MCQ
Read the following mathematical statements carefully :

$I.$ Adifferentiable function $' f '$ with maximum at $x = c$ ==> $ f "(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)\,dx} = \int\limits_0^T {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 $h \rightarrow 0$ for $h > 0.$ Now indicate the correct alternative.

  • 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

Correct option: A.
exactly one statement is correct.
a
$I.$ consider the function $f (x) = - x^4 , f (x) = - 4 x^3 $ $\&$ $f"(x) = - 12 x^2.$

Here $f (x)$ has a maxima at $x = 0$ but $f" (0) = 0$ ==> 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)\,dx} $, let $x + a = y$ ;

$\int\limits_a^{a + T} {f(y)\,dy} = \int\limits_a^0 {f(y)\,dy} + \int\limits_0^T {f(y)\,dy} + \int\limits_T^{a + T} {f(y)\,dy} $

consider $\int\limits_T^{a + T} {f(y)\,dy} $;

$y = T + v$ $\int\limits_0^a {f(v + T)\,dv} = \int\limits_0^a {f(v)\,dv} $.

Hence $\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx} $

==> True.

$IV. $ The statement can be true only if $ 'f'$ is continuous at $x = c .$ Consider the function $f (x) = \left[ {\begin{array}{*{20}{c}}{x\,\,}&{if\,\,\,\,\,\,x\,\, > \,\,0}\\{1\,\,}&{if\,\,\,\,\,\,x\,\, = \,\,0}\\{ - \,x}&{if\,\,\,\,\,\,x\,\, < \,\,0}\end{array}} \right.$

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

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