Application of Derivatives NCERT Solutions

NCERT Exercise 6.2

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

Solution:

Here `f(x)=3x +17`

Let `x_1,x_2 in R` such that `x_1 < x_2`

Now `3 x_1 < 3 x_2`

`=> 3x_1 + 17 < 3x_2 + 17`

`=> 3x_1 + 17 < 3x_2 + 17`

`f(x_1)< f(x_2)` when `x_1 < x_2 ` for all `x_1,x_2 in R`

Thus `f(x)` is strictly increasing function on `R`

Question: (2) Show that the function given by `f(x)=e^(2x) ` is strictly increasing on `R`.

Solution:

Here `f(x) = e^(2x)`

`d/(dx)f(x) = d/(dx) [e^2x]`

` f'(x) = 2e^(2x)`

When `x > 0 `

then `2e^(2x)`

`= 2[1+2x+ ((2x)^2)/(2!) + ----]>0`

`:. f'(x) >0`

When ` x=0`

then `2e^(2x) = 2.1 `

` =2.1 = 2=0

`:. f'(x) > 0`

`:. f'(x)> 0`

When `x< 0`

then ` 2e^(2x)= 2/e^(2y)`

`=2/[1+2y+(2y)^2/(2!) + ---]^(>0)`

`:. f'(x)> 0`

Thus `f(x)` is strictly increasing function on `R`.

Question: (3) Show that the function given by `f(x)= sin x`is

(a)strictly increasing in `(0, pi/2)`

(a)strictly increasing in `(pi/2, pi)`

(a)neiyher increasing nor decreasing in `(0, pi)`

Solution:

Here `f(x)=sin x`

`d/(dx)[f(x)]= d/(dx)(sinx)`

`f' (x) =cos x`

Then `cos x > 0 => f' (x)> 0`

`:.f(x)` is strictly increasimg function in `(0,pi/2)`

(b) When `x in(pi/2, pi)`

Then `cos x < 0 => f' (x) , 0 `

`:. f(x) ` is strictly decreasing function in `(pi/2, pi)`

(c) When `x in (pi, 0)`

We see that `f'(x)> 0` in `(0, pi/2)` and `f'(x) < 0` in `(pi/2, pi)`

So`f' (x) ` is positive and negative in `(0,pi)`

Thus `f(x)` is neither increasing nor decreasing in `(0,pi)`

Question: (4) Find the intervals in which the function f given by `f(x)=2x^2-3x` is

(a) strictly increasing

(b) strictly decreasing

Solution:

Here `f(x)= 2x^2-3x`

` :. f'(x)=4x -3 `

`:. f'(x)=0 `gives `4x-3 =0`

`x= 3/4`

`=>x=3/4`

Case (11)When `x< 3/4`

` :.f'(x)`

`= -ve. So,f'(x)>0`

Hence, `f(x)` is strictly increasing in `(3/4, oo)`

Question: (5) Find the intervals in which the function `f` given by `f(x)=2x^3 -3x^2-36x+7`

(a)Strictly increasing

(b)strictly decreasing

Solution:

We have

` f'(x) =2x^3- 3x^2- 36x+7`

`:. f'(x) = 6x^2 - 6x - 36 `

` =6(x^2-x-6)`

`=6[x^2-3x+2x-6]`

`=6[x(x-3)+ 2(x-3)]`

`= 6(x-3)(x+2)`

Now, `f'(x)`=0`

`=> x=-2` or `x=3`

Thus, the critical points are - 2 and 3

Case 1: When ` x,-2`

in this case , ` x -3<0` and `x+2 < 0`

`:.f(x)=6(-)(-)`

`=+v` . So,`f'(x)>0`

`:.f(x)` is increasing in `(-oo, -2)`

Case 11: When `-2 < x < 3`

In this case , x-3 < 0 ` and `x+2 > 0`

`:. f' (x)=6(-)(+)=- ` ve So , `f'(x) < 0

`:.f(x)` is decreasing in ` (-2,3)`

Conclusions :

(i) `f(x)` is increasing in `(-oo,-2) uu (3,oo)`

(ii) `f(x)` is decreasingin `(-2,3)`.

Question:(6) find the intervals in which the following functions are strictly increasing or decreasing

(a)`x^2+2x-5`

(b)`10-6x-2x^2`

(c)`-2x^3-9x^2-12x+1`

(d)`(x+1)^3(x-3))^3`

(e)`f'(x)=(x+1)^2 (x-3)^3`

Solution:

(a)We have `f(x)= x^2+2x-5`

`:. f'(x)=2x+2 = 2(x+1)`

Now,` f,(x)=0`

`=> 2(x+1)=0`

`=> x=-1`

Case 1: When `x < - 1 `

in this case `x+1 <0`

`:. f'(x)= 2(-)=-ve` .So ,`f(x) < 0`

`:. f(x)` is decreaasing in `(-oo, -1)`

Case11: When `x > -1`

In this case `x+1 > 0`

`:. f'(x) =2(+)= +`ve. So,`f' (x) > 0`

`:. f(x)` is strictly increasing in `(-1, oo)`

(b) We have `f(x)=10-6x-2x^2`

`:. f'(x) =-6-4x=-2(3+2x)`

Now `f(x) =0`

`-2(3+2x) =0`

` =>3+2x=0`

` x = -3/2`

Case 1: When `x < (-3)/2 `

In this case `3+2x <0`

`:. f(x) =-2(-)`

`=+ve`. So , `f' (x) > 0`

`:. f(x)` is increasing in `(-oo, -3/2)`

Case 11: When `x > -3/2 `

In this case `3+2x > 0`

`:. f,(x)=-2(+)=-ve. So, `f'(x)<0`

`:. f(x) ` is strictly decreasing in `(-3/2, oo)`

(c)We have `f(x)=-2x^3- 9x^2 - 12 x+1`

`f'(x)= -6x^2- 18 x-12`

`=-6(x^2+3x+2)`

`=-6(x^2+2x+x+2)`

`=-6[x(x+2)+(x+2)]`

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

Now, `f'(x) =0`

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

`=> x+2=0` and ` x+1=0`

`=>x=-2,x=-1`

Case 1: When `x < -2`

In this case

`6(x+2)(x+1)=-6(-)(-)=-ve`

So, `f'(x)<0`

`:. f(x)` is increasing in `(-22, -1)`

Conclusion:

(i)`f(x)` is strictly increasing in `(-2, -1)`.

(ii)`f(x)` is strictly decreasing in

`(-oo,-2)uu (-1,oo)`

(d) We have `f(x)=(x+1)^3(x-3)^2`

`f'(x)=(x+1)^3xx3(x-3)^2` `+(x-3) x^2xx3(x+1)^2`

`=3(x-3)^2(x+1)^2 [x+1+x-3]`

`=3(x-3)^2(x+1)^2 2(x-1)`

`=6(x-3)^2(x+1)^2(x-1)`

clearly `(x-3)^2> 0 ` and `(x+1)^2 >(0)`

Case 1: When `x < 1`

in this case `f(x) = 6(+)(+)(-)=-ve`

So,` f(x)<0`

Thus , `f(x)` is decreasing in `(-oo,1)`

Case 11: When `x> 1`

In this case

`f(x) = 6(+)(+)(-)=-ve`

So, `f(x)<0`

Thus , `f(x)` is stictly increasing in `(1,-oo)`

(e)`f'(x)=(x+1)^3(x-3)^3`

`f'(x)=3(x+1)^3(x-3)^3+3(x+1)(x-3)^3`

`f'(x)=3(x+1)^3(x-3)^3[x+1)(x-3+1+x-3]`

`f'(x)=6(x+1)^3(x-3)^2(x-1)`

For `f(x)` to be increasing we must have

`f'(x)> 0`

`6(x_1)^3 (x-3)^3(x-1)> 0`

`[:.6(x+1)^3(x-3^2 >0]`

`=> x-1 > 0`

`=> x >1`

`=> x` in `(1,oo)`, so `f(x)` is increasing on (1, oo)` for `f(x)` to decreasing we must have

`f(x)<0 `

`6(x+1)^3 (x-3)^3 (x-1)` < 0

`=>x-1 < 0`

`=>x > 1`

`x in (-oo, 1)`

So ` f(x)` is decreasing on `(-oo,1)`

Question: (7) Show that `y=log (1+x)-(2x)/(2+x),x > -1`, is an increasing function of `x` through out its domain.

Solution:

Clearly, `log (1+x)` is defined only when `x> -1`

Now , `f'(x)= 1/(1+x)-((2+x)2-2x)/(2+x)^2`

`=1/(1+x) (2+(2+x-x))/(2+x)^2`

`1/(1+x)- 4/(2+x)^2`

`((2+x)^2-4(1+x))/(1+x)(2+x^2)`

`=(4+4x + x^2 - 4 - 4x)/ (1+x)(2+x)^2`

`=x^2/(1+x)(2+x)^2`

`:. f'(x)=0`

S0 `x^2=0=>x=0`

Case 1: When `-1 x < 0,`

In this case `f'(x)> 0`

`:. f(x)` is increasing in `(-1,0)`

Case 11: When `x > 0`

in this case `f'(x) >0`

`:. f(x)` is increasing in `(0,oo)`

Hence , `f(x)` Is increasing in `(-1,0)uu(0, oo)`

Question: (8) Find the values of `x` for which `y=[x(x-2)]^2` is an increasing function.

Solution:

` y = [x(x-2)]^2`

`=x^2(x-4x+4)`

` y=x^4 - 4x^3+4x^2`

`:. dy/dx=4x^3 - 12 x^2+8x`

-For the function to be increasing

`dy/dx > 0`

`=> 4x^3 - 12 x^2+ 8x > 0 `

`=> 4x(x^2-3x + 2) >0 `

`=> 4x(x-1(x-2)>0`

For `0< x < 1 `

` (dy)/(dx) = (+)(-)(-)=+ ve ` and for `x > 2`

`(dy)/(dx) = (+)(+)(+)= + ve`

Thus the function is increasing for `0 < x < 1 ` and `x` > 2.

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