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POLYNOMIALS

90110 If 2 is the zero of both the polynomials 3x $^{1}$ + mx - 14 and 2x $^{1}$ + nx $^{1}$ + x - 2, then the value of m - 2n is:

1 5

2 -1

3 -9

4 9

Explanation:

9
According to the question, p(2) = 3x $^{1}$ + mx - 14 = 0 ? 3(2) $^{1}$ + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1

POLYNOMIALS

90111 If $\sqrt{5}$ and $- \sqrt{5}$ are two zeroes of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15, then its third zero is:

1 3

2 -3

3 5

4 -5

Explanation:

-3
Let $α = \sqrt{5}$ and $β = - \sqrt{5}$ be the given zeros and $γ$ be the third zero of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15 Then, By using $α + β + γ = \frac{- {Coefficient of x}^{2}}{{Coefficient of x}^{3}}$ $α + β + γ = \frac{- 3}{1}$ $α + β + γ = - 3$ Substituting $α = \sqrt{5}$ and $β = - \sqrt{5}$ in $α + β + γ = - 3$ We get $\sqrt{5} - \sqrt{5} + γ = - 3$ $γ = - 3$ Hence, the correct choice is (b)

POLYNOMIALS

90112 The zeros of the polynomial x $^{1}$ - 2x - 3 are:

1 3, -1

2 3, 1

3 -3, 1

4 -3, -1

Explanation:

3, -1
x $^{1}$ - 2x - 3 = x(x - 3) + (x - 3) $∴$ (x - 3) (x + 1) = 0 ? x = 3 or x = -1

POLYNOMIALS

90113 If $α, β$ are the zeroes of the polynomial f(x) = x $^{1}$ + x + 1, then $\frac{1}{α} + \frac{1}{β} =$

1 0

2 -1

3 1

4 None of these

Explanation:

-1
Since $α$ and $β$ are the zeroes of the quadratic polynomial f(x) = x $^{1}$ + x + 1 $α + β = \frac{- Coefficient of x}{Coefficient of x^{2}} = \frac{- 1}{1} = - 1$ $α \times β = \frac{Constant term}{Coefficient of x^{2}} = \frac{1}{1} = 1$ Now $\frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{- 1}{1} = - 1$ Thus the value of $\frac{1}{α} + \frac{1}{β}$ is -1

POLYNOMIALS

90110 If 2 is the zero of both the polynomials 3x $^{1}$ + mx - 14 and 2x $^{1}$ + nx $^{1}$ + x - 2, then the value of m - 2n is:

1 5

2 -1

3 -9

4 9

Explanation:

9
According to the question, p(2) = 3x $^{1}$ + mx - 14 = 0 ? 3(2) $^{1}$ + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1

POLYNOMIALS

90111 If $\sqrt{5}$ and $- \sqrt{5}$ are two zeroes of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15, then its third zero is:

1 3

2 -3

3 5

4 -5

Explanation:

-3
Let $α = \sqrt{5}$ and $β = - \sqrt{5}$ be the given zeros and $γ$ be the third zero of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15 Then, By using $α + β + γ = \frac{- {Coefficient of x}^{2}}{{Coefficient of x}^{3}}$ $α + β + γ = \frac{- 3}{1}$ $α + β + γ = - 3$ Substituting $α = \sqrt{5}$ and $β = - \sqrt{5}$ in $α + β + γ = - 3$ We get $\sqrt{5} - \sqrt{5} + γ = - 3$ $γ = - 3$ Hence, the correct choice is (b)

POLYNOMIALS

90112 The zeros of the polynomial x $^{1}$ - 2x - 3 are:

1 3, -1

2 3, 1

3 -3, 1

4 -3, -1

Explanation:

3, -1
x $^{1}$ - 2x - 3 = x(x - 3) + (x - 3) $∴$ (x - 3) (x + 1) = 0 ? x = 3 or x = -1

POLYNOMIALS

90113 If $α, β$ are the zeroes of the polynomial f(x) = x $^{1}$ + x + 1, then $\frac{1}{α} + \frac{1}{β} =$

1 0

2 -1

3 1

4 None of these

Explanation:

-1
Since $α$ and $β$ are the zeroes of the quadratic polynomial f(x) = x $^{1}$ + x + 1 $α + β = \frac{- Coefficient of x}{Coefficient of x^{2}} = \frac{- 1}{1} = - 1$ $α \times β = \frac{Constant term}{Coefficient of x^{2}} = \frac{1}{1} = 1$ Now $\frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{- 1}{1} = - 1$ Thus the value of $\frac{1}{α} + \frac{1}{β}$ is -1

