Class 12

Definite Integration Class 12 Commerce Maths 1 Chapter 6 Exercise 6.1 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.1 Questions and Answers.

Std 12 Maths 1 Exercise 6.1 Solutions Commerce Maths

Evaluate the following definite integrals:

Question 1.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)
Solution:

Question 2.
\(\int_{-2}^{3} \frac{1}{x+5} d x\)
Solution:

Question 3.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Solution:

Question 4.
\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)
Solution:

Question 5.
\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Solution:
Let I = \(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Let \(\frac{x}{(x+2)(x+3)}=\frac{A}{x+3}+\frac{B}{x+2}\)
∴ x = A(x + 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
-3 = A(-1) + B(0)
∴ A = 3
Put x + 2 = 0, i.e. x = -2, we get
-2 = A(0) + B(1)
∴ B = -2

Question 6.
\(\int_{1}^{2} \frac{d x}{x^{2}+6 x+5}\)
Solution:

Question 7.
If \(\int_{0}^{a}(2 x+1) d x\) = 2, find the real values of ‘a’.
Solution:
Let I = \(\int_{0}^{a}(2 x+1) d x\)
= \(\left[2 \cdot \frac{x^{2}}{2}+x\right]_{0}^{a}\)
= a² + a – 0
= a² + a
∴ I = 2 gives a² + a = 2
∴ a² + a – 2 = 0
∴ (a + 2)(a – 1) = 0 1
∴ a + 2 = 0 or a – 1 = 0
∴ a = -2 or a = 1.

Question 8.
If \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\) = 11, find ‘a’.
Solution:
Let I = \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\)
= \(\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{a}\)
= \(\left[x^{3}+x^{2}+x\right]_{1}^{a}\)
= (a³ + a² + a) – (1 + 1 + 1)
= a³ + a² + a – 3
∴ I = 11 gives a³ + a² + a – 3 = 11
∴ a³ + a² + a – 14 = 0
∴ (a³ – 8) + (a² + a – 6) = 0
∴ (a – 2)(a² + 2a + 4) + (a + 3)(a – 2) = 0
∴ (a – 2)(a² + 2a + 4 + a + 3) = 0
∴ (a – 2)(a² + 3a + 7) = 0
∴ a – 2 = 0 or a² + 3a + 7 = 0
∴ a = 2 or a = \(\frac{-3 \pm \sqrt{9-28}}{2}\)
The latter two roots are not real.
∴ they are rejected.
∴ a = 2.

Question 9.
\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)
Solution:

Question 10.
\(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\)
Solution:
Let I = \(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\) = \(\int_{1}^{2} \frac{3 x}{(3 x)^{2}-1} d x\)
Put 3x = t
∴ 3 dx = dt
∴ dx = \(\frac{d t}{3}\)
When x = 1, t = 3 × 1 = 3
When x = 2, t = 3 × 2 = 6

Question 11.
\(\int_{1}^{3} \log x d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 6.1 Solutions
• 12th Commerce Maths Exercise 6.2 Solutions
• 12th Commerce Maths Miscellaneous Exercise 6 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Miscellaneous Exercise 5 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Miscellaneous Exercise 5 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 5 Solutions Commerce Maths

(I) Choose the correct alternative from the following:

Question 1.
The value of \(\int \frac{d x}{\sqrt{1-x}}\) is
(a) 20\(\sqrt{1-x}\) + c
(b) -2\(\sqrt{1-x}\) + c
(c) √x + c
(d) x + c
Answer:
(b) -2\(\sqrt{1-x}\) + c

Question 2.
\(\int \sqrt{1+x^{2}} d x\) =
(a) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left(x+\sqrt{1+x^{2}}\right)+c\)
(b) \(\frac{2}{3}\left(1+x^{2}\right)^{3 / 2}+c\)
(c) \(\frac{1}{3}\left(1+x^{2}\right)+c\)
(d) \(\frac{(x)}{\sqrt{1+x^{2}}}+c\)
Answer:
(a) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left(x+\sqrt{1+x^{2}}\right)+c\)

Question 3.
\(\int x^{2}(3)^{x^{3}} d x\) =
(a) \(\text { (3) }^{x^{3}}+c\)
(b) \(\frac{(3)^{x^{3}}}{3 \cdot \log 3}+c\)
(c) \(\log 3(3)^{x^{3}}+c\)
(d) \(x^{2}(3)^{x 3}\)
Answer:
(b) \(\frac{(3)^{x^{3}}}{3 \cdot \log 3}+c\)
Hint:
Put x³ = t

