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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.15

MULTIPLE CHOICE QUESTIONS :

Question 1.

If x^{3} + 6x^{2} + kx + 6 is exactly divisible by (x + 2), then k = ?

(1) -6

(2) -7

(3) -8

(4) 11

Solution:

(4) 11

Hint: P(-2) = (-2)^{3} + 6 (-2)^{2} + k (-2) + 6 = 0

⇒ – 8 + 24 – 2k +6 = 0

⇒ 22 = 2k

⇒ k = 11

Question 2.

The root of the polynomial equation 2x + 3 = 0 is

(1) \(\frac{1}{3}\)

(2) \(-\frac{1}{3}\)

(3) \(-\frac{3}{2}\)

(4) \(-\frac{2}{3}\)

Solution:

(3) \(-\frac{3}{2}\)

Question 3.

The type of the polynomial 4 – 3x^{3} is

(1) constant polynomial

(2) linear polynomial

(3) quadratic polynomial

(4) cubic polynomial.

Solution:

(4) cubic polynomial.

Hint: Polynomial of degree 3 is called cubic.

Question 4.

If x^{51} + 51 is divided by x + 1, then the remainder is

(1) 0

(2) 1

(3) 49

(4) 50

Solution:

(4) 50

Hint: P(-1 = (-1)^{51} + 51 = -1 + 51 = 50

Question 5.

The zero of the polynomial 2x + 5 is

Solution:

(2) \(-\frac{5}{2}\)

Question 6.

The sum of the polynomials p(x) = x^{3} – x^{2} – 2, q(x) = x^{2} – 3x + 1

(1) x^{3} – 3x – 1

(2) x^{3} + 2x^{2} – 1

(3) x^{3} – 2x^{2} – 3x

(4) x^{3} – 2x^{2} + 3x – 1

Solution:

(1) x^{3} – 3x – 1

Hint:

Question 7.

Degree of the polynomial (y^{3} – 2) (y^{3} + 1) is

(1) 9

(2) 2

(3) 3

(4) 6

Solution:

(4) 6

Hint: (y^{3} – 2)(y^{3} + 1) = (y^{3} – 2)(y^{3} – 2) × 1 = y^{6} – 2y^{3} + y^{3} – 2 = y^{6} – y^{3} – 2

Question 8.

Let the polynomials be

(A) -13q^{5} + 4q^{2} + 12q

(B) (x^{2} + 4 ) (x^{2} + 9)

(C) 4q^{8} – q^{6} + q^{2}

(D) \(-\frac{5}{7}\)y^{12} + y^{3} + y^{5}

Then ascending order of their degree is

(1) A, B, D, C

(2) A, B, C, D

(3) B, C, D, A

(4) B, A, C, D

Solution:

(4) B, A, C, D

Hint: Degree of (A), (B) (C) & (D) are respectively be 5, 4, 8, 12

Question 9.

If p (a) = 0 then (x – a) is a ________ of p(x)

(1) divisor

(2) quotient

(3) remainder

(4) factor

Solution:

(4) factor

Question 10.

Zeros of (2 – 3x) is _____

(1) 3

(2) 2

(3) \(\frac{2}{3}\)

(4) \(\frac{3}{2}\)

Solution:

(3) \(\frac{2}{3}\)

Hint: 2 – 3x =0

-3x = – 2

x = \(\frac{2}{3}\)

Question 11.

Which of the following has x – 1 as a factor?

(1) 2x – 1

(2) 3x – 3

(3) 4x – 3

(4) 3x – 4

Solution:

(2) 3x – 3

Hint: p(x) = 3x – 3

P( 1) = 3(1) – 3 = 0

∴ (x – 1) is a factor of p(x)

Question 12.

If x – 3 is a factor of p (x), then the remainder is

(1) 3

(2) -3

(3) p(3)

(4) p(-3)

Solution:

(3) p(3)

Question 13.

(x + y)(x^{2} – xy + y^{2}) is equal to

(1) (x + y)^{3}

(2) (x – y)^{3}

(3) x^{3} + y^{3}

(4) x^{3} – y^{3}

Solution:

(3) x^{3} + y^{3}

Question 14.

