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This mock test of MCQ : Polynomials - 1 for Class 10 helps you for every Class 10 entrance exam.
This contains 10 Multiple Choice Questions for Class 10 MCQ : Polynomials - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

If 5 is a zero of the quadratic polynomial, x^{2} - kx - 15 then the value of k is

Solution:

x=5

putting x in equation

p(x)=x^{2}-kx-15=0

=(5)^{2}-k(5)-15=0

=25-5k-15=0

=10-5k=0

=-5k=-10

=k=-10/-5

=k=2

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:

Zero of p(x)

Let p(x) = ax + b

Put x = k

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

⇒ -5/2

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

QUESTION: 4

If one of the zeroes of the quadratic polynomial (k - 1)x^{2} + kx + 1 is -3, then the value of k is

Solution:

(k - l)x^{2} + kx +1

One zero is - 3, so it must satisfy the equation and make it zero

∴ (k - 1) (-3)^{2} + k(-3) + 1 - 0

⇒ 9k - 9 - 3k + 1 = 0

⇒ 6k - 8 = 0

⇒

QUESTION: 5

If one of the zeros of a quadratic polynomial of the form x^{2} + ax + b is the negative of the other, then it

Solution:

Let p(x) = x^{2} + ax + b.

Put a = 0, then, p(x) = x^{2} + b = 0

⇒ x^{2 }= -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) = x^{2} + ax+ b

and by given condition the zeroes area and – α.

Sum of the zeroes = α- α = a

= a = 0

f(x) = x^{2} + b, which cannot be linear,

and product of zeroes = α .(- α) = b

⇒ -α^{2} = b

which is possible when, b < 0.

Hence, it has no linear term and the constant term is negative.

QUESTION: 6

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

Solution:

x^{2} + (a + 1)x + b

∵ x = 2 is a zero and x = - 3 is another zero

∴ (2)^{2} + (a + 1)^{2} + b = 0

and (-3)^{2} + (a + 1) (-3) + b = 0

⇒ 4 + 2a + 2 + b = 0 and 9 - 3a - 3 + b = 0

⇒ 2a + b = -6 ...(i) and -3a + b = -6 ..(ii)

Solving (i) and (ii), we get 5a = 0

⇒ a = 0 and b = -6.

QUESTION: 7

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

Solution:

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

∴ Sum of zeroes = -b/a

⇒ ...(i)

and product of zeroes = c/a

⇒ ...(ii)

From Eqs. (i) and (ii)

a = 1, b = -3 and c = -10

∴ p(x) = ax^{2} + bx + c = 1.x^{2} - 3x - 10

= x^{2} - 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) = kx^{2} - 3kx - 10k [where, k is a real number]

⇒ [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 ax^{2} + bx + c, a≠0, the graph of the Corresponding equation y = ax^{2} + bx + c has one of the two shapes either open upwards like u or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas.

Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (a) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial.

QUESTION: 9

If one root of the polynomial p(y) = 5y^{2} + 13y + m is reciprocal of other, then the value of m is

Solution:

Let the roots be α and 1/α. Then α(1/α) = m/5 or 1 = m/5 or m = 5

QUESTION: 10

If p(x) = ax^{2} + bx + c, then - b/a is equal to

Solution:

Sum of zeroes = -b/a

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