Math Twelve

Application of Derivatives NCERT Solutions

NCERT Exercise:6.5

Question(1)Find the maximum and minimum values if any of the following functions given by

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

(ii)`f(x)=9x^2 +12x+2`

(iii)`f(x)=-(x-1)2+10`

(iv)`g(x)=x^3+1`

Solution:

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

`f(x)=(2x-1)xx2 = 4(2x-1)`

`f'(x)=8`

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

`:. 4(2x-1)=0=> x=1/2`

At `x=1/2`

`f(1/2)=8>0`

`x=1/2 ` is a point of minima.

`:.` Minimum value = `f(1/2)`

`=(2xx 1/2 - 1)^2 +3`

` =0 + 3 =3`

(ii)`f(x)=9x^2 +12x+2`

Solution:

(ii)Here `f(x)=9^2+12x+2`

`f(x)=18x+12`

`f(x)= 18`

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

`:. 18x+12=0`

`=> x = - 12/18 = - 2/3`

At `x- 2/3`

` f' (-2/3)= 18>0`

`x= (-2/3)` is a point of minima.

Minimum value `=f(- 2/3)`

`=9(-2/3)^2+ 12 xx - 2/3 +2`

` 9 xx 4/9 - 8+2 `

`4-8 + 2=2`

(iii)`f(x)=-(x-1)2+10`

Solution:

(iii)Here `f(x)= - (x-1)^2 +10`

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

`f(x)=-2`

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

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

`=> x-1=0`

`=> x=1`

At `x=1`

`f(1)=- 2<0`

`:. x=1` is point of maxima.

`:.` Maximum value `= f(1)=-(1-1)^2 +10 `

`=0+10 =10`

(iv)`g(x)=x^2+1`

Solution:

(iv)Here`g(x) = x^3+1`

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

` g'(x)= 6x`

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

`:. 3x^2 = 0`

`=> x=0`

At `x = 0`

` g' (0) = 6xx0 =0`

`:. x=0` is a point of neither maxima nor minima. It is a point of inflection .

Question(2)Find the maximum and minimum values if any of the following functionsgiven by

(i)`f(x)= |x+2|-1`

(ii)`g(x)=-|x+1| +3`

(iii)`h(x)=sin 2x +5`

(iv)` f(x)= |sin 4x + 3|`

(V) ` h(x)=x+1, x in(-1,1)`

(i)`f(x)= |x+2|-1`

Solution:

(i)Here`f(x)= |x+2|-1`

`|x+2| >= 0` For all `x in R`

`|x+2| - 1>= 0-1`

` |x+2| -1 >= -1`

`:. f(x) >= -1`

Now `-1 ` is the value of `f(x)` at ` x=-2`

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

Thus `f(x)` has minimum value `-1` at `x=2` but `f(x)` has no maximum value.

(ii)`g(x)=-|x+1| +3`

Solution:

(ii)Here`g(x)`

`=- |x+1|+3`

`|x+1 >= 0 ` for all `x in R`

` |x+1| <=0`

`=> - |x+1| + 3 <= 0 + 3`

`-|x+1| + 3 <=3`

`:.g(x)<= 3`

N0w 3 is a value of `g(x)` at x=-1`

`:. g(x) <= g( -1)`

Thus `g(x)` has maximum value 3 at `x=1` but `g(x)` has no minimum value.

(iii)`h(x)=sin 2x +5`

Solution:

(iii)Here`h(x)`

`sin 2x +5`

`-1<= sin 2x <= 1` for all `x in R`

`=> -1+5 <= sin 2x + 5 <= 1 + 5`

`=> 4<= sin 2x + 5 <= 6`

Thus maximum value of ` h(x)` is 6 and minimum value of ` h (x) ` is 4`

(iv)` f(x)= |sin 4x + 3|`

Solution:

(iv)Here`f(x) = |sin 4x + 3|`

`-1 <= sin 4x <= 1` for all `x in R`

`=> -1 +3 <= sin 4x + 3 <= 1 + 3`

`=> 2<= sin 4x + 3 <= 4`

`=> 2<= |sin 4x + 3| <= 4`

Thus maximum value of `f(x)` is 4 and minimum value of `f(x) ` is 2

(V) ` h(x)=x+1, x in(-1,1)`

Solution:

(v)Here`h(x)=`

`x + 1, x in (-1, 1)`

THe given function is strictly increasing in `(-1,1)`

THe given function has minimum value in the left of `-1` and maximum value in the right of 1. Thus `h(x)` has neither maximum nor minimum value in `(-1,1)`

Question(3)Find the maximum and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i)`f(x)=x^2`

(ii)`g(x)=x^3-3x`

(iii)`h(x)-sin x + cos x, 0< x < pi/2`

(iv)`f(x)=sin x -cos x,0< x < 2pi`

(v)`f(x)-x^3-6x^2+9x+15`

(vi)`g(x) =x/2 +2/x, x>0`

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

(viii)`f(x) = xsqrt (1-x), x>0`

(i)`f(x)=x^2`

Solution:

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

`f(x) =2x`

`f(x)=2`

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

`:. 2x = 0`

`=>x=0`

At `x=0`

`f"(0)=2>0`

`:. x=0 ` is a point of local minima.

Minimum value at `x=0` is

`f(0)=(0)^2=0`

(ii)`g(x)=x^3-3x`

Solution:

(ii)Here`g(x)= x^3-3x`

`g'(x)=3x^2-3`

`g"(x)=6x`

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

`:. 3x^2-3 = 0`

`=>x^2=1 `

`=> x =+- 1`

At `x=1 `

` g"(1)= 6xx1 =6>0`

`:. x=1 `is a point of minima

Minimum value `g(1)`

`= (1)^3-3xx1`

`= 1-3 =-2`

At `x = - 1`

`g"(-1)=6xx - 1`

`=-6<0`

`:. x = - 1` is a point of maxima.

`:.` Maximum value `g(-1)= (-1)^3- 3 xx -1`

`=-1+3=2`

(iii)`h(x)-sin x + cos x, 0< x < pi/2`

Solution:

(iii)Here` h(x)=sin x + cos x`

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

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

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

`:. cos x- sin x =0`

`=> sin x= cos x`

`=> tan x=1`

` => x=pi/4`

At `x=pi/4`

`h"(pi/4)`

`=-sin pi/4 - cos pi/4`

`=-1/sqrt2`

`=-2/sqrt2 =-sqrt2 < 0`

`At x=pi/4`

`h-(pi/4)=-sin pi/4- cos pi/4`

`=01/sqrt2 - 1/sqrt2 =-2/sqrt2`

`-sqrt2 <0`

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

Maximum value `=h(pi/4) `

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

`1sqrt 2+ 1/sqrt2`

`2/sqrt2= sqrt2`

(iv)`f(x)=sin x -cos x,0< x < 2pi`

Solution:

(iv)Here` 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 `

`tanx=-1`

`x= (3pi)/4 or (2pi)/ 4`

At `x = (3pi)/4`

`f"((3pi)/4)=-sin (3pi)/4 + cos (3pi)/4`

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

`=-sin pi/4 - cos pi/4 `

`-=1/sqrt2 - 1/sqrt 2`

`=-2/sqrt 2 =-sqrt 2 <0`

`:. x= (3pi)/4 ` is a point of maxima

Maximum value =`f((3pi)/4)`

`=sin((3pi)/4)- cos((3pi)/4)`

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

`sin pi/ 4 + cos pi/4`

`1/sqrt2+ 1sqrt2`

`=2/sqrt2 = sqrt2`

At `x=(7pi)/4`

`f" ((7pi)/4)=-sin (7pi/4)+ cos (7pi)/4`

`=-sin (2pi -pi/4) + cos (2x - pi/4)`

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

`= 1/sqrt2 + 1/sqrt2`

`=2/sqrt2 = sqrt2 >0`

`:. x= ((7pi)/4)` is a point of minima:.

