These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 1.

Using differentials, find the approximate value of each of the following.

a. \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\)

b. \(\text { (33) }^{-\frac{1}{5}}\)

Solution:

∆y is approximately equal to dy

∴ (1) ⇒ (33)^{\(\frac { -1 }{ 5 }\)} ≈ 2^{-1} – 0.003

= 0.5 – 0.003 = 0.497

Hence (33)^{\(\frac { -1 }{ 5 }\)} ≈ 0.497

Question 2.

Show that the function given by f(x) = \(\frac { log x }{ x }\) has maximum at x = e.

Solution:

For maxima, f'(x) = 0

i.e., \(\frac{1-\log x}{x^{2}}\) = 0 ⇒ 1 – log x = 0

⇒ log x = 1 ⇒ x = e

At x = e, f”(e) = \(\frac{-3+2 \log e}{e^{3}}\) < 0,

∴ f has a 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:

∆ ABC be an isosceles triangle with AB = AC

Let AB = x, BC = b, \(\frac { dx }{ dt }\) = – 3cm/sec at

Draw AD ⊥ BC. Then D is the midpoint of BC.

i.e., BD = \(\frac { b }{ 2 }\)

Hence area is decreasing at the rate of \(\sqrt{3}\) bcm²/sec.

Question 4.

Find the equation of the normal to curve y² = 4x at the point (1,2).

Solution:

y² = 4x

Differentiating w.r.t. x,

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

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

At(1, 2), \(\frac { dy }{ dx }\) = \(\frac { 2 }{ 2 }\) = 1

Slope of the tangent at (1, 2) = \(\frac { -1 }{ 1 }\) = – 1

Slope of the normal at(1, 2)

Equation of the tangent at (1, 2) is

y – 2 = 1(x – 1)

y – x – 1 = 0

Equation of the normal at (1, 2) is

y – 2 = 1(x – 1)

x + y – 3 = 0

Question 5.

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

Solution:

x = acosθ + a0sinθ

y = asinθ – aθ cosθ

Differentiating w.r.t θ, dx

Slope of the tangent at the point 0 is \(\frac { sinθ }{ cosθ }\)

∴ Slope of the normal at the point θ is \(\frac { – cosθ }{ sinθ }\)

∴ Equation of the normal at the point θ is

i.e., xcosθ + ysin θ = a, which is the equation of the straight line in the normal form. Hence ‘a’ is the distance of the normal line from the origin. Thus the normal at any point 0 to the curve is at a constant distance from the origin.

Question 6.

Find the intervals in which the function f given by

f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}, 0 \leq x \leq 2 \pi\) (i) increasing (ii) decreasing.

Solution:

Since f is continuous

i. f is increasing: 0 ≤ x ≤ \(\frac { π }{ 2 }\) and \(\frac { 3π }{ 2 }\) ≤ 2π

ii. is decreasing: \(\frac { π }{ 2 }\) ≤ x ≤ \(\frac { 3π }{ 2 }\)

Question 7.

Find the intervals in which the function f given by f(x) = x³ + \(\frac{1}{x^{3}}\), x ≠ 0 is

i. increasing

ii. decreasing

Solution:

The critical points are x = – 1, x = 0 and x =1. x = – 1, x = 0, x = 1 divide Rinto disjoint open intervals as

(-∞, -1), (-1, 0),(0, 1) and (1, ∞)

i. f is increasing: x < – 1 and x > 1

ii. f is decreasing: – 1 < x < 0 and 0 < x < 1

Question 8.

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

Solution:

Consider the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) = 1 , a > b

Let A be one end point of the major axis and P(x, y) be a point on the ellipse. Draw PM perpendicular to x axis and extend it to meet the ellipse at Q.

Then the coordinate of Q is (x, – y)

∆APQ is an isosceles triangle

Let S be the area of ∆APQ

S = \(\frac { 1 }{ 2 }\)PQ.AM = \(\frac { 1 }{ 2 }\)(2y)(x + a)

S = y(x + a) = \(\frac { b }{ 2 }\)\(\sqrt{a^{2}-x^{2}}\)(x + a)

Since S is positive, S² is maximum when S is maximum

When x = – a, the point P coincides with A which is not possible.

∴ x = – \(\frac { a }{ 2 }\)

When x = \(\frac { a }{ 2 }\), \(\frac{d^{2}}{d x^{2}}\left(\mathrm{~S}^{2}\right)\)

= \(\frac{-12 b^{2}}{a^{2}} \frac{a}{2}\left(\frac{a}{2}+a\right)\) < 0

∴ S² is maximum when x = \(\frac { a }{ 2 }\)

i.e., S is maximum when x = \(\frac { a }{ 2 }\)

The maximum value of S

Question 9.