POLYNOMIALS

90110 If 2 is the zero of both the polynomials 3x $^{1}$ + mx - 14 and 2x $^{1}$ + nx $^{1}$ + x - 2, then the value of m - 2n is:

1 5

2 -1

3 -9

4 9

Explanation:

9
According to the question, p(2) = 3x $^{1}$ + mx - 14 = 0 ? 3(2) $^{1}$ + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1

POLYNOMIALS

90111 If $\sqrt{5}$ and $- \sqrt{5}$ are two zeroes of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15, then its third zero is:

1 3

2 -3

3 5

4 -5

Explanation:

-3
Let $α = \sqrt{5}$ and $β = - \sqrt{5}$ be the given zeros and $γ$ be the third zero of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15 Then, By using $α + β + γ = \frac{- {Coefficient of x}^{2}}{{Coefficient of x}^{3}}$ $α + β + γ = \frac{- 3}{1}$ $α + β + γ = - 3$ Substituting $α = \sqrt{5}$ and $β = - \sqrt{5}$ in $α + β + γ = - 3$ We get $\sqrt{5} - \sqrt{5} + γ = - 3$ $γ = - 3$ Hence, the correct choice is (b)

POLYNOMIALS

90112 The zeros of the polynomial x $^{1}$ - 2x - 3 are:

1 3, -1

2 3, 1

3 -3, 1

4 -3, -1

Explanation:

3, -1
x $^{1}$ - 2x - 3 = x(x - 3) + (x - 3) $∴$ (x - 3) (x + 1) = 0 ? x = 3 or x = -1

POLYNOMIALS

90113 If $α, β$ are the zeroes of the polynomial f(x) = x $^{1}$ + x + 1, then $\frac{1}{α} + \frac{1}{β} =$

1 0

2 -1

3 1

4 None of these

Explanation:

-1
Since $α$ and $β$ are the zeroes of the quadratic polynomial f(x) = x $^{1}$ + x + 1 $α + β = \frac{- Coefficient of x}{Coefficient of x^{2}} = \frac{- 1}{1} = - 1$ $α \times β = \frac{Constant term}{Coefficient of x^{2}} = \frac{1}{1} = 1$ Now $\frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{- 1}{1} = - 1$ Thus the value of $\frac{1}{α} + \frac{1}{β}$ is -1

POLYNOMIALS

90110 If 2 is the zero of both the polynomials 3x $^{1}$ + mx - 14 and 2x $^{1}$ + nx $^{1}$ + x - 2, then the value of m - 2n is:

1 5

2 -1

3 -9

4 9

Explanation:

9
According to the question, p(2) = 3x $^{1}$ + mx - 14 = 0 ? 3(2) $^{1}$ + m × 2 - 14 = 0 ? 12 + 2m - 14 = 0 ? m = 1

POLYNOMIALS

90111 If $\sqrt{5}$ and $- \sqrt{5}$ are two zeroes of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15, then its third zero is:

1 3

2 -3

3 5

4 -5

Explanation:

-3
Let $α = \sqrt{5}$ and $β = - \sqrt{5}$ be the given zeros and $γ$ be the third zero of the polynomial x $^{1}$ + 3x $^{1}$ - 5x - 15 Then, By using $α + β + γ = \frac{- {Coefficient of x}^{2}}{{Coefficient of x}^{3}}$ $α + β + γ = \frac{- 3}{1}$ $α + β + γ = - 3$ Substituting $α = \sqrt{5}$ and $β = - \sqrt{5}$ in $α + β + γ = - 3$ We get $\sqrt{5} - \sqrt{5} + γ = - 3$ $γ = - 3$ Hence, the correct choice is (b)

POLYNOMIALS

90112 The zeros of the polynomial x $^{1}$ - 2x - 3 are:

1 3, -1

2 3, 1

3 -3, 1

4 -3, -1

Explanation:

3, -1
x $^{1}$ - 2x - 3 = x(x - 3) + (x - 3) $∴$ (x - 3) (x + 1) = 0 ? x = 3 or x = -1

POLYNOMIALS

90113 If $α, β$ are the zeroes of the polynomial f(x) = x $^{1}$ + x + 1, then $\frac{1}{α} + \frac{1}{β} =$

1 0

2 -1

3 1

4 None of these

Explanation:

-1
Since $α$ and $β$ are the zeroes of the quadratic polynomial f(x) = x $^{1}$ + x + 1 $α + β = \frac{- Coefficient of x}{Coefficient of x^{2}} = \frac{- 1}{1} = - 1$ $α \times β = \frac{Constant term}{Coefficient of x^{2}} = \frac{1}{1} = 1$ Now $\frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{- 1}{1} = - 1$ Thus the value of $\frac{1}{α} + \frac{1}{β}$ is -1

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