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Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives – Miscellaneous Exercise on Chapter 6 | Set 1

  • Last Updated : 06 Apr, 2021

Question 1. Using differentials, find the approximate value of each of the following:

(i) (17/81)1/4                                              

(ii) 33-1/5

Solution:

(i) (17/81)1/4 

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Let y = x1/4, x = 16/81 and △x = 1/81



△y = (x + △x)1/4 – x1/4

= (17/81)1/4 – (16/81)1/4

= (17/81)1/4 – (2/3)

So, 

(17/81)1/4 = (2/3) + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{4x^{\frac{3}{4}}}(△x)



\frac{1}{4(\frac{16}{81})^{\frac{3}{4}}}(\frac{1}{81})

= 27/32 × 1/81

= 1/96

= 0.010

Hence, the approximate value of (17/81)1/4 = 2/3 + 0.010 = 0.677

(ii) 33-1/5

Let y = x-1/5, x = 32 and △x = 1

△y = (x + △x)-1/5 – x-1/5

= (33)-1/5 – (32)-1/5

= (33)-1/5 – 1/2



So, 

(33)-1/5 = 1/2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-1}{5x^{\frac{6}{5}}}(△x)

\frac{-1}{5(32)^{\frac{6}{5}}}(1)

= -1/320

= -0.003

Hence, the approximate value of (33)-1/5 = 1/2 – 0.003 = 0.497

Question 2. Show that the function given by f(x) = log x/x has maximum at x = e.

Solution:



The given function is f(x) = log x/x

f'(x)=\frac{x(\frac{1}{x})-\log x}{x^2}=\frac{1-\log x}{x^2}

Now, f'(x) = 0

1 – log x = 0

log x = 1

log x = log e

x = e

Now, f''(x)=\frac{x^2(-\frac{1}{x})-(1-\log x)(2x)}{x^4}

 =\frac{-x-2x(1-\log x}{x^3}

 =\frac{-3+3\log x}{x^3}



f''(e)=\frac{-3+2\log e}{e^3}=\frac{-3+2}{e^3}=\frac{-1}{e^3}<0

Therefore, by second derivatives test, f is the maximum at x = e.

Question 3. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?  

Solution:

Given an isosceles triangle with fixed base b.

Let the other two sides be of length x.

Now its given that,

dx/dy = -3cm/s

Now semi-perimeter(s) = x + x + b/2

s = x + b/2

Area [by heron’s formula] = \sqrt{s(s-x)(s-x)(s-b)}

A=(s-x).\sqrt{s(s-b)}

A=\frac{b}{2}\sqrt{(x+\frac{b}{2})(x-\frac{b}{2})}

A=\frac{b}{2}\sqrt{x^2}\frac{-b^2}{2}

A=\frac{b}{4}\sqrt{4x^2-b^2}

To find: dA/dt = ?

DA/dt = ?

\frac{dA}{dt}=\frac{b}{4}.\frac{1}{2\sqrt{4x^2-b^2}}.8x.\frac{dx}{dt}

\frac{dA}{dt}=\frac{b}{4}.\frac{1}{2b\sqrt{3}}.8b.(-3)

dA/dt = -√3b cm2/s

Hence, the area is decreasing at the rate = √3b cm2/s



Question 4. Find the equation of the normal to curve x2 = 4y which passes through the point (1, 2).

Solution:

Given area: x2 = 4y

On Differentiating both sides with respect to y,

2x(dx/dy) = 4

dx/dy = 2/x

Slope = -1/m = -2/x

By point slope form equation of normal will be,

y – 2 = -1(x – 1)

x + y = 3 is the required equation of normal.

Question 5. Show that the normal at any point θ to the curve x = acosθ + aθsinθ, y = asinθ  – aθ cosθ is at a constant distance from the origin.

