# Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.2

Last Updated : 04 Apr, 2024

### Question 1. Show that the function given by f (x) = 3x + 17 is increasing on R.

Solution:

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function.  (vice-versa is not true)

Given: f(x) = 3x + 17

f'(x) = 3 > 0        -(Always greater than zero)

Hence, 3x + 17 is strictly increasing on R.

### Question 2. Show that the function is given by f (x) = e2x is increasing on R.

Solution:

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function.  (vice-versa is not true)

Given: f(x) = e2x

fâ€™(x) = 2e2x > 0

Hence, f(x) = e2x is strictly increasing on âˆž

### (iii) neither increasing nor decreasing in (0, Ï€)

Solution:

Given: f(x) = sin x

So, fâ€™(x) = d/dx(sin x) = cos x

(i) Now in (0, Ï€/2), fâ€™(x) = cos x > 0 (positive in first quadrant)

Hence, f(x) = sin x is strictly increasing in (0, Ï€/2).

(ii) In (Ï€/2, Ï€), fâ€™(x) = cos x < 0           -(negative in second quadrant)

Hence, f(x) = sin x is strictly decreasing in (Ï€/2,Ï€)

(iii) As we know that fâ€™(x) = cos x is positive in interval(0, Ï€/2)

and fâ€™(x) = cos x is negative in interval (Ï€/2, Ï€)

So, it is neither increasing nor decreasing.

### (ii) decreasing

Solution:

Given: f(x) = 2x2 – 3x

f'(x) = [Tex]\frac{d(2x^2-3x)}{dx}  [/Tex] = 4x – 3           -(1)

= x = 3/4

So the intervals are (-âˆž, 3/4) and (3/4, âˆž)

(i) Interval (3/4, âˆž) let take x = 1

So, from eq(1) f'(x) > 0

Hence, f is strictly increasing in interval (3/4, âˆž)

(ii) Interval (-âˆž, 3/4) let take x = 0.5

So, from eq(1) f'(x) < 0

Hence, f is strictly decreasing in interval (-âˆž, 3/4)

### (ii) decreasing

Solution:

Given: f(x) = 2x3 – 3x2 – 36x + 7

f'(x) = [Tex]\frac{d(2x^3-3x^2-36x+7)}{dx}  [/Tex] = 6x2 – 6x – 36         -(1)

f'(x) = 6(x2 – x – 6)

On putting f'(x) = 0, we get

6(x2 – x – 6) = 0

(x2 – x – 6) = 0

x = -2, x = 3

So, the intervals are (-âˆž, -2), (-2, 3), and (3, âˆž)

For (-âˆž, -2) interval, take x = -3

From eq(1), we get

f'(x) = (+)(-)(-) = (+) > 0

So, f is strictly increasing in interval (-âˆž, -2)

For (-2, 3) interval, take x = 2

From eq(1), we get

f'(x) = (+)(+)(-) = (-) < 0

So, f is strictly decreasing in interval (-2, 3)

For (3, âˆž)interval, take x = 4

From eq(1), we get

f'(x) = (+)(+)(+) = (+) > 0

So, f is strictly increasing in interval (3, âˆž)

(i) f is strictly increasing in interval (-âˆž, -2) and (3, âˆž)

(ii) f is strictly decreasing in interval (-2, 3)

### (v) (x + 1)3 (x – 3)3

Solution:

(i) f(x) = x2 + 2x – 5

f'(x) = 2x + 2         -(1)

On putting f'(x) = 0, we get

2x + 2 = 0

x = -1

So, the intervals are (-âˆž, -1) and (-1, âˆž)

For (-âˆž, -1) interval take x = -2

From eq(1), f'(x) = (-) < 0

So, f is strictly decreasing

For (-1, âˆž) interval take x = 0

From eq(1), f'(x) = (+) > 0

So, f is strictly increasing

(ii) f(x) = 10 – 6x – 2x2

f'(x) = -6 – 4x

On putting f'(x) = 0, we get

-6 – 4x = 0

x = -3/2

So, the intervals are (-âˆž, -3/2) and (-3/2, âˆž)

For (-âˆž, -3/2) interval take x = -2

From eq(1), f'(x) = (-)(-) = (+) > 0

So, f is strictly increasing

For (-3/2, âˆž) interval take x = -1

From eq(1), f'(x) = (-)(+) = (-) < 0

So, f is strictly decreasing

(iii) f(x) = -2x3 – 9x2 – 12x + 1

f'(x) = -6x2 – 8x – 12

On putting f'(x) = 0, we get

-6x2 – 8x – 12 = 0

-6(x + 1)(x + 2) = 0

x = -1, x = -2

So, the intervals are (-âˆž, -2), (-2, -1), and (-1, âˆž)

For (-âˆž, -2) interval take x = -3

From eq(1), f'(x) = (-)(-)(-) = (-) < 0

So, f is strictly decreasing

For (-2, -1) interval take x = -1.5

From eq(1), f'(x) = (-)(-)(+) = (+) > 0

So, f is strictly increasing

For (-1, âˆž) interval take x = 0

From eq(1), f'(x) = (-)(+)(+) = (-) < 0

So, f is strictly decreasing

(iv) f(x) = 6 – 9x – x

f'(x) = -9 – 2x

On putting f'(x) = 0, we get

-9 – 2x = 0

x = -9/2

So, the intervals are (-âˆž, -9/2) and (-9/2, âˆž)

For f to be strictly increasing, f'(x) > 0

– 9 – 2x > 0

x > -9/2

So f is strictly increasing in interval (-âˆž, -9/2)

For f to be strictly decreasing, f'(x) < 0

-9 – 2x < 0

x < -9/2

So f is strictly decreasing in interval (-9/2, âˆž)

(v) f(x) = (x + 1)3 (x – 3)3

f'(x) = (x + 3)3.3(x – 3)3 + (x – 3)3.3(x + 1)2

f'(x) = 6(x – 3)2(x + 1)2(x – 1)

Now, the factor of (x – 3)2 and (x + 1)2 are non-negative for all x

For f to be strictly increasing, f'(x) > 0

(x – 1) > 0

x > 1

So, f is strictly increasing in interval (1, âˆž)

For f to be strictly decreasing, f'(x) < 0

(x – 1) < 0

x < 1

So, f is strictly decreasing in interval (-âˆž, 1)

### Question 7. Show that y = log(1 + x) – [Tex]\frac{2x}{2+x}  [/Tex], is an increasing function of x throughout its domain.

Solution:

f(x) = log(1+x)[Tex]\frac{-2x}{2+x},  x>-1[/Tex]

[Tex]\frac{dy}{dx}=f'(x)=\frac{1}{1+x}-\frac{(2x+1).2-2x(1)}{(2x+1)^2}     [/Tex]

f'(x)=[Tex]\frac{1}{1+x}-\frac{4+2x-2x}{(2x+1)^2}[/Tex]

[Tex]f'(x)=\frac{1}{1+x}-\frac{4}{(2+x)^2}[/Tex]

[Tex]f'(x)=\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)}[/Tex]

[Tex]f'(x)=\frac{x^2}{(x+1)(x+2)^2}[/Tex]

So, the domain of the given function is x > -1

Now, x2 > 0, (x + 2)2 â‰¥ 0, x + 1 > 0

From the above equation f'(x) â‰¥ 0 âˆ€ x in the domain(x > -1) and f is an increasing function.

### Question 8. Find the values of x for which y = [x(x â€“ 2)]2 is an increasing function.

Solution:

Given: y = f(x) = [x(x – 2)]2 = x2(x – 2x)2

= x4 – 4x3 + 4x2

f'(x) = 4x3 – 12x2 + 8x

f'(x) = 4x(x – 2)(x – 1)

x = 0, x = 1, x = 2

So, (âˆž, 0], [0, 1], [1, 2], [2,âˆž)

For (âˆž, 0], let x = -1

So, f'(x) = (-)(-)(-) = (-) â‰¤ 0

f(x) is decreasing

For [0, 1], let x = 1/2

So, f'(x) = (+)(-)(-) = (+) â‰¥ 0

f(x) is increasing

Similarly, for [1, 2], f(x) is decreasing

For [2,âˆž), f(x) is increasing

So, f(x) is increasing in interval [0, 1] and [2,âˆž)

### Question 9. Prove that y = [Tex]\frac{4sin\theta}{2+cos\theta}-\theta    [/Tex] is an increasing function of Î¸ in[0, Ï€/2].

Solution:

y = f(Î¸) = [Tex]\frac{4\sinÎ¸}{(2+\cosÎ¸)^2}-Î¸[/Tex]

[Tex]f'(Î¸)=\frac{(2+\cosÎ¸).4\cosÎ¸-4\sinÎ¸(-\sinÎ¸)}{(2+\cosÎ¸)^2}     [/Tex]

[Tex]f'(Î¸)=\frac{4.(2\cosÎ¸+\cos^2Î¸+\sin^2Î¸)}{(2+\cosÎ¸)^2}-1[/Tex]

[Tex]f'(Î¸)=\frac{8\cosÎ¸+4}{(\cosÎ¸+2)^2}[/Tex]

[Tex]f'(Î¸)=\frac{8\cosÎ¸+4-(2+\cosÎ¸)^2}{(2+\cosÎ¸)^2}[/Tex]

[Tex]f'(Î¸)=\frac{4\cosÎ¸-\cos^2Î¸}{(2+\cosÎ¸)^2}[/Tex]

Now 0 â‰¤ Î¸ â‰¤ Ï€/2, and we have 0 â‰¤ cosÎ¸ â‰¤ 1,

So, 4 – cosÎ¸ > 0

Therefore f'(Î¸) â‰¥ 0 for 0 â‰¤ Î¸ â‰¤ Ï€/2

Hence, f'(x) = [Tex]\frac{4\sinÎ¸}{2+\cosÎ¸}-Î¸    [/Tex] is a strictly increasing in the interval (Î¸, Ï€/2).

### Question 10. Prove that the logarithmic function is increasing on (0, âˆž).

Solution:

Given: f(x) = log(x)         -(logarithmic function)

f'(x) = 1/x âˆ€ x in (0, âˆž)

Therefore,  x > 0, so, 1/x > 0

Hence, the logarithmic function is strictly increasing in interval (0, âˆž)

### Question 11. Prove that the function f given by f(x) = x2 â€“ x + 1 is neither strictly increasing nor decreasing on (â€“ 1, 1).

Solution:

Given: f(x) = x2 – x + 1

f'(x) = 2x – 1

For strictly increasing, f'(x) > 0

2x – 1 > 0

x > 1/2

So, f(x) function is increasing for x > 1/2 in the interval (1/2, 1)          -(Given interval is (-1, 1)

Similarly, for decreasing f'(x) < 0

2x – 1 < 0

x < 1/2

So, f(x) function is increasing for x < 1/2 in the interval (-1, 1/2)          -(Given interval is (-1, 1)

Hence, the function f(x) = x2 – x + 1 is neither strictly increasing nor decreasing.

### (A) cos x    (B) cos 2x    (C) cos 3x    (D) tan x

Solution:

(A) f(x) = cos x

f'(x) = -sin x

Now in (0, Ï€/2) interval, sin x is positive(because it is second quadrant)

So, -sin x < 0

âˆ´ f'(x) < 0

f(x) = cos x is strictly decreasing on(0, Ï€/2).

(B) f(x) = cos 2x

f'(x) = -2 sin 2x

Now in (0, Ï€/2) interval, sin x is positive(because it is second quadrant)

-sin 2x < 0

âˆ´ f'(x) < 0,

f(x) = cos 2x is strictly decreasing on(0, Ï€/2).

(C) f(x) = cos 3x

f'(x) = -3sin 3x

Let 3x = t

So in sin 3x = sin t

When t âˆˆ(0, Ï€), sin t + >0 or 3x âˆˆ (0, Ï€)

But when Ï€/3 < x < Ï€/2

Ï€ < 3x < 3Ï€/2

Here sin 3x < 0

So, in x âˆˆ (0, Ï€/3),

f'(x) = -3sin 3x < 0 & in xâˆˆ(Ï€/3, Ï€/2), f'(x) = -3sin 3x > 0

f'(x) is changing signs, hence f(x) is not strictly decreasing.

(D) f(x) = tan x

f'(x) = sec2x

Now in x âˆˆ (0, Ï€/2), sec2x > 0

Hence, f(x) is strictly increasing on(0, Ï€/2).

So, option (A) and (B) are decreasing on (0, Ï€/2).

### (A) (0, 1)    (B) Ï€/2, Ï€     (C) 0, Ï€/2     (D) None of these

Solution:

f(x) = x100 + sin x – 1

f'(x) = 100x99 + cos x

(A) In (0, 1) interval, x > 0, so 100x99 > 0

and for cos x: (0, 1Â°) = (0, 0.57Â°) > 0

Hence, f(x)is strictly increasing in interval(0, 1)

(B) In (Ï€/2, Ï€) interval,

For 100x99: x âˆˆ (Ï€/2, Ï€) = (11/7, 22/7) = (1.5, 3.1) > 1

So, x99 > 1. Hence 100x99 > 100

For Cos x: (Ï€/2, Ï€) in second quadrant and in second quadrant cos x is negative, so the value is in be -1 and 0.