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

Solution:

Let x and y be the length and breadth of the base of the tank (in metre). Let C be the cost of building the tank.

Volume of the tank = 8m³ (x)(y).(2) = 8 ⇒ xy = 4

∴ y = \(\frac { 4 }{ x }\)

Area of base = xy

Area of 4 sides = 2(x + y)(2) = 4(x + y)

Cost of construction

But x = – 2 is not possible, since length of the base cannot be negative

∴ x = 2 When x = 2,

\(\frac{d^{2} \mathrm{C}}{d x^{2}}=\frac{180 \times 8}{8}\) > 0

The minimum cost of construction,

C = 280 + 180(2 + \(\frac { 4 }{ 2 }\)) = ₹ 1000

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 r be the radius and x be the side of the square.

Then 2πr + 4x = k

∴ r = \(\frac{k-4 x}{2 \pi}\) … (1)

Let S be the sum of areas of circle and square

Question 11.

A window is in the form of a rectangle sur-mounted 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:

Length of the window = 2x,

Height of the rectangular portion = y

i. Perimeter of the window = 10 m

2x + 2y + πx = 10

(π + 2)x + 2y = 10

ii. Let A be the area of the window

iii. Differentiating A w.r.t. x,

∴ A’ = 10 – 2 πx – 4x + π x

A’ = 10 – π x – 4x

A” = – π – 4 = – (π + 4)

For maxima, A’ = 0

10 – πx – 4x = 0

x(π + 4) = 10

x = \(\frac { 10 }{ π + 4 }\)

When x = \(\frac { 10 }{ π + 4 }\), A” = – (π + 4) < 0

∴ A is maximum when x = \(\frac { 10 }{ π + 4 }\)

2y = 10 – (π + 2)x

= 10 – (π + 2)\(\frac { 10 }{ π + 4 }\)

= \(\frac{10 \pi+40-10 \pi-20}{\pi+4}\) = \(\frac { 20 }{ π + 4 }\)

∴ y = \(\frac { 10 }{ π + 4 }\)

∴ Length of the window = 2x = \(\frac { 20 }{ π + 4 }\)

Length of larger side of window = y = \(\frac { 10 }{ π + 4 }\).

Question 12.

A point on the hypotenuse of a triangle is at distance a and b from the side of the tri¬angle. Show that the minimum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).

Solution:

Let ∆ABC be right angled at B and P be a point on the hypotenuse AC at a distance of a from BC and b from AB.

∴ PQ = a and PR = b

Let ∠A = θ then ∠C = 90 – θ

From right triangles ARP and PQC, we get

AP = \(\frac { b }{ sin θ }\) = b cosec θ

PC = \(\frac { a }{ sin(90 – θ) }\) = \(\frac { a }{ cos θ }\) = asec θ

Let l be the length of the hypotenuse

∴ l = AP + PC = asecθ + bcosecθ,

0 < θ < \(\frac { π }{ 2 }\)

\(\frac { dl }{ dθ }\) = asecθ tanθ – bcosecθ cotθ

\(\frac{d^{2} l}{d \theta^{2}}\) = a[secθ.sec²θ + tanθ.secθ tanθ] – b[cosecθ(-cosec²θ) + cot θ(-cosecθ cotθ)] = a[sec³θ + secθtan²θ] + b[cosec³θ + cosecθ cot²θ]

For minima

\(\frac { dl }{ dθ }\) = 0 ⇒ asecθ tanθ – bcosecθ cotθ = 0

⇒ a secθ tanθ = b cosecθ cotθ

Minimum length of the hypotenuse = \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).

Question 13.

Find the points at which the function given by f(x) = (x – 2)^{4} (x + 1)³ has

i. local maxima

ii. local minima

iii. point of in flexion

Solution:

f(x) = (x – 2)^{4} (x + 1)³

f’(x) = (x – 2)^{4}[3(x + 1)²] + (x+ 1)³[4(x – 2)³]

= (x – 2)³(x + 1)²[3x – 6 + 4x + 4]

= (x – 2)³(x + 1)²(7x – 2)

f’(x) = 0 ⇒ (x – 2)³(x + 1)²(7x – 2) = 0

⇒ x = – 1, and x = \(\frac { 2 }{ 7 }\) x = 2

Let us apply first derivative test.

i. local maxima is at x = \(\frac { 2 }{ 7 }\)

ii local minima ¡s at x = 2

iii. point of inflexion is at x = – 1

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) = cos² x + sin x, x ∈ [0, π]

f'(x) = – 2cosx sinx + cosx = cosx (1 – 2sinx)

f'(x) = 0 ⇒ cosx (1 – 2sinx) = 0

⇒ cosx = 0 or sin x = \(\frac { 1 }{ 2 }\)

⇒ x = \(\frac { π }{ 2 }\) or x = \(\frac { π }{ 6 }\) and x = \(\frac { 5π }{ 6 }\)

∴ x = \(\frac { π }{ 6 }\), \(\frac { π }{ 2 }\), \(\frac { 5π }{ 6 }\)

f(0) = cos²0 + sin0 = 1² + 0 = 1

Question 15.

Show that the altitude of the right circular cone of maximum volume that can be in-scribed in a sphere of radius r is \(\frac { 4r }{ 3 }\)

Solution:

Let R be the radius and A be the height of the right circular cone inscribed in a sphere of radius r.

i.e., AB = h, OA = OC = r and BC = R

In right triangle OBC, OB = h – r

and OC² = OB² + BC²

i.e., r² = (h – r)² + R²

r² = h² – 2 hr + r² + R²

∴ R² = 2hr – h²

Volume of the cone V = \(\frac { 1 }{ 3 }\) πR²h

∴ When h = \(\frac { 4r }{ 3 }\), the volume is maximum.

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:

Let c_{1}, c_{2} ∈ [a , b\ such that c_{1}, c_{2}.

Since f'(x) > 0 in (a, b), f is continuous in [a, b] and differentiable in (a, b).

∴ By mean value theorem, there exists c ∈ (c_{1}, c_{2}) such that

Since c_{1} and c_{2} are arbitrary, we get f(x) is increasing on (a, b).

Question 17.

Show that the height of the cylinder of maxi-mum volume that can be inscribed in a sphere of radius R is \(\frac{2 \mathrm{R}}{\sqrt{3}}[/latex . Also find the maximum volume.

Solution:

Let y be the radius and x be the height of the cylinder.

∴ V is maximum when x = [latex]\frac{2 \mathrm{R}}{\sqrt{3}}\)

The height of cylinder with maximum volume is \(\frac{2 \mathrm{R}}{\sqrt{3}}\)

∴ (1) → Maximum volume,

V = \(\frac{\pi}{4}\left(4 \mathrm{R}^{2} \cdot \frac{2 \mathrm{R}}{\sqrt{3}}-\frac{8 \mathrm{R}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{R}^{3}}{3 \sqrt{3}}\)

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 a is one-third that of the cone and the greatest volume of cylinder is \(\frac { 4 }{ 27 }\)πh³ tan²α.

Solution:

Let y be the radius and x be the height of the right circular cylinder inscribed in a cone of height h and semivertical angle α.

Then AB h, CD = y, BC = x

∴ AC = h – x

From right triangle ACD, tan α = \(\frac { CD }{ AC }\)

= \(\frac { y }{ h – x }\)

⇒ y = (h – x)tan α

Volume of the cylinder V = πy²x

= π(h – x)² tan² α.x

∴ V is maximum when the height of the cylinder is one third of the height of the cone.

(1) → Maximum volume,

Question 19.

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

a. 1 m³/h

b. 0.1 m³/h

c. 1.1 m³/h

d. 0.5 m³/h

Solution:

a. 1 m³/h

Let h be the height of the wheat at time t and V be the volume at that instant

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. \(\frac { 22 }{ 7 }\)

b. \(\frac { 6 }{ 7 }\)

c. \(\frac { 7 }{ 6 }\)

d. \(\frac { – 6 }{ 7 }\)

Solution:

b. \(\frac { 6 }{ 7 }\)

x = t² + 3t – 8 and y = 2t² – 2t – 5

Put x = 2 and y = – 1, we get

t² + 3t – 8 = 2 and 2t² – 2t – 5 = – 1

i.e., t² + 3t – 10 = 0 and 2t² – 2t – 4 = 0

i.e, (t + 5) (t – 2) = 0 and 2(t – 2) (t+ 1) = 0

i.e., t = -5 or t = 2 and t = 2 or t = – 1

Hence t = 2

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. \(\frac { 1 }{ 2 }\)

Solution:

y² = 4x

Differentiating w.r.t x,

2y\(\frac { dy }{ dx }\) = 4 ⇒ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ y }\)

∴ Slope of the tangent, m = \(\frac { 2 }{ y }\)

∴ y = mx + 1 becomes y = \(\frac { 2 }{ y }\) x + 1

y² = 2x + y

i.e., y² = \(\frac { y² }{ 2 }\) + y since y² = 4x

\(\frac { y² }{ 2 }\) = y ⇒ y² = 2y

⇒ y(y – 2) = 0

y ≠ 0 since slope = \(\frac { 2 }{ y }\)

∴ y = 2

∴ slope m = \(\frac { 2 }{ y }\) = \(\frac { 2 }{ 2 }\) = 1

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 = 0

Solution:

b. x – y = 0

2y + x² = 3

Differentiating w.r.t x,

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

\(\frac { dy }{ dx }\) = – x

Slope of the tangent at (1, 1)

= \(\frac { dy }{ dx }\) at (1, 1) = – 1

∴ Slope of the normal at (1, 1) = \(\frac { – 1 }{ – 1 }\) = 1

Equation of the normal at (1, 1) is

y – 1 = 1(x – 1)

⇒ x – y = 0

Question 23.

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

a. x + y = 3

b. x – y = 3

c. x + y = 1

d. x – y = 1

Solution:

a. x + y = 3

x² = Ay

Differentiating w.r.t. x,

2x = 4\(\frac { dy }{ dx }\)

∴\(\frac { dy }{ dx }\) = \(\frac { x }{ 2 }\)

Slope of the tangent = \(\frac { x }{ 2 }\)

∴ Slope of the normal = \(\frac { – 2 }{ x }\)

Let (x_{1}, y_{1}) be a point on the curve

∴ Slope of normal at (x_{1}, y_{1}) = \(\frac{-2}{x_{1}}\)

Equation of the normal at (x_{1}, y_{1}) is

y – y_{1} = \(\frac{-2}{x_{1}}\)(x – x_{1})

The normal passes through the point (1, 2)

∴ (2 – y_{1}) = \(\frac{-2}{x_{1}}\)(1 – x_{1})

Equation of the normal at(x_{1}, y_{1}) is

y – y_{1} = \(\frac{-2}{x_{1}}\)(1 – x_{1})

2 – y_{1} = \(\frac{2}{x_{1}}\) + 2

∴ y_{1} = \(\frac{2}{x_{1}}\)

Since (x_{1}, y_{1}) is a point on x² = 4y, we get

x_{1}² = 4y_{1}

i.e., x_{1}³ = \(\frac{4 \times 2}{x_{1}}\), sjnce y_{1} = \(\frac{2}{x_{1}}\)

x_{1}² = 8 ⇒ x, = 2

When x_{1} = 2, y_{1} = \(\frac { 2 }{ 2 }\) = 1

∴ Equation of the normal at (2, 1) is

y – 1 = \(\frac { – 2 }{ 2 }\)(x – 2)

y – 1 = – (x – 2)

i.e., x + y = 3

Question 24

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

a. \(\left(4, \pm \frac{8}{3}\right)\)

b. \(\left(4, \pm \frac{-8}{3}\right)\)

c. \(\left(4, \pm \frac{3}{8}\right)\)

d. \(\left(\pm 4, \frac{3}{8}\right)\)

Solution:

9y² = x³

Let (x_{1}, y_{1}) be the point at which the nor¬mal make equal intercepts on the coordinate axes.

Differentiating 9y³ = x³ w.r.t. x, we get

18y\(\frac { dy }{ dx }\) = 3x²

\(\frac { dy }{ dx }\) = \(\frac{x^{2}}{6 y}\)

Slope of the tangent at (x_{1}, y_{1}) = \(\frac{x_{1}^{2}}{6 y_{1}}\)

∴ Slope of normal at (x_{1}, y_{1}) = \(\frac{-6 y_{1}}{x_{1}^{2}}\)

Equation of the normal at (x_{1}, y_{1}) is

which is in the intercept form.

Since the intercepts of the normal are equal,we get

But x_{1} ≠ 0 since x_{1} = 0 makes the intercepts meaningless.

∴ x_{1} = 4

When x_{1}, 9y²_{1} = x³_{1} becomes

9y²_{1} = 4³ = 64

y²_{1} = \(\frac { 64 }{ 9 }\) ⇒ y_{1} = ≠ \(\frac { 8 }{ 3 }\)

∴ The required points are (4, \(\frac { 8 }{ 3 }\))