GO Classes Test Series 2023 | Calculus | Test 1 | Question: 8

Answer 1

• A. The exp function is $1$ at $x = 0.$
• B. Take $f(x) = x$
• C. It is a point, where the second derivative change sign.
• D. By the intermediate number theorem.

Comments

For Option D, IVT is only applicable if $f''(x)$ is continuous in $(0,1)$. So shouldn’t option D be incorrect?

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Yes , Complete context for option D is missing.

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@IchVonViralto A differentiable function is always continuous, therefore option D is true.

Answer 2

Option 1:- 

Given ,$f(x)=e^{-x^{2}}-1$

Now , $x=0$ , $f(x)=e^{0}-1=1-1=0$ So , $f(x)$ has a root at $x=0$ .

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Option 2:-

The counter example given by the goclass answer will works as $f(x)=x$ .

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Option 3:-

Let  $f(x)=x-1$ , So ,$f(x)$ has root at $x=1$ .

Now , $f’(x)=1$ it doesn’t change sign at $x=1$ it is constant function. So it is also false .

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Option 4:-

Inflection point :- It is a point in function where the function changes from being concave to convex or vice versa .

Mathematically we can write at inflection point $f’’(x)=0$ .

It is given ,$f''(0)< 0$ and $f''(1)> 0$  . Let ,if we consider $f’’(x)=g(x)$

So , $g(0) <0$ and $g(1)>0$ So we can clearly see the sign of $g(x)$ is change between $[0,1]$ . Now as $f’’(x)$ is exists which implies $g(x)$ is continuous as $f’’(x)=g(x)$ .

So , $g(x)$ is continuous in $[0,1]$ and sign of $g(x)$  change from positive to negative which implies $g(x)$ has a root in $(0,1)$ .

So say at $x_{0}$ is the root of $g(x)$.

So, $g(x_{0})=0$

So,$f’’(x_{0})=0$ 

So , $f(x)$ has an inflection point at $x=x_{0}$ . So option $(D)$ is true .

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So correct option is $A$ and $D$.

Comments

thanks for the explanation of option D!!

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My doubt is, how you are concluding $f(x)$ is twice differentiable in interval $(0,1)$?