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Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives – Miscellaneous Exercise on Chapter 6 | Set 2

  • Last Updated : 07 Nov, 2021

Chapter 6 Application of Derivatives – Miscellaneous Exercise on Chapter 6 | Set 1

Question 12. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}

Solution:

Class 12 NCERT Mathematics Part 1 Chapter 6 Miscellaneous Exercise

Given, a triangle ABC

Let, PE = a & PD = b

In the △ABC, ∠B = 90

Let ∠C = θ, so, ∠ DPA = θ

DP|| BC.

Now in △ADDP,

cosθ = DP/AP = b/AP 

AP = b/cosθ

In △EPC,

sinθ = EP/CP = a/CP

CP = a/sin θ               

Now AC = h = PA + PC

h = \frac{b}{\cos θ}+\frac{a}{\sin θ}

h(θ) = b sec θ + a cosec θ

Put h'(θ) = \frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}

\frac{b}{\cos θ}.\frac{\sin θ}{\cos θ}=\frac{a}{\sin θ}.\frac{\cos θ}{\sin θ}

b sin3θ = a cos 3θ

tan3θ = a/b

tanθ = (a/b)1/3 

secθ = \frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}

cosecθ = \frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}

hmax b.\frac{\sqrt{b^{2/3}+a^{2/3}}}{b^{1/3}}+a.\frac{\sqrt{b^{2/3}+a^{2/3}}}{a^{1/3}}

hmax = (b2/3+a2/3)3/2

Question 13. Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has 

(i) local maxima 

(ii) local minima 

(iii) point of inflexion 

Solution:

f(x) = (x – 2)4(x + 1)3

On differentiating w.r.t x, we get

f'(x) = 4(x – 2)3(x + 1)3 + 3(x + 1)2(x – 2)4

Put f'(x) = 0

(x – 2)3(x + 1)2 [4(x + 1) + 3(x – 2)] = 0

(x – 2)3(x + 1)2(7x – 2) = 0

Now,

Around x = -1, sign does not change, i.e

x = -1 is a point of inflation

Around x = 2/7, sign changes from +ve to -ve i.e.,

x = 2/7 is a point of local maxima.

Around x = 2, sign changes from -ve to +ve i.e.,

x = 2 is a point of local minima

Question 14. Find the absolute maximum and minimum values of the function f given by f(x) = cos2 x + sin x, x ∈ [0, π] 

Solution:

f(x) = cos2x + sin x; x ϵ [0, π]

On differentiating w.r.t x, we get

f'(x) = 2cos x(-sin x) + cos x = cos x – sin2x

Put f'(x) = 0

cos x(1 – 2sin x) = 0

cos x = 0; sin x = 1/2

In x ϵ[0, π] if cos x = 0, then x = π/2

and if sin x = 1/2, then x = π/6 & 5π/6

Now, f”(x) = -sin x – 2 cos2x

f”(π/2) = -1 + 2 = 1 > 0

x = π/2 is a point of local minima f(π/2) = 1

f”(π/6) = \frac{-1}{2}-2.\frac{1}{2}=\frac{-3}{2}<0

x = π/6 is a point of local maxima f(π/6) = 5/4

f''(\frac{5π}{6})=\frac{-1}{2}-2.(\frac{-1}{2})>0

x = 5π/6​ is a point of local minima f(5π/6) = 5/4

Global/Absolute maxima = ma{f(0), f(π/6), f(π)}

= max{1, 5/4, 1}

= 5/4 = Absolute maxima value

Global/Absolute minima = min{f(0), f(π/2), f(π/6), f(π)}

= min{1, 1, 5/4, 1}

= 1 = Absolute minima value

Question 15. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π/3

Solution:

Let ABC be the cone

and o is the centre of the sphere.

AO = BO = CO = R

AO = h = height of cone

BD = CD = r = radius of cone.

∠DOC = θ                  -(Properties of circle)

In △ DOC,

OD = R cosθ & CD = Rsinθ,

r = R sin θ 

AD = AO + OD = R + Rcosθ 

h = R(1 + cosθ)

Now, the volume of the cone is 

V = \frac{1}{3}πr^2h

v(θ) = \frac{1}{3}.πR^2\sin{-2}θ.R(1+\cosθ)

v'(θ)=\frac{}{}[\sin{-2}θ(-\sinθ)+(1+\cosθ)(2\sinθ\cosθ)]

Put v(θ) = 0

sinθ[2cosθ + 2cos2θ − sin2θ] = 0

sinθ[2cosθ + 2cos2θ − 1] = 0

sinθ(3cosθ − 1)(1 + cosθ) = 0

sinθ = 0, cos = 1/3​, cosθ = −1

If sinθ = 0, then volume will be 0.

