118063
The maximum number of solution of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is
1 1
2 2
3 4
4 3
Explanation:
D Given, \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) \(8^{\sin ^2 x}+8^{1-\sin ^2 x}=9\) \(8^{\sin ^2 x}+8 / 8^{\sin ^2 x}=9\) Let, \(8^{\sin ^2 x}=t\) \(t+8 / t=9\) \(t^2+8-9 t=0\) \(t^2-8 t-t+8=0\) \(t(t-8)-1(t-8)=0 \quad\) \((t-8)(t-1)=0\) \(8^{\sin ^2 x}=8^0\) \(\sin ^2 x=0\) \(x=0\) \(8^{\sin ^2 x}=8^1\) \(\sin ^2 x=1\) \(\sin ^2 x= \pm 1\) \(x= \pm \pi / 2,3 \pi / 2\) \(\mathrm{x}=\pi, \pi / 2,3 \pi / 2\) Hence, the maximum number of solutions of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is 3 .
COMEDK-2011
Complex Numbers and Quadratic Equation
118064
The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
1 2
2 1
3 0
4 3
Explanation:
C Given, equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) Then, \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{x}^2+9-6 \mathrm{x}=0\) \(3 \mathrm{x}^2-12 \mathrm{x}+14=0\) Now, \(D=b^2-4 a c\) \(D=144-4 \times 3 \times 14\) \(\mathrm{D}=-24\) Here, \(\mathrm{D}\lt 0\) So, equation have no real solutions.
BITSAT-2020
Complex Numbers and Quadratic Equation
118065
If \(x, y\) and \(z\) are real numbers then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
1 positive
2 non-positive
3 zero
4 non-negative
Explanation:
D \(: \mathrm{u}=\mathrm{x}^2+4 \mathrm{y}^2+9 \mathrm{z}^2-6 \mathrm{zy}-3 \mathrm{zx}-2 \mathrm{xy}\) \(=\mathrm{x}^2+(2 \mathrm{y})^2+(3 \mathrm{z})^2-(2 \mathrm{y})(3 \mathrm{z})-(3 \mathrm{z})(\mathrm{x})-\mathrm{x}(2 \mathrm{y}) \geq 0\) \({\left[\because \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]}\) \(\text { As } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \in \mathrm{R}, \mathrm{So}, \mathrm{u} \geq 0\) \(\text { Hence, } \mathrm{u} \text { is always positive or non-negative. }\) As \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z} \in \mathrm{R}\), So, \(\mathrm{u} \geq 0\) Hence, \(u\) is always positive or non-negative.
BITSAT-2020
Complex Numbers and Quadratic Equation
118066
If the roots of the equation \(x^2+2 a x+b=0\) are real and differ by at most \(2 \mathrm{~m}, \mathrm{~m} \neq 0\) then \(b\) lies in the interval
B Let the roots be \(\alpha\) and \(\beta\), Then \(\alpha+\beta=-2 \mathrm{a}\) and \(\alpha \beta=\mathrm{b}\) According to the question \(|\alpha-\beta| \leq 2 \mathrm{~m} \Rightarrow(\alpha-\beta)^2 \leq 4 \mathrm{~m}^2\) \(\Rightarrow(\alpha+\beta)^2-4 \alpha \beta \leq 4 \mathrm{~m}^2\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b} \leq 4 \mathrm{~m}^2 \Rightarrow \mathrm{b} \geq \mathrm{a}^2-\mathrm{m}^2\) Again, the roots are real and distinct \(\therefore \mathrm{D}>0\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b}>0 \Rightarrow \mathrm{b}\lt \mathrm{a}^2\). On combining we get, \(\mathrm{a}^2-\mathrm{m}^2 \leq \mathrm{b}\lt \mathrm{a}^2 \therefore \mathrm{b} \in\left[\mathrm{a}^2-\mathrm{m}^2, \mathrm{a}^2\right)\)
BITSAT-2007
Complex Numbers and Quadratic Equation
118067
If the equation \(x^2+2(k+1) x+9 k-5=0\) has only negative roots, then-
1 \(\mathrm{k} \leq 0\)
2 \(\mathrm{k} \geq 0\)
3 \(\mathrm{k} \geq 6\)
4 \(\mathrm{k} \leq 6\)
Explanation:
C Let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2(\mathrm{k}+1) \mathrm{x}+9 \mathrm{k}-5\). Let \(\alpha, \beta\) be the roots of \(f(x)=0\). The equation \(f(x)=0\) will have both negative roots, if- (i) Disc. \(\geq 0\) (ii) \(\alpha\lt 0, \beta\lt 0\), i.e. \((\alpha+\beta)\lt 0\) and (iii) \(\alpha \cdot \beta>0\) Now, \(\mathrm{D} \geq 0 \Rightarrow 4(\mathrm{k}+1)^2-36 \mathrm{k}+20 \geq 0\) \(\Rightarrow \mathrm{k}^2-7 \mathrm{k}+6 \geq 0 \Rightarrow(\mathrm{k}-1)(\mathrm{k}-6) \geq 0\) \(\Rightarrow \mathrm{k} \leq 1\) or \(\mathrm{k} \geq 6\) \((\alpha+\beta)\lt 0 \Rightarrow-2(\mathrm{k}+1)\lt 0\) \(\Rightarrow \mathrm{k}+1>0 \Rightarrow \mathrm{k}>-1\) and, \(\alpha \cdot \beta>0 \Rightarrow 9 \mathrm{k}-5>0 \Rightarrow \mathrm{k}>5 / 9\) From (i), (ii), (iii), we get \(\mathrm{k} \geq 6\).
118063
The maximum number of solution of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is
1 1
2 2
3 4
4 3
Explanation:
D Given, \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) \(8^{\sin ^2 x}+8^{1-\sin ^2 x}=9\) \(8^{\sin ^2 x}+8 / 8^{\sin ^2 x}=9\) Let, \(8^{\sin ^2 x}=t\) \(t+8 / t=9\) \(t^2+8-9 t=0\) \(t^2-8 t-t+8=0\) \(t(t-8)-1(t-8)=0 \quad\) \((t-8)(t-1)=0\) \(8^{\sin ^2 x}=8^0\) \(\sin ^2 x=0\) \(x=0\) \(8^{\sin ^2 x}=8^1\) \(\sin ^2 x=1\) \(\sin ^2 x= \pm 1\) \(x= \pm \pi / 2,3 \pi / 2\) \(\mathrm{x}=\pi, \pi / 2,3 \pi / 2\) Hence, the maximum number of solutions of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is 3 .
COMEDK-2011
Complex Numbers and Quadratic Equation
118064
The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
1 2
2 1
3 0
4 3
Explanation:
C Given, equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) Then, \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{x}^2+9-6 \mathrm{x}=0\) \(3 \mathrm{x}^2-12 \mathrm{x}+14=0\) Now, \(D=b^2-4 a c\) \(D=144-4 \times 3 \times 14\) \(\mathrm{D}=-24\) Here, \(\mathrm{D}\lt 0\) So, equation have no real solutions.
BITSAT-2020
Complex Numbers and Quadratic Equation
118065
If \(x, y\) and \(z\) are real numbers then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
1 positive
2 non-positive
3 zero
4 non-negative
Explanation:
D \(: \mathrm{u}=\mathrm{x}^2+4 \mathrm{y}^2+9 \mathrm{z}^2-6 \mathrm{zy}-3 \mathrm{zx}-2 \mathrm{xy}\) \(=\mathrm{x}^2+(2 \mathrm{y})^2+(3 \mathrm{z})^2-(2 \mathrm{y})(3 \mathrm{z})-(3 \mathrm{z})(\mathrm{x})-\mathrm{x}(2 \mathrm{y}) \geq 0\) \({\left[\because \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]}\) \(\text { As } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \in \mathrm{R}, \mathrm{So}, \mathrm{u} \geq 0\) \(\text { Hence, } \mathrm{u} \text { is always positive or non-negative. }\) As \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z} \in \mathrm{R}\), So, \(\mathrm{u} \geq 0\) Hence, \(u\) is always positive or non-negative.
BITSAT-2020
Complex Numbers and Quadratic Equation
118066
If the roots of the equation \(x^2+2 a x+b=0\) are real and differ by at most \(2 \mathrm{~m}, \mathrm{~m} \neq 0\) then \(b\) lies in the interval
B Let the roots be \(\alpha\) and \(\beta\), Then \(\alpha+\beta=-2 \mathrm{a}\) and \(\alpha \beta=\mathrm{b}\) According to the question \(|\alpha-\beta| \leq 2 \mathrm{~m} \Rightarrow(\alpha-\beta)^2 \leq 4 \mathrm{~m}^2\) \(\Rightarrow(\alpha+\beta)^2-4 \alpha \beta \leq 4 \mathrm{~m}^2\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b} \leq 4 \mathrm{~m}^2 \Rightarrow \mathrm{b} \geq \mathrm{a}^2-\mathrm{m}^2\) Again, the roots are real and distinct \(\therefore \mathrm{D}>0\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b}>0 \Rightarrow \mathrm{b}\lt \mathrm{a}^2\). On combining we get, \(\mathrm{a}^2-\mathrm{m}^2 \leq \mathrm{b}\lt \mathrm{a}^2 \therefore \mathrm{b} \in\left[\mathrm{a}^2-\mathrm{m}^2, \mathrm{a}^2\right)\)
BITSAT-2007
Complex Numbers and Quadratic Equation
118067
If the equation \(x^2+2(k+1) x+9 k-5=0\) has only negative roots, then-
1 \(\mathrm{k} \leq 0\)
2 \(\mathrm{k} \geq 0\)
3 \(\mathrm{k} \geq 6\)
4 \(\mathrm{k} \leq 6\)
Explanation:
C Let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2(\mathrm{k}+1) \mathrm{x}+9 \mathrm{k}-5\). Let \(\alpha, \beta\) be the roots of \(f(x)=0\). The equation \(f(x)=0\) will have both negative roots, if- (i) Disc. \(\geq 0\) (ii) \(\alpha\lt 0, \beta\lt 0\), i.e. \((\alpha+\beta)\lt 0\) and (iii) \(\alpha \cdot \beta>0\) Now, \(\mathrm{D} \geq 0 \Rightarrow 4(\mathrm{k}+1)^2-36 \mathrm{k}+20 \geq 0\) \(\Rightarrow \mathrm{k}^2-7 \mathrm{k}+6 \geq 0 \Rightarrow(\mathrm{k}-1)(\mathrm{k}-6) \geq 0\) \(\Rightarrow \mathrm{k} \leq 1\) or \(\mathrm{k} \geq 6\) \((\alpha+\beta)\lt 0 \Rightarrow-2(\mathrm{k}+1)\lt 0\) \(\Rightarrow \mathrm{k}+1>0 \Rightarrow \mathrm{k}>-1\) and, \(\alpha \cdot \beta>0 \Rightarrow 9 \mathrm{k}-5>0 \Rightarrow \mathrm{k}>5 / 9\) From (i), (ii), (iii), we get \(\mathrm{k} \geq 6\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118063
The maximum number of solution of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is
1 1
2 2
3 4
4 3
Explanation:
D Given, \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) \(8^{\sin ^2 x}+8^{1-\sin ^2 x}=9\) \(8^{\sin ^2 x}+8 / 8^{\sin ^2 x}=9\) Let, \(8^{\sin ^2 x}=t\) \(t+8 / t=9\) \(t^2+8-9 t=0\) \(t^2-8 t-t+8=0\) \(t(t-8)-1(t-8)=0 \quad\) \((t-8)(t-1)=0\) \(8^{\sin ^2 x}=8^0\) \(\sin ^2 x=0\) \(x=0\) \(8^{\sin ^2 x}=8^1\) \(\sin ^2 x=1\) \(\sin ^2 x= \pm 1\) \(x= \pm \pi / 2,3 \pi / 2\) \(\mathrm{x}=\pi, \pi / 2,3 \pi / 2\) Hence, the maximum number of solutions of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is 3 .
COMEDK-2011
Complex Numbers and Quadratic Equation
118064
The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
1 2
2 1
3 0
4 3
Explanation:
C Given, equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) Then, \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{x}^2+9-6 \mathrm{x}=0\) \(3 \mathrm{x}^2-12 \mathrm{x}+14=0\) Now, \(D=b^2-4 a c\) \(D=144-4 \times 3 \times 14\) \(\mathrm{D}=-24\) Here, \(\mathrm{D}\lt 0\) So, equation have no real solutions.
BITSAT-2020
Complex Numbers and Quadratic Equation
118065
If \(x, y\) and \(z\) are real numbers then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
1 positive
2 non-positive
3 zero
4 non-negative
Explanation:
D \(: \mathrm{u}=\mathrm{x}^2+4 \mathrm{y}^2+9 \mathrm{z}^2-6 \mathrm{zy}-3 \mathrm{zx}-2 \mathrm{xy}\) \(=\mathrm{x}^2+(2 \mathrm{y})^2+(3 \mathrm{z})^2-(2 \mathrm{y})(3 \mathrm{z})-(3 \mathrm{z})(\mathrm{x})-\mathrm{x}(2 \mathrm{y}) \geq 0\) \({\left[\because \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]}\) \(\text { As } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \in \mathrm{R}, \mathrm{So}, \mathrm{u} \geq 0\) \(\text { Hence, } \mathrm{u} \text { is always positive or non-negative. }\) As \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z} \in \mathrm{R}\), So, \(\mathrm{u} \geq 0\) Hence, \(u\) is always positive or non-negative.
BITSAT-2020
Complex Numbers and Quadratic Equation
118066
If the roots of the equation \(x^2+2 a x+b=0\) are real and differ by at most \(2 \mathrm{~m}, \mathrm{~m} \neq 0\) then \(b\) lies in the interval
B Let the roots be \(\alpha\) and \(\beta\), Then \(\alpha+\beta=-2 \mathrm{a}\) and \(\alpha \beta=\mathrm{b}\) According to the question \(|\alpha-\beta| \leq 2 \mathrm{~m} \Rightarrow(\alpha-\beta)^2 \leq 4 \mathrm{~m}^2\) \(\Rightarrow(\alpha+\beta)^2-4 \alpha \beta \leq 4 \mathrm{~m}^2\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b} \leq 4 \mathrm{~m}^2 \Rightarrow \mathrm{b} \geq \mathrm{a}^2-\mathrm{m}^2\) Again, the roots are real and distinct \(\therefore \mathrm{D}>0\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b}>0 \Rightarrow \mathrm{b}\lt \mathrm{a}^2\). On combining we get, \(\mathrm{a}^2-\mathrm{m}^2 \leq \mathrm{b}\lt \mathrm{a}^2 \therefore \mathrm{b} \in\left[\mathrm{a}^2-\mathrm{m}^2, \mathrm{a}^2\right)\)
BITSAT-2007
Complex Numbers and Quadratic Equation
118067
If the equation \(x^2+2(k+1) x+9 k-5=0\) has only negative roots, then-
1 \(\mathrm{k} \leq 0\)
2 \(\mathrm{k} \geq 0\)
3 \(\mathrm{k} \geq 6\)
4 \(\mathrm{k} \leq 6\)
Explanation:
C Let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2(\mathrm{k}+1) \mathrm{x}+9 \mathrm{k}-5\). Let \(\alpha, \beta\) be the roots of \(f(x)=0\). The equation \(f(x)=0\) will have both negative roots, if- (i) Disc. \(\geq 0\) (ii) \(\alpha\lt 0, \beta\lt 0\), i.e. \((\alpha+\beta)\lt 0\) and (iii) \(\alpha \cdot \beta>0\) Now, \(\mathrm{D} \geq 0 \Rightarrow 4(\mathrm{k}+1)^2-36 \mathrm{k}+20 \geq 0\) \(\Rightarrow \mathrm{k}^2-7 \mathrm{k}+6 \geq 0 \Rightarrow(\mathrm{k}-1)(\mathrm{k}-6) \geq 0\) \(\Rightarrow \mathrm{k} \leq 1\) or \(\mathrm{k} \geq 6\) \((\alpha+\beta)\lt 0 \Rightarrow-2(\mathrm{k}+1)\lt 0\) \(\Rightarrow \mathrm{k}+1>0 \Rightarrow \mathrm{k}>-1\) and, \(\alpha \cdot \beta>0 \Rightarrow 9 \mathrm{k}-5>0 \Rightarrow \mathrm{k}>5 / 9\) From (i), (ii), (iii), we get \(\mathrm{k} \geq 6\).
