A) \[\pm 1\]
B) \[\frac{1}{2}\]
C) \[\pm 2\]
D) none of these
Correct Answer: A
Solution :
let \[q=p+d,r=p+2d,s=p+3d\] |
\[\therefore \] |
Applying \[R\to {{R}_{1}}+{{R}_{3}}-2{{R}_{2}},\]we get |
\[=2[\left( p+d+\sin x \right)\left( p+3d+\sin x \right)\]\[-{{\left( p+2d+\sin x \right)}^{2}}=-2{{d}^{2}}\]\[Given\,\int\limits_{0}^{2}{f\left( x \right)dx=-4\Rightarrow \int\limits_{0}^{2}{\left( -2{{d}^{2}} \right)dx=-4}}\]\[\Rightarrow {{d}^{2}}=1\Rightarrow d=\pm 1\] |
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