Question 4.
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=p \int \frac{4 x+6}{2 x^{2}+6 x+5} d x\) + \(\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}\), then p = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{4}\)
(d) 2
Answer:
(c) \(\frac{1}{4}\)
Hint:
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=\int \frac{\frac{1}{4}(4 x+6)+\frac{1}{2}}{2 x^{2}+6 x+5} d x\)

Question 5.
\(\int \frac{d x}{\left(x-x^{2}\right)}\) = ________
(a) log x – log(1 – x) + c
(b) log(1 – x²) + c
(c) -log x + log(1 – x) + c
(d) log(x – x²) + c
Answer:
(a) log x – log(1 – x) + c

Question 6.
\(\int \frac{d x}{(x-8)(x+7)}\) = __________
(a) \(\frac{1}{15} \log \left|\frac{x+2}{x-1}\right|+c\)
(b) \(\frac{1}{15} \log \left|\frac{x+8}{x+7}\right|+c\)
(c) \(\frac{1}{15} \log \left|\frac{x-8}{x+7}\right|+c\)
(d) (x – 8)(x – 7) + c
Answer:
(c) \(\frac{1}{15} \log \left|\frac{x-8}{x+7}\right|+c\)

Question 7.
\(\int\left(x+\frac{1}{x}\right)^{3} d x\) = _________
(a) \(\frac{1}{4}\left(x+\frac{1}{x}\right)^{4}+c\)
(b) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x-\frac{1}{2 x^{2}}+c\)
(c) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x+\frac{1}{x^{2}}+c\)
(d) \(\left(x-x^{-1}\right)^{3}+c\)
Answer:
(b) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x-\frac{1}{2 x^{2}}+c\)
Hint:
\(\left(x+\frac{1}{x}\right)^{3}=x^{3}+3 x+\frac{3}{x}+\frac{1}{x^{3}}\)

Question 8.
\(\int\left(\frac{e^{2 x}+e^{-2 x}}{e^{x}}\right) d x\)
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)
(b) \(e^{x}+\frac{1}{3 e^{3 x}}+c\)
(c) \(e^{-x}+\frac{1}{3 e^{3 x}}+c\)
(d) \(e^{-x}-\frac{1}{3 e^{3 x}}+c\)
Answer:
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)

Question 9.
∫(1 – x)^-2 dx = ___________
(a) (1 + x)^-1 + c
(b) (1 – x)^-1 + c
(c) (1 – x)^-1 – 1 + c
(d) (1 – x)^-1 + 1 + c
Answer:
(b) (1 – x)^-1 + c

Question 10.
\(\int \frac{\left(x^{3}+3 x^{2}+3 x+1\right)}{(x+1)^{5}} d x\) = _______
(a) \(\frac{-1}{x+1}+c\)
(b) \(\left(\frac{-1}{x+1}\right)^{5}+c\)
(c) log(x + 1) + c
(d) log|x + 1|⁵ + c
Answer:
(a) \(\frac{-1}{x+1}+c\)
Hint:
x³ + 3x² + 3x + 1 = (x + 1)³

(II) Fill in the blanks.

Question 1.
\(\int \frac{5\left(x^{6}+1\right)}{x^{2}+1}\)dx = x⁴ + ___x³ + 5x + c.
Answer:
\(-\frac{5}{3}\)
Hint:
x⁶ + 1 = (x² + 1)(x⁴ – x² + 1)

Question 2.
\(\int \frac{x^{2}+x-6}{(x-2)(x-1)} d x\) = x + ______ + c
Answer:
4 log|x – 1|
Hint:
x² + x – 6 = (x + 3)(x – 2)

Question 3.
If f'(x) = \(\frac{1}{x}\) + x and f(1) = \(\frac{5}{2}\) then f(x) = log x + \(\frac{x^{2}}{2}\) + _______
Answer:
2
Hint:

Question 4.
To find the value of \(\int \frac{(1+\log x) d x}{x}\) the proper substitution is __________
Answer:
1 + log x = t

Question 5.
\(\int \frac{1}{x^{3}}\left[\log x^{x}\right]^{2} d x\) = p(log x)³ + c, then p = _______
Answer:
\(\frac{1}{3}\)
Hint:
\(\frac{1}{x^{3}}\left(\log x^{x}\right)^{2}=\frac{1}{x^{3}}(x \log x)^{2}=\frac{(\log x)^{2}}{x}\)

