DifferentiationHard
Question
If y = logu |cos 4x| + |sin x|, where u = sec 2x, then 
at x = - 
is equal to :
at x = - is equal to :Options
A.
B.
C.
D.None of these
Solution
At x = - 
cos(4x) = cos
= - cos 
⇒ |cos 4x| = - cos 4x
and sin x = sin
= - sin 
⇒ |sin x| = - sin x.
∴ y = logu | cos 4x| + |sin x| ⇒ y =
- sin x .....(i)
⇒ (y + sin x) ln sec (2x) = ln (- cos 4x)
Diff. w.r.t. x,
(y1 + cosx) ln sec(2x) + (y + sin x)
(y1 + cosx) ln sec(2x) + (y + sin x) 2 tan (2x) = - 4 tan 4x ... (2)
Put x = -
in (1), y = 
Put this in (2) and x = -
, we get
ln 2 + 
(-2√3) = - 4√3
⇒ y1 =
cos(4x) = cos
= - cos ⇒ |cos 4x| = - cos 4x
and sin x = sin
= - sin ⇒ |sin x| = - sin x.∴ y = logu | cos 4x| + |sin x| ⇒ y =
- sin x .....(i)⇒ (y + sin x) ln sec (2x) = ln (- cos 4x)
Diff. w.r.t. x,
(y1 + cosx) ln sec(2x) + (y + sin x)
(y1 + cosx) ln sec(2x) + (y + sin x) 2 tan (2x) = - 4 tan 4x ... (2)
Put x = -
in (1), y = Put this in (2) and x = -
, we get ln 2 + (-2√3) = - 4√3⇒ y1 =
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