Class 9 Maths MCQs Polynomial with Solutions - Future Study Point

Class 9 Maths MCQs Polynomial with Solutions

class 9 maths MCQ Polynomial

Class 9 Maths MCQs -Chapter 2 Polynomial with Solutions

class 9 maths MCQ Polynomial

Class 9 Maths MCQs Polynomial with Solutions are created for the purpose of helping the class 10 students for the preparation of Term 1 CBSE Board exam 2021. The practice of these MCQs are required to every students because the pattern of the CBSE Board exam has changed and all the questions in the question paper are of MCQs types . All the questions of chapter 2 Polynomial are selected here as per the CBSE sample paper of maths published by CBSE.The solutions of these MCQs of the chapter Polynomial are created here as per the CBSE norms and explained here in easiest way so every students can understand the solutions.

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Q1. The value of the polynomial P(x) = 5xยณ – 3xยฒ + 7x – 10,when x = -1 is:

(a) 7ย  ย  ย (b) 17ย  ย  ย  ย (c) -18ย  ย  ย  (d) -25

Ans. The given polynomial P(x) =5xยณ – 3xยฒ + 7x – 10

The value of the polynomial when x =-1

P(-1) = 5(-1)ยณ -3(-1)ยฒ +7(-1) -10 = -5 -3 -7 -10 =-25

Q2.One of theย  zeroes of the polynomial P(x) = xยฒ – 1 are :

(a) -1ย  ย  ย  ย (b) -2ย  ย  ย  ย (c) 0ย  ย  ย  (d) -3

Ans. (a) -1

The given polynomial is P(x) = xยฒ – 1

P(x) = xยฒ – 1 = 0

(x +1)(x -1) = 0

x = -1 and x = 1

Q3.The product of ( x + 1/x)(x -1/x)(xยฒ + 1/xยฒ) is :

(a) (x – 1/x)ยณย  ย  ย  ย (b) xย  + 1/xยณย  ย  ย  ย (c) ย x4-1/ x4ย  ย  (d) ย x4+ 1/ x4

Ans. The given expression is

(x + 1/x)(x -1/x)(xยฒ + 1/xยฒ)

= (xยฒ – 1/xยฒ)(xยฒ+1/xยฒ) [i.e aยฒ -bยฒ = (a +b)(a – b)]

= ( x4-1/ x4) [ i.e aยฒ – bยฒ = (a + b)( a – b)]

Q4.Ifย  x/yย  + y/x = -1 (x,y โ‰  0), then the value of xยณ – yยณย is

(a) 1ย  (b) (x -y)(xยฒ – yยฒ +1)ย  ย  ย (c) 0ย  ย (d) -1/2

Ans.(c) 0

The given equation is

x/y + y/x = -1

(xยฒ + yยฒ)/xy = -1

xยฒ + yยฒ = -xy….(i)

Cubing both sides

The value of xยณ – yยณ = (x -y)(xยฒ +yยฒ + xy)

Putting the value from equation (i)

xยณ – yยณ = ( x – y)( -xy +xy) =0

Q5. The value of 625ยฒ – 575ยฒย ย is

(a) 50000ย  ย  ย  ย  ย (b)30000ย  ย  ย  ย  ย  ย (c)60000ย  ย  ย  ย  ย  ย  (d) 40000

Ans.

625ยฒ – 575ยฒ = (625 + 575)(625 – 575) = 1200 ร— 50 =60000

Q6. If xยฒ + kx + 6 = ( x +1)( x – 3),then value of k is

(a) 1ย  ย  ย  ย  ย  ย  (b) -1ย  ย  ย  ย  ย  (c) 2ย  ย  ย  ย  ย  (d) -1

Ans. The given equation is

xยฒ + kx + 6 = (x +2)( x +3)

xยฒ + kx + 6 = xยฒ + (2+3)x + 2ร—3 = xยฒ + 5x +6

On comparing LHS and RHS

k = 5

Q7. The coefficient of xยณ in ( 2xยฒ + 2x)(5xยณ – 9xยฒ) is:

(a) -15ย  ย  ย  ย (b) -18ย  ย  ย  ย (c) 27ย  ย  ย  ย  (d) -28

Ans. (2xยฒ + 2x)(5xยณ -9xยฒ) for getting the coefficient of xยณ,multiplying 2x and -9xยฒ i.e 2xร—-9xยฒ = -18xยณ

Hence the coefficient of xยณ is -18

Q8. Find the remainder when the polynomial 3xยณ – 8xยฒ + 17x + 4 is divided by x:

(a) 2ย  ย  ย  ย  (b) -4ย  ย  ย  ย  ย (c) 4ย  ย  ย  ย  (d) 0

Ans.(c) 4

The given polynomial is 3xยณ – 8xยฒ + 17x + 4

Putting the value x = 0

3(0)ยณ – 8(0)ยฒ +17(0) + 4 = 4

Q9.If 2x + y + 3z = 0, then the value ofย  8xยณ + yยณ + 27zยณ is :

(a) 18xyzย  ย  ย  ย  (b) -18xyzย  ย  ย  ย  (c) 36xyzย  ย  ย  ย  (d) -36xyz

Ans.(a) 18xyz

The given equation is

2x + y + 3x = 0

Applying the identity

aยณ + bยณ +cยณ – 3abc = (a + b +c)(aยฒ + bยฒ + cยฒ – ab – bc -ac)

(2x)ยณ + yยณ + (3z)ยณ – 3ร—2xร—3z = ( 2x + y +3z)(4xยฒ + yยฒ + 9zยฒ -2xy – 3yz – 6xz)

Putting the value of 2x + y +3z = 0

8xยณ + yยณ + 27zยณ = 0

Q10. If area of the rectangle is 4xยฒ + 4x – 3,then its possible diensions are :

(a) 2x +1, 2x +3ย  ย  (b) 3x -1, 2x +1ย  ย  (c) 2x -3,2x +5ย  ย  ย (d) 2x +3,2x -1

Ans.

4xยฒ + 4x – 3

=4xยฒ + 6x – 2x – 3

=2x( 2x + 3) -1(2x + 3)

=(2x + 3)(2x – 1)

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