`:.` Minimum value`f((7pi)/4)`

`= sin (7pi)/4 - cos 7pi/4`

`=- sin (2pi-pi/4) - cos (2pi- pi/4)`

`= - sin pi/4 - cos pi/4`

` = 1/sqrt/4 - cos pi/4 `

`=-1/sqrt2 - 1/sqrt2`

` =-2/sqrt2=-sqrt2`

(v)`f(x)-x^3-6x^2+9x+15`

Solution:

(v)Here`f(x)=x^3-6x^2+9x+15`

`f(x)=3x^2-12x+9`

`f"(x)=6x-12`

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

` :. 3x^2-12x+9=0`

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

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

` => 3[x(x-3)-1(x-3)]=0`

`=> 3(x-3)(x-1)=0`

`=> x=3 or x=1`

At `x=3`

`:. f"(3) = 6xx3-12`

`= 18 -12 =6 > 0`

`x =3 ` is a point of minima

`:.` Minimum value `=f(3)`

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

` +9 xx 3 + 15`

At `x=1`

`f"(1) = 6xx1 - 12`

` = 6 - 12 =- < 0`

`x = 1` is a point of mixima,/p>

` :.` Maximum value `= f(1)`

` = (1)^3- 6(1)^2 + 9 xx 1 + 15`

`=1-6+9+15=19`

(vi)`g(x) =x/2 +2/x, x>0`

Solution:

(vi)Here`g(x)= x/2 + 2/x`

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

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

`g"(x) = 4/x^3`

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

`:.(x^2-4)/(2x^2)=0`

`=>x^2-4`

`=>x^2=4`

`=> x=+-2`

`=> x=2`

`(because x>0)`

At `x=2`

`g"(2)=4/(2)^3`

`= 4/8 = 1/2 > 0`

`:. x= 2` is a point of minima.

Minimum value `=g(2)= 2/2+2/2 `

`=1+1 = 2`

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

Solution:

(vii)Here `g(x)1/(x^2+2)`

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

`g"(x)= ((x^2+2)^2 (-2)-(-2x).2(x^2+2).2x)/((x^2+2)^4)`

`=(-2(x^2+2)^2+ 8x^2 (x^2+2))/((x^2+2)^4)`

` =((x^2+2)[-2x^2-4+8x^2])/((x^2+2)^4)`

` (6x^2-4)/(x^2+2)^3`

`(2(3x^3-2))/(x^2+2)`

For maxima or minima ` g(x)=0`

`:. (-2x)/((x^2+2)^2)=0`

`=> - 2x =0 => x=0`

At `x = 0`

`g"(0)=(2[3(0)^2-2])/([(0)^2+2]^3)`

`=-4/8 = - 1/2 <0`

`x=0` is a point of maxima.

Maximum value ` g(0)=1/(0)^2+2 =1/2`

(viii)`f(x) = xsqrt (1-x), x>0`

Solution:

(viii)Here `f(x) =xsqrt (1-x)`

`f'(x) = x .1/(2sqrt(1-x) xx (-1)`

` + sqrt(1-x xx 1)`

`-x/(2sqrt(1-x))+ sqrt(1-x)`

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

=`(2-3x)/(2sqrt(1-x))`

`f"(x)= (2sqrt(1-x)(-3)-(2-3x).2.\1/(2sqrt(1-x)) xx(-1))/(4(1-x))`

`=(-6(1-x)+(2-3x))/(4(1-x)(sqrt(1-x)) `

` (-6+6x+2-3x)/(4(1-x)^(3/2))`

`(3x-4)/(4(1-x)^(3/2))`

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

`:. (2-3x)/(2 sqrt(1-x))=0 `

` =>2-3x=0 `

`=>x=2/3`

At `x=2/3`

` f"(2/3)`

`(3 xx 2/3 -4)/(4(1-2/3)^(2/2))`

`=-2/(4(1/3)^(3/2) <0`

`:. x= 2/3` is a point of maxima.

Maximum value `= f(2/3)`

`2/3 sqrt 1- 2/3`

`= 2/3 xx sqrt (1/3)`

` = 2/(3sqrt3)`

`=(2sqrt3)/9`

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