Application of DerivativeHard
Question
Let the function g: (- ∞, ∞) → 
be given by g(u) = 2tan-1(eu) - 
. Then, g is
be given by g(u) = 2tan-1(eu) - . Then, g isOptions
A.even and is strictly increasing in (0, ∞)
B.odd and is strictly decreasing in (-∞, ∞)
C.odd and is strictly increasing in (-∞, ∞)
D.neither even nor odd, but is strictly increasing in (-∞, ∞)
Solution
g(u) = 2 tan-1(eu)-
= 2tan-1 eu - tan-1 eu - cot-1 eu = tan-1 eu - cot-1 eu
g(-x) = -g(x)
⇒ g(x) is odd
and g′(x) > 0 ⇒ increasing.
= 2tan-1 eu - tan-1 eu - cot-1 eu = tan-1 eu - cot-1 eu
g(-x) = -g(x)
⇒ g(x) is odd
and g′(x) > 0 ⇒ increasing.
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
If α be the angle of intersection between the curves y = ax and y = bx, then tanα is equal to-...The equation of the normal to the curve x = at2, y = 2at at ′t′ point is -...Let Tr be the rth term of an A.P. for r = 1, 2, 3, ........ If for some positive integers m, n we have Tm = & Tn = , the...The equation of normal to the curve x2/3 + y2/3 = a2/3 at the point (a, 0) is -...If f(x) satisfies the requirements of Lagrange′s mean value theorem on [0, 2] and if f(0) = 0 and f′(x) X...