Chapter 6 Application of Derivatives Ex 6.2 Solutions
Question - 1 : - Show that the function given by f(x)= 3x + 17 is strictly increasing on R.
Answer - 1 : - Let
be any two numbers in R.
Then, we have:
Hence, f is strictlyincreasing on R.
Question - 2 : - Show that the function given by f(x)= e2x isstrictly increasing on R.
Answer - 2 : - Letbe any two numbers in R.
Then, we have:
Hence, f is strictlyincreasing on R.
Question - 3 : - Show that the function given by f(x)= sin x is
(a) strictly increasing in
(b) strictlydecreasing in 
(c) neither increasing nor decreasing in (0, π)
Answer - 3 : -
The given function is f(x)= sin x.
(a) Since for each
we have
Hence, f isstrictly increasing in
(b) Since for each
we have 
Hence, f isstrictly decreasing in
(c) From the results obtained in (a) and(b), it is clear that f is neither increasing nor decreasingin (0, π).
Question - 4 : - Find the intervals in which thefunction f given by f(x) = 2x3 −3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing
Answer - 4 : -
The given function is f(x)= 2x3 − 3x2 − 36x +7.
x =− 2, 3
The points x = −2 and x =3 divide the real line into three disjoint intervals i.e.,
In intervals
is positive while in interval
(−2, 3), 
is negative.Hence, the given function (f) isstrictly increasing in intervals
, while function (f) is strictlydecreasing in interval (−2, 3).
Question - 5 : - Find the intervals in which the following functions arestrictly increasing or decreasing:
(a) x2 +2x − 5 (b) 10 − 6x − 2x2
(c) −2x3 −9x2 − 12x + 1 (d) 6 − 9x − x2
(e) (x + 1)3 (x −3)3
Answer - 5 : -
(a) We have,
Now,
x = −1
Point x = −1 divides the real lineinto two disjoint intervals i.e., 
In interval
∴f isstrictly decreasing in interval
Thus, f is strictlydecreasing for x < −1.
In interval
∴ f is strictly increasing in interval
Thus, f is strictlyincreasing for x > −1.
(b) We have,
f(x)= 10 − 6x − 2x2
The point
divides the real line intotwo disjoint intervals i.e.,
In interval
i.e., when
f'(x)=-6-4x>0.
∴ f is strictly increasing for
In interval
i.e., when
∴ f is strictly decreasing for
(c) We have,
f(x)= −2x3 − 9x2 − 12x +1
Points x =−1 and x = −2 divide the real line into three disjoint intervalsi.e.,
In intervals
i.e., when x < −2and x > −1,
∴ f is strictly decreasing for x <−2 and x > −1.
Now, in interval (−2, −1) i.e., when −2 < x <−1,
∴ f is strictly increasing for 
(d) We have,
The point
divides the real line intotwo disjoint intervals i.e., 
In interval
i.e., for
∴ f is strictly increasing for
In interval
i.e., for
∴ f is strictly decreasing for
(e) We have,
f(x)= (x + 1)3 (x − 3)3
The points x =−1, x = 1, and x = 3 divide the real lineinto four disjoint intervals i.e.,
, (−1, 1), (1, 3), and
In intervalsand (−1, 1), 
∴ f is strictly decreasing inintervalsand (−1, 1).
In intervals (1, 3) and, 
∴ f is strictly increasing inintervals (1, 3) and
Question - 6 : - Show that
, is an increasing function of x throughoutits domain.
Answer - 6 : -
We have,
Hence, function f isincreasing throughout this domain.
Question - 7 : - Find the values of x forwhich
is an increasing function.
Answer - 7 : -
We have,
The points x =0, x = 1, and x = 2 divide the real line intofour disjoint intervals i.e., 
In intervals
∴ y is strictly decreasing inintervals
However, in intervals (0,1) and (2, ∞), 
∴ y is strictly increasing in intervals(0, 1) and (2, ∞).
y isstrictly increasing for 0 < x < 1 and x >2.
Question - 8 : - Prove that 
is an increasingfunction of θ in
Answer - 8 : -
We have,
Since cos θ ≠ 4, cos θ =0.
Now,
In interval
we have cos θ >0. Also, 4 > cos θ ⇒ 4− cos θ >0
Therefore, y isstrictly increasing in interval
Also, the given function iscontinuous at 
Hence, y isincreasing in interval
Question - 9 : - Prove that the logarithmic function is strictly increasingon (0, ∞).
Answer - 9 : -
It is clear that for x >0
Hence, f(x) = log x isstrictly increasing in interval (0, ∞).
Prove that the function f givenby f(x) = x2 − x +1 is neither strictly increasing nor strictly decreasing on (−1, 1).
Answer - 10 : -
The given function is f(x)= x2 − x + 1.
The point
divides the interval (−1,1) into two disjoint intervals i.e.,
Now, in interval
Therefore, f isstrictly decreasing in interval
However, in interval
Therefore, f isstrictly increasing in interval
Hence, f is neitherstrictly increasing nor decreasing in interval (−1, 1).