A) \[-2ab{{v}^{2}}\]
B) \[2b{{v}^{3}}\]
C) \[-2a{{v}^{3}}\]
D) \[2a{{v}^{2}}\]
Correct Answer: C
Solution :
[c] Given, \[t=a{{x}^{2}}+bx\] |
Differentiating w.r.t. t |
\[\frac{d}{dt}=2ax\frac{dx}{dt}+b\frac{dx}{dt}\] |
\[v=\frac{dx}{dt}=\frac{1}{(2ax+b)}\] |
Again differentiating w.r.t. t |
\[\frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{-2a}{{{(2ax+b)}^{2}}}\frac{dx}{dt}\] |
\[\therefore f=\frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{-2a}{{{(2ax+b)}^{2}}}.\frac{1}{(2ax+b)}\] |
or \[f=\frac{-2a}{{{(2ax+b)}^{3}}}\] |
\[\therefore f=-2a{{v}^{3}}\] |
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