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Question

# Consider the following statements S1: If f(x)=∣∣x2−5|x|+6∣∣, then f′(−52)=0 S2: The slope of tangent at x=1 on the curve y=sin−1(cosπx) is π S3: If x=f(t),y=g(t), then d2xdy2=f′′(t)g′′(t) Which of the following is/are correct about the truth value of above statements

A
S1 is true
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B
S2 is false
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C
all are true
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D
S3 is false
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Solution

## The correct option is D S3 is falseS1:f(x)=|x2+5x+6|, for x<0 f(x)=−(x2+5x+6), if −3≤x≤−2 f′(x)=−2x−5 f′(−52)=0 S2:y=sin−1(cosπx) =π2−cos−1(cosπx)=⎧⎪ ⎪⎨⎪ ⎪⎩π2−πx,0≤x≤1πx−3π2,1<x≤2 Function is not differentiable at x=1. Hence, no tangent can be drawn at this point. S3:dxdt=f′(t),dydt=g′(t) ⇒dxdy=dxdt⋅dtdy=f′(t)g′(t) ⇒d2xdy2=ddy(dxdy)=ddt(dxdy)⋅dtdy ⇒d2xdy2=f′(t)g′′(t)−f′′(t)g′(t)(f′(t))3

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