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

Content of this article has been merged with Chapter 6 Application of Derivatives – Miscellaneous Exercise as per the revised syllabus of NCERT.

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 

Solution:

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 = 

h(θ) = b sec θ + a cosec θ

Put h'(θ) = 

b sin³θ = a cos ³θ

tan3θ = a/b

tanθ = (a/b)^1/3

secθ = 

cosecθ = 

h_max = 

h_max = (b^2/3+a^2/3)^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) = 

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

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 = 

v(θ) = 

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()

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–          ……….(1)

Now, volume of cylinder = πr²h

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

V = π().h      

V(h) = 

On differentiating both side we get

V ‘(h) = 

Now, put V'(h) = 0

πR² = 

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:

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

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

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, 

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

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

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 = 

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)       (B) 

(C)       (D)

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

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

 According to the question it is given that the normal 

make equal intercepts with the axes.

So,

          …………(1)

The point (x₁, y₁)lie on the curve,

      …………(2)

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

From eq(2), we get

Hence, the required points are 

So, option A is correct.