Question

Read the following mathematical statements carefully:
I. Adifferentiable function ' $f^{\prime }$ with maximum at $x=c\Rightarrow f^{\prime \prime }(c)<0$.
II. Antiderivative of a periodic function is also a periodic function.
III. If $f$ has a period $T$ then for any $a\in R.\underset{0}{\overset{T}{\int }}f(x)dx=\underset{0}{\overset{T}{\int }}f(x+a)dx$
IV. If $f(x)$ has a maxima at $x=c$, then ' $f$ ' is increasing in $(c-h,c)$ and decreasing in $(c,c+h)$ as $h\to 0$ for $h>0$.
Now indicate the correct alternative.

• A. exactly one statement is correct.
• B. exactly two statements are correct.
• C. exactly three statements are correct.
• D. All the four statements are correct.

Answer 1

Answer: (A)

I. consider the function $f(x)=-x^4,f^{\prime }(x)=-4x^3\&f^{\prime \prime }(x)=-12x^2$.
Here $f(x)$ has a maxima at $x=0$ but $f^{\prime \prime }(0)=0\Rightarrow$ False.
II. $f(x)=\cos x+1$ is periodic with period $2\pi$ but $\int (\cos x+1)dx=\sin x+x$ is not periodic.
III. $\underset{0}{\overset{T}{\int }}f(x+a)dx$, let $x+a=y;\underset{a}{\overset{a+T}{\int }}f(y)dy=\underset{a}{\overset{0}{\int }}f(y)dy+\underset{0}{\overset{T}{\int }}f(y)dy+\underset{T}{\overset{a+T}{\int }}f(y)dy$
consider $\underset{T}{\overset{a+T}{\int }}f(y)dy;y=T+v$
${\int }_0^{a}f(v+T)dv=\underset{0}{\overset{a}{\int }}f(v)dv$.
Hence $\underset{0}{\overset{T}{\int }}f(x)dx=\underset{0}{\overset{T}{\int }}f(x+a)dx\Rightarrow$ True.
IV. The statement can be true only if $f^{\prime }{f}^{\prime }$ is continuous at $x=c$. Consider the function

This has a maxima at $x=0$ however this
does not satisfy conditions stated in the problem $\Rightarrow$ False