In this question paper, some of the important one mark, two and five marks questions from the chapter Algebra are covered. The questions are prepared from the book back and previous year questions.
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Questions:

Generic 3

8 x 3 = 24

Let p(n) be the statement "n²+ n is even". If P(k) is true, then show that P(k+1) is true.

If p(n) is the statement "12n + 3" is a multiple of 5, then show that P (3) is false, whereas P(6) is true.

In how many ways can n prizes be given to n boys, when a boy may receive any number of prizes?

If the letters of the word are arranged as in dictionary, find the rank of the word "AGAIN".

Solve : \(\frac { (2x+1)! }{ (x+2)! } .\frac { (x-1)! }{ (2x-1)! } =\frac { 3 }{ 5 } \)

Resolve into partial factors:\(\frac { x+4 }{ ({ x }^{ 2 }-4)(x+1) } \)

Expand the following by using binomial theorem.\({ \left( x+\frac { 1 }{ { x }^{ 2 } } \right) }^{ 6 }\)

Expand the following by using binomial theorem.\(\left( x+\frac { 1 }{ y } \right) ^{ 7 }\)

Generic 5

3 x 5 = 15

Resolve into partial fractions for the following : \(\frac{x+2}{(x-1)(x+3)^2}\)

10.

Resolve into partial fractions for the following : \(\frac{x^2-6 x+2}{x^2(x+2)}\)

11.

Resolve into partial fractions for the following : \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)

Multiple Choice Question

10 x 1 = 10

12.

Thirteen guests has participated in a dinner. The number of handshakes happened in the dinner is __________.

(a)

715

(b)

(c)

286

(d)

13.

The number of ways to arrange the letters of the word "CHEESE" is ______.

(a)

120

(b)

240

(c)

720

(d)

14.

The value of (5C₀ + 5C₁) + (5C₁ + 5C₂) + (5C₂ + 5C₃) + (5C₃ + 5C₄) + (5C₄ + 5C₅ ) is ________.

(a)

2⁶-2

(b)

2⁵-1

(c)

2⁸

(d)

2⁷

15.

There are 10 true or false questions in an examination. Then these questions can be answered in ______.

(a)

240 ways

(b)

120 ways

(c)

1024 ways

(d)

100 ways

16.

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is _________.

(a)

(b)

(c)

(d)

17.

For all n > 0, nC₁ + nC₂ + nC₃ + ... +nC_n is equal to _______

(a)

(b)

2ⁿ- 1

(c)

n²

(d)

n² - 1

18.

If n is a positive integer, then the number of terms in the expansion (x + a)ⁿ is _______.

(a)

(b)

n + 1

(c)

n-1

(d)

19.

The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for n \(\in\) N is ________.

(a)

(b)

(c)

(d)

20.

The number of ways selecting 4 players out of 5 is _______.

(a)

(b)

(c)

(d)

21.

The value of n, when nP₂ = 20 is _______.

(a)

(b)

(c)

(d)

Generic 2

3 x 2 = 6

22.

In how many ways can 10 beads of different colours form a necklace?

23.

How many permutations can be made out of the letters of the word "TRIANGLE" beginning with T?

24.

Show that 10P₃ = 9 P₃ + 3. 9P₂

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Answers Key:

Generic 3

8 x 3 = 24

P(n): "n² + n is even"
Given that P(k) is true.
\(\Rightarrow \) k² + k is even
\(\Rightarrow \) k² + k = 2\(\lambda \) for some \(\lambda \) ....(1)
To prove that P (k + 1) is true.
P (k + 1) : (k + 1)²+ (k + 1) is even
Consider (k + 1)² + (k + 1)
= k²+2k+ 1 +k+ 1
= k²+2k+ 1 +k+ 1
= (k²+ k) + (2 k + 2)
= 2\(\lambda \) + 2 (k + 1) [from (1)]
= even + even
= Sum of two even numbers is always an even number .
∴ P (k + 1) is true.

P(n): "12n + 3" is a multiple of 5.
P(3): 12 (3) + 3 36 + 3 = 39 is not a multiple of 5.
∴ P (3) is false
P(6) = 12(6) + 3 = 72 + 3 = 75 = 5 x 15 is a multiple of 5.
∴ P (6) is true

No. of ways of giving I prize = n
No. of ways of giving II prize = n
.............................................
Similarly no. of ways of giving n^th prize = n
Therefore By fundamental principle of counting, the required number of ways

In "AGAIN" the letters in ascending order are A, A, G, I, N
∴ The first word is AAGIN
Number of words beginning with A A
= No of ways of arranging G, 1, N = 3! = 6
The next word begin with AG and it is AGAIN
∴ No. of words before AGAIN = 6.
∴ Rank of word AGAIN = 7

\({{(2x+1)!}\over{(x+2)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\) \({{(2x+1)(2x)(2x-1)!}\over{(x+2)(x+1)(x-1)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\) \({{(2x+1)(2)}\over{(x+2)(x+1)}}={{3}\over{5}}\)
\(\Rightarrow\) 10 (2x+1) = 3 (x+2) (x+1)
\(\Rightarrow\) 20x+10=3 (x²+3x+2)
\(\Rightarrow\) 20x+10+3x²-9x-6=0
\(\Rightarrow\) -3x²+11x+4 = 0
\(\Rightarrow\) -3x²-11x-4=0
\(\Rightarrow\) (x-4)(3x+1)=0
\(\Rightarrow\) x = 4 or x \(={{-1}\over{3}}\)
Since \(x={{-1}\over{3}}\) is not possible, x = 4.

\({{x+4}\over{(x^2-4)(x+1)}}={{x+4}\over{(x+2)(x-2)(x+1)}}={{A}\over{x+2}}+{{B}\over{x-2}}+{{C}\over{x+1}}\)
\(\Rightarrow\)  \({{x+4}\over{(x^2-4)(x+1)}}=\frac { { { A(x-2)(x+1)B+(x+2)(x+1)+C(x+2)(x-2) } } }{ (x+2)(x-2)(x+1) } xx\)
\(\Rightarrow\) x + 4 = A (x - 2) (x + 1) + B (x + 2) (x + 1) C (x +2)(x - 2) ...(1)
Putting x = 2 in (1) we get,
6 = B (4) (3)  \(\Rightarrow\boxed{B={{1}\over{2}}}\)
Putting x = -1 in (1) we get,
3 = C (1)(-3)  \(\Rightarrow\quad \boxed{C=-1}\)
Putting x = 0 in (1) we get,
4 = - 2A + 2B - 4C
\(\Rightarrow\) \(4=-2A+2\left( {{1}\over{2}} \right)-4(-1)\) \(\left[ \because B={{1}\over{2}},C=-1 \right]\)
\(\Rightarrow\) \(4=-2A+1+4\ \ \Rightarrow\ 4=-2A+5\ \Rightarrow -1\) =-2A
\(\Rightarrow\) \(\boxed{A={{1}\over{2}}}\)
\(\therefore\) \({{x+4}\over{(x^2-4)(x+1)}}={{{{1}\over{2}}}\over{x+2}}+{{{{1}\over{2}}}\over{x-2}}-{{1}\over{}x+1}={{1}\over{2(x+2)}}+{{1}\over{2(2x-2)}}-{{1}\over{x+1}}\)

\(\left(x+\frac{1}{x^2}\right)^6 = 6 C_0 x^3+6 C_1 x^5\left(\frac{1}{x^2}\right)+6 C_2 x^4\left(\frac{1}{x^2}\right)+6 C_3 x^3\left(\frac{1}{x^2}\right)^3 +6 C_4 x^2\left(\frac{1}{x^2}\right)^4+6 C_5 x\left(\frac{1}{x^2}\right)^5+6 C_6\left(\frac{1}{x^2}\right)^6 \)
\(=x^6+6 x^3+15 x^4\left(\frac{1}{x^4}\right)+20 x^3\left(\frac{1}{x^6}\right)+15 x^2\left(\frac{1}{x^8}\right)+6 x\left(\frac{1}{x^{10}}\right)+\frac{1}{x^{12}}\)
\(=x^6+6 x^3+15+\frac{20}{x^3}+\frac{15}{x^6}+\frac{6}{x^9}+\frac{1}{x^{12}}\)

\(\left(x+\frac{1}{y}\right)^7 =7 C_0 x^7+7 C_1 x^6\left(\frac{1}{y}\right)+7 C_2 x^5\left(\frac{1}{y}\right)^2 +7 C_4 x^3\left(\frac{1}{y}\right)^4+7 C_5 x^2\left(\frac{1}{y}\right)^5 +7 C_6 x\left(\frac{1}{y}\right)^6+7 C_7\left(\frac{1}{y}\right)^7\)
\(=x^7+\frac{7 x^6}{y}+\frac{21 x^5}{y^2}+\frac{35 x^4}{y^3} +\frac{35 x^3}{y^4}+\frac{21 x^2}{y^5}+\frac{7 x}{y^6}+\frac{1}{y^7}\)

Generic 5

3 x 5 = 15

\(\frac{x+2}{(x-1)(x+3)^2}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^2}\)
\(=\frac{A(x+3)^2+B(x-1)(x+3)+C(x-1)}{(x-1)(x+3)^2}\)
\(\Rightarrow\) x + 2 = A ( x + 3 )² + B ( x - 1 ) ( x + 3 ) + C( x - 1 ) ....(1)
Putting x = 1 in (1) we get,
\(3=A(4)^2 \Rightarrow A=\frac{3}{16}\)
Puttingx = -3 in (1)we get
\(-1=C(-3-1)\)
\(C=\frac{1}{4}\)
Equate co-efficient of x² on both sides of (1)
\(0 =A+B \Rightarrow B=-A=\frac{-3}{16} \)
\(\frac{x+2}{(x-1)(x+3)^2} =\frac{3}{16(x-1)}-\frac{3}{16(x+3)}+\frac{1}{4(x+3)^2}\)

10.

\(\frac { { x }^{ 2 }-6x+2 }{ { x }^{ 2 }(x+2) } =\frac { A }{ x } +\frac { B }{ { x }^{ 2 } } +\frac { C }{ x+2 } \)
⇒ \(\frac { { x }^{ 2 }-6x+2 }{ { x }^{ 2 }(x+2) } =\frac { Ax(x+2)+B(x+2)+Cx^{ 2 } }{ { x }^{ 2 }(x+2) } \)
⇒ x² - 6x + 2 = Ax (x + 2) + B (x + 2)+ Cx² ..(1)
x = -2 in (1) we get,
4 + 12+ 2 = \(C(-2)^2\) = C(4)
⇒ 18 = 4C ⇒ C = \(\frac { 18 }{ 4 } =\frac { 9 }{ 2 } \)
x = 0 in (1) we get,
2 = B (2) ⇒ B = 1
Equating co-efficient of x² on both sides of (1)
1 = A + C ⇒ = A + \(\frac { 9 }{ 2 } \) [∵ C=\(\frac { 9 }{ 2 } \)]
⇒ A = \(1-\frac { 9 }{ 2 } =\frac { 2-9 }{ 2 } =\frac { -7 }{ 2 } \)
\(\frac{x^2-6 x+2}{x^2(x+2)}=\frac{-7}{2 x}+\frac{1}{x^2}+\frac{9}{2(x+2)}\).

11.

\(\frac { { x }^{ 2 }-3 }{ (x+2)({ x }^{ 2 }+1) } =\frac { A }{ x+2 } +\frac { Bx+C }{ { x }^{ 2 }+1 } \)
[since x² + 1 cannot be factorised into linear factors]
⇒ \(\frac { { x }^{ 3 }-3 }{ (x+2)({ x }^{ 2 }+1) } =\frac { A({ x }^{ 2 }+1)+(Bx+C)(x+2) }{ (x+2)({ x }^{ 2 }+1) } \)
⇒ x² - 3 = A(x² + 1) + (Bx + C) (x + 2) ...(1)
Putting x = -2 in (1) we get
4 - 3 = A(4 + 1) ⇒ 1 ⇒ 5A ⇒ \(A=\frac{1}{5}\)
Equate co-efficient of x² on both sides of (1)
1 = A + B
\(1=\frac{1}{5}+B\)
\(B=1-\frac{1}{5}=\frac{4}{5}\)
Putting x = 0 in (1) we get,
\(-3=A+2 C\)
\(-3=\frac{1}{5}+2 C\)
\(2 C=-3-\frac{1}{5}=-\frac{16}{5}\)
\(C=-\frac{8}{5}\)
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{\frac{4 x}{5}-\frac{8}{5}}{x^2+1}\)
\(=\frac{1}{5(x+2)}+\frac{4(x-2)}{5\left(x^2+1\right)}\).

Multiple Choice Question

10 x 1 = 10

12.

(b)
78

13.

The number of ways = 6!/3! = 120

14.

Since sum of all binomial coefficients is 2n
5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅+ (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (2⁵ - (5C₀ + 5C₅)
= 32 + 32 - (1 + 1)
= 64 - 2 = 2⁶ - 2

15.

210 = 1024 (2.x 2... 10 terms)

16.

The number of parallelograms = 4C₂ x 3C₂

17.

nC₀+ nC₁ + nC₂ + nC₃ + ... +nC_n = 2n and nC₀ = 1

18.

(b)
n + 1

19.

Since if n = 1 then (1) (2) (3) (4) = 24 is divisible by = 24

20.

No. of ways = 5C₄ = 5C₁ = 5

21.

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

Generic 2

3 x 2 = 6

22.

Number of ways of forming necklace
\(\frac { (10-1)! }{ 2! } =\frac { 9! }{ 2 } =181440\)

23.

The first place can be filled in only one way namely T and the remaining 7 letters can be arranged in 7! ways.
\(\therefore\) Total number of arrangements = 1 x 7! = 5040.

24.

LHS 10P₃ = 10 x 9 x 8 = 720
RHS 9P₃ + 3. 9P₂ = 9 x 8 x 7 + 3 x 9 x 8
= 9 x 8 (7 + 3) = 72 (10) = 720
LHS= RHS Hence proved.