If x = a cos t, y = a sin t,then d2ydx2 at t=π4 is
A
a2√2
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B
−a2√2
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C
2√2a
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D
−2√2a
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Solution
The correct option is D−2√2a Clearly x2+y2=a2 and y(π4)=a√2,x(π4)=a√2. Differentiating we get, 2x+2yy1=0⇒y1=−xy,soy1(π4)=−1 Now x+yy1=0⇒1+y21+yy2=0 ⇒y2(π4)=−1+(y1(π4))2y(π4)=−2√2a