Question

Let g(x) a polynomial of degree 3 passing through origin and have a local maximum at $x=-\frac{1}{2\sqrt{2}}$. Also g’(x) has a local minimum at x = 0 and g(1) = 5.

Let f(x) = sgn(x) (where sgn(x) denotes signum function of x), then which of the following statement is incorrect for f(g(x))?

• A. Area enclosed by f(g(x)) with x-axis between ordinates $x=-\alpha$ to $x=\alpha$ is $2\alpha$
• B. ${\int }_0^{\alpha }dx=2\alpha$
• C. f(g(x)) is many one function
• D. f(g(x)) is a periodic function

Answer 1

The correct option is D

f(g(x)) is a periodic function

$g"\left(x\right)=AX\Rightarrow g^{\prime }\left(x\right)=\frac{A{x}^2}{2}+B\because g^{\prime }(-\frac{1}{2\sqrt{2}})=0\Rightarrow \frac{A}{16}+B=0\Rightarrow B=-\frac{A}{16}\Rightarrow g^{\prime }\left(x\right)=\frac{A{x}^2}{2}-\frac{A}{16}\Rightarrow g^{\prime }\left(x\right)=\frac{A{x}^3}{6}-\frac{Ax}{16}+C\because g\left(0\right)=0andg\left(1\right)=5\Rightarrow c=0,A=48g\left(x\right)=8x^3-3x$
so, $f\left(g\right(x\left)\right)=sgn(8x^3-3x)$

$=\{\begin{array}{ccc}1& if& xϵ(-\infty ,-\sqrt{\frac{3}{8}})\\ 1& if& xϵ(0,\sqrt{\frac{3}{8}})\\ 1& if& xϵ(0,\sqrt{\frac{3}{8}})\\ 1& if& xϵ(0,\sqrt{\frac{3}{8}},\infty )\end{array}$

clearly f(g(x)) is not a periodic function
area enclosed between ordinates $x=-\alpha tox=\alpha is=2{\int }_0^{\alpha }dx=2\alpha$