Question

The line $y=a$ intersects the curve $y=g(x)$, atleast at two points. If $\underset{2}{\overset{x}{\int }}g(t)dt=$$\frac{x^2}{2}+\underset{x}{\overset{2}{\int }}{t}^{2}g(t)dt$ then possible value of $\alpha$ is/are -

• A. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• B. $\left[-\frac{1}{2},\frac{1}{2}\right]$
• C. $\left(-\frac{1}{2},\frac{1}{2}\right)-\left\{0\right\}$
• D. $\left\{-\frac{1}{2},0.\frac{1}{2}\right\}$

Answer 1

Answer: (C)

$\underset{2}{\overset{x}{\int }}g(t)dt=$$\frac{x^2}{2}+\underset{x}{\overset{2}{\int }}{t}^{2}g(t)dt$
Differentiating w.r.t. x, we get
$g(x)=x+(-x^2(g(x))$
$\Rightarrow g\left(x\right)=\frac{x}{1+x^2}$

Clearly from graph, $\alpha \in \left(-\frac{1}{2},\frac{1}{2}\right)-\left\{0\right\}$