2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem

Students can Download Basic Maths Question Bank Chapter 4 Binomial Theorem Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem

2nd PUC Basic Maths Binomial Theorem Two Marks Questions and Answers

Question 1.
Find the 5^th term in \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\)
Answer:

Question 2.
Find the 6th term in \((\sqrt{x}-\sqrt{y})^{17}\)
Answer:

Question 3.
Find the 8th term in \(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\)
Answer:

Question 4.
Find the 7th term \(\left(3 x^{2}-\frac{y}{3}\right)^{9}\)
Answer:

Question 5.
Find the 10^th term in the Expansion of \(\left(\frac{a}{b}-\frac{2 b}{a^{2}}\right)^{12}\)
Answer:

2nd PUC Basic Maths Binomial Theorem Three or Four Marks Questions and Answers

A Binomial Expansion Calculator that doesn’t require any scripting in your browser. Just enter your values and compute.

Question 1.
Expand (1 – 2x)^s using binomial theorem. x = 1, a = – 2x, n = 5.
Answer:
(1 + (-2x))⁵ = 1⁵ + ⁵C₁ . I⁴ (-2x) + ⁵C₂. I³ (-2x)² + ⁵C₃. I²(-2x)³ + ⁵C₄ . 1(-2x)⁴ + ⁵C₅(-2x)⁵
= 1+5 (-2x) + 10(-2x)² + 10 (-2x)³+ 5 (-2x)⁴ + (-2x)⁵
= 1 – 10x + 10 C4x² + 10(-8x³) + 5 (16x⁴) + (-32x⁵)
= 1 – 10x + 40x²– 80x³ + 80x⁴– 32x⁵

Question 2.
Expand
Answer:

Question 3.
Simplify : \((2+\sqrt{5})^{4}+(2-\sqrt{5})^{4}\)
Answer:

Question 4.
Simplify : \((3+\sqrt{2})^{4}-(3-\sqrt{2})^{4}\) using Binomial theorem.
Answer:

Question 5.
Simplify  \((\sqrt{2}+1)^{6}-(\sqrt{2}-1)^{6}\)
Answer:

Question 6.
Expand (0.99)⁵ by Binomial theorem correctly decimal places.
Answer:
(0.99)⁵ = (1- 0.01)⁵=1⁵+5C₁.1⁴.(- 0.01) + 5C₂.1³(-0.01)²+5C₃.1²(-0.01)³+5C₄.1¹(-0.01)⁴+  ⁵C₅ ( – 0:01 )⁵

Question 7.
Expand (0.98)^s by Binomial theorem.
Answer:
(0.98)⁵ = (1 – 0.02)⁵
= 1⁵ + 5C_r 1⁴ (- 0.02) + 5C_r 1³( – 0.02)² + 5C_r 1²( – 0.02)³ + 5C₄. 1(- 0.02)⁴ + 5C₅(- 0.02)⁵ = 1 – 5(0.02) + 10(0.0004) – 10 (0.000008) + 5(0.00000016) – 0.0000000032
= 1-0.1 + 0.004 – 0.00008 + 0.00000080 – 0.0000000032
= 1.0040008 – 0.1000800032 = 0.9039.

Question 8.
Find the value of (1.02)⁴ using the binomial theorem.
Answer:
(1.02)⁴ =(1 +0.02)⁴
= 1⁴ + 4C₁.1³(0.02) + 4C₂.1²(0.02)² + 4C_r 1(0.02)³ + 4C₄(0.02)⁴
= 1 +4 (0.02) + 6 (0.0004) + 4 (0.000008) + 1 .(0.00000016)
= 1 + 0.08 + 0.0024 + 0.000032 + 0.00000016
= 1.08243216.

Question 9.
Write down the middle term in the expansion of \(\left(2 x^{2}-\frac{1}{x}\right)^{12}\)
Answer:
Here n = 12,
∴ There are 13 terms. The middle terms is \(\frac{12}{2}+1\) = 7^th terms
To find 7th term put r = 6

Question 10.
Find the middle term of the Expansion of \(\left(\frac{1}{x^{2}}+\frac{x}{2}\right)^{10}\)
Answer:
Here n = 10, There are 11 terms.
∴  Middle term = \(\frac{10}{2}\)+1 = 6 terms
To find 6th term put r = 5

Expanding binomials calculator using the binomial expansion method step-by-step.

Question 11.
Find the 5^th term in the expansion of \(\left(2-\frac{1}{x}\right)^{3}\)
Answer:

Question 12.
Find the 6th term in the expansion of \(\left(3+\frac{2}{x}\right)^{10}\)
Answer:

Question 13.
Find the value of (10.1)⁵ up to 4 decimal places, using binomial theorem
Answer:
(10 + 0.1)⁵= 10⁵ + 5C_r10⁴.(0.1) + 5C₂.10³. (0.1)² + 5C_r10².(0.1)³ + 5C₄.10.(0.1)⁴+5C₅(0.1)⁵

Question 14.
S.T there is no term containing x³ in the expression \(\left(3 x-\frac{1}{2 x}\right)^{8}\)
Answer:

To find the term containg x³ equate the power of x to 3
∴ 8 – 2r = 3 ⇒ 5 = 2r = s.r = r = 5 /2
Since r is a fraction there is no term containing x³.

Question 15.
S.T. There is no term independent of x in the expansion of \(\left(2 x-\frac{1}{x^{2}}\right)^{10}\)
Answer:

To find the terms independent of x equate the power of x to O, we get 10 – 3r = 0
⇒ 10 = 3r ⇒ r = \(\frac{10}{3}\) is a fraction. Hence there is no terms independent of x.

Question 16.
Find the co-efficient of x¹ in the expression of \(\left(x^{2}+\frac{2}{x}\right)^{11}\)
Answer:

To find the coeffecint of x¹ equate the power of x to 7
∴ 22 – 3r = 7 ⇒ 22 -1 = 3r ⇒ 15 – 3r  ⇒ r = 5.
∴ Co-efficient of x⁷ is ¹¹C₅.2⁵.

Question 17.
\(\left(\frac{a}{x}+b x\right)^{12}\)
Answer:
∵  n = 12, we have 13 terms,
∴ 7^th term is the middle tem
To find 7th term put r = 6

Question 18.
\(\left(\frac{a}{3}+\frac{b}{2}\right)^{8}\)
Answer:
Here n = 8,
∵  we have only one middle term, i.e. 5^th terms.
To find 5^th term put r = 4

Question 19.
\((3+\sqrt{2})^{6}-(3-\sqrt{2})^{6}\)
Answer:

Question 20.
Find x if the 17th and 18th terms of the expansion (2 + x)⁵⁰ are equal.
Answer:
17^th term = T_{16 + 1} = ⁵⁰C₁₇ . 2^{50 – 16}. x¹⁶ = ⁵⁰C₁₆ . 2³⁴. x¹⁶
18^th term = T_{17 + 1} = ⁵⁰C₁₇ . 5^{50 – 17}. x¹⁷ = ^50C₁₇ . 2³³ . x¹⁷

Question 21.
If 2^nd 3^rd and 4^th terms of (x + y)ⁿ are 108,54,12 respectively. Find the value of x,y and n.
Answer:

Question 22.
If the coefficient of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal then find r.
Answer:
T_r+1 = ¹⁸Cr . (1)^{18 – r} . x^r
= ¹⁸Cr. x^r
Here The coefficient of (r + 1 )^th term is ¹⁸C_r
Therefore the coefficient of (2r + 4))^th term is ¹⁸C_2r+J
Therefore the coefficient of (r – 2))^th term is ¹⁸C_{r – 3}

Question 23.
Find the middle terms in the expansion of
(i) \(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
∵ n = 15, we have 2 middle terms i.e., 8^th and 9^th terms,
(a) To find 8^th term put r = 7

Question 24.
\(\left(\frac{x}{2}+\frac{3}{x^{2}}\right)^{19}\)
Answer:
n = 19, (odd no.)
We have two middle terms i.e., 10^th and 11^th terms,

Question 25.
\(\left(2 x^{2}+\frac{1}{\sqrt{x}}\right)^{11}\)
Answer:
n = 19, (odd no.). We have two middle terms i.e., 10^th and 11^th terms,

Question 26.
Find the coefficients of
(a) \(x^{11} \text { in }\left(x+\frac{2}{x^{2}}\right)^{17}\)
Answer:

Question 27.
Find the coefficient of y² in \(\left(7 y^{2}-\frac{2}{y}\right)^{12}\)
Answer:

Question 28.
Find the coefficient of y¹¹ in \(\left(\sqrt{x}-\frac{2}{x}\right)^{17}\)
Answer:

Question 29.
To find coefficient of x^-11, equate the power of x, to -11
Answer:

Question 30.
Find the term independent of x in.
(i) \(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Answer:

To find the term independent of x, equal the power of x to 0
18 – 3r = 0 ⇒ r = 6

Question 31.
\(\left(x^{3}-\frac{3}{x^{2}}\right)^{15}\)
Answer:

To find the term independent of x, equale the power of x to 0
45 – 5r = 0 ⇒ r = 9

Question 32.
\(\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}\)
Answer:

To find the terms independent of x , Equale the power of x to 0

Question 33.
The first 3 terms in (1 + ax)n where n is a positive integer are 1,6x, 6×2. Find the values of a and n.
Answer:

Question 34.
In the expansion of (3 + Kx)⁹, the coefficients of x² and x³ are Equal. Find K.
Answer:

Question 35.
Find the ratio of the coefficients of x⁴ in the two expansion (1 + x)⁷ and (1 + x)¹⁰.
Answer: