Continuity and DifferentiabilityHard
Question
The slope(s) of common tangent(s) to the curves y = e-x & y = e-x sin x can be -
Options
A.-e-π/2
B.-e-π
C.π/2
D.1
Solution
y = e-x & y = e-xsinx
y′ = -e-x ...(i) & y′ = -e-x(sin x - cos x) ...(ii)
equating (i) & (ii)
e-x(1 - sin x + cos x) = 0
e-x ≠ 0 ⇒1 - sin x + cos x = 0
⇒ 2cos2
= 2 sin cos
⇒ 2 cos
= 0 ⇒ x = 
, π
slope can be -e-π/2 & -e-π.
y′ = -e-x ...(i) & y′ = -e-x(sin x - cos x) ...(ii)
equating (i) & (ii)
e-x(1 - sin x + cos x) = 0
e-x ≠ 0 ⇒1 - sin x + cos x = 0
⇒ 2cos2
= 2 sin cos ⇒ 2 cos
= 0 ⇒ x = , πslope can be -e-π/2 & -e-π.
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