118256
The quadratic equation \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite sign, Then,
1 a \(\leq 0\)
2 \(0\lt \) a \(\lt 4\)
3 \(4 \leq\) a \(\lt 8\)
4 \(a \geq 8\)
Explanation:
B Given, \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) Let \(\alpha\) and \(\beta\) be the roots of the equation \(\alpha+\beta=\frac{a^3+8 a-1}{2}\) \(\alpha \cdot \beta=\frac{a^2-4 a}{2}\) Roots are opposite sign, then, \(\alpha . \beta\lt 0\) \(\frac{a^2-4 a}{2}\lt 0\) \(a(a-4)\lt 0\)Hence, \(0\lt \mathrm{a}\lt 4 \quad \mathrm{a} \in(0,4)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118257
The equations \(x^2+x+a=0\) and \(x^2+a x+1=0\) have a common real root
1 for no value of a
2 for exactly one value of a
3 for exactly two vales of a
4 for exactly three values of a
Explanation:
B Let \(\alpha\) be the common roots, \(\alpha^2+\alpha+a=0\) and \(\alpha^2+a \alpha+1=0\) \(\frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1} \quad(a \neq 1)\) Eliminating \(\alpha\), we get, \((a-1)^2=\left(1-a^2\right)(a-1)\) \((a-1)=1-a^2\) \(a^2+a-2=0\) \(a^2+2 a-a-2=0\) \(a(a+2)-1(a+2)=0\) \((a+2)(a-1)=0\) \(a=-2 \quad(\because a \neq 1)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118340
If \(k \in R\) then roots of \((x-2)(x-3)=k^2\) are always
1 real and distinct
2 real and equal
3 complex number
4 rational numbers
Explanation:
A Given, \((x-2)(x-3)=k^2, k \in R\) \(x^2-3 x-2 x+6-k^2=0\) \(x^2-5 x+6-k^2=0\) Now, discriminate \(\mathrm{D}=(-5)^2-4 \times 1 \times\left(6-\mathrm{k}^2\right)\) \(=25-24+4 \mathrm{k}^2\) \(=1+4 \mathrm{k}^2>0\)Hence, the roots are real and distinct.
TS EAMCET-05.05.2018
Complex Numbers and Quadratic Equation
118213
If one root of the equation \(\mathbf{x}^2+p x+12=0\) is 4 and the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, the value of \(q\) is
1 \(49 / 4\)
2 4
3 3
4 12
Explanation:
A Since 4 is one of the roots of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\), Therefore \(\mathrm{x}=4\) satisfies the equation. \((4)^2+\mathrm{p}(4)+12=0\) \(\mathrm{p}=-7\) \(\therefore \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) Now, It is given that \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) has equal roots. \(\therefore D=0\) \(\Rightarrow b^2-4 a c=0\) \(49-4 q=0\) \(q=\frac{49}{4}\)
BCECE-2012]
Complex Numbers and Quadratic Equation
118214
If \(p\) and \(q\) are roots of the quadratic equation \(\mathbf{x}^2+\mathbf{m x}+\mathbf{m}^2+\mathbf{a}=\mathbf{0}\), then the value of \(\mathbf{p}^2+\mathbf{q}^2+\) pq is
1 0
2 a
3 \(-\mathrm{a}\)
4 \(\pm \mathrm{m}^2\)
Explanation:
C \(\mathrm{p}\) and \(\mathrm{q}\) are roots of the equation :- \(\therefore \mathrm{p}+\mathrm{q} =-\mathrm{m}\) \(\mathrm{pq} =\left(\mathrm{m}^2+\mathrm{a}\right)\) Now, \(p q =\left(m^2+a\right)\) \(p^2+q^2+p q =(p+q)^2-p q\) \(=(-m)^2-\left(m^2+a\right)\) \(=m^2-m^2-a\) \(=-a\)
118256
The quadratic equation \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite sign, Then,
1 a \(\leq 0\)
2 \(0\lt \) a \(\lt 4\)
3 \(4 \leq\) a \(\lt 8\)
4 \(a \geq 8\)
Explanation:
B Given, \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) Let \(\alpha\) and \(\beta\) be the roots of the equation \(\alpha+\beta=\frac{a^3+8 a-1}{2}\) \(\alpha \cdot \beta=\frac{a^2-4 a}{2}\) Roots are opposite sign, then, \(\alpha . \beta\lt 0\) \(\frac{a^2-4 a}{2}\lt 0\) \(a(a-4)\lt 0\)Hence, \(0\lt \mathrm{a}\lt 4 \quad \mathrm{a} \in(0,4)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118257
The equations \(x^2+x+a=0\) and \(x^2+a x+1=0\) have a common real root
1 for no value of a
2 for exactly one value of a
3 for exactly two vales of a
4 for exactly three values of a
Explanation:
B Let \(\alpha\) be the common roots, \(\alpha^2+\alpha+a=0\) and \(\alpha^2+a \alpha+1=0\) \(\frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1} \quad(a \neq 1)\) Eliminating \(\alpha\), we get, \((a-1)^2=\left(1-a^2\right)(a-1)\) \((a-1)=1-a^2\) \(a^2+a-2=0\) \(a^2+2 a-a-2=0\) \(a(a+2)-1(a+2)=0\) \((a+2)(a-1)=0\) \(a=-2 \quad(\because a \neq 1)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118340
If \(k \in R\) then roots of \((x-2)(x-3)=k^2\) are always
1 real and distinct
2 real and equal
3 complex number
4 rational numbers
Explanation:
A Given, \((x-2)(x-3)=k^2, k \in R\) \(x^2-3 x-2 x+6-k^2=0\) \(x^2-5 x+6-k^2=0\) Now, discriminate \(\mathrm{D}=(-5)^2-4 \times 1 \times\left(6-\mathrm{k}^2\right)\) \(=25-24+4 \mathrm{k}^2\) \(=1+4 \mathrm{k}^2>0\)Hence, the roots are real and distinct.
TS EAMCET-05.05.2018
Complex Numbers and Quadratic Equation
118213
If one root of the equation \(\mathbf{x}^2+p x+12=0\) is 4 and the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, the value of \(q\) is
1 \(49 / 4\)
2 4
3 3
4 12
Explanation:
A Since 4 is one of the roots of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\), Therefore \(\mathrm{x}=4\) satisfies the equation. \((4)^2+\mathrm{p}(4)+12=0\) \(\mathrm{p}=-7\) \(\therefore \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) Now, It is given that \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) has equal roots. \(\therefore D=0\) \(\Rightarrow b^2-4 a c=0\) \(49-4 q=0\) \(q=\frac{49}{4}\)
BCECE-2012]
Complex Numbers and Quadratic Equation
118214
If \(p\) and \(q\) are roots of the quadratic equation \(\mathbf{x}^2+\mathbf{m x}+\mathbf{m}^2+\mathbf{a}=\mathbf{0}\), then the value of \(\mathbf{p}^2+\mathbf{q}^2+\) pq is
1 0
2 a
3 \(-\mathrm{a}\)
4 \(\pm \mathrm{m}^2\)
Explanation:
C \(\mathrm{p}\) and \(\mathrm{q}\) are roots of the equation :- \(\therefore \mathrm{p}+\mathrm{q} =-\mathrm{m}\) \(\mathrm{pq} =\left(\mathrm{m}^2+\mathrm{a}\right)\) Now, \(p q =\left(m^2+a\right)\) \(p^2+q^2+p q =(p+q)^2-p q\) \(=(-m)^2-\left(m^2+a\right)\) \(=m^2-m^2-a\) \(=-a\)
118256
