Application of Derivatives NCERT Solutions

NCERT Exercise 6.5:Q: 4-9

Question(4) Prove that the following functions do not have maxima or minima.

(i)`f(x) = e^x`

(ii)`g(x)=log x`

`(iii)h(x)=+x^2+x+1`

(i)`f(x) = e^x`

Solution:

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

`f'(x)= e^x`

Domain of `e^x` is R.

Now `f'(x) ` has no value of `x` in R at which `f'(x) =0 `, so there exist no critical point in R.

Thus `f(x)`= e^x does not have point of maxima of minima.

(ii)`g(x)=log x`

Solution:

(ii)Here `g(x) = log x`

g"(x)=1/x`

Domain of log `x` is `(0,oo)`

Now `g (x) ` has no value of `x` in `(0,oo)` at which `g' (x) =0`, So there exists no critical point in ` (0,oo)`

Thus , `f(x) = log x ` does not have point of maxima or minima.

(iii)`h(x)=+x^2+x+1`

Solution:

(ii)Here `h(x)= x^3 + x^2 + 1`

`h'(x) = 3x^2 + 2x +1`

For maxima or minima, `f'(x) =0`

`:. 3x^2 + 2x +1 =0`

`x = (-2+-sqrt(4-12))/6 `

`=(-2+-sqrt (-8))/6`

These exists no real value of `x` for which `f(x)=0`. So there exists no critical point.Thus `h(x)= x^3+x^2 + +1` does not have point of maxima pr minima.

Question (5) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) ` f(x)=x^3, x in [-2,2]`

(ii)` f(x)= sinx + cos x. x in[0,pi]`

(iii)`f(x)=4x -1/ 2 x^2 ,xin[-2, 9/2]`

(iv)`f(x)=(x-1)^2+3. x in [-3.1]`

(i) ` f(x)=x^3, x in [-2,2]`

Solution:

(i)Here `f(x)= x^3`

`f'(x) =3x^2`

For maxima or minima.`f'(x)=0`

`:. 3x^2=0 `

`=> x =0`

At ` x- 2 `

`f(-2) = (-2)^3=-8`

At `x=0`

`f(0)=(0)^3=0`

At`x=2 `

`f(2)=(2)^3=8`

Thus absolute maximum value is `8` at ` x= 2` and absolute minimum value is `-8` at `x=-2`

(ii)` f(x)= sinx + cos x. x in[0,pi]`

Solution:

(ii)Here `f(x) = sin x+cos x`

`f'(x) = cos x -sin x`

For maxima or minima, `f(x) =0`

`:. cos x- sin x =0`

`=> sin x = cos x`

`=> tan x=1`

`=> x= pi/4`

At `x=0`

`f(0)= sin (0) + cos (0) `

`0+1=1`

At ` x= pi/ 4`

`f (pi/4) `

`=sin (pi/4)+ cos (pi/4)`

`1/sqrt 2 + 1/sqrt2 `

`2/sqrt2 = sqrt2`

At ` f(x) = sin pi + cos pi =0-1 =-1`

Thus absolute maximum value is `sqrt2` at `x = pi/4 ` and absolute minimum value is `- 1` at `x=pi`

(iii)`f(x)=4x -1/ 2 x^2 ,xin[-2, 9/2]`

Solution:

(iii)Here `f(x) = 4x- 1/2 x^2`

`f(x)=4-x`

For maxima or minima,` f"(x) =0`

`:. 4-x = 0`

`=> x=4`

At `x=-2 `

`f(-2) = 4xx-2-1/2 xx(-2)^2`

`=-8- 4/2 =-8-2=-10`

`f4=4xx4 - 1/2 xx (4)^2`

`16 - (16)/2 = 16 -8=8`

At `x= 9/2`

`f(9/2)= 4xx 9/2 - 1/2 xx ( 9/2)^2 `

`= 18 - 81/8= 63/8`

Thus absolute maximum value is `8` at `x = 4` and absolute minimum value is `-10` at `x=-2`

(iv)`f(x)=(x-1)^2+3. x in [-3.1]`

Solution:

