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Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.2 Questions and Answers.

Std 11 Maths 1 Exercise 5.2 Solutions Commerce Maths

Question 1.
Find the slope of each of the following lines which pass through the points:
(a) (2, -1), (4, 3)
(b) (-2, 3), (5, 7)
(c) (2, 3), (2, -1)
(d) (7, 1), (-3, 1)
Solution:
(a) Let A = (x₁, y₁) = (2, -1) and B = (x₂, y₂) = (4, 3).
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-(-1)}{4-2}\)
= \(\frac{4}{2}\)
= 2

(b) Let C = (x₁, y₁) = (-2, 3) and D = (x₂, y₂) = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{7-3}{5-(-2)}\)
= \(\frac{4}{7}\)

(c) Let E = (2, 3) = (x₁, y₁) and F = (2, -1) = (x₂, y₂)
Since x₁ = x₂ = 2
∴ The slope of EF is not defined. ……[EF || y-axis]

(d) Let G = (7, 1) = (x₁, y₁) and H = (-3, 1) = (x₂, y₂) say.
Since y₁ = y₂
∴ The slope of GH = 0 …..[GH || x-axis]

Question 2.
If the X and Y-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ the line L intersects X-axis at (2, 0) and Y-axis at (0, 3).
i.e. the line L passes through (2, 0) = (x₁, y₁) and (0, 3) = (x₂, y₂) say.
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{0-2}\)
= \(\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
Slope of the line = tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is 45°.
Solution:
Given, inclination (θ) = 45°
Slope of the line = tan θ = tan 45° = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the slope of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
i.e. the line passes through (3, 0) = (x₁, y₁) and (0, 3) = (x₂, y₂) say.
Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{0-3}\)
= -1

Question 6.
Without using Pythagoras theorem, show that points A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right-angled triangle.
Solution:
Given, A(4, 4) = (x₁, y₁), B(3, 5) = (x₂, y₂), C(-1, -1) = (x₃, y₃)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}=-1\)
Slope of BC = \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1\)
Slope of AB × slope of AC = -1 × 1 = -1
∴ side AB ⊥ side AC
∴ ΔABC is a right angled triangle, right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured clockwise.
Solution:

Since, the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction.
∴ Inclination of the line (θ) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= -cot 45° …….[tan(90 + θ°) = -cot θ]
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2, 1), and R(4, 5) are collinear.
Solution:
Given, points P(k, -1), Q(2, 1), and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Check:
For collinear points P, Q, R,
Slope of PQ = Slope of QR = Slop of PR
For k = 1, if the given points are collinear, then our answer is correct.
P(1, -1), Q(2, 1) and R(4, 5)
Slope of PQ = \(\frac{1-(-1)}{2-1}=\frac{2}{1}=2\)
Slope of QR = \(\frac{5-1}{4-2}=\frac{4}{2}=2\)
Slope of PQ = Slope of QR
∴ The given points are collinear.
Thus, our answer is correct.

Maharashtra State Board 11th Commerce Maths

• Locus and Straight Line Ex 5.1 11th Commerce Maths
• Locus and Straight Line Ex 5.2 11th Commerce Maths
• Locus and Straight Line Ex 5.3 11th Commerce Maths
• Locus and Straight Line Ex 5.4 11th Commerce Maths
• Locus and Straight Line Miscellaneous Exercise 5 11th Commerce Maths
• Determinants Ex 6.1 11th Commerce Maths
• Determinants Ex 6.2 11th Commerce Maths
• Determinants Ex 6.3 11th Commerce Maths
• Determinants Miscellaneous Exercise 6 11th Commerce Maths

Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.1 Questions and Answers.

Std 11 Maths 1 Exercise 5.1 Solutions Commerce Maths

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3) and B(2, 1).
PA = PB
∴ PA² = PB²
∴ (x – 1)² + (y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
∴ -2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(-5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(-5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² – 4y + 4 = x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
∴ The required equation of locus is 9x – y – 6 = 0

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
∴ x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3y² + 4x – 24y + 32 = 0
∴ The required equation of locus is 3x² + 3y² + 4x – 24y + 32 = 0

Question 4.
If A(4, 1) and B(5, 4), find the equation of the locus of point P if PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus.
Given, A(4, 1), B(5, 4) and PA² = 3PB²
∴ (x – 4)² + (y – 1)² = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² – 30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 4), B(5, 8) and PA² – PB² = 13
∴ [(x – 2)² + (y – 4)²] – [(x – 5)² + (y – 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) = 13
∴ 6x + 8y – 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y – 41 = 0

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends a right angle at P. (∠APB = 90°)
Solution:
Let P(x. y) be any point on the required locus.
Given, A(1, 6) and B(3, 5), ∠APB = 90°
∴ ΔAPB is a right-angled triangle.

By Pythagoras theorem,
AP² + PB² = AB²
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 – 5)²
∴ x² – 2x + 1 + y² – 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0

Question 7.
If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 3 …..(i)
(a) Given, A(x, y) = A(1, 3)
x = X + 2 and y = Y + 3 …..[From (i)]
∴ 1 = X + 2 and 3 = Y + 3
∴ X = -1 and Y = 0
∴ the new co-ordinates of point A are (-1, 0).

(b) Given, B(x, y) = B(2, 5)
x = X + 2 andy = Y + 3 ……[From (i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ the new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points (a) C(5, 4) (b) D(3, 3)
Solution:
Origin is shifted to (1, 3) = (h, k)
Let the new co-ordinates be (X, Y)
x = X + h and y = Y + k
∴ x = X + 1 and 7 = Y + 3 …..(i)
(a) Given, C(X, Y) = C(5, 4)
∴ x = X + 1 andy = Y + 3 …..[From(i)]
∴ x = 5 + 1 = 6 and y = 4 + 3 = 7
∴ the old co-ordinates of point C are (6, 7).