The quadratic equation \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite sign, Then,
1 a \(\leq 0\)
2 \(0\lt \) a \(\lt 4\)
3 \(4 \leq\) a \(\lt 8\)
4 \(a \geq 8\)
Explanation:
B Given, \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) Let \(\alpha\) and \(\beta\) be the roots of the equation \(\alpha+\beta=\frac{a^3+8 a-1}{2}\) \(\alpha \cdot \beta=\frac{a^2-4 a}{2}\) Roots are opposite sign, then, \(\alpha . \beta\lt 0\) \(\frac{a^2-4 a}{2}\lt 0\) \(a(a-4)\lt 0\)Hence, \(0\lt \mathrm{a}\lt 4 \quad \mathrm{a} \in(0,4)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118257
The equations \(x^2+x+a=0\) and \(x^2+a x+1=0\) have a common real root
1 for no value of a
2 for exactly one value of a
3 for exactly two vales of a
4 for exactly three values of a
Explanation:
B Let \(\alpha\) be the common roots, \(\alpha^2+\alpha+a=0\) and \(\alpha^2+a \alpha+1=0\) \(\frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1} \quad(a \neq 1)\) Eliminating \(\alpha\), we get, \((a-1)^2=\left(1-a^2\right)(a-1)\) \((a-1)=1-a^2\) \(a^2+a-2=0\) \(a^2+2 a-a-2=0\) \(a(a+2)-1(a+2)=0\) \((a+2)(a-1)=0\) \(a=-2 \quad(\because a \neq 1)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118340
If \(k \in R\) then roots of \((x-2)(x-3)=k^2\) are always
1 real and distinct
2 real and equal
3 complex number
4 rational numbers
Explanation:
A Given, \((x-2)(x-3)=k^2, k \in R\) \(x^2-3 x-2 x+6-k^2=0\) \(x^2-5 x+6-k^2=0\) Now, discriminate \(\mathrm{D}=(-5)^2-4 \times 1 \times\left(6-\mathrm{k}^2\right)\) \(=25-24+4 \mathrm{k}^2\) \(=1+4 \mathrm{k}^2>0\)Hence, the roots are real and distinct.
TS EAMCET-05.05.2018
Complex Numbers and Quadratic Equation
118213
If one root of the equation \(\mathbf{x}^2+p x+12=0\) is 4 and the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, the value of \(q\) is
1 \(49 / 4\)
2 4
3 3
4 12
Explanation:
A Since 4 is one of the roots of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\), Therefore \(\mathrm{x}=4\) satisfies the equation. \((4)^2+\mathrm{p}(4)+12=0\) \(\mathrm{p}=-7\) \(\therefore \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) Now, It is given that \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) has equal roots. \(\therefore D=0\) \(\Rightarrow b^2-4 a c=0\) \(49-4 q=0\) \(q=\frac{49}{4}\)
BCECE-2012]
Complex Numbers and Quadratic Equation
118214
If \(p\) and \(q\) are roots of the quadratic equation \(\mathbf{x}^2+\mathbf{m x}+\mathbf{m}^2+\mathbf{a}=\mathbf{0}\), then the value of \(\mathbf{p}^2+\mathbf{q}^2+\) pq is
1 0
2 a
3 \(-\mathrm{a}\)
4 \(\pm \mathrm{m}^2\)
Explanation:
C \(\mathrm{p}\) and \(\mathrm{q}\) are roots of the equation :- \(\therefore \mathrm{p}+\mathrm{q} =-\mathrm{m}\) \(\mathrm{pq} =\left(\mathrm{m}^2+\mathrm{a}\right)\) Now, \(p q =\left(m^2+a\right)\) \(p^2+q^2+p q =(p+q)^2-p q\) \(=(-m)^2-\left(m^2+a\right)\) \(=m^2-m^2-a\) \(=-a\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118256
The quadratic equation \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite sign, Then,
1 a \(\leq 0\)
2 \(0\lt \) a \(\lt 4\)
3 \(4 \leq\) a \(\lt 8\)
4 \(a \geq 8\)
Explanation:
B Given, \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) Let \(\alpha\) and \(\beta\) be the roots of the equation \(\alpha+\beta=\frac{a^3+8 a-1}{2}\) \(\alpha \cdot \beta=\frac{a^2-4 a}{2}\) Roots are opposite sign, then, \(\alpha . \beta\lt 0\) \(\frac{a^2-4 a}{2}\lt 0\) \(a(a-4)\lt 0\)Hence, \(0\lt \mathrm{a}\lt 4 \quad \mathrm{a} \in(0,4)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118257
The equations \(x^2+x+a=0\) and \(x^2+a x+1=0\) have a common real root
1 for no value of a
2 for exactly one value of a
3 for exactly two vales of a
4 for exactly three values of a
Explanation:
B Let \(\alpha\) be the common roots, \(\alpha^2+\alpha+a=0\) and \(\alpha^2+a \alpha+1=0\) \(\frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1} \quad(a \neq 1)\) Eliminating \(\alpha\), we get, \((a-1)^2=\left(1-a^2\right)(a-1)\) \((a-1)=1-a^2\) \(a^2+a-2=0\) \(a^2+2 a-a-2=0\) \(a(a+2)-1(a+2)=0\) \((a+2)(a-1)=0\) \(a=-2 \quad(\because a \neq 1)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118340
If \(k \in R\) then roots of \((x-2)(x-3)=k^2\) are always
1 real and distinct
2 real and equal
3 complex number
4 rational numbers
Explanation:
A Given, \((x-2)(x-3)=k^2, k \in R\) \(x^2-3 x-2 x+6-k^2=0\) \(x^2-5 x+6-k^2=0\) Now, discriminate \(\mathrm{D}=(-5)^2-4 \times 1 \times\left(6-\mathrm{k}^2\right)\) \(=25-24+4 \mathrm{k}^2\) \(=1+4 \mathrm{k}^2>0\)Hence, the roots are real and distinct.
TS EAMCET-05.05.2018
Complex Numbers and Quadratic Equation
118213
If one root of the equation \(\mathbf{x}^2+p x+12=0\) is 4 and the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, the value of \(q\) is
1 \(49 / 4\)
2 4
3 3
4 12
Explanation:
A Since 4 is one of the roots of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\), Therefore \(\mathrm{x}=4\) satisfies the equation. \((4)^2+\mathrm{p}(4)+12=0\) \(\mathrm{p}=-7\) \(\therefore \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) Now, It is given that \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) has equal roots. \(\therefore D=0\) \(\Rightarrow b^2-4 a c=0\) \(49-4 q=0\) \(q=\frac{49}{4}\)
BCECE-2012]
Complex Numbers and Quadratic Equation
118214
If \(p\) and \(q\) are roots of the quadratic equation \(\mathbf{x}^2+\mathbf{m x}+\mathbf{m}^2+\mathbf{a}=\mathbf{0}\), then the value of \(\mathbf{p}^2+\mathbf{q}^2+\) pq is
1 0
2 a
3 \(-\mathrm{a}\)
4 \(\pm \mathrm{m}^2\)
Explanation:
C \(\mathrm{p}\) and \(\mathrm{q}\) are roots of the equation :- \(\therefore \mathrm{p}+\mathrm{q} =-\mathrm{m}\) \(\mathrm{pq} =\left(\mathrm{m}^2+\mathrm{a}\right)\) Now, \(p q =\left(m^2+a\right)\) \(p^2+q^2+p q =(p+q)^2-p q\) \(=(-m)^2-\left(m^2+a\right)\) \(=m^2-m^2-a\) \(=-a\)
118256
The quadratic equation \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite sign, Then,
1 a \(\leq 0\)
2 \(0\lt \) a \(\lt 4\)
3 \(4 \leq\) a \(\lt 8\)
4 \(a \geq 8\)
Explanation:
B Given, \(2 x^2-\left(a^3+8 a-1\right) x+a^2-4 a=0\) Let \(\alpha\) and \(\beta\) be the roots of the equation \(\alpha+\beta=\frac{a^3+8 a-1}{2}\) \(\alpha \cdot \beta=\frac{a^2-4 a}{2}\) Roots are opposite sign, then, \(\alpha . \beta\lt 0\) \(\frac{a^2-4 a}{2}\lt 0\) \(a(a-4)\lt 0\)Hence, \(0\lt \mathrm{a}\lt 4 \quad \mathrm{a} \in(0,4)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118257
The equations \(x^2+x+a=0\) and \(x^2+a x+1=0\) have a common real root
1 for no value of a
2 for exactly one value of a
3 for exactly two vales of a
4 for exactly three values of a
Explanation:
B Let \(\alpha\) be the common roots, \(\alpha^2+\alpha+a=0\) and \(\alpha^2+a \alpha+1=0\) \(\frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1} \quad(a \neq 1)\) Eliminating \(\alpha\), we get, \((a-1)^2=\left(1-a^2\right)(a-1)\) \((a-1)=1-a^2\) \(a^2+a-2=0\) \(a^2+2 a-a-2=0\) \(a(a+2)-1(a+2)=0\) \((a+2)(a-1)=0\) \(a=-2 \quad(\because a \neq 1)\)
WB JEE-2012
Complex Numbers and Quadratic Equation
118340
If \(k \in R\) then roots of \((x-2)(x-3)=k^2\) are always
1 real and distinct
2 real and equal
3 complex number
4 rational numbers
Explanation:
A Given, \((x-2)(x-3)=k^2, k \in R\) \(x^2-3 x-2 x+6-k^2=0\) \(x^2-5 x+6-k^2=0\) Now, discriminate \(\mathrm{D}=(-5)^2-4 \times 1 \times\left(6-\mathrm{k}^2\right)\) \(=25-24+4 \mathrm{k}^2\) \(=1+4 \mathrm{k}^2>0\)Hence, the roots are real and distinct.
TS EAMCET-05.05.2018
Complex Numbers and Quadratic Equation
118213
If one root of the equation \(\mathbf{x}^2+p x+12=0\) is 4 and the equation \(\mathbf{x}^2+\mathbf{p x}+\mathbf{q}=\mathbf{0}\) has equal roots, the value of \(q\) is
1 \(49 / 4\)
2 4
3 3
4 12
Explanation:
A Since 4 is one of the roots of the equation \(\mathrm{x}^2+\mathrm{px}+12=0\), Therefore \(\mathrm{x}=4\) satisfies the equation. \((4)^2+\mathrm{p}(4)+12=0\) \(\mathrm{p}=-7\) \(\therefore \mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\) \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) Now, It is given that \(\mathrm{x}^2-7 \mathrm{x}+\mathrm{q}=0\) has equal roots. \(\therefore D=0\) \(\Rightarrow b^2-4 a c=0\) \(49-4 q=0\) \(q=\frac{49}{4}\)
BCECE-2012]
Complex Numbers and Quadratic Equation
118214
If \(p\) and \(q\) are roots of the quadratic equation \(\mathbf{x}^2+\mathbf{m x}+\mathbf{m}^2+\mathbf{a}=\mathbf{0}\), then the value of \(\mathbf{p}^2+\mathbf{q}^2+\) pq is
1 0
2 a
3 \(-\mathrm{a}\)
4 \(\pm \mathrm{m}^2\)
Explanation:
C \(\mathrm{p}\) and \(\mathrm{q}\) are roots of the equation :- \(\therefore \mathrm{p}+\mathrm{q} =-\mathrm{m}\) \(\mathrm{pq} =\left(\mathrm{m}^2+\mathrm{a}\right)\) Now, \(p q =\left(m^2+a\right)\) \(p^2+q^2+p q =(p+q)^2-p q\) \(=(-m)^2-\left(m^2+a\right)\) \(=m^2-m^2-a\) \(=-a\)