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

Question 1.
The proper substitution for \(\int x\left(x^{x}\right)^{x}(2 \log x+1) d x \text { is }\left(x^{x}\right)^{x}=t\)
Answer:
True

Question 2.
If ∫x e^2x dx is equal to e^2x f(x) + c where c is constant of integration, then f(x) is \(\frac{(2 x-1)}{2}\).
Answer:
False

Question 3.
If ∫x f(x) dx = \(\frac{f(x)}{2}\), then f(x) = \(e^{x^{2}}\).
Answer:
True

Question 4.
If \(\int \frac{(x-1) d x}{(x+1)(x-2)}\) = A log|x + 1| + B log|x – 2|, then A + B = 1.
Answer:
True

Question 5.
For \(\int \frac{x-1}{(x+1)^{3}} e^{x} d x\) = e^x f(x) + c, f(x) = (x + 1)².
Answer:
False

(IV) Solve the following:

1. Evaluate:

(i) \(\int \frac{5 x^{2}-6 x+3}{2 x-3} d x\)
Solution:

(ii) \(\int(5 x+1)^{\frac{4}{9}} d x\)
Solution:

(iii) \(\int \frac{1}{2 x+3} d x\)
Solution:

(iv) \(\int \frac{x-1}{\sqrt{x+4}} d x\)
Solution:

(v) If f'(x) = √x and f(1) = 2, then find the value of f(x).
Solution:
By the definition of integral

(vi) ∫|x| dx if x < 0
Solution:
∫|x| dx = ∫-x dx …..[∵ x < 0]
= -∫x dx
= \(-\frac{x^{2}}{2}\) + c

2. Evaluate:

(i) Find the primitive of \(\frac{1}{1+e^{x}}\)
Solution:
Let I be the primitive of \(\frac{1}{1+e^{x}}\)

(ii) \(\int \frac{a e^{x}+b e^{-x}}{\left(a e^{x}-b e^{-x}\right)^{2}} d x\)
Solution:

(iii) \(\int \frac{1}{2 x+3 x \log x} d x\)
Solution:

(iv) \(\int \frac{1}{\sqrt{x}+x} d x\)
Solution:

(v) \(\int \frac{2 e^{x}-3}{4 e^{x}+1} d x\)
Solution:

3. Evaluate:

(i) \(\int \frac{d x}{\sqrt{4 x^{2}-5}} d x\)
Solution:

(ii) \(\int \frac{d x}{3-2 x-x^{2}} d x\)
Solution:

(iii) \(\int \frac{d x}{9 x^{2}-25}\)
Solution:

(iv) \(\int \frac{e^{x}}{\sqrt{e^{2 x}+4 e^{x}+13}} d x\)
Solution:

(v) \(\int \frac{d x}{x\left[(\log x)^{2}+4 \log x-1\right]}\)
Solution:

(vi) \(\int \frac{d x}{5-16 x^{2}}\)
Solution:

(vii) \(\int \frac{d x}{25 x-x(\log x)^{2}}\)
Solution:

(viii) \(\int \frac{e^{x}}{4 e^{2 x}-1} d x\)
Solution:

4. Evaluate:

(i) ∫(log x)² dx
Solution:

(ii) \(\int e^{x} \frac{1+x}{(2+x)^{2}} d x\)
Solution:

(iii) ∫x e^2x dx
Solution:

(iv) ∫log(x² + x) dx
Solution:

(v) \(\int e^{\sqrt{x}} d x\)
Solution:

(vi) \(\int \sqrt{x^{2}+2 x+5} d x\)
Solution:

(vii) \(\int \sqrt{x^{2}-8 x+7} d x\)
Solution:

5. Evaluate:

(i) \(\int \frac{3 x-1}{2 x^{2}-x-1} d x\)
Solution:

(ii) \(\int \frac{2 x^{3}-3 x^{2}-9 x+1}{2 x^{2}-x-10} d x\)
Solution:

(iii) \(\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.6 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.6 Questions and Answers.

