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# Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.2 | Set 1

• Last Updated : 06 Apr, 2021

### 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) =  = 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) =  = 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) – , is an increasing function of x throughout its domain.

Solution:

f(x) = log(1+x)

f'(x)=

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 =  is an increasing function of Î¸ in[0, Ï€/2].

Solution:

y = f(Î¸) =

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) =  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)

### Chapter 6 Application of Derivatives – Exercise 6.2 | Set 2

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