

Now since $f$ is twice differentiable on $(-\infty, \infty)$, it implies $f'$ is continuous on $(-\infty, \infty)$. Now consider the interval $[x,y]$, in this interval $f'$ is continuous and $f'$ is differentiable on $(x,y)$. it has satisfied the hypothesis of LMVT now apply it,

So from mean valule theorem it implies that  $\exists c\in(x,y)$ such that  $f''(c) = \dfrac{f'(y)-f'(x)}{b-a} = \dfrac{0}{b-a} = 0$. which is a contradiction since $f''(x) >0$ for every $x \in \mathbb{R}$.

By proof by contradiction, There does not exist $x$ and $y$, $x\neq y$  such that $f'(x) = f'(y) = 0$.

And even it's true that There does not exist $x$ and $y$, $x\neq y$  such that $f'(x) = f'(y)$.

SO option B is always True.

 

C. For a contradiction assume, global minima occurs at two different point $x,y$.

First derivative theorem:- If $f$ has local maxima/minima at $x=c$, c is an interior point and $f'(c) $ exists then $f'(c) = 0$ .(proof is very simple, try to do it.)

Since $x,y$ are interior points and $f$ has local minima at those points and derivative exists those points $\implies$ $f'(x) = f'(y) = 0$.

we have already proved in option B, that There does not exist $x$ and $y$, $x\neq y$  such that $f'(x) = f'(y) = 0$.

So by proof by contradiction $f$ has atmost one global minima. SO  C is always True.

 

D.

For a contradiction assume, local minima occurs at two different point $x,y$.

First derivative theorem:- If $f$ has local maxima/minima at $x=c$, c is an interior point and $f'(c) $ exists then $f'(c) = 0$ .(proof is very simple, try to do it.)

Since $x,y$ are interior points and $f$ has local minima at those points and derivative exists those points $\implies$ $f'(x) = f'(y) = 0$.

we have already proved in option B, that There does not exist $x$ and $y$, $x\neq y$  such that $f'(x) = f'(y) = 0$.

So by proof by contradiction $f$ has atmost one  local minima.

D is always True.

 

The answer is $\mathrm{B,C,D}$.

 

Another wasy to answer this question is , $f''(x)> 0, \forall x  \in R$, so $f'$ is strictly increasing function, so $f'(x) = 0$ either occurat exactly one point or it doen't occur at all. So by reasoning we can conclude that B , C , D are true.
 

Comments

@sailesh may i know why did you flaged my answer as low quality.

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f''(x) > 0 => f is convex not concave

 

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convex is nothing but concave up, i just missed to write "up" their.

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all that effort and you just tagged as low quality. shame man

Answer 2

f'' > 0, the function is convex.

We don't know about the derivative f'.

f' may be +ve -ve or 0

A is False. We are not guaranteed to have a local minima

B is True. Even if we have a local minima, we can't have more than 1 local minima

C is True. We can have at 0 or 1 global minima

D is True. Global minima and local minima of convex funcs are the same

Answer 3

first all we need to understand options carefully,, 

option a say , f has a local minima , which means, no matter what, f need to have exactly one minima

Option B states that there cannot exist two distinct critical points. In other words, f′(x)=0 can occur at at most one point or zero.

Option C states that f has at most one global minimum. This includes both cases exactly one global minimum and no global minimum.

Option D states that f has at most one local minimum. This includes both cases: exactly one local minimum and no local minimum.

 

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>> It is given that f''(x)>0, for all x $\in$ R ,, from this statement we can say that f'(x) is strictly increasing on R,, now here three cases arises

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case 1 : f''(x) starts increasing only from above x axis which say that f'(x) >0 for all x $\in$ R,, from this statement we can say that f(x) is strictly incresing function ,, so f(x) has neither minima nor maxima
 

conclusion

>>  we do not have minima so option a is false

>>  we don't have a global minimia so option c is true

>>  we don't have a local minimia so option d is true

>>  Since f′(x)≠0  for every x, there do not exist distinct points x,y with f′(x)=f′(y)=0. Hence B is true.

 

 

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case 2 : f''(x) starts increasing only below x-axis which say that f'(x) < 0 for all x $\in$ R,, from this statement we can say that f(x) is now a decreasing function ,, so f(x) has neither minima nor maxima
 

conclusion:

>>  we do not have minima so option a is false

>>  we don't have global minimia so option c is true

>>  we don't have local minimia so option d is true

>> Since f′(x)≠0  for every x, there do not exist distinct points x,y with f′(x)=f′(y)=0. Hence B is true.

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case 3 : f''(x) starts increasing from below x-axis , crosses zero on x-axis exactly once and continues increasing  above x-axsis which say that f'(x)<0 before the zero crossing 

f′(x)=0 at exactly one point, f′(x)>0 after the zero crossing.,, Therefore, f(x) first decreases, attains exactly one minimum when f′(x)=0, and then increases.
 

conclusion

>>  we have exactly one minima so option a is true (because only once zero crossing )

>>  we have eaxctly one global minimia so option c is true

>>  we have eaxctly one local minimia so option d is true

>>  Since there is exactly one critical point, if f′(x)=f′(y)=0, then necessarily x=y."

 

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Now based on above all three cases we say that 

option a is true only in case 3, not in all cases (so not always true), 

option b is true in all three cases, so option b is always true,

option c is true in all three cases, so option c is always true,

option d is true in all three cases, so option d is always true,