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 $a \in R . \int\limits_0^T f(x) d x=\int\limits_0^T f(x+a) d x$
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 $h \rightarrow 0$ 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 $f(x)=-x^4, f^{\prime}(x)=-4 x^3 \& f^{\prime \prime}(x)=-12 x^2$.
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. $ \int\limits_0^T f(x+a) d x$, let $x+a=y ; \int\limits_a^{a+T} f(y) d y=\int\limits_a^0 f(y) d y+\int\limits_0^T f(y) d y+\int\limits_T^{a+T} f(y) d y$
consider $\int\limits_T^{a+T} f(y) d y ; y=T+v$
$\int_0^a f(v+T) d v=\int\limits_0^a f(v) d v$.
Hence $\int\limits_0^T f(x) d x=\int\limits_0^T f(x+a) d x \Rightarrow$ 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