Hence, f(x)is strictly increasing in interval (Ï€/2, Ï€)

(C) In (0, Ï€/2) interval, both cos x > 0 and 100x99 > 0

So f'(x) > 0

Hence, f(x)is strictly increasing in interval (0, Ï€/2)

So, the correct option is (D).

### Question 14. For what values of a the function f given by f(x) = x2 + ax + 1 is increasing on (1, 2)?

Solution:

Given: f(x) = x2 + ax + 1

f'(x) = 2x + a

Now, x âˆˆ (1, 2), 2x âˆˆ (2, 4)

2x + a âˆˆ (2 + a, 4 + a)

For f(x) to be strictly increasing, f'(x) > 0

If the minimum value of f'(x) > 0 then

f'(x) on its entire domain will be > 0.

f'(x)min > 0

2 + a > 0

a > -2

### Question 15. Let I be any interval disjoint from [â€“1, 1]. Prove that the function f given by [Tex]f(x)=x+\frac{1}{x} [/Tex] is increasing on I.

Solution:

Clearly the maximum interval I is R-(-1,1)

Now, f(x) = [Tex]x+\frac{x}{1}[/Tex]

f'(x) = [Tex]1-\frac{1}{x^2}=\frac{x^2-1}{x^2}[/Tex]

It is given that I be any interval disjoint from [â€“1, 1]

So, for every x âˆˆ I either x < -1 or x > 1

So, for x < -1, f'(x) is positive.

So, for x < 1, f'(x) is positive.

Hence, f'(x) > 0 âˆ€ x âˆˆ I, so, f(x) is strictly increasing on I.

### Question 16. Prove that the function f given by f(x) = log sin x is increasing on (0, Ï€/2) and decreasing on (Ï€/2, Ï€).

Solution:

f(x) = log sin x

f'(x) = [Tex]\frac{1}{\sin x}.\cos x=\cot x  [/Tex]

Interval (0, Ï€/2), it is first quadrant, here cot x is positive.

So, f'(x) = cot x is positive (i.e., cot x > 0)

Hence, f(x) is strictly increasing in interval (0, Ï€/2)

Interval (Ï€/2, Ï€), it is second quadrant, here cot x is negative.

So, f'(x) = cot x is negative (i.e., cot x < 0)

Hence, f(x) is strictly decreasing in interval (Ï€/2, Ï€)

### Question 17. Prove that the function f given by f(x) = log|cos x| is decreasing on (0, Ï€/2) and increasing on (Ï€/2, Ï€).

Solution:

f(x) = log cos x

f'(x) = 1/cos x (-sin x) = -tan x

Interval (0, Ï€/2), it is first quadrant, here tan x is positive.

So, f'(x) = -tan x is negative(i.e., tan x < 0)

Hence, f(x) is strictly decreasing in interval (0, Ï€/2)

Interval (Ï€/2, Ï€), it is second quadrant, here tan x is negative.

So, f'(x) = -tan x is positive (i.e., tan x > 0)

Hence, f(x) is strictly increasing in interval (Ï€/2, Ï€)

### Question 18. Prove that the function given by f(x) = x3 – 3x2 + 3x – 100 is increasing in R.

Solution:

f(x) = x3 – 3x2 + 3x – 100

f'(x) = 3x2 – 6x + 3

f'(x) = 3(x2 – 2x + 1)

f'(x) = 3(x – 1)2 â‰¥ 0 âˆ€ x in R

So f(x) is strictly increasing in R.

### (A) (â€“ âˆž, âˆž)    (B) (â€“ 2, 0)    (C) (2, âˆž)     (D) (0, 2)

Solution:

Given, f(x) = x2e-x

f'(x) = x2(-e-x) + e-x.2x

f'(x) = e-x(2x – x2)

f'(x) = e-x.x(2 – x)

For f(x) to be increasing, f'(x) â‰¥ 0

So, f'(x) â‰¥ 0

e-x.x.(2 – x) â‰¥ 0

x.(2 – x) â‰¥ 0

x(x – 2) â‰¥ 0

x âˆˆ [0, 2]

So, the f(x) is strictly increasing in interval (0, 2). Correct option in D.

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