Let the function y=f(x) be given by x=t5−5t3−20t+7 and y=4t3−3t2−18t+3, where tϵ(−2,2). Then f′(x) at t=1 is ?
A
52
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B
25
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C
75
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D
none of these
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Solution
The correct option is B25 Given, x=t5−5t3−20t+7 dxdt=5t4−15t2−20 (dxdt)t=1=−30 Also, given y=4t3−3t2−18t+3 dydt=12t2−6t−18 (dydt)t=1=−12 So, (dydx)t=1=25