Solution:

Given curve,

x = acosθ + aθsin θ

y = asinθ – aθcos θ

Now -dx/dy = slope of normal =\frac{\frac{-dx}{dθ}}{\frac{dy}{dθ}}               -(1)

\frac{dx}{dθ}   = -asinθ + asinθ + aθcosθ

\frac{dx}{dθ}   = aθcosθ          -(2)

 \frac{dx}{dθ}   = acosθ + aθsinθ – acosθ

 \frac{dx}{dθ}   = aθsinθ                -(3)

\frac{-dx}{dy}=\frac{-aθ\cos θ}{aθ\sin θ}         -(From 1, 2 & 3)

-dx/dy = -cotθ



Now using point slope from, equation of normal will be, 

y-a\sin θ+aθ\cos θ=\frac{-\cos θ}{\sin θ}(x-a\cos θ-aθ\sinθ)

ysinθ – asin-2θ + aθcosθsinθ = -xcosθ + acos2θ + aθsinθcosθ

ysinθ + ysinθ a = 0

d=\frac{|0+0-a|}{\sqrt{\cos^2θ+\sin^2θ}}=a=   constant.

Hence proved

Question 6. Find the intervals in which the function f given by f(x)=\frac{4\sin x-2x-x\cos x}{2t\cos x}   is 

(i) increasing (ii) decreasing  

Solution:

f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}

f'(x)=\frac{(2+\cos x)(4\cos x-2+x\sin x-\cos x)+(4\sin x-2x-x-\cos x)(\sin x)}{(2+\cos x)^2}

f'(x)=\frac{4\cos x-cos^2x}{(2+\cos x)^2}



(i) For f(x) tp be increasing f'(x) ≥ 0

\frac{4\cos x-cos^2x}{(2+\cos x)^2}>0

 \cos x(4-\cos x)>0

Now, 4 – cos x > 0           -(because 4 – cos x ≥ 3)

So, cos x > 0

Hence, f(x) is increasing for 0 < x < x/2 and 3π/2 < x < 2π

(ii) For f(x)to be decreasing,

f'(x) < 0

\frac{4\cos x-\cos^2x}{(\cos x+2)^2}<0

cosx(4 − cosx) < 0

cosx < 0

Hence, f(x) is decreasing for π/2 < x < 3π/2

Question 7. Find the intervals in which the function f given by f(x)=x^3+\frac{1}{x^3},x≠0   is 

(i)increasing            (ii)decreasing  

Solution:

f(x) = x^3+\frac{1}{x^3},x≠ 0

f'(x) =3x^2-\frac{1}{3x^4}

(i) For f(x) to be increasing,

f'(x) > 0

3x^2-\frac{1}{3x^4}>0

\frac{9x^6-1}{3x^4}>0

9x6 > 1       



x>(\frac{1}{9})^{\frac{1}{6}}   or x∈((\frac{1}{9}^{\frac{1}{6}}),∞)

(ii) For f(x) to be decreasing,

f'(x) < 0 3x^2-\frac{1}{3x^4}<0

9x6 < 1               

x<(\frac{1}{9})^{\frac{1}{6}}   or x∈(∞,(\frac{1}{9}^{\frac{1}{6}}))

Question 8. Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1   with its vertex at one end of the major axis. 

Solution:

Given ellipse: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

Its major axis is the x-axis

Using parametric form of ellipse, x = acosθ, y = bsinθ,

If coordinates of A are (acosθ, bsinθ)

Then B’S coordinates will be (acosθ, -bsin θ).

Now, OC = a, OD = acos θ, so CD = a(1 + cos θ)

AB = |AD| + |BD| = 2b sin θ

Area of △ABC = 1/2.AB.CD

= 1/2.2bsin θ.a(1 + cos θ)

 △(θ) = ab.sinθ.(1 + sin θ)

For maxima/minima, put △'(θ) = 0

△'(θ) = ab[cosθ[1 + cosθ] + sinθ[-sinθ]]

△'(θ) = ab[2cos2θ + cosθ – 1] = 0

2cos2θ + cosθ – 1 = 0

2cos2θ + 2cosθ – cosθ – 1 = 0

2cosθ(cosθ + 1) – 1(cosθ + 1) = 0

(2cosθ – 1).(cosθ + 1) = 0

cosθ = 1/2 or cosθ = -1

If cosθ = -1, then sinθ = 0 & △(θ) = 0

But if cosθ = 1/2, sinθ = √3/2 & △(θ) = ab.\frac{\sqrt{3}}{2}(1+\frac{1}{2})

△(θ)_{max}=\frac{2\sqrt{3}ab}{4}

Question 9. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2m and volume is 8m3. If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square meter for sides. What is the cost of least expensive tank?

Solution:

Given:

Depth of tank = 2m



Volume = 8m3

Let the length be equal to x & width be to y

The base area will be equal to x.y.

Area of sides will be equal to; 2x, 2y, 2x, 2y

Now, volume = x.y.2 = 2xy = 8m3

so, xy = 4m2          -(1)

y = 4/x

Total cost = 70.base + 45.(sides)

c = 70xy + 45(2x + 2y + 2x + 2y)

c = 70.4 + 45.4(x + y)          -(xy = 4)

c(x) = 180 – \\frac{180.4}{x^2}=0

1-\frac{4}{x^2}=0

x2 = 4                        

x = ±2, x = 2,           (Rejecting -ve value)

y = 4/x = 4/2 = 2

Now cost c(x) = 280 + 180(x + 4/x)

c = 280 + 180(2 + 2)

c = 1000 rupees

Question 10. The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Solution:

Let the sides of square be x & radius of circle be r.

Perimeter of square = 4x

Circumference of circle = 2πr

Now given that, 4x + 2πr = k          -(1)

x = \frac{k-2πr}{4}

Area of square = x2

Area of circle = πr2

Sum of areas = x2 + πr2

s(r)=(\frac{k-2πr}{4})^2+πr^2

Put s'(r) = 0

s(r)=2(\frac{k-2πr}{4})(\frac{-π}{2})+2πr=0             -(From eq(1))

πr=\frac{π}{2}\frac{k-2πr}{4}

8πr = kπ – 2π2r

8r = k – 2πr

8r = (4x + 2πr) – 2πr           -(k = 4x + 2πr)

8r = 4x

x = 2r

Hence, proved that the sides of the square is double the radius of the circle.

Question 11. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. 

Solution:

Let the length of the rectangle = x

the breadth of the rectangle = y

and the radius of the semicircle = x/2

So given that total perimeter of the window = 10m

P = πx/2 + x + 2y = 10

x(1 + π/2) + 2y = 10

2y = 10 – x(1 + π/2)

y = 5 – x(1/2 – π/4)          -(1)

Now, the area of the window

A=\frac{πx^2}{2}+xy           -(2)

From eq(1) put the value of y in eq(2), we get

A=x.[5 - x(\frac{1}{2} - \frac{π}{4})]+\frac{πx^2}{2}



= 5x – x2(1/2 + π/4) + πx2/8

On differentiating we get

A’ = 5 – 2x(1/2 + π/4) + 2xπ/8

= 5 – x(1 + π/2) + xπ/4

Put A’ = 0

 5 – x(1 + π/2) + xπ/4 = 0

-x(1 + π/2) + xπ/4 = -5  

x(-1 – π/2 + π/4) = -5

x(-1 – π/2 + π/4) = -5

x(1 + π/4) = 5

x = 5/ (1 + π/4) 

x = 20/ π + 4

Hence, the length of the rectangle = 20/ π + 4

Now put the value of x in eq(1)

y = 5 – (20/ π + 4)(1/2 – π/4)      

y = 10/π + 4

Hence, breadth of the rectangle = 10/π + 4

and the radius of the semicircle = x/2 = \frac{\frac{20}{π + 4}}{2}  = 10/π + 4

Chapter 6 Application of Derivatives – Miscellaneous Exercise on Chapter 6 | Set 2




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