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Question

The slope of the tangent to the curve x = 3t2 + 1, y = t3 −1 at x = 1 is

(a) 1/2
(b) 0
(c) −2
(d) ∞

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Solution

(b) 0

Given:x=3t2+1 y=t3-1x=1Now,3t2+1=13t2=0t=0dxdt=6t and dydt=3t2dydx=dydtdxdt=3t26t=t2Thus, we getSlope of the tangent=dydxt=0=02=0

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