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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

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 â–³ABC, âˆ B = 90

Let âˆ C = Î¸, so, âˆ  DPA = Î¸

DP|| BC.

cosÎ¸ = DP/AP = b/AP

AP = b/cosÎ¸

In â–³EPC,

sinÎ¸ = EP/CP = a/CP

CP = a/sin Î¸

Now AC = h = PA + PC

h =

h(Î¸) = b sec Î¸ + a cosec Î¸

Put h'(Î¸) =

b sin3Î¸ = a cos 3Î¸

tan3Î¸ = a/b

tanÎ¸ = (a/b)1/3

secÎ¸ =

cosecÎ¸ =

hmax

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

(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 âˆˆ [0, Ï€]

Solution:

f(x) = cos2x + sin x; x Ïµ [0, Ï€]

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 Ïµ[0, Ï€] if cos x = 0, then x = Ï€/2

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

Now, f”(x) = -sin x – 2 cos2x

f”(Ï€/2) = -1 + 2 = 1 > 0

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

f”(Ï€/6) =

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

x = 5Ï€/6â€‹ is a point of local minima f(5Ï€/6) = 5/4

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

= max{1, 5/4, 1}

= 5/4 = Absolute maxima value

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

= 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 4Ï€/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.

âˆ DOC = Î¸                  -(Properties of circle)

In â–³ DOC,

OD = R cosÎ¸ & CD = RsinÎ¸,

r = R sin Î¸

AD = AO + OD = R + RcosÎ¸

h = R(1 + cosÎ¸)

Now, the volume of the cone is

V =

v(Î¸) =

Put v(Î¸) = 0

sinÎ¸[2cosÎ¸ + 2cos2Î¸ âˆ’ sin2Î¸] = 0

sinÎ¸[2cosÎ¸ + 2cos2Î¸ âˆ’ 1] = 0

sinÎ¸(3cosÎ¸ âˆ’ 1)(1 + cosÎ¸) = 0

sinÎ¸ = 0, cos = 1/3â€‹, cosÎ¸ = âˆ’1

If sinÎ¸ = 0, then volume will be 0.

If cosÎ¸ = -1, then sinÎ¸ = 0 & again volume will be 0.

But if cosÎ¸ = 1/3; sinÎ¸ = 2âˆš2/3 and

Volume, v = 32/81â€‹Ï€R3, which is maximum.

Height, h = R(1 + cosÎ¸) = R()

h = 4r/3

Hence proved

Question 16. Let f be a function defined on [a, b] such that fâ€²(x) > 0, for all x âˆˆ (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/âˆš3. Also, find the maximum volume.

Solution:

In â–³ABC,

AC2 = BC2 + AB2

4R2 = 4r2 + h2

r2 = R2          ……….(1)

Now, volume of cylinder = Ï€r2h

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

V = Ï€().h

V(h) =

On differentiating both side we get

V ‘(h) =

Now, put V'(h) = 0

Ï€R2

Now the maximum volume of cylinder = Ï€[R2. 2R/âˆš3 – 1/4.4R2/3.2R/âˆš3]

= 4Ï€R3/ 3âˆš3

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 Î± is one-third that of the cone and the greatest volume of the cylinder is 4/27Ï€h3tan2Î±.

Solution:

Let,

XQ = r

XO = h’

AO = h

OC = r’

âˆ XAQ = Î±

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 = Ï€r’2h’

v = Ï€r’2(h(r – r’)/r)

= Ï€(h(rr’2 – r’3)/r)

On differentiating we get

v’ = Ï€h/r(2rr’ – 3r’2)

Again on differentiating we get

v” = Ï€h/r(2r – 6r’) ………(1)

Now put v’ = 0

Ï€h/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 = Ï€h/r[r. 4r2/9 – 8r2/27]

= Ï€hr2[4/27]

= 4/27Ï€h(h tanÎ±)2

= 4/27Ï€h3 tan2Î±

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,

Rate of increase of volume = 314m3/h

ie   dv/dt = 314m3/h

Now, the volume of cylinder = Ï€r2h

v = Ï€.(10)2.h

v = 100Ï€h

On differentiating w.r.t t, we get

dv/dt = 100Ï€

So option A is correct

Question 20. The slope of the tangent to the curve x = t2 + 3t â€“ 8, y = 2t2 â€“ 2t â€“ 5 at the point (2,â€“ 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 â€“ y = 0

(C) x + y +1 = 0        (D) x â€“ 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 â€“ y = 3           (C) x + y = 1       (D) x â€“ 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

(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.

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