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(2) \(\frac{-7}{4}\)

Let Raman’s age = x 

Let the sum of his two sons age = y 

Now x = 3y ⇒ x – 3y = 0 … (1) 

After 5 years, 

Step (1) 

x + 5 = 2(y + 10) 

x + 5 = 2y + 20 

x – 2y = 20 – 5 

x – 2y = 15 

Step (2) 

From equation (1) x = 3y 

Step (3) 

Substitute x = 3y in (2) 

3y – 2y = 15

y = 15 

Step (4) 

Substitute y = 15 in (1) 

x = 3y = 3 x 15 

x = 45 

∴ Raman’s age is 45 years.

(3) 1, 2, -15 

p(-5) = a (-5²) + b (-5) + c = 25a – 5b + c = 0 ... (1)

p(3) = a (3²) + bc + 3 + c = 9 + 3b + c = 0 … (2) 

25a – 5b = 9a + 3b 

25a – 9a = 3b + 5b 

16a = 8 b 

\(\frac{a}{b}=\frac{8}{16}=\frac{1}{2}\)

Substitute a = 1, b = 2 in (1) 

25(1) – 5 (2) = -c 

25 – 10 = 15 = – c 

c = -15

(4) 50 

P(-1) = (-1)⁵¹ + 51 = -1 + 51 = 50

(2) \(-\frac{5}{2}\)

(2) x = \(\frac{1}{3}\)

x = \(\frac{1}{3}\) gives 3(\(\frac{1}{3}\)) - 1 = 1 - 1 = 0

(4) cubic polynomial. 

Polynomial of degree 3 is called cubic.

Answer is (3) 3

(1) \(\frac{5}{2}\)

Zero is given by 2x - 5 = 0 

x = \(\frac{5}{2}\)

(2) quadratic polynomial

A polynomial is quadratic of its highest power of x is 2

(1) linear polynomial

A polynomial is linear if its degree is 1

(3) 3x + 5 = \(\frac{2}{3}\)

x + \(\frac{1}{x}\) = 2 ⇒ x² – 2x + 1 = 0; x(x – 1) = 2 ⇒ x² - x – 2 = 0

(2) -2 

2x + 3y = m, x = 2, y = – 2

m = 2(2) + 3(-2) 

= 4 – 6 = -2

(2) (4, 2) 

2x – y = 6 

2(4) – 2 = 8 – 2 = 6 = RHS

(4) 13

2x + 3y = k 

2(2) + 3(3) = 4 + 9 = 13

(3) a = 0, b = 0, c ≠ 0 

a = 0, b = 0, c ≠ 0 ⇒ (0)x + (0) y + c = 0 

False

(1) k = 3

4x + 6y = 1 

6y = -4x + 1 

y = \(\frac{-4}{6}x+\frac{1}{6}\) … (1) 

2x + ky – 7 = 0 

ky = -2x + 7 

y = \(\frac{-2}{k}x+\frac{7}{k}\) ... (2) 

Since the lines (1) and (2) parallel m₁ = m₂

\(\frac{-4}{6}=\frac{-2}{k}\)

k = -2 x \(\frac{-6}{4}\) = 3

(2) 0x + 0y + c = 0 

a and b both can not be zero

(b) 78

To handshakes, we need two guests. 

Number of selecting 2 guests from 13 is 13C₂

= \(\frac{13\times12}{2\times1}\) = 78

5y² (x²y³ – 2x⁴y + 10x²) – [(-2)(xy)² (y³ + 7x²y + 5)]

= [5y² (x²y³) – 5y² (2x⁴y) + 5y² (10x²)] – [(-2)x²y² (y³ + 7x²y + 5)]

= (5y⁵5x² – 10x⁴y³ + 50x²y²)

= 5x²y⁵ – 10x⁴y³ + 50x²y² – [(-2x²y⁵) – 14x⁴y³ – 10x²y²]

= 5x²y⁵ – 10x⁴y³ + 50x²)y² + 2x² y⁵ + 14x⁴)y³ + 10x²)y²

= (5 + 2)x²y⁵ + (-10 + 14)x⁴y³ + (+50 + 10)x²y²

= 7x²y⁵ + 4x⁴y³ + 60x²y²

Length of the rectangular hall be l 
Given, breadth =b  and length =3b−4
∴l=3b−4

Let    P(x) = 4x² -12x + 9

=  4x² - 6x + 9

or,  p(x)  =  2x(2x - 3)  -3(2x  - 3)

 :.    0  =  (2x - 3) (2x - 3)

The zeros are    3 / 2, 3 / 2

:.    x = 3 / 2, 3 / 2

Hence, zeroes of the polynomial  are 3 / 2, 3 / 2.

