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The equation of the normal to the curve `y=x+sin x cos x " at " x=(pi)/(2),` isA. `x=2`B. `x=pi`C. `x+pi=0`D. `2x=pi` |
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Answer» Correct Answer - D When `x=(pi)/(2)`, we have `y=(pi)/(2) + "sin" (pi)/(2)"cos"(pi)/(2)=(pi)/(2)` So, the coordinates of the point are `(pi//2,pi//2)` Now, `y=x+sin x cos x ` `rArr (dy)/(dx)=1+cos^(2)x-sin^(2)x rArr ((dy)/(dx))_(x=(n)/(2))=1+0-1=0` This means that the tangent at `(pi//2,pi//2)` is parallel to x-axis. So, the normal is parallel to y-axis and hence its equation is `x=(pi)/(2)`. |
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