Question

The line $y = a$ intersects the curve $y = g (x)$, atleast at two points. If $\int\limits^{x}_{{2}}g(t)dt=$$\frac{x^{2}}{2}+\int\limits^{2}_{{x}}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)

$\int\limits^{x}_{{2}}g(t)dt=$$\frac{x^{2}}{2}+\int\limits^{2}_{{x}}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, $\alpha\in\left(-\frac{1}{2}, \frac{1}{2}\right)-\left\{0\right\}$