I applied L-Hospital’s rule and got a ifferent answer which is not in the options. MY answer={(2a*alpha+b^2 }/2;

Question

Vikas TU · Answer

Dear student You have applies lhospital wrongly , It will be – sin( ax^2 + bx + c) *(2ax +b) / ( 2x – 2a) apply differentiation again. {(2a) (-sin(ax^2 + bx + c) -cos(ax^2 + bx +c) (2ax +b)^2 } / 2 Now put value of a. Good Luck

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I applied L-Hospital’s rule and got a ifferent answer which is not in the options. MY answer={(2a*alpha+b^2 }/2;

I applied L-Hospital’s rule and got a ifferent answer which is not in the options.

MY answer={(2a*alpha+b^2 }/2;

1 Answers

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I applied L-Hospital’s rule and got a ifferent answer which is not in the options. MY answer={(2a*alpha+b^2 }/2;

I applied L-Hospital’s rule and got a ifferent answer which is not in the options. MY answer={(2a*alpha+b^2 }/2;

1 Answers

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I applied L-Hospital’s rule and got a ifferent answer which is not in the options.

MY answer={(2a*alpha+b^2 }/2;