(b) Given, D(X, Y) = D(3, 3)
∴ x = X + 1 and y = Y + 3 …..[From (i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ the old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates (5, 14) change to (8, 3) by the shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, (x,y) = (5, 14), (X, Y) = (8, 3)
Since, x = X + h and y = Y + k
∴ 5 = 8 + h and 14 = 3 + k
∴ h = -3 and k = 11
∴ the co-ordinates of the point, where the origin is shifted are (-3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point O'(2, 2), the direction of axes remaining the same:
(a) 3x – y + 2 = 0
(b) x² + y² – 3x = 7
(c) xy – 2x – 2y + 4 = 0
Solution:
Given, (h, k) = (2, 2)
Let (X, Y) be the new co-ordinates of the point (x, y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 2
(a) Substituting the values of x and y in the equation 3x – y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6 – Y – 2 + 2 = 0
∴ 3X – Y + 6 = 0, which is the new equation of locus.

(b) Substituting the values of x and y in the equation x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new equation of locus.

(c) Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4 – 2Y – 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

Maharashtra State Board 11th Commerce Maths

• Locus and Straight Line Ex 5.1 11th Commerce Maths
• Locus and Straight Line Ex 5.2 11th Commerce Maths
• Locus and Straight Line Ex 5.3 11th Commerce Maths
• Locus and Straight Line Ex 5.4 11th Commerce Maths
• Locus and Straight Line Miscellaneous Exercise 5 11th Commerce Maths
• Determinants Ex 6.1 11th Commerce Maths
• Determinants Ex 6.2 11th Commerce Maths
• Determinants Ex 6.3 11th Commerce Maths
• Determinants Miscellaneous Exercise 6 11th Commerce Maths

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.2 Questions and Answers.

Std 12 Maths 1 Exercise 4.2 Solutions Commerce Maths

Question 1.
Test whether the following functions are increasing and decreasing:
(i) f(x) = x³ – 6x² + 12x – 16, x ∈ R
Solution:
f(x) = x³ – 6x² + 12x – 16
∴ f'(x) = \(\frac{d}{d x}\)(x³ – 6x² + 12x – 16)
= 3x² – 6 × 2x + 12 × 1 – 0
= 3x² – 12x + 12
= 3(x² – 4x + 4)
= 3(x – 2)² > 0 for all x ∈ R, x ≠ 2
∴ f'(x) > 0 for all x ∈ R – {2}
∴ f is increasing for all x ∈ R – {2}.

(ii) f(x) = x – \(\frac{1}{x}\), x ∈ R, x ≠ 0
Solution:
f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{d}{d x}\left(x-\frac{1}{x}\right)\)
= 1 – \(\left(-\frac{1}{x^{2}}\right)\)
= 1 + \(\frac{1}{x^{2}}\) > 0 for all x ∈ R, x ≠ 0
∴ f'(x) > 0 for all x ∈ R, where x ≠ 0
∴ f is increasing for all x > R, where x ≠ 0.

(iii) f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0
Solution:
f(x) = \(\frac{7}{x}\) – 3
∴ f'(x) = \(\frac{d}{d x}\left(\frac{7}{x}-3\right)=7\left(-\frac{1}{x^{2}}\right)-0\)
= \(-\frac{7}{x^{2}}\) < 0 for all x ∈ R, x ≠ 0
∴ f'(x) < 0 for all x ∈ R, where x ≠ 0.
∴ f is decreasing for all x ∈ R, where x ≠ 0.

Question 2.
Find the values of x, such that f(x) is increasing function:
(i) f(x) = 2x³ – 15x² + 36x + 1
Solution:
f(x) = 2x³ – 15x² + 36x + 1
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² + 36x + 1)
= 2 × 3x² – 15 × 2x + 36 × 1 + 0
= 6x² – 30x + 36
= 6(x² – 5x + 6)
f is increasing, if f'(x) > 0
i.e. if 6(x² – 5x + 6) > 0
i.e. if x² – 5x + 6 > 0
i.e. if x² – 5x > -6
i.e. if x – 5x + \(\frac{25}{4}\) > -6 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{1}{4}\)
i.e. if x – \(\frac{5}{2}\) > \(\frac{1}{2}\) or x – \(\frac{5}{2}\) < –\(\frac{1}{2}\)
i.e. if x > 3 or x < 2
i.e. if x ∈ (-∞, 2) ∪ (3, ∞)
∴ f is increasing, if x ∈ (-∞, 2) ∪ (3, ∞).

(ii) f(x) = x² + 2x – 5
Solution:
f(x) = x² + 2x – 5
∴ f'(x) = \(\frac{d}{d x}\)(x² + 2x – 5)
= 2x + 2 × 1 – 0
= 2x + 2
f is increasing, if f'(x) > 0
i.e. if 2x + 2 > 0
i.e. if 2x > -2
i.e. if x > -1, i.e. x ∈ (-1, ∞)
∴ f is increasing, if x > -1, i.e. x ∈ (-1, ∞)

(iii) f(x) = 2x³ – 15x² – 144x – 7
Solution:
f(x) = 2x³ – 15x² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\) (2x³ – 15x² – 144x – 7)
= 2 × 3x² – 15 × 2x – 144 × 1 – 0
= 6x² – 30x – 144
= 6(x² – 5x – 24)
f is increasing if, f'(x) > 0
i.e. if 6(x² – 5x – 24) > 0
i.e. if x² – 5x – 24 > 0
i.e. if x² – 5x > 24
i.e. if x² – 5x + \(\frac{25}{4}\) > 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{121}{4}\)
i.e. if \(x-\frac{5}{2}>\frac{11}{2} \text { or } x-\frac{5}{2}<-\frac{11}{2}\)
i.e. if x > 8 or x < -3
i.e. if x ∈ (-∞, -3) ∪ (8, ∞)
∴ f is increasing, if x ∈ (-∞, -3) ∪ (8, ∞).

Question 3.
Find the values of x such that f(x) is decreasing function:
(i) f(x) = 2x³ – 15x² – 144x – 7
Solution:
f(x) = 2x³ – 15x² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² – 144x – 7)
= 2 × 3x² – 15 × 2x – 144 × 1 – 0
= 6x² – 30x – 144
= 6(x² – 5x – 24)
f is decreasing, if f'(x) < 0
i.e. if 6(x² – 5x – 24) < 0
i.e. if x² – 5x – 24 < 0
i.e. if x² – 5x < 24
i.e. if x² – 5x + \(\frac{25}{4}\) < \(\frac{121}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{121}{4}\)
i.e. if \(-\frac{11}{2}<x-\frac{5}{2}<\frac{11}{2}\)
i.e. if \(-\frac{11}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{11}{2}+\frac{5}{2}\)
i.e. if -3 < x < 8
∴ f is decreasing, if -3 < x < 8.

(ii) f(x) = x⁴ – 2x³ + 1
Solution:
f(x) = x⁴ – 2x³ + 1
∴ f'(x) = \(\frac{d}{d x}\)(x⁴ – 2x³ + 1)
= 4x³ – 2 × 3x² + 0
= 4x³ – 6x²
f is decreasing, if f'(x) < 0
i.e. if 4x³ – 6x² < 0
i.e. if x²(4x – 6) < 0
i.e. if 4x – 6 < 0 …….[∵ x² > 0]
i.e. if x < \(\frac{3}{2}\)
i.e. -∞ < x < \(\frac{3}{2}\)
∴ f is decreasing, if -∞ < x < \(\frac{3}{2}\).

(iii) f(x) = 2x³ – 15x² – 84x – 7
Solution:
f(x) = 2x³ – 15x² – 84x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x³ – 15x² – 84x – 7)
= 2 × 3x² – 15 × 2x – 84 × 1 – 0
= 6x² – 30x – 84
= 6(x² – 5x – 14)
f is decreasing, if f'(x) < 0
i.e. if 6(x² – 5x – 14) < 0
i.e. if x² – 5x – 14 < 0
i.e. if x² – 5x < 14
i.e. if x – 5x + \(\frac{25}{4}\) < 14 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{81}{4}\)
i.e. if \(-\frac{9}{2}<x-\frac{5}{2}<\frac{9}{2}\)
i.e. if \(-\frac{9}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{9}{2}+\frac{5}{2}\)
i.e. if -2 < x < 7
∴ f is decreasing, if -2 < x < 7.

12th Commerce Maths Solution Book Pdf 

• Applications of Derivatives Ex 4.1 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.2 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.3 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.4 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Miscellaneous Exercise 4 Class 12 Commerce Maths Textbook Solutions

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.

Std 12 Maths 1 Exercise 4.1 Solutions Commerce Maths

Question 1.
Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x² – x + 1 at (1, 3)
Solution:
y = 3x² – x + 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\) (3x² – x + 1)
= 3 × 2x – 1 + 0
= 6x – 1
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}\) = 6(1) – 1
= 5
= slope of the tangent at (1, 3).
∴ the equation of the tangent at (1, 3) is
y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0.
The slope of the normal at (1, 3) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}\)
∴ the equation of the normal at (1, 3) is
y – 3 = \(-\frac{1}{5}\)(x – 1)
∴ 5y – 15 = -x + 1
∴ x + 5y – 16 = 0
Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.

(ii) 2x² + 3y² = 5 at (1, 1)
Solution:
2x² + 3y² = 5
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = \(\frac{-2}{3}\)(x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}\)
∴ the equation of the normal at (1, 1) is
y – 1 = \(\frac{3}{2}\)(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.

(iii) x² + y² + xy = 3 at (1, 1)
Solution:
x² + y² + xy = 3
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}\)
= \(\frac{-1}{-1}\)
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.

Question 2.
Find the equations of the tangent and normal to the curve y = x² + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Solution:
Let P(x₁, y₁) be the point on the curve y = x² + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Differentiating y = x² + 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(x² + 5) = 2x + 0 = 2x
\(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}\)
= slope of the tangent at (x₁, y₁)
Let m₁ = 2x₁
The slope of the line 4x – y + 1 = 0 is
m₂ = \(\frac{-4}{-1}\) = 4
Since, the tangent at P(x₁, y₁) is parallel to the line 4x – y + 1 = 0,
m₁ = m₂
∴ 2x₁ = 4
∴ x₁ = 2
Since, (x₁, y₁) lies on the curve y = x² + 5, y₁ = \(x_{1}^{2}\) + 5
∴ y₁ = (2)² + 5 = 9 ……[x₁ = 2]
∴ the coordinates of the point are (2, 9) and the slope of the tangent = m₁ = m₂ = 4.
∴ the equation of the tangent at (2, 9) is
y – 9 = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x – y + 1 = 0
Slope of the normal = \(\frac{-1}{m_{1}}=-\frac{1}{4}\)
∴ the equation of the normal at (2, 9) is
y – 9 = \(-\frac{1}{4}\)(x – 2)
∴ 4y – 36 = -x + 2
∴ x + 4y – 38 = 0
Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

