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-

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

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

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

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

The line \[y=\alpha \] intersects the curve \[y=g(x)\],atleast at two

1.	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.	\[\left\{ -\frac{1}{2},0,\frac{1}{2} \right\}\]
Answer» D. \[\left\{ -\frac{1}{2},0,\frac{1}{2} \right\}\]

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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-

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