Application of DerivativeHard
Question
For function f(x) = esinx - e-sinx, which of the following is/are correct -
Options
A.Equation of normal at x = π is x - 2y = π
B.f(x) is strictly increasing ∀ x ∈
; n ∈ I
C.f(x) is strictly decreasing ∀ x ∈
; n ∈ I
D.x = nπ, n∈I is either point of local maxima or minima of f(x).
Solution
y = f(x) = esinx - e-sinx
⇒ f′(x) = cosx(esinx + e-sinx)
f′(π) = - 2
⇒ slope of normal at x = π is
⇒ Equation of normal at x = π
⇒ x = 2y = π
sign of f′(x)


x = nπ is neither point of local maxima nor minima
⇒ f′(x) = cosx(esinx + e-sinx)
f′(π) = - 2
⇒ slope of normal at x = π is
⇒ Equation of normal at x = π
⇒ x = 2y = π
sign of f′(x)


x = nπ is neither point of local maxima nor minima
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
For the curve represented parametrically by the equation, x = 2 ln cot t + 1 and y = tan t + cot t...If the line ax + by + c = 0 is a normal to the curve xy = 1, then-...The length of perpendicular drawn from the origin to the normal at any point θ of the curve x = a cos3 θ, y = ...Suppose p is the first of n(n > 1) AM′s between two positive numbers a and b, then value of p is -...Let f(x) and g(x) be differentiable for 0 ≤ x ≤ 1, such that f(0) = 2, g(0) = 0, f (1) = 6. Let there exist ...