The value of a and b such that

Question

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

a) - 1, -2                    b) 1, 2

c) -1,2                       d) 1,-2

Comments

Where is b in limit ?

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thanks for notice. i left it. please check it now.

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How about using lim{x->0} sinx/x = 1 ?

Answer 1

Option (A) is correct.

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

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

applying L'Hospital rule again ,we get,

 = > lim_x->0 (2a*cos2x - b*sec²x) / 12x²  =  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)

lim_x->0 (b*cos2x - b*sec²x) / 12x²  =  1/2    [0/0 form], applying L'Hospital rule again ,we get,

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

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

lim_x->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.

Comments

but what about the cos2x and $sec^2x$ (involved with the coefficients)in the numerator can they be removed directly as you have removed in above answer.

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Put x->0 in N^r ,it evaluates to 2b[-2 -1]. Do further simplification.

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Plz explain how you are using Sin(2x) formula while in question sin^2x is mentioned

Answer 2

so, we can say

-4a-b=6

So ans will be (A) only