Std 12 Maths 1 Exercise 5.6 Solutions Commerce Maths

Evaluate:

Question 1.
\(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Let \(\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\)
∴ 2x + 1 = A(x – 2) + B(x + 1)
Put x + 1 = 0, i.e. x = -1, we get
2(-1) + 1 = A(-3) + B(0)
∴ A = \(\frac{1}{3}\)
Put x – 2 = 0, i.e. x = 2, we get
2(2) + 1 = A(0) + B(3)
∴ B = \(\frac{5}{3}\)

Question 2.
\(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Let \(\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}\)
∴ 2x + 1 = A(x – 1)(x – 4) + Bx(x – 4) + Cx(x – 1)
Put x = 0, we get
2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1)
∴ 1 = 4A
∴ A = \(\frac{1}{4}\)
Put x – 1 = 0, i.e. x = 1, we get
2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0)
∴ 3 = -3B
∴ B = -1
Put x – 4 = 0, i.e x = 4, we get
2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3)
∴ 9 = 12C
∴ C = \(\frac{3}{4}\)

Question 3.
\(\int \frac{x^{2}+x-1}{x^{2}+x-6} d x\)
Solution:

∴ 1 = A(x – 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
1 = A(-5) + B (0)
∴ A = \(\frac{-1}{5}\)
Put x – 2 = 0, i.e. x = 2, we get
1 = A(0) + B(5)
∴ B = \(\frac{1}{5}\)

Question 4.
\(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Solution:
Let I = \(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Let \(\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}\)
∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²
Put x – 1 = 0, i.e. x = 1, we get
1 = A(0)(3) + B(3) + C(0)
∴ B = \(\frac{1}{3}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 = A (-3)(0) + B(0) + C(9)
∴ C = \(-\frac{2}{9}\)
Put x = -1, we get,
-1 = A(-2)(1) + B(1) + C(4)
But B = \(\frac{1}{3}\) and C = \(-\frac{2}{9}\)

Question 5.
\(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Solution:
Let I = \(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Let \(\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}\)
∴ 3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)²
Put x + 1 = 0, i.e. x = -1, we get
3(-1) – 2 = A(0)(2) + B(2) + C(0)
∴ -5 = 2B
∴ B = \(-\frac{5}{2}\)
Put x + 3 = 0, i.e. x = -3, we get
3(-3) – 2 = A(-2)(0) + B(0) + C(4)
∴ -11 = 4C
∴ C = \(-\frac{11}{4}\)
Put x = 0, we get
3(0) – 2 = A(1)(3) + B(3) + C(1)
∴ -2 = 3A + 3B + C
But B = \(-\frac{5}{2}\) and C = \(-\frac{11}{4}\)

Question 6.
\(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
Solution:
Let I = \(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
= \(\int \frac{x^{4}}{x^{5}\left(x^{5}+1\right)} d x\)

Question 7.
\(\int \frac{1}{x\left(x^{n}+1\right)} d x\)
Solution:

Question 8.
\(\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x\)
Solution:

Let \(\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\)
∴ 5x² + 20x + 6 = A(x + 1)² + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i.e. x = -1, we get
5(1) + 20(-1) + 6 = A(0) + B(-1)(0) + C(-1)
∴ -9 = -C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = -1

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.5 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.5 Questions and Answers.

Std 12 Maths 1 Exercise 5.5 Solutions Commerce Maths

Evaluate the following.

Question 1.
∫x log x
Solution:

Question 2.
∫x² e^4x dx
Solution:

Question 3.
∫x² e^3x dx
Solution:

Question 4.
\(\int x^{3} e^{x^{2}} d x\)
Solution:

Question 5.
\(\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x\)
Solution:

Question 6.
\(\int e^{x} \frac{x}{(x+1)^{2}} d x\)
Solution:

Question 7.
\(\int e^{x} \frac{x-1}{(x+1)^{3}} d x\)
Solution:

Question 8.
\(\int e^{x}\left[(\log x)^{2}+\frac{2 \log x}{x}\right] d x\)
Solution:

Question 9.
\(\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] d x\)
Solution:

Question 10.
\(\int \frac{\log x}{(1+\log x)^{2}} d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.4 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.4 Questions and Answers.

Std 12 Maths 1 Exercise 5.4 Solutions Commerce Maths

Evaluate the following.

Question 1.
\(\int \frac{1}{4 x^{2}-1} d x\)
Solution:

Question 2.
\(\int \frac{1}{x^{2}+4 x-5} d x\)
Solution:

Question 3.
\(\int \frac{1}{4 x^{2}-20 x+17} d x\)
Solution:

Question 4.
\(\int \frac{x}{4 x^{4}-20 x^{2}-3} d x\)
Solution:

Question 5.
\(\int \frac{x^{3}}{16 x^{8}-25} d x\)
Solution:

Question 6.
\(\int \frac{1}{a^{2}-b^{2} x^{2}} d x\)
Solution:

Question 7.
\(\int \frac{1}{7+6 x-x^{2}} d x\)
Solution:

Question 8.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x^{2}+4 x+29}} d x\)
Solution:

Question 10.
\(\int \frac{1}{\sqrt{3 x^{2}-5}} d x\)
Solution:

Question 11.
\(\int \frac{1}{\sqrt{x^{2}-8 x-20}} d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.3 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.3 Questions and Answers.

Std 12 Maths 1 Exercise 5.3 Solutions Commerce Maths

Evaluate the following:

Question 1.
\(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Solution:
Let I = \(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Put, Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[\(\frac{d}{d t}\)(4e^2t – 5)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[4e^2t × 2 – 0]
∴ 3e^2t + 5 = (4A + 8B) e^2t – 5A
Equating the coefficient of e^2t and constant on both sides, we get
4A + 8B = 3
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)

Question 2.
\(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Solution:
Let I = \(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 20 – 12e^x = A(3e^x – 4) + B[\(\frac{d}{d x}\)(3e^x – 4)]
∴ 20 – 12e^x = A(3e^x – 4) + B(3e^x – 0)
∴ 20 – 12e^x = (3A + 3B)e^x – 4A
Equating the coefficient of ex and constant on both sides, we get
3A + 3B = -12 ……(1)
and -4A = 20
∴ A = -5
from (1), 3(-5) + 3B = -12
∴ 3B = 3
∴ B = 1
∴ 20 – 12e^x = -5(3e^x – 4) + (3e^x)

Question 3.
\(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Solution:
Let I = \(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^x + 4 = A(2e^x – 8) + B[\(\frac{d}{d x}\)(2e^x – 8)]
∴ 3e^x + 4 = A(2e^x – 8) + B(2e^x – 0)
∴ 3e^x + 4 = (2A + 2B)e^x – 8A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 3 ……..(1)
and -8A = 4
∴ A = \(-\frac{1}{2}\)
∴ from (1), 2(\(-\frac{1}{2}\)) + 2B = 3
∴ 2B = 4
∴ B = 2

Question 4.
\(\int \frac{2 e^{x}+5}{2 e^{x}+1} d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.2 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.2 Questions and Answers.

Std 12 Maths 1 Exercise 5.2 Solutions Commerce Maths

Evaluate the following.

Question 1.
\(\int x \sqrt{1+x^{2}} d x\)
Solution:

Question 2.
\(\int \frac{x^{3}}{\sqrt{1+x^{4}}} d x\)
Solution:

Question 3.
\(\int\left(e^{x}+e^{-x}\right)^{2}\left(e^{x}-e^{-x}\right) d x\)
Solution:

Question 4.
\(\int \frac{1+x}{x+e^{-x}} d x\)
Solution:

Question 5.
∫(x + 1)(x + 2)⁷(x + 3) dx
Solution:
Let I = ∫(x + 1)(x + 2)⁷(x + 3) dx
= ∫(x + 2)⁷ (x + 1)(x + 3) dx
= ∫(x + 2)⁷ [(x + 2) – 1][(x + 2) + 1] dx
= ∫(x + 2)⁷ [(x + 2)² – 1] dx
= ∫[(x + 2)⁹ – (x + 2 )⁷] dx
= ∫(x + 2 )⁹ dx – ∫(x + 2)⁷ dx
= \(\frac{(x+2)^{10}}{10}\) – \(\frac{(x+2)^{8}}{8}\) + c

Question 6.
\(\int \frac{1}{x \log x} d x\)
Solution:
Put log x = t
∴ \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{d x}{x \cdot \log x}=\int \frac{1}{\log x} \cdot \frac{1}{x} d x\)
= ∫\(\frac{1}{t}\) dt
= log |t| + c
= log|log x| + c.

Question 7.
\(\int \frac{x^{5}}{x^{2}+1} d x\)
Solution:

Question 8.
\(\int \frac{2 x+6}{\sqrt{x^{2}+6 x+3}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x}+x} d x\)
Solution:

Question 10.
\(\int \frac{1}{x\left(x^{6}+1\right)} d x\)
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Integration Class 12 Commerce Maths 1 Chapter 5 Exercise 5.1 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.1 Questions and Answers.

Std 12 Maths 1 Exercise 5.1 Solutions Commerce Maths

Question 1.
Evaluate \(\int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x\)
Solution:

Question 2.
Evaluate \(\int\left(1+x+\frac{x^{2}}{2 !}\right) d x\)
Solution:

Question 3.
Evaluate \(\int \frac{3 x^{3}-2 \sqrt{x}}{x} d x\)
Solution:

Question 4.
Evaluate ∫(3x² – 5)² dx
Solution:
∫(3x² – 5)² dx
= ∫(9x⁴ – 30x² + 25) dx
= 9∫x⁴ dx – 30∫x² dx + 25∫1 dx
= 9(\(\frac{x^{5}}{5}\)) – 30(\(\frac{x^{3}}{3}\)) + 25x + c
= \(\frac{9x^{5}}{5}\) – 10x³ + 25x + c.

