90086
If one zero of 3x\(^{1}\) + 8x + k be the reciprocal of the other, then k = ?
1 \(3\)
2 \(-3\)
3 \(\frac{1}{3}\)
4 \(\frac{-1}{3}\)
Explanation:
\(3\) Let \(\alpha\) and \(\frac{1}{\alpha}\) be the roots of 3x\(^{1}\) + 8x + k. Then, we have \(\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}\) \(\Rightarrow1=\frac{\text{k}}{3}\) \(\Rightarrow\text{k}=3\)
POLYNOMIALS
90087
If \(\alpha,\ \beta,\ \gamma\) are the zeros of the polynomial x\(^{1}\) - 6x\(^{1}\) - x + 30, then the value of \((\alpha\beta+\beta\gamma+\gamma\alpha)\) is:
1 -1
2 1
3 -5
4 30
Explanation:
-1 Here, p(x) = x\(^{1}\) - 6x\(^{1}\) - x + 30 Comparing the given polynomial with \(\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,\) we get: \((\alpha\beta+\beta\gamma+\gamma\alpha)=-1\)
POLYNOMIALS
90088
The zeroes of a polynomial x\(^{1}\) + 5x - 24 are:
1 Both negative
2 One positive and one negative
3 Both positive
4 Both equal
Explanation:
One positive and one negative x\(^{1}\) + 5x - 24 = x\(^{1}\) + 8x - 3x - 24 = x (x + 8) - 3 (x + 8) = 0 = (x + 8) (x - 3) = 0 = x + 8 = 0 or x - 3 = 0 ? x = -8 or x = 3
POLYNOMIALS
90089
The zeroes of a polynomial x\(^{1}\) - 7x + 12 are:
1 One positive and one negative
2 Both equal
3 Both positive
4 Both negative
Explanation:
Both positive x\(^{1}\) - 7x + 12 = x\(^{1}\) - 4x - 3x + 12 = 0 = x(x - 4) - 3 (x - 4) = 0 = (x - 4) (x - 3) = 0 \(\therefore\) x - 4 = 0 or x - 3 = 0 ? x = 4 or x = 3
90086
If one zero of 3x\(^{1}\) + 8x + k be the reciprocal of the other, then k = ?
1 \(3\)
2 \(-3\)
3 \(\frac{1}{3}\)
4 \(\frac{-1}{3}\)
Explanation:
\(3\) Let \(\alpha\) and \(\frac{1}{\alpha}\) be the roots of 3x\(^{1}\) + 8x + k. Then, we have \(\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}\) \(\Rightarrow1=\frac{\text{k}}{3}\) \(\Rightarrow\text{k}=3\)
POLYNOMIALS
90087
If \(\alpha,\ \beta,\ \gamma\) are the zeros of the polynomial x\(^{1}\) - 6x\(^{1}\) - x + 30, then the value of \((\alpha\beta+\beta\gamma+\gamma\alpha)\) is:
1 -1
2 1
3 -5
4 30
Explanation:
-1 Here, p(x) = x\(^{1}\) - 6x\(^{1}\) - x + 30 Comparing the given polynomial with \(\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,\) we get: \((\alpha\beta+\beta\gamma+\gamma\alpha)=-1\)
POLYNOMIALS
90088
The zeroes of a polynomial x\(^{1}\) + 5x - 24 are:
1 Both negative
2 One positive and one negative
3 Both positive
4 Both equal
Explanation:
One positive and one negative x\(^{1}\) + 5x - 24 = x\(^{1}\) + 8x - 3x - 24 = x (x + 8) - 3 (x + 8) = 0 = (x + 8) (x - 3) = 0 = x + 8 = 0 or x - 3 = 0 ? x = -8 or x = 3
POLYNOMIALS
90089
The zeroes of a polynomial x\(^{1}\) - 7x + 12 are:
1 One positive and one negative
2 Both equal
3 Both positive
4 Both negative
Explanation:
Both positive x\(^{1}\) - 7x + 12 = x\(^{1}\) - 4x - 3x + 12 = 0 = x(x - 4) - 3 (x - 4) = 0 = (x - 4) (x - 3) = 0 \(\therefore\) x - 4 = 0 or x - 3 = 0 ? x = 4 or x = 3
90086
If one zero of 3x\(^{1}\) + 8x + k be the reciprocal of the other, then k = ?
1 \(3\)
2 \(-3\)
3 \(\frac{1}{3}\)
4 \(\frac{-1}{3}\)
Explanation:
\(3\) Let \(\alpha\) and \(\frac{1}{\alpha}\) be the roots of 3x\(^{1}\) + 8x + k. Then, we have \(\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}\) \(\Rightarrow1=\frac{\text{k}}{3}\) \(\Rightarrow\text{k}=3\)
POLYNOMIALS
90087
If \(\alpha,\ \beta,\ \gamma\) are the zeros of the polynomial x\(^{1}\) - 6x\(^{1}\) - x + 30, then the value of \((\alpha\beta+\beta\gamma+\gamma\alpha)\) is:
1 -1
2 1
3 -5
4 30
Explanation:
-1 Here, p(x) = x\(^{1}\) - 6x\(^{1}\) - x + 30 Comparing the given polynomial with \(\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,\) we get: \((\alpha\beta+\beta\gamma+\gamma\alpha)=-1\)
POLYNOMIALS
90088
The zeroes of a polynomial x\(^{1}\) + 5x - 24 are:
1 Both negative
2 One positive and one negative
3 Both positive
4 Both equal
Explanation:
One positive and one negative x\(^{1}\) + 5x - 24 = x\(^{1}\) + 8x - 3x - 24 = x (x + 8) - 3 (x + 8) = 0 = (x + 8) (x - 3) = 0 = x + 8 = 0 or x - 3 = 0 ? x = -8 or x = 3
POLYNOMIALS
90089
The zeroes of a polynomial x\(^{1}\) - 7x + 12 are:
1 One positive and one negative
2 Both equal
3 Both positive
4 Both negative
Explanation:
Both positive x\(^{1}\) - 7x + 12 = x\(^{1}\) - 4x - 3x + 12 = 0 = x(x - 4) - 3 (x - 4) = 0 = (x - 4) (x - 3) = 0 \(\therefore\) x - 4 = 0 or x - 3 = 0 ? x = 4 or x = 3
NEET Test Series from KOTA - 10 Papers In MS WORD
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POLYNOMIALS
90086
If one zero of 3x\(^{1}\) + 8x + k be the reciprocal of the other, then k = ?
1 \(3\)
2 \(-3\)
3 \(\frac{1}{3}\)
4 \(\frac{-1}{3}\)
Explanation:
\(3\) Let \(\alpha\) and \(\frac{1}{\alpha}\) be the roots of 3x\(^{1}\) + 8x + k. Then, we have \(\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}\) \(\Rightarrow1=\frac{\text{k}}{3}\) \(\Rightarrow\text{k}=3\)
POLYNOMIALS
90087
If \(\alpha,\ \beta,\ \gamma\) are the zeros of the polynomial x\(^{1}\) - 6x\(^{1}\) - x + 30, then the value of \((\alpha\beta+\beta\gamma+\gamma\alpha)\) is:
1 -1
2 1
3 -5
4 30
Explanation:
-1 Here, p(x) = x\(^{1}\) - 6x\(^{1}\) - x + 30 Comparing the given polynomial with \(\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,\) we get: \((\alpha\beta+\beta\gamma+\gamma\alpha)=-1\)
POLYNOMIALS
90088
The zeroes of a polynomial x\(^{1}\) + 5x - 24 are:
1 Both negative
2 One positive and one negative
3 Both positive
4 Both equal
Explanation:
One positive and one negative x\(^{1}\) + 5x - 24 = x\(^{1}\) + 8x - 3x - 24 = x (x + 8) - 3 (x + 8) = 0 = (x + 8) (x - 3) = 0 = x + 8 = 0 or x - 3 = 0 ? x = -8 or x = 3
POLYNOMIALS
90089
The zeroes of a polynomial x\(^{1}\) - 7x + 12 are:
1 One positive and one negative
2 Both equal
3 Both positive
4 Both negative
Explanation:
Both positive x\(^{1}\) - 7x + 12 = x\(^{1}\) - 4x - 3x + 12 = 0 = x(x - 4) - 3 (x - 4) = 0 = (x - 4) (x - 3) = 0 \(\therefore\) x - 4 = 0 or x - 3 = 0 ? x = 4 or x = 3
