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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.
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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:

p(x) = x^{2}- 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)x^{2} + kx + 1 is -3, then the value of k is:

Solution:

p(x) = (k - 1)x^{2} + 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 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

Given condition the zeroes are α and – α.

Sum of the zeroes (-a) = α - α = 0

Product of zeroes (b) = α .(- α) = - α^{2}

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:

p(x) = x^{2} + (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) = x^{2} + (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) = ax^{2} + 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) = 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]

⇒ p(x) = x^{2}/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 ax^{2} + 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) = 5y^{2} + 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) = ax^{2} + bx + c, then - b/a is equal to

Solution:

Sum of zeroes = -b/a

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