(a + b – c)^{2} is equal to

(1) (a – b + c)^{2}

(2) (-a – b + c)^{2}

(3) (a + b + c)^{2}

(4) (a – b – c)^{2}

Solution:

(2) (-a – b + c)^{2}

Hint: (a + b – c)^{2} = [- (- a – b + c)]^{2} = (-a – b + c)^{2}

Question 15.

In an expression ax^{2} + bx + c the sum and product of the factors respectively,

(1) a, bc

(2) b, ac

(3) ac, b

(4) bc, a

Solution:

(2) b, ac

Question 16.

If (x + 5) and (x – 3) are the factors of ax^{2} + bx + c, then values of a, b and c are

(1) 1, 2, 3

(2) 1, 2, 15

(3) 1, 2, -15

(4) 1, -2, 15

Solution:

(3) 1, 2, -15

Hint: p(-5) = a (-5^{2}) + b (-5) + c = 25a – 5b + c = 0 ……… (1)

p( 3) = a (3^{2}) + 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

Question 17.

Cubic polynomial may have a maximum of

(1) 1

(2) 2

(3) 3

(4) 4

Solution:

(3) 3

Question 18.

Degree of the constant polynomial is

(1) 3

(2) 2

(3) 1

(4) 0

Solution:

(4) 0

Question 19.

Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = – 2.

(1) 2

(2) -2

(3) 10

(4) 0

Solution:

(2) -2

Hint: 2x + 3y = m, x = 2, y = – 2

m = 2(2) + 3(-2)

= 4 – 6 = -2

Question 20.

Which of the following is a linear equation?

(1) x + \(\frac{1}{x}\) = 2 x

(2) x(x – 1) = 2

(3) 3x + 5 = \(\frac{2}{3}\)

(4) x^{3} – x = 5

Solution:

(3) 3x + 5 = \(\frac{2}{3}\)

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

Question 21.

Which of the following is a solution of the equation 2x – y = 6?

(1) (2, 4)

(2) (4, 2)

(3) (3, -1)

(4) (0, 6)

Solution:

(2) (4, 2)

Hint: 2x – y = 6

2(4) – 2 = 8 – 2 = 6 = RHS

Question 22.

If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is

(1) 12

(2) 6

(3) 0

(4) 13

Solution:

(4) 13

Hint: 2x + 3y = k

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

Question 23.

Which condition does not satisfy the linear equation ax + by + c = 0

(1) a ≠ 0,b = 0

(2) a = 0, b ≠ 0

(3) a = 0, b = 0, c ≠ 0

(4) a ≠ 0, b ≠ 0

Solution:

(3) a = 0, b = 0, c ≠ 0

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

Question 24.

Which of the following is not a linear equation in two variable

(1) ax + by + c = 0

(2) 0x + 0y + c = 0

(3) 0x + by + c = 0

(4) ax + 0y + c = 0

Solution:

(2) 0x + 0y + c = 0

Hint: a and b both can not be zero

Question 25.

The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 represents parallel lines is

(1) k = 3

(2) k = 2

(3) k = 4

(4) k = -3

Solution:

(1) k = 3

Hint: 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 m_{1} = m_{2}

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

k = -2 × \(\frac{-6}{4}\) = 3

Question 26.

A pair of linear equations has no solution then the graphical representation is

Solution:

(2)

Hint: Parallel lines have no solution

Question 27.

If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) where a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 then the given pair of linear equation has ______ solution(s)

(1) no solution

(2) two solutions

(3) unique

(4) infinite

Solution:

(3) unique

Hint: \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\); unique solution

Question 28.

If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) where a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 then the given pair of linear equation has

(1) no solution

(2) two solutions

(3) infinite

(4) unique

Solution:

(1) no solution

Hint: \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\): parallel

Question 29.

GCD of any two prime numbers is _______

(1) -1

(2) 0

(3) 1

(4) 2

Solution:

(3) 1

Question 30.

The GCD of x^{4} – y^{4} and x^{2} – y^{2} is

(1) x^{4} – y^{4}

(2) x^{2} – y^{2}

(3) (x + y)^{2}

(4) (x + y)^{4}

Solution:

(2) x^{2} – y^{2}

Hint:

x^{4} – y^{4} = (x^{2} )^{2} – (y^{2})^{2} = (x^{2} + y^{2}) (x^{2} – y^{2})

x^{2} – y^{2} = x^{2} – y^{2}

G.C.D. is = x^{2} – y^{2}