Question 5.
Evaluate \(\int \frac{1}{x(x-1)} d x\)
Solution:

Question 6.
If f'(x) = x² + 5 and f(0) = -1, then find the value of f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(x² + 5) dx
= ∫x² dx + 5∫1 dx
= \(\frac{x^{3}}{3}\) + 5x + c
Now, f(0) = -1 gives
f(0) = 0 + 0 + c = -1
∴ c = -1
∴ from (1), f(x) = \(\frac{x^{3}}{3}\) + 5x – 1.

Question 7.
If f(x) = 4x³ – 3x² + 2x + k, f(0) = -1 and f(1) = 4, find f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(4x³ – 3x² + 2x + k) dx
= 4∫x³ dx – 3∫x² dx + 2∫x dx + k∫1 dx
= 4(\(\frac{x^{4}}{4}\)) – 3(\(\frac{x^{3}}{3}\)) + 2(\(\frac{x^{2}}{2}\)) + kx + c
∴ f(x) = x⁴ – x³ + x² + kx + c
Now, f(0) = 1 gives
f(0) = 0 – 0 + 0 + 0 + c = 1
∴ c = 1
∴ from (1), f(x) = x⁴ – x³ + x² + kx + 1
Further f(1) = 4 gives
f(1) = 1 – 1 + 1 + k + 1 = 4
∴ k = 2
∴ from (2), f(x) = x⁴ – x³ + x² + 2x + 1.

Question 8.
If f(x) = \(\frac{x^{2}}{2}\) – kx + 1, f(0) = 2 and f(3) = 5, find f(x).
Solution:
By the definition of integral

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions 
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Exercise 5.4 Solutions
• 12th Commerce Maths Exercise 5.5 Solutions
• 12th Commerce Maths Exercise 5.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

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

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

Std 12 Maths 1 Miscellaneous Exercise 4 Solutions Commerce Maths

(I) Choose the correct alternative:

Question 1.
The equation of tangent to the curve y = x² + 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 x² + y² = 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) = x³ – 3x² + 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) = 3x³ – 9x² – 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) = x³ – 3x² + 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 = 3x² + 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 = c² at (ct, \(\frac{c}{t}\)), where t is a parameter.
Solution:
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 = \(\frac{1}{t}\), 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 = x³ – x² – 1 at the point whose abscissa is -2.
Solution:
y = x³ – x² – 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\)(x³ – x² – 1)
= 3x² – 2x – 0
= 3x² – 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(x₁, y₁) 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 x²y is 4.
Solution:
x + y = 3
∴ y = 3 – x
∴ x²y = x²(3 – x) = 3x² – x³
Let f(x) = 3x² – x³
Then f'(x) = \(\frac{d}{d x}\)(3x² – x³)
= 3 × 2x – 3x²
= 6x – 3x²
and f”(x) = \(\frac{d}{d x}\)(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.

Question 6.
Examine the function f for maxima and minima, where f(x) = x³ – 9x² + 24x.
Solution:
f(x) = x³ – 9x² + 24x
∴ f'(x) = \(\frac{d}{d x}\)(x³ – 9x² + 24x)
= 3x² – 9 × 2x + 24 × 1
= 3x² – 18x + 24
and f”(x) = \(\frac{d}{d x}\)(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.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Exercise 4.2 Solutions
• 12th Commerce Maths Exercise 4.3 Solutions
• 12th Commerce Maths Exercise 4.4 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.4 Questions and Answers.

Std 12 Maths 1 Exercise 4.4 Solutions Commerce Maths

Question 1.
The demand function of a commodity at price P is given as D = 40 – \(\frac{5 P}{8}\). Check whether it is an increasing or decreasing function.
Solution:
D = 40 – \(\frac{5 P}{8}\)
∴ \(\frac{d D}{d P}=\frac{d}{d P}\left(40-\frac{5 P}{8}\right)\)
= 0 – \(\frac{5}{8}\) × 1
= \(\frac{-5}{8}\)
Hence, the given function is decreasing function.