(i) (x + a)ⁿ = nC₀ xⁿa⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + … + nC_rx^n-r a^r + … + nC_n aⁿ 

(101)⁴ = (100 + 1)⁴ = 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4C₃ (100)¹ (1)³ + 4C₄ (1)⁴ 

= 1 x (100000000) + 4 x (1000000) + 6 x (10000) + 4 x 100 + 1 x 1 

= 100000000 + 4000000 + 60000 + 400 + 1 

= 10,40,60,401

(ii) (x + a)ⁿ = nC₀ xⁿa⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + … + nC_rx^n-r a^r + … + nC_n aⁿ  

(999)⁵ = (1000 – 1)⁵ = 5C₀ (1000)⁵ – 5C₁ (1000)⁴ (1)¹ + 5C₂ (1000)³ (1)² – 5C₃ (1000)² (1)³ + 5C₄ (1000) (1)⁴ – 5C₅ (1)⁵ 

= 1(1000)⁵ – 5(1000)⁴ – 10(1000)³ – 10(1000)² + 5(1000) – 1 

= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1 

= 995009990004999

Given that ⁿP₄ = 12(ⁿP₂) 

n(n – 1) (n – 2) (n – 3) = 12n(n – 1) 

Cancelling n(n – 1) on both sides we get 

(n – 2) (n – 3) = 4 x 3 

We have product of consecutive number on both sides with decreasing order. 

n – 2 = 4 

∴ n = 6

5 boys can be seated among themselves in ⁵P₅ = 5! Ways. After this arrangement, we have to arrange the three girls in such a way that in between two girls there atleast one boy. So the possible places girls can be placed with the x symbol given below. 

x B x B x B x B x B x 

∴ There are 6 places to seated by 3 girls which can be done ⁶P₃ ways. 

∴ Total number of ways = 5! x ⁶P₃ 

= 120 x (6 x 5 x 4) 

= 120 x 120 

= 14400

Given that each number starts with 35. We need a 6 digit number. So we have to fill only one’s place, 10’s place, 100th place, and 1000th places. We have to use 10 digits.

In these digits, 3 and 5 should not be used as a repetition of digits is not allowed. Except for these two digits, we have to use 8 digits. One’s place can be filled by any of the 8 digits in different ways, 10’s place can be filled by the remaining 7 digits in 7 different ways.

100th place can be filled by the remaining 6 different ways and 1000th place can be filled by the remaining 5 digits in 5 different ways.

∴ Number of 6 digit telephone numbers = 8 x 7 x 6 x 5 = 1680

The required numbers are lesser than 1000. They are one digit, two-digit or three-digit numbers. 

There are five numbers to be used without repetition. 

One digit number: One-digit numbers are 5. Two-digit number: 10th place can be filled by anyone of the digits by 5 ways and 1’s place can be 4 filled by any of the remaining four digits in 4 ways. 

∴ Two-digit number are 5 x 4 = 20. 

Three-digit number: 100th place can be filled by any of the 5 digits, 10th place can be filled by 4 digits and one’s place can be filled by 3 digits.

∴ Three digit numbers are = 5 x 4 x 3 = 60 

∴ Total numbers = 5 + 20 + 60 = 85.

Let P(n) denote the statement

1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Put n = 1 

LHS = 1(1 + 1) = 2 

RHS = \(\frac{1(1+1)(1+2)}{3}\) = \(\frac{1(2)(3)}{3}\) = 2

∴ P(1) is true. 

Now assume that the statement be true for n = k 

(i.e.,) assume P(k) be true 

(i.e.,) assume 1.2 + 2.3 + 3.4 + … + k(k + 1) = \(\frac{k(k+1)(k+2)}{3}\) be true

To prove: P(k + 1) is true 

(i.e.,) to prove: 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k + 1) (k + 2) 

= \(\frac{(k+1)(k+2)(k+3)}{3}\)

Consider 1.2 + 2.3 + 3.4 + ... + k(k + 1) + (k + 1) (k + 2) 

= [1.2 + 23 + … + k(k + 1)] + (k + 1) (k + 2)

= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)

= \(\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\)

= \(\frac{(k+1)(k+2)(k+3)}{3}\)

∴ P(k + 1) is true. 

Thus if P(k) is true, P(k + 1) is true. 

By the principle of Mathematical ‘induction, P(n) is true for all n ∈ N.

1.2 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Let P(n) denote the statement 2ⁿ > n for all n ∈ N 

i.e., P(n): 2ⁿ > n for n ≥ 1 

Put n = 1, P(1): 2¹ > 1 which is true. 