If cosθ = -1, then sinθ = 0 & again volume will be 0.

But if cosθ = 1/3; sinθ = 2√2/3 and 

Volume, v = 32/81​πR3, which is maximum.

Height, h = R(1 + cosθ) = R(1+\frac{1}{3}    )

h = 4r/3

Hence proved

Question 16. Let f be a function defined on [a, b] such that f′(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b). 

Solution:

Given that on [a, b] f'(x) > 0, for all x in interval I. 

So let us considered x1, x2 belongs to I with x1 < x2 

To prove: f(x) is increasing in (a, b)

According to the Lagrange’s Mean theorem

f(x2) – f(x1)/ x2 – x1 = f'(c)

f(x2) – f(x1) = f'(c)(x2 – x1)

Where x1 < c < x2 

As we know that x1 < x2 

so x1 < x2 > 0

It is given that f'(x) > 0

so, f'(c) > 0

Hence, f(x2) – f(x1) > 0

f(x2) < f(x1)

Therefore, for every pair of points x1, x2 belongs to I with x1 < x2 

f(x2) < f(x1)

f(x) is strictly increasing in I

Question 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also, find the maximum volume.

Solution:

In △ABC, 

AC2 = BC2 + AB2

4R2 = 4r2 + h2

r2 = R2\frac{h^2}{4}                ……….(1)

Now, volume of cylinder = πr2h

Put the value ov r2 from eq(1), we get

V = π(R^2\frac{-h^2}{4}   ).h      

V(h) = πR^2h-\frac{πh^3}{4}

On differentiating both side we get

V ‘(h) = πR^2h-\frac{3πh^3}{4}

Now, put V'(h) = 0

πR2 \frac{3}{4}πh^2

h=\frac{2R}{\sqrt{3}}

Now the maximum volume of cylinder = π[R2. 2R/√3 – 1/4.4R2/3.2R/√3]

= 4πR3/ 3√3

Question 18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α is one-third that of the cone and the greatest volume of the cylinder is 4/27πh3tan2α.

Solution:

Class 12 NCERT Mathematics Part 1 Chapter 6 Miscellaneous Exercise

Let, 

XQ = r

XO = h’

AO = h

OC = r’

∠XAQ = α

In triangle AXQ and AOC = XQ/OC = AX/AO

So, r’/r = h-h’/h

hr’ = r(h-h’)

hr’ = rh – rh’

rh’ = rh – hr’

rh’ = h(r – r’)

h’ = h(r – r’)/r

The volume of cylinder = πr’2h’

v = πr’2(h(r – r’)/r)

= π(h(rr’2 – r’3)/r)

On differentiating we get

v’ = πh/r(2rr’ – 3r’2)

Again on differentiating we get

v” = πh/r(2r – 6r’) ………(1)

Now put v’ = 0

 πh/r(2rr’ – 3r’2) = 0

(2rr’ – 3r’2) = 0

2r’r = 3r’2

r’ = 2r/3

So, v is maximum at r’ = 2r/3

The maximum volume of cylinder = πh/r[r. 4r2/9 – 8r2/27]

= πhr2[4/27]

= 4/27πh(h tanα)2

= 4/27πh3 tan2α

Question 19. A cylindrical tank of a radius 10 m is being filled with wheat at the rate of 314 cubic meters per hour. Then the depth of the wheat is increasing at the rate of 

(A) 1 m/h      (B) 0.1 m/h      (C) 1.1 m/h      (D) 0.5 m/h 

Solution:

Given,

Radius of cylinder = 10m   [radius is fixed]

Rate of increase of volume = 314m3/h

ie   dv/dt = 314m3/h

Now, the volume of cylinder = πr2h

v = π.(10)2.h

v = 100πh

On differentiating w.r.t t, we get

dv/dt = 100π\frac{dh}{dt}

\frac{dh}{dt}=\frac{1}{100π}.\frac{dv}{dt}=\frac{1}{314}.314

\frac{dh}{dt}=1m/h        

So option A is correct

Question 20. The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) is

(A) 22/7     (B) 6/7     (C) 7/6      (D) -6/7

Solution:

Given that the slope of the tangent to the curve x = t2 + 2t – 8 and y = 2t2 – 2t – 5

On differentiating we get

\frac{dy}{dx}=2t+3  ;\frac{dy}{dt}=4t-2

Now, when x = 2,

t2 + 3 – 8 = 2

t2 + 3 – 10 = 0

t2 – 2t + 5t – 10 = 0

(t – 2)(t + 5) = 0

Here, t = 2, t = -5   ……….(1)

When y = -1

2t2 – 2t – 5 = -1

2t2 – 2t – 4 = 0

t2 – t – 2 = 0

(t + 1)(t – 2) = 0

t = -1 or t = 2  ……….(2)

From eq(1) & eq(2) satisfies both,

Now, \frac{dy}{dx}=slope=\frac{dy}{dt}

\frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}

So, option B is the correct.

Question 21.The line y = mx + 1 is a tangent to the curve y2 = 4x if the value of m is

(A) 1       (B) 2      (C) 3       (D)1/2

Solution:

The curve if y2 = 4x  …….(1)

On differentiating we get

2y\frac{dy}{dx}=4

\frac{dy}{dx}=\frac{2}{y}

The slope of the tangent to the given curve at point(x, y)

\frac{dy}{dx}=\frac{2}{y}

m = 2/y

y = 2/m

The equation of line is y = mx + 1

Now put the value of y, we get the value of x

2/m = mx + 1

x = 2 – m/m

Now put the value of y and x in eq(1), we get

(2/m)2 = 4(2 – m/m)

m = 1

Hence, the option A is correct

Question 22. The normal at the point (1, 1) on the curve 2y + x2 = 3 is 

(A) x + y = 0             (B) x – y = 0 

(C) x + y +1 = 0        (D) x – y = 1 

Solution:

The equation of curve 2y + x2 = 3

On differentiating w.r.t x, we get

2\frac{dy}{dx}+2x=0

dy/dx = -x

The slope of the tangent to the given curve at point(1, 1)

dy/dx = -x = -1

m = -1

And slope of normal = 1

Now the equation of normal 

(y -1) = 1(x – 1)

x – y = 0

So, B option is correct

Question 23. The normal to the curve x2 = 4y passing (1, 2) is 

(A) x + y = 3            (B) x – y = 3           (C) x + y = 1       (D) x – y = 1

Solution:

The equation of curve is x2 = 4y …….(1)

On differentiating w.r.t x, we get

2x = 4\frac{dy}{dx}

\frac{dy}{dx}=\frac{x}{2}

The slope of normal at (x, y)

-dx/dy = -2/x = m

The slope at given point(1, 2)

m = (y – 2)/(x – 1)

-2/x = (y – 2)/(x – 1)

y = 2/x

Now put the value of y in eq(1)

x2 = 4(2/x)

x = 2

and y = 1

So the point is (2, 1)

Now the slope of normal at point(2, 1) = -2/2 = -1

The equation of the normal is

(y – 1) = -1(x – 2)

x + y = 3

So option A is correct

Question 24. The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are

(A) (4,±\frac{8}{3})            (B) (4,\frac{-8}{3})

(C)(4,±\frac{3}{8})             (D) (±4,\frac{3}{8})

Solution:

Given equation 9y2 = x3

On differentiating w.r.t x, we get

18y dy/dx = 3x2

dy/dx = 3x2/18y 

dy/dx = x2/6y 

Now, the slope of the normal to the given curve at point (x1, y1) is

-1=\frac{6y_1}{x^2_1}

Hence, the equation of the normal to the curve at point (x1, y1) is

y-y_1=\frac{-6y_1}{x^2_1}(x-x_1)

x^2_1y-x_1^2y=-6xy_1+6x_1y_1

\frac{6xy_1}{6x_1y_1+x_1^2y_1}+\frac{x_1^2y}{6x_1y_1+x^2_1y_1}

\frac{\frac{x}{x_1(6+x_1)}}{6}+\frac{\frac{y}{y_1(6+x_1)}}{x_1}=1

 According to the question it is given that the normal 

make equal intercepts with the axes.

So,

\frac{x_1(6+x_1)}{6}+\frac{y_1(6+x_1)}{x_1}

x_1^2=6y_1                …………(1)

The point (x1, y1)lie on the curve,

9_1^2=x_1^22            …………(2)

From eq(1) and (2), we get

9(\frac{x_1^2}{6})^2=x_1^3=\frac{x_1^4}{4}=x^3_1=x_1=4

From eq(2), we get

9y_1^2=(4)^3=64

y_1^2=\frac{64}{9}

y_1=±\frac{8}{3}

Hence, the required points are (4,±\frac{8}{3})

So, option A is correct.


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