Question 2.
The price P for demand D is given as P = 183 + 120D – 3D², find D for which price is increasing.
Solution:
P = 183 + 120D – 3D²
∴ \(\frac{d P}{d D}=\frac{d}{d D}\)(183 + 120D – 3D²)
= 0 + 120 × 1 – 3 × 2D
= 120 – 6D
If price P is increasing, then \(\frac{d P}{d D}\) > 0
∴ 120 – 6D > 0
∴ 120 > 6D
∴ D < 20
Hence, the price is increasing when D < 20.

Question 3.
The total cost function for production of x articles is given as C = 100 + 600x – 3x². Find the values of x for which the total cost is decreasing.
Solution:
The cost function is given as
C = 100 + 600x – 3x²
∴ \(\frac{d C}{d D}=\frac{d}{d D}\)(100 + 600x – 3x²)
= 0 + 600 × 1 – 3 × 2x
= 600 – 6x
If the total cost is decreasing, then \(\frac{d C}{d x}\) < 0
∴ 600 – 6x < 0
∴ 600 < 6x
∴ x > 100
Hence, the total cost is decreasing for x > 100.

Question 4.
The manufacturing company produces x items at the total cost of ₹(180 + 4x). The demand function for this product is P = (240 – x). Find x for which
(i) revenue is increasing
(ii) profit is increasing.
Solution:
(i) Let R be the total revenue.
Then R = P.x = (240 – x)x
∴ R = 240x – x²
∴ \(\frac{d R}{d D}=\frac{d}{d D}\)(240x – x²)
= 240 × 1 – 2x
= 240 – 2x
R is increasing, if \(\frac{d R}{d x}\) > 0
i.e. if 240 – 2x > 0
i.e. if 240 > 2x
i.e. if x < 120
Hence, the revenue is increasing, if x < 120.

(ii) Profit π = R – C
∴ π = (240x – x²) – (180 + 4x)
= 240x – x² – 180 – 4x
= 236x – x² – 180
∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(236x – x² – 180)
= 236 × 1 – 2x – 0
= 236 – 2x
Profit is increasing, if \(\frac{d \pi}{d x}\) > 0
i.e. if 236 – 2x > 0
i.e. if 236 > 2x
i.e. if x < 118
Hence, the profit is increasing, if x < 118.

Question 5.
For manufacturing x units, labour cost is 150 – 54x and processing cost is x². Price of each unit is p = 10800 – 4x². Find the values of x for which
(i) total cost is decreasing
(ii) revenue is increasing.
Solution:
(i) Total cost C = labour cost + processing cost
∴ C = 150 – 54x + x²
∴ \(\frac{d C}{d x}=\frac{d}{d x}\)(150 – 54x + x²)
= 0 – 54 × 1 + 2x
= -54 + 2x
The total cost is decreasing, if \(\frac{d C}{d x}\) < 0
i.e. if -54 + 2x < 0
i.e. if 2x < 54
i.e. if x < 27
Hence, the total cost is decreasing, if x < 27.

(ii) The total revenue R is given as
R = p.x
R = (10800 – 4x²) x
R = 10800x – 4x³
∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(10800x – 4x³)
= 10800 × 1 – 4 × 3x²
= 10800 – 12x²
The revenue is increasing, if \(\frac{d R}{d x}\) > 0
i.e. if 10800 – 12x² > 0
i.e. if 10800 > 12x²
i.e. if x² < 900
i.e. if x < 30 ……[∵ x > 0]
Hence, the revenue is increasing, if x < 30.

Question 6.
The total cost of manufacturing x articles is C = 47x + 300x² – x⁴. Find x, for which average cost is
(i) increasing
(ii) decreasing.
Solution:
The total cost is given as C = 47x + 300x² – x⁴
∴ the average cost is given by

(i) CA is increasing, if \(\frac{d C_{A}}{d x}\) > 0
i.e. if 300 – 3x² > 0
i.e. if 300 > 3x²
i.e. if x² < 100
i.e. if x < 10 …..[∵ x > 0]
Hence, the average cost is increasing, if x < 10.

(ii) CA is decreasing, if \(\frac{d C_{A}}{d x}\) < 0
i.e. if 300 – 3x² < 0
i.e. if 300 < 3x²
i.e. if x² > 100
i.e. if x > 10 ……[∵ x > 0]
Hence, the average cost is decreasing, if x > 10.

Question 7.
(i) Find the marginal revenue, if the average revenue is 45 and the elasticity of demand is 5.
Solution:
Given R_A = 45 and η = 5
Now, R_m = \(R_{A}\left(1-\frac{1}{\eta}\right)\)
= 45(1 – \(\frac{1}{5}\))
= 45(\(\frac{4}{5}\))
= 36
Hence, the marginal revenue = 36.

(ii) Find the price, if the marginal revenue is 28 and elasticity of demand is 3.
Solution:
Given R_m = 28 and η = 3

Hence, the price = 42.

(iii) Find the elasticity of demand, if the marginal revenue is 50 and price is ₹ 75.
Solution:
Given R_m = 50 and R_A = 75

Hence, the elasticity of demand = 3.

Question 8.
If the demand function is D = \(\frac{p+6}{p-3}\), find the elasticity of demand at p = 4.
Solution:
The demand function is

Elasticity of demand is given by

Hence, the elasticity of demand at p = 4 is 3.6

Question 9.
Find the price for the demand function D = \(\frac{2 p+3}{3 p-1}\), when elasticity of demand is \(\frac{11}{14}\).
Solution:
The demand function is

Elasticity of demand is given by

Question 10.
If the demand function is D = 50 – 3p – p² elasticity of demand at (i) p = 5 (ii) p = 2. Comment on the result.
Solution:
The demand function is D = 50 – 3p – p²
∴ \(\frac{d D}{d p}=\frac{d}{d p}\)(50 – 3p – p²)
= 0 – 3 × 1 – 2p
= -3 – 2p
Elasticity of demand is given by

(i) When p = 5, then

Since, η >1, the demand is elastic.
(ii) When p = 2, then

Since, 0 < η < 1, the demand is inelastic.

Question 11.
For the demand function D = 100 – \(\frac{p^{2}}{2}\), find the elasticity of demand at (i) p = 10 (ii) p = 6 and comment on the results.
Solution:
The demand function is

The elasticity of demand is given by

(i) When p = 10, then

Since, η > 1, the demand is elastic.
(ii) When p = 6, then

Since, 0 < η < 1, the demand is inelastic.

Question 12.
A manufacturing company produces, x items at a total cost of ₹(40 + 2x). Their price is given as p = 120 – x. Find the value of x for which
(i) revenue is increasing
(ii) profit is increasing
(iii) Also find an elasticity of demand for price 80.
Solution:
(i) The total revenue R is given by
R = p.x = (120 – x)x
∴ R = 120x – x²
∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(120x – x²)
= 120 × 1 – 2x
= 120 – 2x
If the revenue is increasing, then \(\frac{d R}{d x}\) > 0
∴ 120 – 2x > 0
∴ 120 > 2x
∴ x < 60
Hence, the revenue is increasing when x < 60.

(ii) Profit π = R – C
= (120x – x²) – (40 + 2x)
= 120x – x² – 40 – 2x
= 118x – x² – 40
∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(118x – x² – 40)
= 118 × 1 – 2x – 0
= 118 – 2x
If the profit is increasing, then \(\frac{d \pi}{d x}\) > 0
∴ 118 – 2x > 0
∴ 118 > 2x
∴ x < 59
Hence, the profit is increasing when x < 59.

(iii) p = 120 – x
∴ x = 120 – p
∴ \(\frac{d x}{d p}=\frac{d}{d p}\)(120 – p)
= 0 – 1
= -1
Elasticity of demand is given by

Question 13.
Find MPC, MPS, APC and APS, if the expenditure E_c of a person with income I is given as
E_c = (0.0003)I² + (0.075)I, when I = 1000.
Solution:
E_c = (0.0003)I² + (0.075)I
MPC = \(\frac{d E_{c}}{d I}=\frac{d}{d I}\)[(0.0003)I2 + (0.075)I]
= (0.0003)(2I) + (0.075)(1)
= (0.0006)I + 0.075
When I = 1000, then
MPC = (0.0006)(1000) + 0.075
= 0.6 + 0.075
= 0.675.
∴ MPC + MPS = 1
∴ 0.675 + MPS = 1
∴ MPS = 1 – 0.675 = 0.325
Now, APC = \(\frac{E_{c}}{I}=\frac{(0.0003) I^{2}+(0.075) I}{I}\)
= (0.0003)I + (0.075)
When I = 1000, then
APC = (0.0003)(1000) + 0.075
= 0.3 + 0.075
= 0.375
∵ APC + APS = 1
∴ 0.375 + APS = 1
∴ APS = 1 – 0.375 = 0.625
Hence, MPC = 0.675, MPS = 0.325, APC = 0.375, APS = 0.625.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Exercise 4.2 Solutions
• 12th Commerce Maths Exercise 4.3 Solutions
• 12th Commerce Maths Exercise 4.4 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions