JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y = a\cos (\log x) + b\sin (\log x)

where

a,b

are parameters, then

{x^2}y'' + xy'

is equal to

View answer and explanation

Correct option: B

- y

Option B (

-y

) is correct because after differentiating twice and substituting, the expression

{x^2}y'' + xy'

simplifies to

-y

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