Class 10  >  Mathematics (Maths) Class 10  >  RS Aggarwal Test: Polynomials Download as PDF

RS Aggarwal Test: Polynomials


Test Description

10 Questions MCQ Test Mathematics (Maths) Class 10 | RS Aggarwal Test: Polynomials

RS Aggarwal Test: Polynomials for Class 10 2022 is part of Mathematics (Maths) Class 10 preparation. The RS Aggarwal Test: Polynomials questions and answers have been prepared according to the Class 10 exam syllabus.The RS Aggarwal Test: Polynomials MCQs are made for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RS Aggarwal Test: Polynomials below.
Solutions of RS Aggarwal Test: Polynomials questions in English are available as part of our Mathematics (Maths) Class 10 for Class 10 & RS Aggarwal Test: Polynomials solutions in Hindi for Mathematics (Maths) Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt RS Aggarwal Test: Polynomials | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Mathematics (Maths) Class 10 for Class 10 Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
RS Aggarwal Test: Polynomials - Question 1

If 5 is a zero of the quadratic polynomial, x2 - kx - 15 then the value of k is:

Detailed Solution for RS Aggarwal Test: Polynomials - Question 1

p(x) = x2- kx - 15

Given: p(5) = 0
⇒ (5)2- k(5) - 15 = 0
⇒ 25 - 5k - 15 = 0
⇒ 5k = 10
⇒ k = 10/5 = 2

Thus, Value of k is 2

RS Aggarwal Test: Polynomials - Question 2

If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as:

Detailed Solution for RS Aggarwal Test: Polynomials - Question 2

Let p(x) = ax + b
p(k) = ak + b = 0
∴ k is zero of p(x)

RS Aggarwal Test: Polynomials - Question 3

The zero of the polynomial p(x) = 2x + 5 is:

Detailed Solution for RS Aggarwal Test: Polynomials - Question 3

Given, p(x) = 2x+5
For zero of the polynomial, put p(x) = 0
⇒ 2x + 5 = 0
⇒ x = -5/2

Thus, zero of the polynomial p(x) is -5/2. 

RS Aggarwal Test: Polynomials - Question 4

If one of the zeroes of the quadratic polynomial (k - 1)x2 + kx + 1 is -3, then the value of k is:

Detailed Solution for RS Aggarwal Test: Polynomials - Question 4

p(x) = (k - 1)x2 + kx +1

One zero of polynomial is - 3 i.e. p(-3) = 0
⇒ (k - 1) (-3)2 + k(-3) + 1 = 0
⇒ 9k - 9 - 3k + 1 = 0
⇒ 6k - 8 = 0
⇒ k = 8/6 = 4/3

RS Aggarwal Test: Polynomials - Question 5

If one of the zeros of a quadratic polynomial of the form x2 + ax + b is the negative of the other, then it

Detailed Solution for RS Aggarwal Test: Polynomials - Question 5

Let p(x) = x2 + ax + b.
Put a = 0 then,  p(x) = x2 + b = 0
⇒  x= -b
⇒  x =  ±-b [∴ b < 0]
Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other,
then it has no linear term i.e., a = 0 and the constant term is negative i.e., b< 0.

Alternate Method
Let   f(x) = x2 + ax+ b
Given condition the zeroes are α and – α.
Sum of the zeroes (-a) = α - α = 0
Product of zeroes (b) = α .(- α) = - α2
Hence, it has no linear term and the constant term is negative.

RS Aggarwal Test: Polynomials - Question 6

If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then

Detailed Solution for RS Aggarwal Test: Polynomials - Question 6

p(x) = x2 + (a + 1)x + b
Given: zeros of polynomial is 2, -3
For x = 2
(2)2 + (a + 1)2 + b = 0
⇒ 4 + 2a + 2 + b = 0
⇒ 2a + b = -6 ...(i)

For x = -3
(-3)2 + (a + 1) (-3) + b = 0
⇒ 9 - 3a - 3 + b = 0
⇒ -3a + b = -6 ..(ii)

Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = -6

Alternative method:

p(x) = x2 + (a + 1)x + b
Given: zeros of polynomial is 2, -3

Sum of zeros: -(a+1) = 2-3
⇒ a = 0
Product of zeros: b = 2.(-3) = -6 

RS Aggarwal Test: Polynomials - Question 7

The number of polynomials having zeros as - 2 and 5 is:

Detailed Solution for RS Aggarwal Test: Polynomials - Question 7

 Let p (x) = ax2 + bx + c be the required polynomial whose zeroes are -2 and 5.

∴ Sum of zeroes = -b/a
⇒  - 2 + 5 = 3 = -(-3)/1  ...(i)
and product of zeroes = c/a
⇒  -2 x 5 = -10/1 ...(ii)
From Eqs. (i) and (ii)
a = 1, b = -3 and c = -10
∴ p(x) = ax2 + bx + c = 1.x2 - 3x - 10
= x2 - 3x - 10
But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.
∴ p(x) = kx2 - 3kx - 10k [where, k is a real number]
⇒ p(x) = x2/k - 3x/k -10/k [where, k is a non-zero real number]

Hence, the required number of polynomials are infinite i.e., more than 3.

RS Aggarwal Test: Polynomials - Question 8

Which of the following is NOT the graph of a quadratic polynomial ?

Detailed Solution for RS Aggarwal Test: Polynomials - Question 8

For any quadratic polynomial ax2 + bx + c, a≠0, graph for the corresponding equation:

  • Has one of the two shapes either open upwards like ∪ or open downwards like  ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. 
  • The curve of a quadratic polynomial crosses the X-axis on at most two points.

Thus, option A is correct.

RS Aggarwal Test: Polynomials - Question 9

If one root of the polynomial p(y) = 5y2 + 13y + m is reciprocal of other, then the value of m is

Detailed Solution for RS Aggarwal Test: Polynomials - Question 9

Let the roots be α and 1/α.

Product of roots = α(1/α) = 1
⇒ m/5 = 1
⇒ m = 5

RS Aggarwal Test: Polynomials - Question 10

If p(x) = ax2 + bx + c, then - b/a is equal to 

Detailed Solution for RS Aggarwal Test: Polynomials - Question 10

Sum of zeroes = -b/a

53 videos|369 docs|138 tests
Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about RS Aggarwal Test: Polynomials Page
In this test you can find the Exam questions for RS Aggarwal Test: Polynomials solved & explained in the simplest way possible. Besides giving Questions and answers for RS Aggarwal Test: Polynomials, EduRev gives you an ample number of Online tests for practice