Application of DerivativeHard
Question
If f(x)
and g(x)
, where 0 < x ≤ 1 then in this interval
and g(x), where 0 < x ≤ 1 then in this interval Options
A.both f(x) and g(x) are increasing functions
B.both f(x) and g(x) are decreasing functions
C.f(x) is an increasing functions
D.g(x) is an increasing functions
Solution
Let g(x) 
, if 0 < x ≤ 1
Now, g(x) is continuous on [0, 1] and differentiable on ]0,1[
For 0 < x < 1
g′(x)
Again, H(x) = tan x - x sec2 x, 0 ≤ x ≤ 1
Now, H(x) is continuous on [0, 1] and differentiable on ]0,1[.
For 0 < x <1
H(x) = tan x - x sec2 x,0 ≤ x ≤ 1
⇒ H′(x) = sec2 x - sec2 x - 2x sec2 x tan x
= - 2x sec2 x tan x < 0
∴ H(x) is decreasing function on [0, 1]. Thus
H(x) < H (0) for 0 < x < 1
⇒ H(x) < 0 for 0 < x <1
⇒ g′(x) < 0 for 0 < x < 1
⇒ g(x) is decreasing function on [0, 1 ].
Therefore, g(x) =
is a decreasing function on 0 < x ≤ 1
Also, g(x) < g(0) for 0 < x ≤ 1
⇒ < 1 for o < x ≤ 1
⇒ x < tan x fopr 0 < x ≤ 1
Next, let f(x)
Now, f is continuous on [0, 1] and differentiable on ]0, 1[.
For 0 < x ≤ 1
f′(x)
> 0 for o < x < 1
⇒ f(x) increases on [0, 1].
Thus, f(x)
increases on 0 < x ≤ 1
Therefore, (c) is the answer.
, if 0 < x ≤ 1 Now, g(x) is continuous on [0, 1] and differentiable on ]0,1[
For 0 < x < 1
g′(x)
Again, H(x) = tan x - x sec2 x, 0 ≤ x ≤ 1
Now, H(x) is continuous on [0, 1] and differentiable on ]0,1[.
For 0 < x <1
H(x) = tan x - x sec2 x,0 ≤ x ≤ 1
⇒ H′(x) = sec2 x - sec2 x - 2x sec2 x tan x
= - 2x sec2 x tan x < 0
∴ H(x) is decreasing function on [0, 1]. Thus
H(x) < H (0) for 0 < x < 1
⇒ H(x) < 0 for 0 < x <1
⇒ g′(x) < 0 for 0 < x < 1
⇒ g(x) is decreasing function on [0, 1 ].
Therefore, g(x) =
is a decreasing function on 0 < x ≤ 1Also, g(x) < g(0) for 0 < x ≤ 1
⇒
< 1 for o < x ≤ 1 ⇒ x < tan x fopr 0 < x ≤ 1
Next, let f(x)
Now, f is continuous on [0, 1] and differentiable on ]0, 1[.
For 0 < x ≤ 1
f′(x)
> 0 for o < x < 1 ⇒ f(x) increases on [0, 1].
Thus, f(x)
increases on 0 < x ≤ 1 Therefore, (c) is the answer.
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
The slope of the normal to the curve x = a(θ − sin θ), y = a(1− cos θ) at point θ = `...Let f(x) = xm/n for x ∈ R where m and n are integers, m even and n odd and 0 < m < n. Then...The equation of tangent to the curve = 2 at the point (a, b) for all values of n is-...The sum of the intercepts made by a tangent to the curve √x + √y = 4 at point (4, 4) on coordinate axes is-...If f(x) is differentiable in the interval [2, 5], where f(2) = and f(5) = , then there exists a number c, 2 < c < ...