Question 3.
Find the equations of the tangent and normal to the curve y = 3x² – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Solution:
Let P(x₁, y₁) be the point on the curve y = 3x² – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Differentiating y = 3x² – 3x – 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(3x² – 3x – 5)
= 3 × 2x – 3 × 1 – 0
= 6x – 3
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3\)
= slope of the tangent at (x₁, y₁)
Let m₁ = 6x₁ – 3
The slope of the line 3x – y + 1 = 0
m₂ = \(\frac{-3}{-1}\) = 3
Since, the tangent at P(x₁, y₁) is parallel to the line 3x – y + 1 = 0,
m₁ = m₂
∴ 6x₁ – 3 = 3
∴ 6x₁ = 6
∴ x₁ = 1
Since, (x₁, y₁) lies on the curve y = 3x² – 3x – 5,
\(y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5\), where x₁ = 1
= 3(1)² – 3(1) – 5
= 3 – 3 – 5
= -5
∴ the coordinates of the point are (1, -5) and the slope of the tangent = m₁ = m₂ = 3.
∴ the equation of the tangent at (1, -5) is
y – (-5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x – y – 8 = 0
Slope of the normal = \(-\frac{1}{m_{1}}=-\frac{1}{3}\)
∴ the equation of the normal at (1, -5) is
y – (-5) = \(-\frac{1}{3}\)(x – 1)
∴ 3y + 15 = -x + 1
∴ x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.

12th Commerce Maths Solution Book Pdf 

• Applications of Derivatives Ex 4.1 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.2 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.3 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Ex 4.4 Class 12 Commerce Maths Textbook Solutions
• Applications of Derivatives Miscellaneous Exercise 4 Class 12 Commerce Maths Textbook Solutions

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Miscellaneous Exercise 4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Miscellaneous Exercise 4 Questions and Answers.

Std 11 Maths 1 Miscellaneous Exercise 4 Solutions Commerce Maths

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:

Question 2.
For a G.P. a = \(\frac{4}{3}\) and t₇ = \(\frac{243}{1024}\), find the value of r.
Solution:

Question 3.
For a sequence, if t_n = \(\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (t_n) is a G.P., if \(\frac{5^{n-2}}{7^{n-3}}\) = constant, for all n ∈ N.

∴ the sequence is a G.P. with common ratio = \(\frac{5}{7}\)
∴ first term = t₁ = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^{2}}{5}=\frac{49}{5}\)

Question 4.
Find three numbers in G.P., such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the first condition,


∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.

Question 5.
Find 4 numbers in G. P. such that the sum of the middle 2 numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\).
According to the second condition,
\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)
∴ a⁴ = 1
∴ a = 1
According to the first condition,

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be
\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)
According to the first condition,

Question 7.
For a sequence, S_n = 4(7ⁿ – 1), verify whether the sequence is a G.P.
Solution:

Question 8.
Find 2 + 22 + 222 + 2222 + …… upto n terms.
Solution:
S_n = 2 + 22 + 222 +….. upto n terms
= 2(1 + 11 + 111 +…… upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +…… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since, 10, 100, 1000, …… n terms are in G.P.
with a = 10, r = \(\frac{100}{10}\) = 10

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…..
Solution:
0.6, 0.66, 0.666, 0.6666, ……
∴ t₁ = 0.6
t₂ = 0.66 = 0.6 + 0.06
t₃ = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
t_n = 0.6 + 0.06 + 0.006 + …… upto n terms.
The terms are in G.P.with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ t_n = the sum of first n terms of the G.P.

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\).
Solution:

Question 11.
Find \(\sum_{\mathbf{r}=1}^{\mathbf{n}} \mathbf{r}(\mathbf{r}-\mathbf{3})(\mathbf{r}-\mathbf{2})\).
Solution:

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that,

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}{(r+1)^{2}}\)
Solution:

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + …… upto n terms.
Solution:
2, 4, 6, … are in A.P.
∴ rth term = 2 + (r – 1)2 = 2r
6, 9, 12, … are in A.P.
∴ rth term = 6 + (r – 1) (3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 +…… upto n terms

= n(n + 1) (2n + 1 + 3)
= 2n(n + 1)(n + 2)

Question 15.
Find 12² + 13² + 14² + 15² + …… + 20².
Solution:
12² + 13² + 14² + 15² + …… + 20²
= (1² + 2² + 3² + 4² + ……. + 202) – (1² + 2² + 3² + 4² + …… + 11²)

= 2870 – 506
= 2364

Question 16.
Find (50² – 49²) + (48² – 47²) + (46² – 45²) + …… + (2² – 1²).
Solution:
(50² – 49²) + (48² – 47²) + (46² – 45²) + …… + (2² – 1²)
= (50² + 48² + 46² + …… + 2²) – (49² + 47² + 45² + …… + 1²)
= \(\sum_{r=1}^{25}(2 r)^{2}-\sum_{r=1}^{25}(2 r-1)^{2}\)

= 1300 – 25
= 1275

Question 17.
In a G.P., if t₂ = 7, t₄ = 1575, find r.
Solution:

Question 18.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.
∴ \(\frac{k}{k-1}=\frac{k+2}{k}\)
∴ k² = k² + k – 2
∴ k – 2 = 0
∴ k = 2

Question 19.
If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.
∴ t_n = \(\text { a. } R^{n-1}\)
∴ x = \(\text { a. } R^{p-1}\), y = \(\text { a. } R^{q-1}\), z = \(\text { a. } R^{r-1}\)

Maharashtra State Board 11th Commerce Maths

• Sequences and Series Ex 4.1 11th Commerce Maths
• Sequences and Series Ex 4.2 11th Commerce Maths
• Sequences and Series Ex 4.3 11th Commerce Maths
• Sequences and Series Ex 4.4 11th Commerce Maths
• Sequences and Series Ex 4.5 11th Commerce Maths
• Sequences and Series Miscellaneous Exercise 4 11th Commerce Maths

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.5 Questions and Answers.

Std 11 Maths 1 Exercise 4.5 Solutions Commerce Maths

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

Question 2.
Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\).
Solution:

Question 3.
Find \(\sum_{r=1}^{n} \frac{1+2+3+\ldots+r}{r}\).
Solution:

Question 4.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+\ldots+r^{3}}{r(r+1)}\).
Solution:
We know that,

Question 5.
Find the sum 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms.
Solution:
5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + ….. upto n terms
Now, 5, 7, 9, 11, … are in A.P.
rth term = 5 + (r – 1) (2) = 2r + 3
7, 9, 11,. … are in A.P.
rth term = 7 + (r – 1) (2) = 2r + 5
∴ 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms

Question 6.
Find the sum 2² + 4² + 6² + 8² + …… upto n terms.
Solution:
2² + 4² + 6² + 8² + …… upto n terms
= (2 × 1)² + (2 × 2)² + (2 × 3)² + (2 × 4)² + ……

Question 7.
Find (70² – 69²) + (68² – 67²) + (66² – 65²) + ……. + (2² – 1²)
Solution:
Let S = (70² – 69²) + (68² – 67²) + …… +(2² – 1²)
∴ S = (2² – 1²) + (4² – 3²) + …… + (70² – 69²)
Here, 2, 4, 6,…, 70 is an A.P. with rth term = 2r
and 1, 3, 5,….., 69 in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2.
∴ rth term = 1 + (r – 1)2 = 2r – 1
3, 5, 7, 9, … are in A.P. with a = 3 and d = 2
∴ rth term = 3 + (r – 1)2 = 2r + 1
and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2
∴ rth term = 5 + (r – 1)2 = 2r + 3
∴ 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… upto n terms

= n(n + 1)[2n(n + 1) + 4n + 2 – 1] – 3n
= n(n + l)(2n² + 6n + 1) – 3n
= n(2n³ + 8n² + 7n + 1 – 3)
= n(2n³ + 8n² + 7n – 2)

Question 9.
Find n, if \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots+\text { upto } n \text { terms }}{1+2+3+4+\ldots+\text { upto } n \text { terms }}\) = \(\frac{100}{3}\)
Solution:

Question 10.
If S₁, S₂, and S₃ are the sums of first n natural numbers, their squares, and their cubes respectively, then show that:
9\(S_{2}^{2}\) = S₃(1 + 8S₁).
Solution:

Maharashtra State Board 11th Commerce Maths

• Sequences and Series Ex 4.1 11th Commerce Maths
• Sequences and Series Ex 4.2 11th Commerce Maths
• Sequences and Series Ex 4.3 11th Commerce Maths
• Sequences and Series Ex 4.4 11th Commerce Maths
• Sequences and Series Ex 4.5 11th Commerce Maths
• Sequences and Series Miscellaneous Exercise 4 11th Commerce Maths

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.4 Questions and Answers.

Std 11 Maths 1 Exercise 4.4 Solutions Commerce Maths

Question 1.
Verify whether the following sequences are H.P.
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)
(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)
Solution:
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Here, the reciprocal sequence is 3, 5, 7, 9, …
∴ t₁ = 3, t₂ = 5, t₃ = 7, …..
∵ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 2, constant
∴ The reciprocal sequence is an A.P.
∴ the given sequence is H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)
Here, the reciprocal sequence is 3, 6, 9, 12 …
∴ t₁ = 3, t₂ = 6, t₃ = 9, t₄ = 12, …
∵ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 3, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.

(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)
Here, the reciprocal sequence is 7, 9, 11, 13, 15, ……
∴ t₁ = 7, t₂ = 9, t₃ = 11, t₄ = 13, …..
∵ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 2, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.

Question 2.
Find the nth term and hence find the 8th term of the following H.P.s:
(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots \ldots \ldots\)
(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots \ldots \ldots \ldots\)
(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \cdots \cdots \cdots\)
Solution:

Question 3.
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\).
Solution:
G.M. = 4, H.M. = \(\frac{16}{5}\)
∵ (G.M.)² = (A.M.) (H.M.)
∴ 16 = A.M. × \(\frac{16}{5}\)
∴ A.M. = 5

Question 4.
Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.
Solution:
A.M. = \(\frac{15}{2}\), G.M. = 6
Now, (G.M.)² = (A.M.) (H.M.)
∴ 6² = \(\frac{15}{2}\) × H.M.
∴ H.M. = 36 × \(\frac{2}{15}\)
∴ H.M. = \(\frac{24}{5}\)

Question 5.
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution:
A.M. = 75, H.M. = 48
(G.M.)² = (A.M.) (H.M.)
∵ (G.M.)² = 75 × 48
∵ (G.M.)² = 25 × 3 × 16 × 3
∵ (G.M.)² = 5² × 4² × 3²
∴ G.M. = 5 × 4 × 3
∴ G.M. = 60

Question 6.
Insert two numbers between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.
Solution:
Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).
∴ \(\frac{1}{7}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{13}\) are in H.P.
∴ 7, H₁, H₂ and 13 are in A.P.
∴ t₁ = a = 7 and t₄ = a + 3d = 13
∴ 7 + 3d = 13
∴ 3d = 6
∴ d = 2
∴ H₁ = t₂ = a + d = 7 + 2 = 9
and H₂ = t₃ = a + 2d = 7 + 2(2) = 11
∴ \(\frac{1}{9}\) and \(\frac{1}{11}\) are the required numbers to be inserted between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.