Assume that P(k) is true for n = k 

i.e., 2^k > k for k ≥ 1 

To prove P(k + 1) is true. 

i.e., to prove 2^{k + 1} > k + 1 for k ≥ 1 

Since 2^k > k 

Multiply both sides by 2 

2 . 2^k > 2k 

2^{k + 1} > k + k 

i.e., 2^{k + 1} > k + 1 (∵ k ≥ 1) 

∴ P(k + 1) is true whenever P(k) is true. 

∴ By principal of mathematical induction P(n) is true for all n ∈ N.

P(n): n(n + 1) (n + 2) is divisible by 6. 

P(1): 1 (2) (3) = 6 is divisible by 6 

∴ P(1) is true. 

Let us assume that P(k) is true for n = k 

That is, k (k + 1) (k + 2) = 6m for some m 

To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2) (k + 3) is divisible by 6. 

P(k + 1) = (k + 1) (k + 2) (k + 3) 

= (k + 1)(k + 2)k + 3(k + 1)(k + 2) 

= 6m + 3(k + 1)(k + 2) 

In the second term either k + 1 or k + 2 will be even, whatever be the value of k. 

Hence second term is also divisible by 6. 

∴ P (k + 1) is also true whenever P(k) is true. 

By Mathematical Induction P (n) is true for all values of n.

Let P(n) denote the statement 4 + 8 + … + 4n = 2n(n + 1) 

i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1) 

Put n = 1, 

P(1): LHS = 4 

RHS = 2 (1)(1 + 1) = 4 

P(1) is true. 

Assume that P(n) is true for n = k

P(k): 4 + 8 + 12 + … + 4k = 2k(k + 1) 

To prove P(k + 1) 

i.e., to prove 4 + 8 + 12 + … + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1) 

4 + 8 + 12 + … + 4k + (4k + 4) = 2(k + 1) (k + 2) 

Consider, 4 + 8 + 12 + … + 4k + (4k + 4) = 2k(k + 1) + (4k + 4) 

= 2k(k + 1) + 4(k + 1) 

= 2k² + 2k + 4k + 4 

= 2k² + 6k + 4 

= 2(k + 1)(k + 2) 

P(k + 1) is also true. 

∴ By Mathematical Induction, P(n) for all value n ∈ N.

Let P(n) denote the statement 3²ⁿ – 1 is divisible by 8 for all n ∈ N 

Put n = 1 

P(1) is the statement 3²⁽¹⁾ – 1 = 3² – 1 = 9 – 1 = 8, which is divisible by 8 

∴ P(1) is true. 

Assume that P(k) is true for n = k. 

i.e., 3^2k – 1 is divisible by 8 to be true. 

Let 3^2k – 1 = 8m 

To prove P(k + 1) is true. 

i.e., to prove 3^{2(k + 1)} – 1 is divisible by 8 

Consider 3^{2(k + 1)} – 1 = 3^{2k + 2} – 1 

= 3^2k.3² – 1

= 3^2k(9) – 1 

= 3^2k(8 + 1) – 1

= 3 x 8 + 3 x 1 – 1 

= 3^2k(8) + 3^2k – 1 

= 3^2k(8) + 8m (∵ 3^2k – 1 = 8m) 

= 8(3^2k + m), which is divisible by 8. 

∴ P(k + 1) is true wherever P(k) is true. 

∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.

(a) 2⁶ – 2

(5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) + (5C₁ + 5C₂ + 5C₃ + 5C₄) 

= 2⁵ + (5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) – (5C₀ + 5C₅) 

= 2⁵ + 2⁵ – (1 + 1) (∵ Adding and subtracting of 5C₀ and 5C₅) 

= 2(2⁵) – 2 (∵ 5C₀ = 5C₅ = 1) 

= 2⁶ – 2

Let P(n) be the proposition that 5²ⁿ – 1 is divisible by 24. 

For n = 1, P(1) is: 5² – 1 = 25 – 1 = 24, 24 is divisible by 24. 

Assume that P(k) is true. 

i.e., 5^2k – 1 is divisible by 24 

Let 5^2k – 1 = 24m 

To prove P(k + 1) is true. 

i.e., to prove 5^{2(k + 1)} – 1 is divisible by 24. 

P(k): 5^2k – 1 is divisible by 24. 

P(k + 1) = 5^{2(k + 1)} – 1 

= 5^2k.5² – 1 

= 5^2k(25) – 1 

= 5^2k(24 + 1) – 1 

= 24.5^2k + 5^2k – 1 

= 24.5^2k + 24m 

= 24 [5^2k + 24] 

Which is divisible by 24 ⇒ P(k + 1) is also true.

Hence by mathematical induction, P(n) is true for all values n ∈ N.

Let P(n) denote the statement aⁿ – bⁿ is divisible by a – b. 

Put n = 1. Then P(1) is the statement: a¹ – b¹ = a – b is divisible by a – b 

∴ P(1) is true. Now assume that the statement be true for n = k 

(i.e.,) assume P(k) be true, (i.e.,) a^k – b^k is divisible by (a – b) be true. 

⇒ \(\frac{a^k-b^k}{a-b}\) = m (say) where m ∈ N

⇒ a^k – b^k = m(a – b) 

⇒ a^k = b^k + m(a – b) … (1) 

Now to prove P(k + 1) is true, (i.e.,) to prove: a^{k + 1} – b^{k + 1} is divisible by a – b 

Consider a^{k + 1} – b^{k + 1} = a^k.a – b^k.b 

= [b^k + m(a – b)] a – b^k.b [∵ a^k = bm + k(a – b)] 

= b^k.a + am(a – b) – b^k.b 

= b^k.a – b^k.b + am(a – b) 

= b^k(a – b) + am(a – b) 

= (a – b) (b^k + am) is divisible by (a – b) 

∴ P(k + 1) is true. 

By the principle of Mathematical induction. P(n) is true for all n ∈ N.

∴ aⁿ – bⁿ is divisible by a – b for n ∈ N.

(c) 1024 ways

For each question, there are two ways of answering it. 

For 10 questions the numbers of ways to answer 

= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 

= 2¹⁰ 

= 1024 ways

(d) 24

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

= 1 x 2 x 3 x 4 

= 24

(d) 5

Given that nC₃ = nC₂ 

We know that if nC_x = nC_y then x + y = n or x = y 

Here 3 + 2 = n 

∴ n = 5

A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon. 

A number of line segments obtained by joining the vertices of a n sided polygon taken two at a time = Number of ways of selecting 2 out of n.

= ⁿC₂

= \(\frac{n(n-1)}{2}\)

Out of these lines, n lines are the sides of the polygon, sides can’t be diagonals.

∴ Number of diagonals of the polygon = \(\frac{n(n-1)}{2}\) - n = \(\frac{n(n-3)}{2}\)

Given that a polygon has 44 diagonals. 

Let n be the number of sides of the polygon.

\(\frac{n(n-3)}{2}\) = 44

⇒ n(n – 3) = 88 

⇒ n² – 3n – 88 = 0 

⇒ (n + 8) (n – 11) 

⇒ n = -8 (or) n = 11 

n cannot be negative. 

∴ n = 11 is number of sides of polygon is 11.

(c) 6

Given nP_r = 720(nC_r)

\(\frac{n!}{(n-r)!}\) = 720 \(\frac{n!}{r!(n-r)!}\)

1 = \(\frac{720}{r!}\)

r! = 720

r! = 6 x 5 x 4 x 3 x 2 x 1

r! = 6! 

r = 6

Answer is (c) nC₂ – n

(a) 2⁵

Number of possible outcomes When a coin is tossed is 2 

∴ When five coins are tossed (same as a coin is tossed five times) 

Possible outcomes = 2 x 2 x 2 x 2 x 2 = 2⁵

When a die is rolled number of possible outcomes is selecting an event from 6 events = ⁶C₁ 

When four dice are rolled number of possible outcomes = ⁶C₁ x ⁶C₁ x ⁶C₁ x ⁶C₁ 

When a die is rolled number of possible outcomes in which ‘2’ does not appear is selecting an event from 5 events = ⁵C₁ 

When four dice are rolled number of possible outcomes in which 2 does not appear = ⁵C₁ x ⁵C₂ x ⁵C₃ x ⁵C₄ 

Therefore the number of possible outcomes in which atleast one die shows 2 

= ⁶C₁ x ⁶C₁ x ⁶C₁ x ⁶C₁  – ⁵C₁ x ⁵C₂ x ⁵C₃ x ⁵C₄ 

= 6 x 6 x 6 x 6 – 5 x 5 x 5 x 5 

= 1296 – 625

= 671

Note: when two dice are rolled number of possible outcomes is 36 and the number of possible outcomes in which 2 doesn’t appear = 25. When two dice are rolled the number of possible outcomes in which atleast one die shows 2 = 36 – 25 = 11. Use the sample space, S = {(1, 1), (1, 2), … (6, 6)}.

True.

Negative of a negative integer is always a positive integer.