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Limits Class 11 Commerce Maths 1 Chapter 7 Miscellaneous Exercise 7 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Miscellaneous Exercise 7 Questions and Answers.

Std 11 Maths 1 Miscellaneous Exercise 7 Solutions Commerce Maths

I.

Question 1.
If \(\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80\) then find the value of n.
Solution:

II. Evaluate the following Limits:

Question 1.
\(\lim _{x \rightarrow a} \frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{(x-2)}{2 x^{2}-7 x+6}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 1}\left[\frac{x^{3}-1}{x^{2}+5 x-6}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 3}\left[\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right]\)
Solution:

Question 6.
\(\lim _{x \rightarrow 4}\left[\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right]\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}-1}{x}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left(1+\frac{x}{5}\right)^{\frac{1}{x}}\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{\log (1+9 x)}{x}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0} \frac{(1-x)^{5}-1}{(1-x)^{3}-1}\)
Solution:

Question 11.
\(\lim _{x \rightarrow 0}\left[\frac{a^{x}+b^{x}+c^{x}-3}{x}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}}\)
Solution:

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\left(2^{x}-1\right) \cdot \log (1+x)}\right]\)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x^{2}}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{x \cdot \log (1+x)}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow 0}\left[\frac{a^{4 x}-1}{b^{2 x}-1}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow 0}\left[\frac{\log 100+\log (0.01+x)}{x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 0}\left[\frac{\log (4-x)-\log (4+x)}{x}\right]\)
Solution:

Question 19.
Evaluate the limit of the function if exist at x = 1 where,
\(f(x)= \begin{cases}7-4 x & x<1 \\ x^{2}+2 & x \geq 1\end{cases}\)
Solution:

Maharashtra State Board 11th Commerce Maths

• Limits Ex 7.1 11th Commerce Maths
• Limits Ex 7.2 11th Commerce Maths
• Limits Ex 7.3 11th Commerce Maths
• Limits Ex 7.4 11th Commerce Maths
• Limits Miscellaneous Exercise 7 11th Commerce Maths
• Continuity Ex 8.1 11th Commerce Maths
• Continuity Miscellaneous Exercise 8 11th Commerce Maths
• Differentiation Ex 9.1 11th Commerce Maths
• Differentiation Ex 9.2 11th Commerce Maths
• Differentiation Miscellaneous Exercise 9 11th Commerce Maths

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.4 Questions and Answers.

Std 11 Maths 1 Exercise 7.4 Solutions Commerce Maths

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{9^{x}-5^{x}}{4^{x}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}+3^{x}-2^{x}-1}{x}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{\log (2+x)-\log (2-x)}{x}\right]\)
Solution:

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{3^{x}+3^{-x}-2}{x^{2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{3+x}{3-x}\right]^{\frac{1}{x}}\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{\log (3-x)-\log (3+x)}{x}\right]\)
Solution:

III. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-b^{2 x}}{\log (1+4 x)}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\left(2^{x}-1\right)^{2}}{\left(3^{x}-1\right) \cdot \log (1+x)}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{15^{x}-5^{x}-3^{x}+1}{x^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{3^{\frac{x}{2}}-3}{3^{x}-9}\right]\)
Solution:

IV. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{(25)^{x}-2(5)^{x}+1}{x^{2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{(49)^{x}-2(35)^{x}+(25)^{x}}{x^{2}}\right]\)
Solution:

Maharashtra State Board 11th Commerce Maths

• Limits Ex 7.1 11th Commerce Maths
• Limits Ex 7.2 11th Commerce Maths
• Limits Ex 7.3 11th Commerce Maths
• Limits Ex 7.4 11th Commerce Maths
• Limits Miscellaneous Exercise 7 11th Commerce Maths
• Continuity Ex 8.1 11th Commerce Maths
• Continuity Miscellaneous Exercise 8 11th Commerce Maths
• Differentiation Ex 9.1 11th Commerce Maths
• Differentiation Ex 9.2 11th Commerce Maths
• Differentiation Miscellaneous Exercise 9 11th Commerce Maths

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.3 Questions and Answers.

Std 11 Maths 1 Exercise 7.3 Solutions Commerce Maths

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^{2}}-\sqrt{6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^{2}}-\sqrt{1+y^{2}}}{y^{2}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 1}\left[\frac{x^{2}+x \sqrt{x}-2}{x-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 4}\left[\frac{x^{2}+x-20}{\sqrt{3 x+4}-4}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-8}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

IV. Evaluate the following limits:

Question 1.
\(\lim _{y \rightarrow 2}\left[\frac{2-y}{\sqrt{3-y}-1}\right]\)
Solution:

Question 2.
\(\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]\)
Solution:

Maharashtra State Board 11th Commerce Maths

• Limits Ex 7.1 11th Commerce Maths
• Limits Ex 7.2 11th Commerce Maths
• Limits Ex 7.3 11th Commerce Maths
• Limits Ex 7.4 11th Commerce Maths
• Limits Miscellaneous Exercise 7 11th Commerce Maths
• Continuity Ex 8.1 11th Commerce Maths
• Continuity Miscellaneous Exercise 8 11th Commerce Maths
• Differentiation Ex 9.1 11th Commerce Maths
• Differentiation Ex 9.2 11th Commerce Maths
• Differentiation Miscellaneous Exercise 9 11th Commerce Maths

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 Notes 

• Chapter 5 Integration Ex 5.1 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.2 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.3 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.4 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.5 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.6 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Miscellaneous Exercise 5 Class 12 Commerce Maths Textbook 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 Notes 

• Chapter 5 Integration Ex 5.1 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.2 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.3 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.4 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.5 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.6 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Miscellaneous Exercise 5 Class 12 Commerce Maths Textbook 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 Notes 

• Chapter 5 Integration Ex 5.1 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.2 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.3 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.4 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.5 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.6 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Miscellaneous Exercise 5 Class 12 Commerce Maths Textbook Solutions

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.2 Questions and Answers.

Std 11 Maths 1 Exercise 7.2 Solutions Commerce Maths

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow 2}\left[\frac{z^{2}-5 z+6}{z^{2}-4}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow-3}\left[\frac{x+3}{x^{2}+4 x+3}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{5 y^{3}+8 y^{2}}{3 y^{4}-16 y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^{3}+2 x^{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{u \rightarrow 1}\left[\frac{u^{4}-1}{u^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^{3}-27}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow-2}\left[\frac{x^{7}+x^{5}+160}{x^{3}+8}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^{3}}{y-4 y^{3}}\right]\)
Solution:

Question 3.
\(\lim _{v \rightarrow \sqrt{2}}\left[\frac{v^{2}+v \sqrt{2}-4}{v^{2}-3 v \sqrt{2}+4}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 3}\left[\frac{x^{2}+2 x-15}{x^{2}-5 x+6}\right]\)
Solution:

Maharashtra State Board 11th Commerce Maths

• Limits Ex 7.1 11th Commerce Maths
• Limits Ex 7.2 11th Commerce Maths
• Limits Ex 7.3 11th Commerce Maths
• Limits Ex 7.4 11th Commerce Maths
• Limits Miscellaneous Exercise 7 11th Commerce Maths
• Continuity Ex 8.1 11th Commerce Maths
• Continuity Miscellaneous Exercise 8 11th Commerce Maths
• Differentiation Ex 9.1 11th Commerce Maths
• Differentiation Ex 9.2 11th Commerce Maths
• Differentiation Miscellaneous Exercise 9 11th Commerce Maths

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 Notes 

• Chapter 5 Integration Ex 5.1 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.2 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.3 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.4 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.5 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Ex 5.6 Class 12 Commerce Maths Textbook Solutions
• Chapter 5 Integration Miscellaneous Exercise 5 Class 12 Commerce Maths Textbook Solutions

Limits Class 11 Commerce Maths 1 Chapter 7 Exercise 7.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

Std 11 Maths 1 Exercise 7.1 Solutions Commerce Maths

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{x+6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{5}-3125}\right]\)
Solution:

Question 4.
If \(\lim _{x \rightarrow 1}\left[\frac{x^{4}-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^{3}-a^{3}}{x-a}\right]\), find all possible values of a.
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]\)
Solution:

Question 2.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]=500\), find all possible values of k.
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

Question 3.
\(\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{2}-25}\right]\)
Solution:

Maharashtra State Board 11th Commerce Maths

• Limits Ex 7.1 11th Commerce Maths
• Limits Ex 7.2 11th Commerce Maths
• Limits Ex 7.3 11th Commerce Maths
• Limits Ex 7.4 11th Commerce Maths
• Limits Miscellaneous Exercise 7 11th Commerce Maths
• Continuity Ex 8.1 11th Commerce Maths
• Continuity Miscellaneous Exercise 8 11th Commerce Maths
• Differentiation Ex 9.1 11th Commerce Maths
• Differentiation Ex 9.2 11th Commerce Maths
• Differentiation Miscellaneous Exercise 9 11th Commerce Maths

Determinants Class 11 Commerce Maths 1 Chapter 6 Miscellaneous Exercise 6 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Miscellaneous Exercise 6 Questions and Answers.

Std 11 Maths 1 Miscellaneous Exercise 6 Solutions Commerce Maths

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
Solution:

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)
Solution:

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Question 2.
Find the value(s) of x, if
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
∴ 1(-10x² + 10x) – 4(5x² + 5) + 20(2x + 2) = 0
∴ -10x² + 10x – 20x² – 20 + 40x + 40 = 0
∴ -30x² + 50x + 20 = 0
∴ 3x² – 5x – 2 = 0 ……[Dividing throughout by (-10)]
∴ 3x² – 6x + x – 2 = 0
∴ 3x(x – 2) + 1(x – 2) = 0
∴ (x – 2) (3x + 1) = 0
∴ x – 2 = 0 or 3x + 1 = 0
∴ x = 2 or x = \(-\frac{1}{3}\)

(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
∴ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
∴ 1(-12) – 2x(-15) + 4x(-3) = 0
∴ -12 + 30x – 12x = 0
∴ 18x = 12
∴ x = \(\frac{2}{3}\)

Question 3.
By using properties of determinants, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0\).
Solution:

Question 4.
Without expanding the determinants, show that

Solution:

Question 5.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
Solution:
Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0
D_x = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40
D_y = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) – 1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80
D_z = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,
x = \(\frac{D_{x}}{D}=\frac{-40}{-40}\) = 1
y = \(\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}\) = 2
z = \(\frac{\mathrm{D}_{z}}{\mathrm{D}}=\frac{-40}{-40}\) = 1
∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

(ii) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-2\), \(\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3\), \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=-1\)
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
The given equations become
p + q + r = -2
p – 2q + r = 3
2p – q + 3r = -1
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(-6 + 1) – 1(3 – 2) + 1(-1 + 4)
= -5 – 1 + 3
= -3
D_p = \(\left|\begin{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
\end{array}\right|\)
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
D_q = \(\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
D_r = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
\end{array}\right|\)
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,
p = \(\frac{\mathrm{D}_{\mathrm{p}}}{\mathrm{D}}=\frac{-5}{-3}=\frac{5}{3}\)
q = \(\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}\) = 2
r = \(\frac{D_{2}}{D}=\frac{-40}{-40}\) = 1
∴ x = \(\frac{3}{5}\), y = \(\frac{-3}{5}\), z = \(\frac{-1}{2}\) are the solutions of the given equations.

(iii) x – y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
Given equations are
x – y + 2z = 1
3x + 4y – 5z = 5
2x – y + 3z = 12
D = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\)
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0
D_x = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7)+ 1(75) +2(-53)
= 49 + 75 – 106
= 18
D_y = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6
D_z = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53)+ 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,
x = \(\frac{\mathrm{D}_{x}}{\mathrm{D}}=\frac{18}{4}=\frac{9}{2}\)
y = \(\frac{D_{y}}{D}=\frac{-6}{4}=\frac{-3}{2}\)
z = \(\frac{D_{z}}{D}=\frac{2}{4}=\frac{1}{2}\)
∴ x = \(\frac{9}{2}\), y = \(\frac{-3}{2}\) and z = \(\frac{1}{2}\) are the solutions of the given equations.

Question 6.
Find the value(s) of k, if the following equations are consistent.
(i) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
Solution:
Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3 i.e. 2x – y – 3 = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0\)
∴ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
∴ 3(-9) – 1 (-3k + 6) – 2(-k – 4) = 0
∴ -27 + 3k – 6 + 2k + 8 = 0
∴ 5k – 25 = 0
∴ k = 5

(ii) kx + 3y + 4 = 0, x + ky + 3 = 0, 3x + 4y + 5 = 0
Solution:
Given equations are
kx + 3y + 4 = 0
x + ky + 3 = 0
3x + 4y + 5 = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{lll}
\mathrm{k} & 3 & 4 \\
1 & \mathrm{k} & 3 \\
3 & 4 & 5
\end{array}\right|=0\)
∴ k(5k – 12) – 3(5 – 9) + 4(4 – 3k) = 0
∴ 5k² – 12k + 12 + 16 – 12k = 0
∴ 5k² – 24k + 28 = 0
∴ 5k² – 10k – 14k + 28 = 0
∴ 5k(k – 2) – 14(k – 2) = 0
∴ (k – 2) (5k – 14) = 0
∴ k – 2 = 0 or 5k – 14 = 0
∴ k = 2 or k = \(\frac{14}{5}\)

Question 7.
Find the area of triangles whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
Solution:
Here, A(x₁, y₁) ≡ A(-1, 2), B(x₂, y₂) ≡ B(2, 4), C(x₃, y₃) ≡ C(0, 0)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-1 & 2 & 1 \\
2 & 4 & 1 \\
0 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-1(4 – 0) – 2(2 – 0) + 1(0 – 0)]
= \(\frac{1}{2}\) (-4 – 4)
= \(\frac{1}{2}\) (-8)
= -4
Since, area cannot be negative.
∴ A(∆ABC) = 4 sq.units

(ii) P(3, 6), Q(-1, 3), R(2, -1)
Solution:
Here, P(x₁, y₁) ≡ P(3, 6), Q(x₂, y₂) ≡ Q(-1, 3), R(x₃, y₃) ≡ R(2, -1)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & 6 & 1 \\
-1 & 3 & 1 \\
2 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(3 + 1) – 6(-1 – 2) + 1(1 – 6)]
= \(\frac{1}{2}\) [3(4) – 6(-3) + 1(-5)]
= \(\frac{1}{2}\) (12 + 18 – 5)
∴ A(∆PQR) = \(\frac{25}{2}\) sq.units

(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
Here, L(x₁, y₁) ≡ L(1, 1), M(x₂, y₂) ≡ M(-2, 2), N(x₃, y₃) ≡ N(5, 4)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(∆LMN) = \(\frac{1}{2}\left|\begin{array}{rrr}
1 & 1 & 1 \\
-2 & 2 & 1 \\
5 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [1(2 – 4) -1(-2 – 5) + 1(-8 – 10)]
= \(\frac{1}{2}\) [1(-2) – 1(-7) + 1(-18)]
= \(\frac{1}{2}\) (-2 + 7 – 18)
= \(\frac{-13}{2}\)
Since, area cannot be negative.
∴ A(∆LMN) = \(\frac{13}{2}\) sq.units

Question 8.
Find the value of k,
(i) if the area of ∆PQR is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
Solution:
Here, P(x₁, y₁) ≡ P(k, 0), Q(x₂, y₂) ≡ Q(4, 0), R(x₃, y₃) ≡ R(0, 2)
A(∆PQR) = 4 sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ ±4 = \(\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
∴ ±4 = \(\frac{1}{2}\) [k(0 – 2) – 0 + 1(8 – 0)]
∴ ±8 = -2k + 8
∴ 8 = -2k + 8 or -8 = -2k + 8
∴ -2k = 0 or 2k = 16
∴ k = 0 or k = 8

(ii) if area of ∆LMN is \(\frac{33}{2}\) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
Here, L(x₁, y₁) ≡ L(3, -5), M(x₂, y₂) ≡ M(-2, k), N(x₃, y₃) ≡ N(1, 4)
A(∆LMN) = \(\frac{33}{2}\) sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
-2 & \mathrm{k} & 1 \\
1 & 4 & 1
\end{array}\right|\)
∴ ±\(\frac{33}{2}\) = \(\frac{1}{2}\) [3(k – 4) – (-5) (-2 – 1) + 1(-8 – k)]
∴ ±33 = 3k – 12 – 15 – 8 – k
∴ 33 = 2k – 35
∴ 2k – 35 = 33 or 2k – 35 = -33
∴ 2k = 68 or 2k = 2
∴ k = 34 or k = 1