Question 7.
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution:
Let the required numbers be G₁ and G₂.
∴ 1, G₁, G₂, -27 are in G.P.
∴ t₁ = 1, t₂ = G₁, t₃ = G₂, t₄ = -27
∴ t₁ = a = 1
t_n = ar^n-1
∴ t₄ = (1) r^4-1
∴ -27 = r³
∴ r³ = (-3)³
∴ r = -3
∴ G₁ = t₂ = ar = 1(-3) = -3
∴ G₂ = t₃ = ar = 1(-3)² = 9
∴ -3 and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P.

Question 8.
Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).
Solution:
Let a, b be the two numbers.


∴ a + b = 13
∴ b = 13 – a …….(iii)
and ab = 36
∴ a(13 – a) = 36 …… [From (iii)]
∴ a² – 13a + 36 = 0
∴ (a – 4)(a – 9) = 0
∴ a = 4 or a = 9
When a = 4, b = 13 – 4 = 9
When a = 9, b = 13 – 9 = 4
∴ the two numbers are 4 and 9.

Question 9.
Find two numbers whose A.M. exceeds G.M. bv 7 and their H.M. by \(\frac{63}{5}\).
Solution:
Let a, b be the two numbers.

∴ a + b = 70
∴ b = 70 – a …..(ii)
∴ G = A – 7 = 35 – 7 = 28 …….[From (i)]
∴ √ab = 28
∴ ab = 28² = 784
∴ a(70 – a) = 784 ……[From (ii)]
∴ 70a – a² = 784
∴ a² – 70a + 784 = 0
∴ a² – 56a – 14a + 784 = 0
∴ (a – 56) (a – 14) = 0
∴ a = 14 or a = 56
When a = 14, b = 70 – 14 = 56
When a = 56, b = 70 – 56 = 14
∴ the two numbers are 14 and 56.

Maharashtra State Board 11th Commerce Maths

• Sequences and Series Ex 4.1 11th Commerce Maths
• Sequences and Series Ex 4.2 11th Commerce Maths
• Sequences and Series Ex 4.3 11th Commerce Maths
• Sequences and Series Ex 4.4 11th Commerce Maths
• Sequences and Series Ex 4.5 11th Commerce Maths
• Sequences and Series Miscellaneous Exercise 4 11th Commerce Maths

Differentiation Class 12 Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Miscellaneous Exercise 3 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 3 Solutions Commerce Maths

(I) Choose the correct alternative:

Question 1.
If y = (5x³ – 4x² – 8x)⁹, then \(\frac{d y}{d x}\) = ___________
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)
(b) 9(5x³ – 4x² – 8x)⁹ (15x² – 8x – 8)
(c) 9(5x³ – 4x² – 8x)⁸ (5x² – 8x – 8)
(d) 9(5x³ – 4x² – 8x)⁹ (5x² – 8x – 8)
Answer:
(a) 9(5x³ – 4x² – 8x)⁸ (15x² – 8x – 8)

Question 2.
If y = \(\sqrt{x+\frac{1}{x}}\), then \(\frac{d y}{d x}\) = ?
(a) \(\frac{x^{2}-1}{2 x^{2} \sqrt{x^{2}+1}}\)
(b) \(\frac{1-x^{2}}{2 x^{2} \sqrt{x^{2}+1}}\)
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
(d) \(\frac{1-x^{2}}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Answer:
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Hint:

Question 3.
If y = \(e^{\log x}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{e^{\log x}}{x}\)
(b) \(\frac{1}{x}\)
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{e^{\log x}}{x}\)

Question 4.
If y = 2x² + 2² + a², then \(\frac{d y}{d x}\) = ?
(a) x
(b) 4x
(c) 2x
(d) -2x
Answer:
(b) 4x

Question 5.
If y = 5^x . x⁵, then \(\frac{d y}{d x}\) = ?
(a) 5^x . x⁴(5 + log 5)
(b) 5^x . x⁵(5 + log 5)
(c) 5^x . x⁴(5 + x log 5)
(d) 5^x . x⁵(5 + x log 5)
Answer:
(c) 5^x . x⁴(5 + x log 5)

Question 6.
If y = \(\log \left(\frac{e^{x}}{x^{2}}\right)\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{2-x}{x}\)
(b) \(\frac{x-2}{x}\)
(c) \(\frac{e-x}{ex}\)
(d) \(\frac{x-e}{ex}\)
Answer:
(b) \(\frac{x-2}{x}\)
Hint:

Question 7.
If ax² + 2hxy + by² = 0, then \(\frac{d y}{d x}\) = ?
(a) \(\frac{(a x+h y)}{(h x+b y)}\)
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)
(c) \(\frac{(a x-h y)}{(h x+b y)}\)
(d) \(\frac{(2 a x+h y)}{(h x+3 b y)}\)
Answer:
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)

Question 8.
If x⁴ . y⁵ = (x + y)^(m+1) and \(\frac{d y}{d x}=\frac{y}{x}\) then m = ?
(a) 8
(b) 4
(c) 5
(d) 20
Answer:
(a) 8
Hint:
If x^p . y^q = (x + y)^p+q, then \(\frac{d y}{d x}=\frac{y}{x}\)
∴ m + 1 = 4 + 5 = 9
∴ m = 8.

