QBManager

POLYNOMIALS

89960 If $α, β$ are the zeros of kx $^{1}$ - 2x + 3k such that $α + β = α β,$ then k = ?

1

\frac{1}{3}

2

\frac{- 1}{3}

3

\frac{2}{3}

4

\frac{- 2}{3}

Explanation:

$\frac{2}{3}$
Here, p(x) = x $^{1}$ - 2x + 3k Comparing the given polynomial with ax $^{1}$ + bx + c, we get: a = 1, b = -2 and c = 3k It is given that $α$ and $β$ are the roots of the polynomial. $∴ α + β = - \frac{b}{a}$ $\Rightarrow α + β = - (\frac{- 2}{1})$ $\Rightarrow α + β = 2 \dots (i)$ Also, $α β = \frac{c}{a}$ $\Rightarrow α β = \frac{5k}{1}$ $\Rightarrow α β = 3k \dots (ii)$ Now, $α + β = α β$ $\Rightarrow 2 = 3k$ [Using (i) and (ii)] $\Rightarrow k = \frac{2}{3}$

POLYNOMIALS

89961 If $α$ and $β$ are zeros of x $^{1}$ + 5x + 8, then the value of $(α + β)$ is:

1 8

2 5

3 -5

4 -8

Explanation:

-5
$x^{2} + 5 x + 8$ $α + β = \frac{Coefficient of x}{Coefficient of x^{2}}$ $= \frac{- 5}{1} = - 5$

POLYNOMIALS

89962 Which of the following is not a polynomial?

1

\sqrt{3} x^{2} - 2 \sqrt{3} x + 5

2

{9x}^{2} - 4x + \sqrt{2}

3

\frac{3}{2} x^{3} + {6x}^{2} - \frac{1}{\sqrt{2}} x - 8

4

x + \frac{3}{x}

Explanation:

$x + \frac{3}{x}$
An expression of the form p(x) = a $_{\1}$ + a $_{\1}$ x + a $_{\1}$ x $^{1}$ + ... + a $_{\1}$ x $^{1}$ , where a $_{\1}$ ? 0, is called a polynomial in x of degree n. Here, a $_{\1}$ , a $_{\1}$ , a $_{\1}$ , ..., a $_{\1}$ are real numbers and each power of x is a non-negative integer. Hence, $x + \frac{3}{x}$ is not a polynomial.

POLYNOMIALS

89963 If the sum of the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 is 6, then the value of k is:

1 2

2 4

3 -2

4 -4

Explanation:

4
Let $α, β$ be the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 and we are given that Then, $α + β + γ = \frac{- Coefficient of x}{{Coefficient of x}^{2}}$ It is given that $α + β + γ = 6$ Substituting $α + β + γ = \frac{3 k}{2},$ we get $\frac{+ 3 k}{2} = 6$ $+ 3 k = 6 \times 2$ $+ 3 k = 12$ $k = \frac{12}{+ 3}$ $k = + 4$ The value of k is 4 Hence, the correct alternative is (b)

POLYNOMIALS

89960 If $α, β$ are the zeros of kx $^{1}$ - 2x + 3k such that $α + β = α β,$ then k = ?

1

\frac{1}{3}

2

\frac{- 1}{3}

3

\frac{2}{3}

4

\frac{- 2}{3}

Explanation:

$\frac{2}{3}$
Here, p(x) = x $^{1}$ - 2x + 3k Comparing the given polynomial with ax $^{1}$ + bx + c, we get: a = 1, b = -2 and c = 3k It is given that $α$ and $β$ are the roots of the polynomial. $∴ α + β = - \frac{b}{a}$ $\Rightarrow α + β = - (\frac{- 2}{1})$ $\Rightarrow α + β = 2 \dots (i)$ Also, $α β = \frac{c}{a}$ $\Rightarrow α β = \frac{5k}{1}$ $\Rightarrow α β = 3k \dots (ii)$ Now, $α + β = α β$ $\Rightarrow 2 = 3k$ [Using (i) and (ii)] $\Rightarrow k = \frac{2}{3}$

POLYNOMIALS

89961 If $α$ and $β$ are zeros of x $^{1}$ + 5x + 8, then the value of $(α + β)$ is:

1 8

2 5

3 -5

4 -8

Explanation:

-5
$x^{2} + 5 x + 8$ $α + β = \frac{Coefficient of x}{Coefficient of x^{2}}$ $= \frac{- 5}{1} = - 5$

POLYNOMIALS

89962 Which of the following is not a polynomial?

1

\sqrt{3} x^{2} - 2 \sqrt{3} x + 5

2

{9x}^{2} - 4x + \sqrt{2}

3

\frac{3}{2} x^{3} + {6x}^{2} - \frac{1}{\sqrt{2}} x - 8

4

x + \frac{3}{x}

Explanation:

$x + \frac{3}{x}$
An expression of the form p(x) = a $_{\1}$ + a $_{\1}$ x + a $_{\1}$ x $^{1}$ + ... + a $_{\1}$ x $^{1}$ , where a $_{\1}$ ? 0, is called a polynomial in x of degree n. Here, a $_{\1}$ , a $_{\1}$ , a $_{\1}$ , ..., a $_{\1}$ are real numbers and each power of x is a non-negative integer. Hence, $x + \frac{3}{x}$ is not a polynomial.

POLYNOMIALS

89963 If the sum of the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 is 6, then the value of k is:

1 2

2 4

3 -2

4 -4

Explanation:

4
Let $α, β$ be the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 and we are given that Then, $α + β + γ = \frac{- Coefficient of x}{{Coefficient of x}^{2}}$ It is given that $α + β + γ = 6$ Substituting $α + β + γ = \frac{3 k}{2},$ we get $\frac{+ 3 k}{2} = 6$ $+ 3 k = 6 \times 2$ $+ 3 k = 12$ $k = \frac{12}{+ 3}$ $k = + 4$ The value of k is 4 Hence, the correct alternative is (b)

POLYNOMIALS

89960 If $α, β$ are the zeros of kx $^{1}$ - 2x + 3k such that $α + β = α β,$ then k = ?

1

\frac{1}{3}

2

\frac{- 1}{3}

3

\frac{2}{3}

4

\frac{- 2}{3}

Explanation:

$\frac{2}{3}$
Here, p(x) = x $^{1}$ - 2x + 3k Comparing the given polynomial with ax $^{1}$ + bx + c, we get: a = 1, b = -2 and c = 3k It is given that $α$ and $β$ are the roots of the polynomial. $∴ α + β = - \frac{b}{a}$ $\Rightarrow α + β = - (\frac{- 2}{1})$ $\Rightarrow α + β = 2 \dots (i)$ Also, $α β = \frac{c}{a}$ $\Rightarrow α β = \frac{5k}{1}$ $\Rightarrow α β = 3k \dots (ii)$ Now, $α + β = α β$ $\Rightarrow 2 = 3k$ [Using (i) and (ii)] $\Rightarrow k = \frac{2}{3}$

POLYNOMIALS

89961 If $α$ and $β$ are zeros of x $^{1}$ + 5x + 8, then the value of $(α + β)$ is:

1 8

2 5

3 -5

4 -8

Explanation:

-5
$x^{2} + 5 x + 8$ $α + β = \frac{Coefficient of x}{Coefficient of x^{2}}$ $= \frac{- 5}{1} = - 5$

POLYNOMIALS

89962 Which of the following is not a polynomial?

1

\sqrt{3} x^{2} - 2 \sqrt{3} x + 5

2

{9x}^{2} - 4x + \sqrt{2}

3

\frac{3}{2} x^{3} + {6x}^{2} - \frac{1}{\sqrt{2}} x - 8

4

x + \frac{3}{x}

Explanation:

$x + \frac{3}{x}$
An expression of the form p(x) = a $_{\1}$ + a $_{\1}$ x + a $_{\1}$ x $^{1}$ + ... + a $_{\1}$ x $^{1}$ , where a $_{\1}$ ? 0, is called a polynomial in x of degree n. Here, a $_{\1}$ , a $_{\1}$ , a $_{\1}$ , ..., a $_{\1}$ are real numbers and each power of x is a non-negative integer. Hence, $x + \frac{3}{x}$ is not a polynomial.

POLYNOMIALS

89963 If the sum of the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 is 6, then the value of k is:

1 2

2 4

3 -2

4 -4

Explanation:

4
Let $α, β$ be the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 and we are given that Then, $α + β + γ = \frac{- Coefficient of x}{{Coefficient of x}^{2}}$ It is given that $α + β + γ = 6$ Substituting $α + β + γ = \frac{3 k}{2},$ we get $\frac{+ 3 k}{2} = 6$ $+ 3 k = 6 \times 2$ $+ 3 k = 12$ $k = \frac{12}{+ 3}$ $k = + 4$ The value of k is 4 Hence, the correct alternative is (b)

POLYNOMIALS

89960 If $α, β$ are the zeros of kx $^{1}$ - 2x + 3k such that $α + β = α β,$ then k = ?

1

\frac{1}{3}

2

\frac{- 1}{3}

3

\frac{2}{3}

4

\frac{- 2}{3}

Explanation:

$\frac{2}{3}$
Here, p(x) = x $^{1}$ - 2x + 3k Comparing the given polynomial with ax $^{1}$ + bx + c, we get: a = 1, b = -2 and c = 3k It is given that $α$ and $β$ are the roots of the polynomial. $∴ α + β = - \frac{b}{a}$ $\Rightarrow α + β = - (\frac{- 2}{1})$ $\Rightarrow α + β = 2 \dots (i)$ Also, $α β = \frac{c}{a}$ $\Rightarrow α β = \frac{5k}{1}$ $\Rightarrow α β = 3k \dots (ii)$ Now, $α + β = α β$ $\Rightarrow 2 = 3k$ [Using (i) and (ii)] $\Rightarrow k = \frac{2}{3}$

POLYNOMIALS

89961 If $α$ and $β$ are zeros of x $^{1}$ + 5x + 8, then the value of $(α + β)$ is:

1 8

2 5

3 -5

4 -8

Explanation:

-5
$x^{2} + 5 x + 8$ $α + β = \frac{Coefficient of x}{Coefficient of x^{2}}$ $= \frac{- 5}{1} = - 5$

POLYNOMIALS

89962 Which of the following is not a polynomial?

1

\sqrt{3} x^{2} - 2 \sqrt{3} x + 5

2

{9x}^{2} - 4x + \sqrt{2}

3

\frac{3}{2} x^{3} + {6x}^{2} - \frac{1}{\sqrt{2}} x - 8

4

x + \frac{3}{x}

Explanation:

$x + \frac{3}{x}$
An expression of the form p(x) = a $_{\1}$ + a $_{\1}$ x + a $_{\1}$ x $^{1}$ + ... + a $_{\1}$ x $^{1}$ , where a $_{\1}$ ? 0, is called a polynomial in x of degree n. Here, a $_{\1}$ , a $_{\1}$ , a $_{\1}$ , ..., a $_{\1}$ are real numbers and each power of x is a non-negative integer. Hence, $x + \frac{3}{x}$ is not a polynomial.

POLYNOMIALS

89963 If the sum of the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 is 6, then the value of k is:

1 2

2 4

3 -2

4 -4

Explanation:

4
Let $α, β$ be the zeros of the polynomial f(x) = 2x $^{1}$ - 3kx $^{1}$ + 4x - 5 and we are given that Then, $α + β + γ = \frac{- Coefficient of x}{{Coefficient of x}^{2}}$ It is given that $α + β + γ = 6$ Substituting $α + β + γ = \frac{3 k}{2},$ we get $\frac{+ 3 k}{2} = 6$ $+ 3 k = 6 \times 2$ $+ 3 k = 12$ $k = \frac{12}{+ 3}$ $k = + 4$ The value of k is 4 Hence, the correct alternative is (b)

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: