Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 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:
Solution & Step-by-Step Answer:
(a) 2x – y = 0
Solution & Step-by-Step Answer:
(a) 2x – y + 5 = 0; 2x – y – 5 = 0
Solution & Step-by-Step Answer:
(c) unitary elastic
Solution & Step-by-Step Answer:
(b) inelastic
Solution & Step-by-Step Answer:
(a) increasing for all x ∈ R, x ≠ 1
Solution & Step-by-Step Answer:
(c) f has maxima at x = -1
(II) Fill in the blanks:
Solution & Step-by-Step Answer:
gradient
Solution & Step-by-Step Answer:
6x – 6 = 6(x – 1)
Solution & Step-by-Step Answer:
14x-3
Solution & Step-by-Step Answer:
27 and 27
Solution & Step-by-Step Answer:
(III) State whether each of the following is True or False:
Solution & Step-by-Step Answer:
True
Solution & Step-by-Step Answer:
False
Solution & Step-by-Step Answer:
True
Solution & Step-by-Step Answer:
True. Hint: Hence, function f(x) is increasing on (, 1).
(IV) Solve the following:
Solution & Step-by-Step Answer:
xy = c2 Differentiating both sides w.r.t. x, we get Hence, equations of tangent and normal are x + t2y – 2ct = 0 and t3x – ty – c(t4 + 1) = 0 respectively.
(ii) y = x2+ 4x at the point whose ordinate is -3.
Solution:
Let P(x1, y1) be the point on the curve
y = x2+ 4x, where y1= -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 = , y = t – , at t = 2.
Solution:
When t = 2, x = and y = 2 – =
Hence, the point P at which we want to find the equations of tangent and normal is (, )
Hence, the equations of tangent and normal are 5x + y – 4 = 0 and x – 5y + 7 = 0 respectively.
(iv) y = x3– x2– 1 at the point whose abscissa is -2.
Solution:
y = x3– x2– 1
∴ (x3– x2– 1)
= 3x2– 2x – 0
= 3x2– 2x
∴ = 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
=
∴ the equation of the normal at (-2, -13) is
y – (-13) = [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.
Solution & Step-by-Step Answer:
Let P(x1, y1) be the foot of the required normal to the curve y = Differentiating y = w.r.t. x, we get ∴ x – 2y – = 0 i.e. 16x – 32y – 57 = 0 Hence, the equation of the normals are 16x – 32y – 41 = 0 and 16x – 32y – 57 = 0.
Solution & Step-by-Step Answer:
f(x) = ∴ f'(x) > 0, for all x ∈ R, x ≠ -1 Hence, the function f is increasing for all x ∈ R, where x ≠ -1.
Solution & Step-by-Step Answer:
f(x) = + 10 ∴ f'(x) < 0 for all x ∈ R, x ≠ 0 Hence, the function f is decreasing for all x ∈ R, where x ≠ 0.
Solution & Step-by-Step Answer:
x + y = 3 ∴ y = 3 – x ∴ x2y = x2(3 – x) = 3x2 – x3 Let f(x) = 3x2 – x3 Then f'(x) = (3x2 – x3) = 3 × 2x – 3x2 = 6x – 3x2 and f”(x) = (6x – 3x2) = 6 × 1 – 3 × 2x = 6 – 6x Now, f'(x) = 0 gives 6x – 3x2 = 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 x2y = (2)2(1) = 4.
Solution & Step-by-Step Answer:
f(x) = x3 – 9x2 + 24x ∴ f'(x) = (x3 – 9x2 + 24x) = 3x2 – 9 × 2x + 24 × 1 = 3x2 – 18x + 24 and f”(x) = (3x2 – 18x + 24) = 3 × 2x – 18 × 1 + 0 = 6x – 18 f'(x) = 0 gives 3x2 – 18x + 24 = 0 ∴ x2 – 6x + 8 = 0 ∴ (x – 2)(x – 4) = 0 ∴ the roots of f'(x) = 0 are x1 = 2 and x2 = 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.