Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Miscellaneous Exercise 4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Miscellaneous Exercise 4

(I) Choose the correct alternative:

Question 1.

The equation of tangent to the curve y = x^{2} + 4x + 1 at (-1, -2) is

(a) 2x – y = 0

(b) 2x + y – 5 = 0

(c) 2x – y – 1 = 0

(d) x + y – 1 = 0

Answer:

(a) 2x – y = 0

Question 2.

The equation of tangent to the curve x^{2} + y^{2} = 5, where the tangent is parallel to the line 2x – y + 1 = 0 are

(a) 2x – y + 5 = 0; 2x – y – 5 = 0

(b) 2x + y + 5 = 0; 2x + y – 5 = 0

(c) x – 2y + 5 = 0; x – 2y – 5 = 0

(d) x + 2y + 5; x + 2y – 5 = 0

Answer:

(a) 2x – y + 5 = 0; 2x – y – 5 = 0

Question 3.

If the elasticity of demand η = 1, then demand is

(a) constant

(b) inelastic

(c) unitary elastic

(d) elastic

Answer:

(c) unitary elastic

Question 4.

If 0 < η < 1, then the demand is

(a) constant

(b) inelastic

(c) unitary elastic

(d) elastic

Answer:

(b) inelastic

Question 5.

The function f(x) = x^{3} – 3x^{2} + 3x – 100, x ∈ R is

(a) increasing for all x ∈ R, x ≠ 1

(b) decreasing

(c) neither increasing nor decreasing

(d) decreasing for all x ∈ R, x ≠ 1

Answer:

(a) increasing for all x ∈ R, x ≠ 1

Question 6.

If f(x) = 3x^{3} – 9x^{2} – 27x + 15, then

(a) f has maximum value 66

(b) f has minimum value 30

(c) f has maxima at x = -1

(d) f has minima at x = -1

Answer:

(c) f has maxima at x = -1

(II) Fill in the blanks:

Question 1.

The slope of tangent at any point (a, b) is called as ___________

Answer:

gradient

Question 2.

If f(x) = x^{3} – 3x^{2} + 3x – 100, x ∈ R, then f”(x) is ___________

Answer:

6x – 6 = 6(x – 1)

Question 3.

If f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0, then f”(x) is ___________

Answer:

14x^{-3}

Question 4.

A rod of 108 m in length is bent to form a rectangle. If area j at the rectangle is maximum, then its dimensions are ___________

Answer:

27 and 27

Question 5.

If f(x) = x . log x, then its maximum value is ___________

Answer:

\(-\frac{1}{e}\)

(III) State whether each of the following is True or False:

Question 1.

The equation of tangent to the curve y = 4xe^{x} at (-1, \(\frac{-4}{e}\)) is y.e + 4 = 0.

Answer:

True

Question 2.

x + 10y + 21 = 0 is the equation of normal to the curve y = 3x^{2} + 4x – 5 at (1, 2).

Answer:

False

Question 3.

An absolute maximum must occur at a critical point or at an endpoint.

Answer:

True

Question 4.

The function f(x) = x.e^{x(1-x)} is increasing on (\(\frac{-1}{2}\), 1).

Answer:

True.

Hint:

Hence, function f(x) is increasing on (\(\frac{-1}{2}\), 1).

(IV) Solve the following:

Question 1.

Find the equations of tangent and normal to the following curves:

(i) xy = c^{2} at (ct, \(\frac{c}{t}\)), where t is a parameter.

Solution:

xy = c^{2}

Differentiating both sides w.r.t. x, we get

Hence, equations of tangent and normal are x + t^{2}y – 2ct = 0 and t^{3}x – ty – c(t^{4} + 1) = 0 respectively.

(ii) y = x^{2} + 4x at the point whose ordinate is -3.

Solution:

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

y = x^{2} + 4x, where y_{1} = -3

Hence, the equations of tangent and normal at

(i) (-3, -3) are 2x + y + 9 = 0 and x – 2y – 3 = 0

(ii) (-1, -3) are 2x – y – 1 = 0 and x + 2y + 7 = 0

(iii) x = \(\frac{1}{t}\), y = t – \(\frac{1}{t}\), at t = 2.

Solution:

When t = 2, x = \(\frac{1}{2}\) and y = 2 – \(\frac{1}{2}\) = \(\frac{3}{2}\)

Hence, the point P at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{3}{2}\))

Hence, the equations of tangent and normal are 5x + y – 4 = 0 and x – 5y + 7 = 0 respectively.

(iv) y = x^{3} – x^{2} – 1 at the point whose abscissa is -2.

Solution:

y = x^{3} – x^{2} – 1

∴ \(\frac{d y}{d x}=\frac{d}{d x}\)(x^{3} – x^{2} – 1)

= 3x^{2} – 2x – 0

= 3x^{2} – 2x

∴ \(\left(\frac{d y}{d x}\right)_{\text {at } x=-2}\) = 3(-2)^{2} – 2(-2) = 16

= slope of the tangent at x = -2

When x = -2, y = (-2)^{3} – (-2)^{2} – 1 = -13

∴ the point P is (-2, -13)

∴ the equation of the tangent at (-2, -13) is

y – (-13) = 16[x – (-2)]

∴ y + 13 = 16x + 32

∴ 16x – y + 19 = 0

The slope of the normal at x = -2

= \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at } x=-2}}=\frac{-1}{16}\)

∴ the equation of the normal at (-2, -13) is

y – (-13) = \(-\frac{1}{16}\)[x – (-2)]

∴ 16y + 208 = -x – 2

∴ x + 16y + 210 = 0

Hence, equations of tangent and normal are 16x – y + 19 = 0 and x + 16y + 210 = 0 respectively.

Question 2.

Find the equation of the normal to the curve y = \(\sqrt{x-3}\) which is perpendicular to the line 6x + 3y – 4 = 0.

Solution:

Let P(x_{1}, y_{1}) be the foot of the required normal to the curve y = \(\sqrt{x-3}\)

Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get

∴ x – 2y – \(\frac{57}{16}\) = 0

i.e. 16x – 32y – 57 = 0

Hence, the equation of the normals are 16x – 32y – 41 = 0 and 16x – 32y – 57 = 0.

Question 3.

Show that the function f(x) = \(\frac{x-2}{x+1}\), x ≠ -1 is increasing.

Solution:

f(x) = \(\frac{x-2}{x+1}\)

∴ f'(x) > 0, for all x ∈ R, x ≠ -1

Hence, the function f is increasing for all x ∈ R, where x ≠ -1.

Question 4.

Show that the function f(x) = \(\frac{3}{x}\) + 10, x ≠ 0 is decreasing.

Solution:

f(x) = \(\frac{3}{x}\) + 10

∴ f'(x) < 0 for all x ∈ R, x ≠ 0

Hence, the function f is decreasing for all x ∈ R, where x ≠ 0.

Question 5.

If x + y = 3, show that the maximum value of x^{2}y is 4.

Solution:

x + y = 3

∴ y = 3 – x

∴ x^{2}y = x^{2}(3 – x) = 3x^{2} – x^{3}

Let f(x) = 3x^{2} – x^{3}

Then f'(x) = \(\frac{d}{d x}\)(3x^{2} – x^{3})

= 3 × 2x – 3x^{2}

= 6x – 3x^{2}

and f”(x) = \(\frac{d}{d x}\)(6x – 3x^{2})

= 6 × 1 – 3 × 2x

= 6 – 6x

Now, f'(x) = 0 gives 6x – 3x^{2} = 0

∴ 3x(2 – x) = 0

∴ x = 0 or x = 2

f”(0) = 6 – 0 = 6 > 0

∴ f has minimum value at x = 0

Also, f”(2) = 6 – 12 = -6 < 0

∴ f has maximum value at x = 2

When x = 2, y = 3 – 2 = 1

∴ maximum value of x^{2}y = (2)^{2}(1) = 4.

Question 6.

Examine the function f for maxima and minima, where f(x) = x^{3} – 9x^{2} + 24x.

Solution:

f(x) = x^{3} – 9x^{2} + 24x

∴ f'(x) = \(\frac{d}{d x}\)(x^{3} – 9x^{2} + 24x)

= 3x^{2} – 9 × 2x + 24 × 1

= 3x^{2} – 18x + 24

and f”(x) = \(\frac{d}{d x}\)(3x^{2} – 18x + 24)

= 3 × 2x – 18 × 1 + 0

= 6x – 18

f'(x) = 0 gives 3x^{2} – 18x + 24 = 0

∴ x^{2} – 6x + 8 = 0

∴ (x – 2)(x – 4) = 0

∴ the roots of f'(x) = 0 are x_{1} = 2 and x_{2} = 4.

(a) f”(2) = 6(2) – 18 = -6 < 0

∴ by the second derivative test,

f has maximum at x = 2 and maximum value of f at x = 2

f(2) = (2) – 9(2)^{2} + 24(2)

= 8 – 36 + 48

= 20

(b) f”(4) = 6(4) – 18 = 6 > 0

∴ by the second derivative test, f has minimum at x = 4

and minimum value of f at x = 4

f(4) = (4)^{3} – 9(4)^{2} + 24(4)

= 64 – 144 + 96

= 16.