(iv)Here `f(x) = (x-1)^2+3`

`f(x) = 2(x-1)`

For maxima or minima, `f' (x) =0`

`:. 2(x-1)` ` = 0 => x=1`

At `x=-3 `

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

`=16+ 3 = 19`

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

`= 0 + 3 = 3`

Thus absolute maximum value is `19` at ` x= -3` and absolute minimum value is `3 ` at `x=1`

Question (6) Find the maximum profit that a company can make, if the profit function is given by `p(x)=41 - 72x- 18x^2`

Solution:

Here `p(x)= 41-72x-18x^2`

`p' (x) =-72 - 36 x`

`=-36(2+x)`

`p"(x) = -36`

For maxima or minima, `p'(x) = 0`

`:. -36 (2+x)= 0`

` => x+2 =0`

`=> x = -2`

At `x=-2`

`p"(-2)= - 36 < 0`

`:. x= -2` is a point of maxima.

`:. ` Maximum profit ` p(-2) = 41 - 72 ` `(-2)-18 (-2)^2`

`= 41+ 144 -72 = 113 ` units`

Question (7) Find both the maximum value and minimum value of ` 3x^4 - 8x^3 + 12 x^2- 48x + 25 ` on the interval `[0,3]`

Solution:

Let `f(x) = 3 x^4- 8x^3` ` +12x^2 - 48x + 25`

`f'(x)= 12(x)^3- 24x^2 ` ` + 24x - 48`

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

`12[x^2(x-2) + 2 ( x-2)] `

`= 12 [x^2(x-2)+ 2 (x-2)]`

`12 (x-2)(x^2+2)`

for maxima or minima,` f' (x) = 0`

`:. 12(x-2) (x^2+2)`

`0 => x - 2 = 0`

` =>x=2`

At `x=0`

Thus maximum value is `25` at `x=0` and minimum value `-39` is at `x=2`

Question (8)At what points in the interval `[0,2pi]` , does the function sin `2x` attain its maximum value?

Solution:

Let `f(x)=sin 2x`

`f'(x) = 2cos 2x`

`f' (x) =-4 sin 2x`

For maxima or minima, `f' (x) =0`

`:. 2 cos 2x = 0`

` => cos 2x =0`

`2x = pi /2, (3pi)/2 , (5pi)/2, (7pi)/2`

`x = Pi/4, (3pi)/4, (3pi)/4, (5pi)/4, (7pi)/4`

`x=0`

`f(0)=sin (2xx0)`

` =sin 0=0`

At `x= pi/4 `

`f(pi/4)= sin (2xx pi/4)`

`=sin pi/2 =1`

At `x=(3pi)/4`

`f((3pi)/4)`

`=sin (2xx (3pi)/4)`

` sin (3pi)/2`

` =(pi + pi/2)`

`-sin pi/2 =-1`

At `x=(5pi)/4`

`f((5pi)/4)`

`=sin(2xx (5pi)/4)`

`=sin (2pi + pi/2)`

`=sin pi/2 =1`

At `x=(7pi)/4`

`f((7pi)/4)=sin (2 xx (7pi)/4) `

`= sin (7pi)/2 `

`=sin (3pi +pi/2)`

`=-sin pi/2 =-1`

At ` x=2pi`

` f(2pi) = sin (2xx 2pi)`

`= sin 4 pi =0`

Thus maximum value is `1` at `x=pi/4 ` and `x = (5pi)/4`

Question (9)What is the maximum value of the function sin `x + cos x`?

Solution:

Let `f(x)= sin x + cos x`

`f'(x)= cos x - sin x`

`f"(x)=-sin x-cos x`

For maxima or minima, `f'(x) =0`

`:. cos x - sin x `

`= 0 => sin x = cos x`

`=> tan x =1 `

`=> x=pi/4`

At `x = pi/4`

`f" (pi/4)`

`= - sin pi/4`

` 1/sqrt2 - 1/sqrt2 `

`-2/sqrt2 = -sqrt2 <0`

`:. = pi/4` is a point of maxima.

`:.` Maximum value ` f(pi/4)`

`=sin pi/4 + cos pi/4`

`1/sqrt2 + 1/sqrt2`

`=2/sqrt2 = sqrt2`

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