118063
The maximum number of solution of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is
1 1
2 2
3 4
4 3
Explanation:
D Given, \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) \(8^{\sin ^2 x}+8^{1-\sin ^2 x}=9\) \(8^{\sin ^2 x}+8 / 8^{\sin ^2 x}=9\) Let, \(8^{\sin ^2 x}=t\) \(t+8 / t=9\) \(t^2+8-9 t=0\) \(t^2-8 t-t+8=0\) \(t(t-8)-1(t-8)=0 \quad\) \((t-8)(t-1)=0\) \(8^{\sin ^2 x}=8^0\) \(\sin ^2 x=0\) \(x=0\) \(8^{\sin ^2 x}=8^1\) \(\sin ^2 x=1\) \(\sin ^2 x= \pm 1\) \(x= \pm \pi / 2,3 \pi / 2\) \(\mathrm{x}=\pi, \pi / 2,3 \pi / 2\) Hence, the maximum number of solutions of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is 3 .
COMEDK-2011
Complex Numbers and Quadratic Equation
118064
The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
1 2
2 1
3 0
4 3
Explanation:
C Given, equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) Then, \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{x}^2+9-6 \mathrm{x}=0\) \(3 \mathrm{x}^2-12 \mathrm{x}+14=0\) Now, \(D=b^2-4 a c\) \(D=144-4 \times 3 \times 14\) \(\mathrm{D}=-24\) Here, \(\mathrm{D}\lt 0\) So, equation have no real solutions.
BITSAT-2020
Complex Numbers and Quadratic Equation
118065
If \(x, y\) and \(z\) are real numbers then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
1 positive
2 non-positive
3 zero
4 non-negative
Explanation:
D \(: \mathrm{u}=\mathrm{x}^2+4 \mathrm{y}^2+9 \mathrm{z}^2-6 \mathrm{zy}-3 \mathrm{zx}-2 \mathrm{xy}\) \(=\mathrm{x}^2+(2 \mathrm{y})^2+(3 \mathrm{z})^2-(2 \mathrm{y})(3 \mathrm{z})-(3 \mathrm{z})(\mathrm{x})-\mathrm{x}(2 \mathrm{y}) \geq 0\) \({\left[\because \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]}\) \(\text { As } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \in \mathrm{R}, \mathrm{So}, \mathrm{u} \geq 0\) \(\text { Hence, } \mathrm{u} \text { is always positive or non-negative. }\) As \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z} \in \mathrm{R}\), So, \(\mathrm{u} \geq 0\) Hence, \(u\) is always positive or non-negative.
BITSAT-2020
Complex Numbers and Quadratic Equation
118066
If the roots of the equation \(x^2+2 a x+b=0\) are real and differ by at most \(2 \mathrm{~m}, \mathrm{~m} \neq 0\) then \(b\) lies in the interval
B Let the roots be \(\alpha\) and \(\beta\), Then \(\alpha+\beta=-2 \mathrm{a}\) and \(\alpha \beta=\mathrm{b}\) According to the question \(|\alpha-\beta| \leq 2 \mathrm{~m} \Rightarrow(\alpha-\beta)^2 \leq 4 \mathrm{~m}^2\) \(\Rightarrow(\alpha+\beta)^2-4 \alpha \beta \leq 4 \mathrm{~m}^2\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b} \leq 4 \mathrm{~m}^2 \Rightarrow \mathrm{b} \geq \mathrm{a}^2-\mathrm{m}^2\) Again, the roots are real and distinct \(\therefore \mathrm{D}>0\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b}>0 \Rightarrow \mathrm{b}\lt \mathrm{a}^2\). On combining we get, \(\mathrm{a}^2-\mathrm{m}^2 \leq \mathrm{b}\lt \mathrm{a}^2 \therefore \mathrm{b} \in\left[\mathrm{a}^2-\mathrm{m}^2, \mathrm{a}^2\right)\)
BITSAT-2007
Complex Numbers and Quadratic Equation
118067
If the equation \(x^2+2(k+1) x+9 k-5=0\) has only negative roots, then-
1 \(\mathrm{k} \leq 0\)
2 \(\mathrm{k} \geq 0\)
3 \(\mathrm{k} \geq 6\)
4 \(\mathrm{k} \leq 6\)
Explanation:
C Let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2(\mathrm{k}+1) \mathrm{x}+9 \mathrm{k}-5\). Let \(\alpha, \beta\) be the roots of \(f(x)=0\). The equation \(f(x)=0\) will have both negative roots, if- (i) Disc. \(\geq 0\) (ii) \(\alpha\lt 0, \beta\lt 0\), i.e. \((\alpha+\beta)\lt 0\) and (iii) \(\alpha \cdot \beta>0\) Now, \(\mathrm{D} \geq 0 \Rightarrow 4(\mathrm{k}+1)^2-36 \mathrm{k}+20 \geq 0\) \(\Rightarrow \mathrm{k}^2-7 \mathrm{k}+6 \geq 0 \Rightarrow(\mathrm{k}-1)(\mathrm{k}-6) \geq 0\) \(\Rightarrow \mathrm{k} \leq 1\) or \(\mathrm{k} \geq 6\) \((\alpha+\beta)\lt 0 \Rightarrow-2(\mathrm{k}+1)\lt 0\) \(\Rightarrow \mathrm{k}+1>0 \Rightarrow \mathrm{k}>-1\) and, \(\alpha \cdot \beta>0 \Rightarrow 9 \mathrm{k}-5>0 \Rightarrow \mathrm{k}>5 / 9\) From (i), (ii), (iii), we get \(\mathrm{k} \geq 6\).
118063
The maximum number of solution of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is
1 1
2 2
3 4
4 3
Explanation:
D Given, \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) \(8^{\sin ^2 x}+8^{1-\sin ^2 x}=9\) \(8^{\sin ^2 x}+8 / 8^{\sin ^2 x}=9\) Let, \(8^{\sin ^2 x}=t\) \(t+8 / t=9\) \(t^2+8-9 t=0\) \(t^2-8 t-t+8=0\) \(t(t-8)-1(t-8)=0 \quad\) \((t-8)(t-1)=0\) \(8^{\sin ^2 x}=8^0\) \(\sin ^2 x=0\) \(x=0\) \(8^{\sin ^2 x}=8^1\) \(\sin ^2 x=1\) \(\sin ^2 x= \pm 1\) \(x= \pm \pi / 2,3 \pi / 2\) \(\mathrm{x}=\pi, \pi / 2,3 \pi / 2\) Hence, the maximum number of solutions of \(8^{\sin ^2 x}+8^{\cos ^2 x}=9\) in \((0,5)\) is 3 .
COMEDK-2011
Complex Numbers and Quadratic Equation
118064
The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
1 2
2 1
3 0
4 3
Explanation:
C Given, equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) Then, \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{x}^2+9-6 \mathrm{x}=0\) \(3 \mathrm{x}^2-12 \mathrm{x}+14=0\) Now, \(D=b^2-4 a c\) \(D=144-4 \times 3 \times 14\) \(\mathrm{D}=-24\) Here, \(\mathrm{D}\lt 0\) So, equation have no real solutions.
