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.
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