question_answer

The line \[y=\alpha \] intersects the curve \[y=g(x)\],atleast at two points. If \[\int\limits_{2}^{x}{g(t)dt=\frac{{{x}^{2}}}{2}+\int\limits_{x}^{2}{{{t}^{2}}g(t)dt}}\]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) \[\left( -\frac{1}{2},\frac{1}{2} \right)-\{0\}\]

D) \[\left\{ -\frac{1}{2},0,\frac{1}{2} \right\}\]

Correct Answer: C

Solution :

[c] \[\int\limits_{2}^{x}{g(t)dt=\frac{{{x}^{2}}}{2}+\int\limits_{x}^{2}{{{t}^{2}}g(t)dt}}\] Differentiating w.r.t. x, we get \[g(x)=x+(-{{x}^{2}}(g(x))\Rightarrow g(x)=\frac{x}{1+{{x}^{2}}}\] Clearly from graph, \[\alpha \in \left( -\frac{1}{2},\frac{1}{2} \right)-\{0\}\]