BITSAT-2020
Complex Numbers and Quadratic Equation
118065
If \(x, y\) and \(z\) are real numbers then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
1 positive
2 non-positive
3 zero
4 non-negative
Explanation:
D \(: \mathrm{u}=\mathrm{x}^2+4 \mathrm{y}^2+9 \mathrm{z}^2-6 \mathrm{zy}-3 \mathrm{zx}-2 \mathrm{xy}\) \(=\mathrm{x}^2+(2 \mathrm{y})^2+(3 \mathrm{z})^2-(2 \mathrm{y})(3 \mathrm{z})-(3 \mathrm{z})(\mathrm{x})-\mathrm{x}(2 \mathrm{y}) \geq 0\) \({\left[\because \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]}\) \(\text { As } \mathrm{x}, \mathrm{y} \text { and } \mathrm{z} \in \mathrm{R}, \mathrm{So}, \mathrm{u} \geq 0\) \(\text { Hence, } \mathrm{u} \text { is always positive or non-negative. }\) As \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z} \in \mathrm{R}\), So, \(\mathrm{u} \geq 0\) Hence, \(u\) is always positive or non-negative.
BITSAT-2020
Complex Numbers and Quadratic Equation
118066
If the roots of the equation \(x^2+2 a x+b=0\) are real and differ by at most \(2 \mathrm{~m}, \mathrm{~m} \neq 0\) then \(b\) lies in the interval
B Let the roots be \(\alpha\) and \(\beta\), Then \(\alpha+\beta=-2 \mathrm{a}\) and \(\alpha \beta=\mathrm{b}\) According to the question \(|\alpha-\beta| \leq 2 \mathrm{~m} \Rightarrow(\alpha-\beta)^2 \leq 4 \mathrm{~m}^2\) \(\Rightarrow(\alpha+\beta)^2-4 \alpha \beta \leq 4 \mathrm{~m}^2\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b} \leq 4 \mathrm{~m}^2 \Rightarrow \mathrm{b} \geq \mathrm{a}^2-\mathrm{m}^2\) Again, the roots are real and distinct \(\therefore \mathrm{D}>0\) \(\Rightarrow 4 \mathrm{a}^2-4 \mathrm{~b}>0 \Rightarrow \mathrm{b}\lt \mathrm{a}^2\). On combining we get, \(\mathrm{a}^2-\mathrm{m}^2 \leq \mathrm{b}\lt \mathrm{a}^2 \therefore \mathrm{b} \in\left[\mathrm{a}^2-\mathrm{m}^2, \mathrm{a}^2\right)\)
BITSAT-2007
Complex Numbers and Quadratic Equation
118067
If the equation \(x^2+2(k+1) x+9 k-5=0\) has only negative roots, then-
1 \(\mathrm{k} \leq 0\)
2 \(\mathrm{k} \geq 0\)
3 \(\mathrm{k} \geq 6\)
4 \(\mathrm{k} \leq 6\)
Explanation:
C Let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2(\mathrm{k}+1) \mathrm{x}+9 \mathrm{k}-5\). Let \(\alpha, \beta\) be the roots of \(f(x)=0\). The equation \(f(x)=0\) will have both negative roots, if- (i) Disc. \(\geq 0\) (ii) \(\alpha\lt 0, \beta\lt 0\), i.e. \((\alpha+\beta)\lt 0\) and (iii) \(\alpha \cdot \beta>0\) Now, \(\mathrm{D} \geq 0 \Rightarrow 4(\mathrm{k}+1)^2-36 \mathrm{k}+20 \geq 0\) \(\Rightarrow \mathrm{k}^2-7 \mathrm{k}+6 \geq 0 \Rightarrow(\mathrm{k}-1)(\mathrm{k}-6) \geq 0\) \(\Rightarrow \mathrm{k} \leq 1\) or \(\mathrm{k} \geq 6\) \((\alpha+\beta)\lt 0 \Rightarrow-2(\mathrm{k}+1)\lt 0\) \(\Rightarrow \mathrm{k}+1>0 \Rightarrow \mathrm{k}>-1\) and, \(\alpha \cdot \beta>0 \Rightarrow 9 \mathrm{k}-5>0 \Rightarrow \mathrm{k}>5 / 9\) From (i), (ii), (iii), we get \(\mathrm{k} \geq 6\).