Question 9.
If x = \(\frac{e^{t}+e^{-t}}{2}\), y = \(\frac{e^{t}-e^{-t}}{2}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{-y}{x}\)
(b) \(\frac{y}{x}\)
(c) \(\frac{-x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(d) \(\frac{x}{y}\)
Hint:

Question 10.
If x = 2at², y = 4at, then \(\frac{d y}{d x}\) = ?
(a) \(-\frac{1}{2 a t^{2}}\)
(b) \(\frac{1}{2 a t^{3}}\)
(c) \(\frac{1}{t}\)
(d) \(\frac{1}{4 a t^{3}}\)
Answer:
(c) \(\frac{1}{t}\)

(II) Fill in the blanks:

Question 1.
If 3x²y + 3xy² = 0 then \(\frac{d y}{d x}\) = …………
Answer:
-1
Hint:
3x²y + 3xy² = 0
∴ 3xy(x + y) = 0
∴ x + y = 0
∴ y = -x
∴ \(\frac{d y}{d x}\) = -1

Question 2.
If x^m . yⁿ = (x+y)^(m+n) then \(\frac{d y}{d x}=\frac{\ldots \ldots}{x}\)
Answer:
y

Question 3.
If 0 = log(xy) + a then \(\frac{d y}{d x}=\frac{-y}{\ldots . .}\)
Answer:
x

Question 4.
If x = t log t and y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y
Hint:
x = t log t = log t^t = log y
∴ 1 = \(\frac{1}{y} \cdot \frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = y

Question 5.
If y = x . log x then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{1}{x}\)

Question 6.
If y = [log(x)]² then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{2(1-\log x)}{x^{2}}\)
Hint:

Question 7.
If x = y + \(\frac{1}{y}\) then \(\frac{d y}{d x}\) = …………
Answer:
\(\frac{y^{2}}{y^{2}-1}\)
Hint:

Question 8.
If y = e^ax, then x.\(\frac{d y}{d x}\) = …………
Answer:
axy

Question 9.
If x = t . log t, y = t^t then \(\frac{d y}{d x}\) = …………
Answer:
y

Question 10.
If y = \(\left(x+\sqrt{x^{2}-1}\right)^{m}\) then \(\sqrt{\left(x^{2}-1\right)} \frac{d y}{d x}\) = ………
Answer:
my
Hint:

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

Question 1.
If f’ is the derivative of f, then the derivative of the inverse of f is the inverse of f’.
Answer:
False

Question 2.
The derivative of log_a x, where a is constant is \(\frac{1}{x \cdot \log a}\).
Answer:
True

Question 3.
The derivative of f(x) = a^x, where a is constant is x . a^x-1
Answer:
False

Question 4.
The derivative of a polynomial is polynomial.
Answer:
True

Question 5.
\(\frac{d}{d x}\left(10^{x}\right)=x \cdot 10^{x-1}\)
Answer:
False

Question 6.
If y = log x, then \(\frac{d y}{d x}=\frac{1}{x}\).
Answer:
True

Question 7.
If y = e², then \(\frac{d y}{d x}\) = 2e.
Answer:
False

Question 8.
The derivative of a^x is a^x. log a.
Answer:
True

Question 9.
The derivative of x^m . yⁿ = (x + y)^(m+n) is \(\frac{x}{y}\)
Answer:
False

(IV) Solve the following:

Question 1.
If y = (6x³ – 3x² – 9x)¹⁰, find \(\frac{d y}{d x}\)
Solution:
Given y = (6x³ – 3x² – 9x)¹⁰

Question 2.
If y = \(\sqrt[5]{\left(3 x^{2}+8 x+5\right)^{4}}\), find \(\frac{d y}{d x}\).
Solution:

Question 3.
If y = [log(log(log x))]², find \(\frac{d y}{d x}\).
Solution:

Question 4.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = 25 + 30x – x².
Solution:

Question 5.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = \(\frac{5 x+7}{2 x-13}\)
Solution:

Question 6.
Find \(\frac{d y}{d x}\) if y = x^x.
Solution:
y = x^x
∴ log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 7.
Find \(\frac{d y}{d x}\) if y = \(2^{x^{x}}\).
Solution:

Question 8.
Find \(\frac{d y}{d x}\), if y = \(\sqrt{\frac{(3 x-4)^{3}}{(x+1)^{4}(x+2)}}\)
Solution:

Question 9.
Find \(\frac{d y}{d x}\) if y = x^x + (7x – 1)^x
Solution:

Question 10.
If y = x³ + 3xy² + 3x²y, find \(\frac{d y}{d x}\).
Solution:
y = x³ + 3xy² + 3x²y
Differentiating both sides w.r.t. x, we get

Question 11.
If x³ + y² + xy = 7, find \(\frac{d y}{d x}\).
Solution:
x³ + y² + xy = 7
Differentiating both sides w.r.t. x, we get

Question 12.
If x³y³ = x² – y², find \(\frac{d y}{d x}\).
Solution:
x³y³ = x² – y²
Differentiating both sides w.r.t. x, we get

Question 13.
If x⁷ . y⁹ = (x + y)¹⁶, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 14.
If x^a . y^b = (x + y)^a+b, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:

Question 15.
Find \(\frac{d y}{d x}\) if x = 5t², y = 10t.
Solution:
x = 5t², y = 10t
Differentiating x and y w.r.t. t, we get

Question 16.
Find \(\frac{d y}{d x}\) if x = e^3t, y = \(e^{\sqrt{t}}\).
Solution:
x = e^3t, y = \(e^{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get

Question 17.
Differentiate log(1 + x²) with respect to a^x.
Solution:
Let u = log(1 + x²) and v = a^x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 18.
Differentiate e^(4x+5) with resepct to 10^4x.
Solution:
Let u = e^(4x+5) and v = 10^4x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

Question 19.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = log x.
Solution:
y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\)

Question 20.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = 2at, x = at².
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

Question 21.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = x² . e^x
Solution:
y = x² . e^x
Differentiating w.r.t. x, we get

= e^x (2x + 2 + x² + 2x)
= e^x (x² + 4x + 2).

Question 22.
If x² + 6xy + y² = 10, then show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\).
Solution:
x² + 6xy + y² = 10 ……..(1)
Differentiating both sides w.r.t. a, we get

Question 23.
If ax² + 2hxy + by² = 0, then show that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution:
ax² + 2hxy + by² = 0 ……..(1)
∴ ax² + hxy + hxy + by² = 0
∴ x(ax + hy) + y(hx + by) = 0
∴ x(ax + hy) = -y(hx + by)
∴ \(\frac{a x+h y}{h x+b y}=-\frac{y}{x}\) …….(2)
Differentiating (1) w.r.t. x, we get

Std 12 Commerce Statistics Part 1 Digest Pdf 

• Differentiation Ex 3.1 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.2 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.3 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.4 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.5 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.6 Class 12 Commerce Maths Textbook Solutions
• Differentiation Miscellaneous Exercise 3 Class 12 Commerce Maths Textbook Solutions

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.3 Questions and Answers.

Std 11 Maths 1 Exercise 4.3 Solutions Commerce Maths

Question 1.
Determine whether the sum to infinity of the following G.P’.s exist. If exists, find it.
(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)
(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)
(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)
(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)
Solution:

Question 2.
Express the following recurring decimals as a rational number.
(i) \(0 . \overline{32}\)
(ii) 3.5
(iii) \(4 . \overline{18}\)
(iv) \(0.3 \overline{45}\)
(v) \(3.4 \overline{56}\)
Solution:
(i) \(0 . \overline{32}\) = 0.323232…..
= 0.32 + 0.0032 + 0.000032 + …..
Here, 0.32, 0.0032, 0.000032, … are in G.P. with a = 0.32 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.
∴ Sum to infinity = \(\frac{a}{1-r}\)

(ii) 3.5 = 3.555… = 3 + 0.5 + 0.05 + 0.005 + …
Here, 0.5, 0.05, 0.005, … are in G.P. with a = 0.5 and r = 0.1
Since, |r| = |0.1| < 1
∴ Sum to infinity exists.

(iii) \(4 . \overline{18}\) = 4.181818…..
= 4 + 0.18 + 0.0018 + 0.000018 + …..
Here, 0.18, 0.0018, 0.000018, … are in G.P. with a = 0.18 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.

(iv) 0.345 = 0.3454545…..
= 0.3 + 0.045 + 0.00045 + 0.0000045 + …..
Here, 0.045, 0.00045, 0.0000045, … are in G.P. with a = 0.045, r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.

(v) \(3.4 \overline{56}\) = 3.4565656 …..
= 3.4 + 0.056 + 0.00056 + 0.0000056 + ….
Here, 0.056, 0.00056, 0.0000056, … are in G.P. with a = 0.056 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.

Question 3.
If the common ratio of a G.P. is \(\frac{2}{3}\) and sum of its terms to infinity is 12. Find the first term.
Solution:
r = \(\frac{2}{3}\), sum to infinity = 12 … [Given]
Sum to infinity = \(\frac{a}{1-r}\)
∴ 12 = \(\frac{a}{1-\frac{2}{3}}\)
∴ a = 12 × \(\frac{1}{3}\)
∴ a = 4

Question 4.
If the first term of a G.P. is 16 and sum of its terms to infinity is \(\frac{176}{5}\), find the common ratio.
Solution:
a = 16, sum to infinity = \(\frac{176}{5}\) … [Given]
Sum to infinity = \(\frac{a}{1-r}\)
∴ \(\frac{176}{5}=\frac{16}{1-r}\)
∴ \(\frac{11}{5}=\frac{1}{1-r}\)
∴ 11 – 11r = 5
∴ 11r = 6
∴ r = \(\frac{6}{11}\)

Question 5.
The sum of the terms of an infinite G.P. is 5 and the sum of the squares of those terms is 15. Find the G.P.
Solution:
Let the required G.P. be a, ar, ar², ar³, …..
Sum to infinity of this G.P. = 5
∴ 5 = \(\frac{a}{1-r}\)
∴ a = 5(1 – r) ……(i)
Also, the sum of the squares of the terms is 15.

Maharashtra State Board 11th Commerce Maths

• Sequences and Series Ex 4.1 11th Commerce Maths
• Sequences and Series Ex 4.2 11th Commerce Maths
• Sequences and Series Ex 4.3 11th Commerce Maths
• Sequences and Series Ex 4.4 11th Commerce Maths
• Sequences and Series Ex 4.5 11th Commerce Maths
• Sequences and Series Miscellaneous Exercise 4 11th Commerce Maths

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Ex 3.6 Questions and Answers.

Std 12 Maths 1 Exercise 3.6 Solutions Commerce Maths

1. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = √x
Solution:

Question 2.
y = x⁵
Solution:

Question 3.
y = x^-7
Solution:

2. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = e^x
Solution:

Question 2.
y = e^(2x+1)
Solution:

Question 3.
y = e^{log x}
Solution:

Std 12 Commerce Statistics Part 1 Digest Pdf 

• Differentiation Ex 3.1 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.2 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.3 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.4 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.5 Class 12 Commerce Maths Textbook Solutions
• Differentiation Ex 3.6 Class 12 Commerce Maths Textbook Solutions
• Differentiation Miscellaneous Exercise 3 Class 12 Commerce Maths Textbook Solutions