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`
Reference: