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Application of Derivatives Ex-6.2 Questions Answers

Subjects

Question 1 :

Show that the function given by f(x)= 3x + 17 is strictly increasing on R.

Answer 1 : Letbe any two numbers in R.

Then, we have:

Hence, 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, 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 =3 divide the real line into three disjoint intervals i.e.,

In intervalsis 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 pointdivides 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 forwhichis 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, ∞).

Question 10 :

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 pointdivides 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).