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$\lim_{x \to 0} \frac{ a sin ^2x + b log ( cos x) }{ x^4} = \frac{1}{2}$

a) - 1, -2                    b) 1, 2

c) -1,2                       d) 1,-2
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Option (A) is correct.

limx->0   (asin2x + b log(cosx))/x4  = 1/2     [0/0 form]   ,applying L'Hospital rule ,we get

= > limx->0   (2a*sinx*cosx  - (b /cosx)*sinx)/ 4x3  = 1/2 => limx->0   (a*sin2x  - b*tanx)/ 4x= 1/2      [0/0 form],

applying L'Hospital rule again ,we get,

 = > limx->0 (2a*cos2x - b*sec2x) / 12x =  1/2

For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.

Hence 2a - b =0 =>  2a = b ------(A)

limx->0 (b*cos2x - b*sec2x) / 12x =  1/2    [0/0 form], applying L'Hospital rule again ,we get,

= >  limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x  =  1/2   => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x  =  1/2

[0/0 form], applying L'Hospital rule again ,we get,

limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24  =  1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2

from (A), we have  ,  2a = b => 2a = -2 => a = -1

Hence a =-1 & b = -2

Hence option (A) is correct.
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