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

### 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

**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 sin

^{3}θ = a cos^{ 3}θtan3θ = 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) = 0Now,

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) = cos^{2} x + sin x, x ∈ [0, π]

**Solution:**

f(x) = cos

^{2}x + 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 = 5

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

sin

θ[2cosθ+ 2cos^{2}θ− sin2θ] = 0sin

θ[2cosθ+ 2cos^{2}θ− 1] = 0sin

θ(3cosθ− 1)(1 + cosθ) = 0sin

θ= 0, cos = 1/3, cosθ= −1If 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

πR, which is maximum.^{3}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,

AC

^{2 }= BC^{2 }+ AB^{2}4R

^{2 }= 4r^{2 }+ h^{2}r

^{2 }= R^{2}– ……….(1)Now, volume of cylinder = πr

^{2}hPut the value ov r

^{2}from eq(1), we getV = π().h

V(h) =

On differentiating both side we get

V ‘(h) =

Now, put V'(h) = 0

πR

^{2 }=Now the maximum volume of cylinder = π[R

^{2}. 2R/√3 – 1/4.4R^{2}/3.2R/√3]= 4πR

^{3}/ 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 4/27*πh*^{3}tan^{2}α*.*

**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’

^{2}h’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}) = 02r’r = 3r’

^{2}r’ = 2r/3

So, v is maximum at r’ = 2r/3

The maximum volume of cylinder = πh/r[r. 4r

^{2}/9 – 8r^{2}/27]= πhr

^{2}[4/27]= 4/27πh(h tanα)

^{2}= 4/27πh

^{3}tan^{2}α

### 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 = 314m

^{3}/hie dv/dt = 314m

^{3}/hNow, the volume of cylinder = πr

^{2}hv = π.(10)

^{2}.hv = 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 = t^{2} + 3t – 8, y = 2t^{2} – 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 = t

^{2 }+ 2t – 8 and y = 2t^{2 }– 2t – 5On differentiating we get

Now, when x = 2,

t

^{2 }+ 3 – 8 = 2t

^{2 }+ 3 – 10 = 0t

^{2 }– 2t + 5t – 10 = 0(t – 2)(t + 5) = 0

Here, t = 2, t = -5 ……….(1)

When y = -1

2t

^{2 }– 2t – 5 = -12t

^{2 }– 2t – 4 = 0t

^{2 }– 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 y^{2} = 4x if the value of m is

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

**Solution:**

The curve if y

^{2 }= 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 + x^{2} = 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 + x

^{2 }= 3On 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 x^{2} = 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 x

^{2 }= 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)

x

^{2 }= 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 9y^{2 }= x^{3}, where the normal to the curve makes equal intercepts with the axes are

### (A) (B)

### (C) (D)

**Solution:**

Given equation 9y

^{2 }= x^{3}On differentiating w.r.t x, we get

18y dy/dx = 3x

^{2}dy/dx = 3x

^{2}/18ydy/dx = x

^{2}/6yNow, the slope of the normal to the given curve at point (x

_{1}, y_{1}) isHence, the equation of the normal to the curve at point (x

_{1}, y_{1}) isAccording to the question it is given that the normal

make equal intercepts with the axes.

So,

…………(1)

The point (x

_{1}, y_{1})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.

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