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# RS Aggarwal Test: Polynomials

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

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This mock test of RS Aggarwal Test: Polynomials for Class 10 helps you for every Class 10 entrance exam. This contains 10 Multiple Choice Questions for Class 10 RS Aggarwal Test: Polynomials (mcq) to study with solutions a complete question bank. The solved questions answers in this RS Aggarwal Test: Polynomials quiz give you a good mix of easy questions and tough questions. Class 10 students definitely take this RS Aggarwal Test: Polynomials exercise for a better result in the exam. You can find other RS Aggarwal Test: Polynomials extra questions, long questions & short questions for Class 10 on EduRev as well by searching above.
QUESTION: 1

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

Solution:

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

QUESTION: 2

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

Solution:

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

QUESTION: 3

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

Solution:

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.

QUESTION: 4

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

Solution:

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

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

Solution:

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.

QUESTION: 6

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

Solution:

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

QUESTION: 7

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

Solution:

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.

QUESTION: 8

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

Solution:

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.

QUESTION: 9

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

Solution:

Let the roots be α and 1/α.

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

QUESTION: 10

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

Solution:

Sum of zeroes = -b/a