118082
The equation \(4^{\left(x^2+2\right)}-9 \cdot 2^{\left(x^2+2\right)}+8=0\) has the solution
1 \(x=1\)
2 \(x=0\)
3 \(\mathrm{x}=\sqrt{2}\)
4 \(x=-\sqrt{2}\)
Explanation:
A Given equation is- \(4^{\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(2^{2\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(\left\{2^{\left(\mathrm{x}^2+2\right)}\right\}^2-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) Let \(2^{\left(\mathrm{x}^2+2\right)}=\mathrm{t}\) Then, \(\mathrm{t}^2-9 \mathrm{t}+8=0\) \(\mathrm{t}^2-8 \mathrm{t}-\mathrm{t}+8=0\) \(\mathrm{t}(\mathrm{t}-8)-1(\mathrm{t}-8)=0\) \((\mathrm{t}-1)(\mathrm{t}-8)=0\) \(t=1\), and 8 Now, When \(t=1\) \(2^{\left(\mathrm{x}^2+2\right)}=1\) \(2^{\left(\mathrm{x}^2+2\right)}=2^0\) \(\mathrm{x}^2+2=0\) \(x^2 \neq-2\) Which is not possible? When \(\mathrm{t}=8\) \(2^{\left(x^2+2\right)}=8\) \(2^{\left(x^2+2\right)}=2^3\) \(x^2+2=3\) \(x^2=1\) \(x= \pm 1\)Therefore, \(\mathrm{x}=1\) and \(\mathrm{x}=-1\)
UPSEE-2014
Complex Numbers and Quadratic Equation
118084
If \(\alpha, \beta\) are the roots of the equation \(6 x^2-5 x+1\) \(=0\), then the value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) is:
1 0
2 \(\pi / 4\)
3 1
4 \(\pi / 2\)
Explanation:
B Given equation is- \(6 x^2-5 x+1=0\) and \(\alpha, \beta\) are roots of the equation- then \(\alpha+\beta=\frac{5}{6}\) \(\alpha \cdot \beta=\frac{1}{6}\) Now, value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) We know that \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{(\alpha+\beta)}{1-\alpha \cdot \beta}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{1}{6}}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}(1)=\tan ^{-1}(\tan \pi / 4)\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\pi / 4\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118085
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+x+1=0\), then the value of \(\alpha^3+\beta^3+\gamma^3\) is:
1 0
2 3
3 -3
4 -1
Explanation:
C Given \( \alpha, \beta, \gamma \text { are the roots of the equation- }\) \(\mathrm{x}^3+\mathrm{x}+1=0\) \(\text { So, } \alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=1\) \(\alpha \cdot \beta \cdot \gamma=-1\) \(\text { We, know that, }\) \(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)+\) \(3 \mathrm{abc}\) \(\text { Now }\) \(\alpha^2+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)\) \(+3 \alpha \beta \gamma\) \(\alpha^3+\beta^3+\gamma^3=0+3 \alpha \beta \gamma\) \(=3(-1)\) \(\alpha^3+\beta^3+\gamma^3=-3\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118086
The number of solutions of the equation \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)=\mathbf{5}^{\mathrm{x}}+\mathbf{5}^{-\mathbf{x}}\), are:
1 no solution
2 one solution
3 two solutions
4 infinitely many solutions
Explanation:
A Let \(5^{\mathrm{x}}=\mathrm{t}\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}=\mathrm{t}+\frac{1}{\mathrm{t}}\) \(\left(\sqrt{\mathrm{t}}-\frac{1}{\sqrt{\mathrm{t}}}\right)^2+2\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}>2\) So, \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)>2\) \(\cos \left(\mathrm{e}^{\mathrm{x}}\right)>1\) \(\because-1\lt \cos \left(\mathrm{e}^{\mathrm{x}}\right)\lt 1\) \(\therefore\) equation (i) is not possible \(\therefore\) There will be no solution for given equation.
UPSEE-2004
Complex Numbers and Quadratic Equation
118087
The real roots of the equation \(x^{2 / 3}+x^{1 / 3}-2=0\) are:
1 1,8
2 \(-1,-8\)
3 \(-1,8\)
4 \(1,-8\)
Explanation:
D Given equation- \(x^{2 / 3}+x^{1 / 3}-2=0\) \(Let x^{1 / 3}=t\) \(Then\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) \((t-1)=0\) \(t=1\) \((t+2)=0\) \(t=-2\) \(\text { when } t=1\) \(x^{1 / 3}=1\) \(x=1\) \(\text { When } t=-2\) \(x^{1 / 3}=-2\) \(x=-8\) \(\text { Therefore root are } 1,-8\) \(hline right.\)Therefore root are \(1,-8\)
118082
The equation \(4^{\left(x^2+2\right)}-9 \cdot 2^{\left(x^2+2\right)}+8=0\) has the solution
1 \(x=1\)
2 \(x=0\)
3 \(\mathrm{x}=\sqrt{2}\)
4 \(x=-\sqrt{2}\)
Explanation:
A Given equation is- \(4^{\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(2^{2\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(\left\{2^{\left(\mathrm{x}^2+2\right)}\right\}^2-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) Let \(2^{\left(\mathrm{x}^2+2\right)}=\mathrm{t}\) Then, \(\mathrm{t}^2-9 \mathrm{t}+8=0\) \(\mathrm{t}^2-8 \mathrm{t}-\mathrm{t}+8=0\) \(\mathrm{t}(\mathrm{t}-8)-1(\mathrm{t}-8)=0\) \((\mathrm{t}-1)(\mathrm{t}-8)=0\) \(t=1\), and 8 Now, When \(t=1\) \(2^{\left(\mathrm{x}^2+2\right)}=1\) \(2^{\left(\mathrm{x}^2+2\right)}=2^0\) \(\mathrm{x}^2+2=0\) \(x^2 \neq-2\) Which is not possible? When \(\mathrm{t}=8\) \(2^{\left(x^2+2\right)}=8\) \(2^{\left(x^2+2\right)}=2^3\) \(x^2+2=3\) \(x^2=1\) \(x= \pm 1\)Therefore, \(\mathrm{x}=1\) and \(\mathrm{x}=-1\)
UPSEE-2014
Complex Numbers and Quadratic Equation
118084
If \(\alpha, \beta\) are the roots of the equation \(6 x^2-5 x+1\) \(=0\), then the value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) is:
1 0
2 \(\pi / 4\)
3 1
4 \(\pi / 2\)
Explanation:
B Given equation is- \(6 x^2-5 x+1=0\) and \(\alpha, \beta\) are roots of the equation- then \(\alpha+\beta=\frac{5}{6}\) \(\alpha \cdot \beta=\frac{1}{6}\) Now, value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) We know that \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{(\alpha+\beta)}{1-\alpha \cdot \beta}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{1}{6}}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}(1)=\tan ^{-1}(\tan \pi / 4)\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\pi / 4\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118085
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+x+1=0\), then the value of \(\alpha^3+\beta^3+\gamma^3\) is:
1 0
2 3
3 -3
4 -1
Explanation:
C Given \( \alpha, \beta, \gamma \text { are the roots of the equation- }\) \(\mathrm{x}^3+\mathrm{x}+1=0\) \(\text { So, } \alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=1\) \(\alpha \cdot \beta \cdot \gamma=-1\) \(\text { We, know that, }\) \(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)+\) \(3 \mathrm{abc}\) \(\text { Now }\) \(\alpha^2+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)\) \(+3 \alpha \beta \gamma\) \(\alpha^3+\beta^3+\gamma^3=0+3 \alpha \beta \gamma\) \(=3(-1)\) \(\alpha^3+\beta^3+\gamma^3=-3\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118086
The number of solutions of the equation \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)=\mathbf{5}^{\mathrm{x}}+\mathbf{5}^{-\mathbf{x}}\), are:
1 no solution
2 one solution
3 two solutions
4 infinitely many solutions
Explanation:
A Let \(5^{\mathrm{x}}=\mathrm{t}\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}=\mathrm{t}+\frac{1}{\mathrm{t}}\) \(\left(\sqrt{\mathrm{t}}-\frac{1}{\sqrt{\mathrm{t}}}\right)^2+2\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}>2\) So, \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)>2\) \(\cos \left(\mathrm{e}^{\mathrm{x}}\right)>1\) \(\because-1\lt \cos \left(\mathrm{e}^{\mathrm{x}}\right)\lt 1\) \(\therefore\) equation (i) is not possible \(\therefore\) There will be no solution for given equation.
UPSEE-2004
Complex Numbers and Quadratic Equation
118087
The real roots of the equation \(x^{2 / 3}+x^{1 / 3}-2=0\) are:
1 1,8
2 \(-1,-8\)
3 \(-1,8\)
4 \(1,-8\)
Explanation:
D Given equation- \(x^{2 / 3}+x^{1 / 3}-2=0\) \(Let x^{1 / 3}=t\) \(Then\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) \((t-1)=0\) \(t=1\) \((t+2)=0\) \(t=-2\) \(\text { when } t=1\) \(x^{1 / 3}=1\) \(x=1\) \(\text { When } t=-2\) \(x^{1 / 3}=-2\) \(x=-8\) \(\text { Therefore root are } 1,-8\) \(hline right.\)Therefore root are \(1,-8\)
118082
The equation \(4^{\left(x^2+2\right)}-9 \cdot 2^{\left(x^2+2\right)}+8=0\) has the solution
1 \(x=1\)
2 \(x=0\)
3 \(\mathrm{x}=\sqrt{2}\)
4 \(x=-\sqrt{2}\)
Explanation:
A Given equation is- \(4^{\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(2^{2\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(\left\{2^{\left(\mathrm{x}^2+2\right)}\right\}^2-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) Let \(2^{\left(\mathrm{x}^2+2\right)}=\mathrm{t}\) Then, \(\mathrm{t}^2-9 \mathrm{t}+8=0\) \(\mathrm{t}^2-8 \mathrm{t}-\mathrm{t}+8=0\) \(\mathrm{t}(\mathrm{t}-8)-1(\mathrm{t}-8)=0\) \((\mathrm{t}-1)(\mathrm{t}-8)=0\) \(t=1\), and 8 Now, When \(t=1\) \(2^{\left(\mathrm{x}^2+2\right)}=1\) \(2^{\left(\mathrm{x}^2+2\right)}=2^0\) \(\mathrm{x}^2+2=0\) \(x^2 \neq-2\) Which is not possible? When \(\mathrm{t}=8\) \(2^{\left(x^2+2\right)}=8\) \(2^{\left(x^2+2\right)}=2^3\) \(x^2+2=3\) \(x^2=1\) \(x= \pm 1\)Therefore, \(\mathrm{x}=1\) and \(\mathrm{x}=-1\)
UPSEE-2014
Complex Numbers and Quadratic Equation
118084
If \(\alpha, \beta\) are the roots of the equation \(6 x^2-5 x+1\) \(=0\), then the value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) is:
1 0
2 \(\pi / 4\)
3 1
4 \(\pi / 2\)
Explanation:
B Given equation is- \(6 x^2-5 x+1=0\) and \(\alpha, \beta\) are roots of the equation- then \(\alpha+\beta=\frac{5}{6}\) \(\alpha \cdot \beta=\frac{1}{6}\) Now, value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) We know that \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{(\alpha+\beta)}{1-\alpha \cdot \beta}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{1}{6}}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}(1)=\tan ^{-1}(\tan \pi / 4)\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\pi / 4\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118085
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+x+1=0\), then the value of \(\alpha^3+\beta^3+\gamma^3\) is:
1 0
2 3
3 -3
4 -1
Explanation:
C Given \( \alpha, \beta, \gamma \text { are the roots of the equation- }\) \(\mathrm{x}^3+\mathrm{x}+1=0\) \(\text { So, } \alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=1\) \(\alpha \cdot \beta \cdot \gamma=-1\) \(\text { We, know that, }\) \(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)+\) \(3 \mathrm{abc}\) \(\text { Now }\) \(\alpha^2+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)\) \(+3 \alpha \beta \gamma\) \(\alpha^3+\beta^3+\gamma^3=0+3 \alpha \beta \gamma\) \(=3(-1)\) \(\alpha^3+\beta^3+\gamma^3=-3\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118086
The number of solutions of the equation \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)=\mathbf{5}^{\mathrm{x}}+\mathbf{5}^{-\mathbf{x}}\), are:
1 no solution
2 one solution
3 two solutions
4 infinitely many solutions
Explanation:
A Let \(5^{\mathrm{x}}=\mathrm{t}\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}=\mathrm{t}+\frac{1}{\mathrm{t}}\) \(\left(\sqrt{\mathrm{t}}-\frac{1}{\sqrt{\mathrm{t}}}\right)^2+2\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}>2\) So, \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)>2\) \(\cos \left(\mathrm{e}^{\mathrm{x}}\right)>1\) \(\because-1\lt \cos \left(\mathrm{e}^{\mathrm{x}}\right)\lt 1\) \(\therefore\) equation (i) is not possible \(\therefore\) There will be no solution for given equation.
UPSEE-2004
Complex Numbers and Quadratic Equation
118087
The real roots of the equation \(x^{2 / 3}+x^{1 / 3}-2=0\) are:
1 1,8
2 \(-1,-8\)
3 \(-1,8\)
4 \(1,-8\)
Explanation:
D Given equation- \(x^{2 / 3}+x^{1 / 3}-2=0\) \(Let x^{1 / 3}=t\) \(Then\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) \((t-1)=0\) \(t=1\) \((t+2)=0\) \(t=-2\) \(\text { when } t=1\) \(x^{1 / 3}=1\) \(x=1\) \(\text { When } t=-2\) \(x^{1 / 3}=-2\) \(x=-8\) \(\text { Therefore root are } 1,-8\) \(hline right.\)Therefore root are \(1,-8\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118082
The equation \(4^{\left(x^2+2\right)}-9 \cdot 2^{\left(x^2+2\right)}+8=0\) has the solution
1 \(x=1\)
2 \(x=0\)
3 \(\mathrm{x}=\sqrt{2}\)
4 \(x=-\sqrt{2}\)
Explanation:
A Given equation is- \(4^{\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(2^{2\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(\left\{2^{\left(\mathrm{x}^2+2\right)}\right\}^2-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) Let \(2^{\left(\mathrm{x}^2+2\right)}=\mathrm{t}\) Then, \(\mathrm{t}^2-9 \mathrm{t}+8=0\) \(\mathrm{t}^2-8 \mathrm{t}-\mathrm{t}+8=0\) \(\mathrm{t}(\mathrm{t}-8)-1(\mathrm{t}-8)=0\) \((\mathrm{t}-1)(\mathrm{t}-8)=0\) \(t=1\), and 8 Now, When \(t=1\) \(2^{\left(\mathrm{x}^2+2\right)}=1\) \(2^{\left(\mathrm{x}^2+2\right)}=2^0\) \(\mathrm{x}^2+2=0\) \(x^2 \neq-2\) Which is not possible? When \(\mathrm{t}=8\) \(2^{\left(x^2+2\right)}=8\) \(2^{\left(x^2+2\right)}=2^3\) \(x^2+2=3\) \(x^2=1\) \(x= \pm 1\)Therefore, \(\mathrm{x}=1\) and \(\mathrm{x}=-1\)
UPSEE-2014
Complex Numbers and Quadratic Equation
118084
If \(\alpha, \beta\) are the roots of the equation \(6 x^2-5 x+1\) \(=0\), then the value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) is:
1 0
2 \(\pi / 4\)
3 1
4 \(\pi / 2\)
Explanation:
B Given equation is- \(6 x^2-5 x+1=0\) and \(\alpha, \beta\) are roots of the equation- then \(\alpha+\beta=\frac{5}{6}\) \(\alpha \cdot \beta=\frac{1}{6}\) Now, value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) We know that \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{(\alpha+\beta)}{1-\alpha \cdot \beta}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{1}{6}}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}(1)=\tan ^{-1}(\tan \pi / 4)\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\pi / 4\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118085
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+x+1=0\), then the value of \(\alpha^3+\beta^3+\gamma^3\) is:
1 0
2 3
3 -3
4 -1
Explanation:
C Given \( \alpha, \beta, \gamma \text { are the roots of the equation- }\) \(\mathrm{x}^3+\mathrm{x}+1=0\) \(\text { So, } \alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=1\) \(\alpha \cdot \beta \cdot \gamma=-1\) \(\text { We, know that, }\) \(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)+\) \(3 \mathrm{abc}\) \(\text { Now }\) \(\alpha^2+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)\) \(+3 \alpha \beta \gamma\) \(\alpha^3+\beta^3+\gamma^3=0+3 \alpha \beta \gamma\) \(=3(-1)\) \(\alpha^3+\beta^3+\gamma^3=-3\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118086
The number of solutions of the equation \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)=\mathbf{5}^{\mathrm{x}}+\mathbf{5}^{-\mathbf{x}}\), are:
1 no solution
2 one solution
3 two solutions
4 infinitely many solutions
Explanation:
A Let \(5^{\mathrm{x}}=\mathrm{t}\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}=\mathrm{t}+\frac{1}{\mathrm{t}}\) \(\left(\sqrt{\mathrm{t}}-\frac{1}{\sqrt{\mathrm{t}}}\right)^2+2\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}>2\) So, \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)>2\) \(\cos \left(\mathrm{e}^{\mathrm{x}}\right)>1\) \(\because-1\lt \cos \left(\mathrm{e}^{\mathrm{x}}\right)\lt 1\) \(\therefore\) equation (i) is not possible \(\therefore\) There will be no solution for given equation.
UPSEE-2004
Complex Numbers and Quadratic Equation
118087
The real roots of the equation \(x^{2 / 3}+x^{1 / 3}-2=0\) are:
1 1,8
2 \(-1,-8\)
3 \(-1,8\)
4 \(1,-8\)
Explanation:
D Given equation- \(x^{2 / 3}+x^{1 / 3}-2=0\) \(Let x^{1 / 3}=t\) \(Then\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) \((t-1)=0\) \(t=1\) \((t+2)=0\) \(t=-2\) \(\text { when } t=1\) \(x^{1 / 3}=1\) \(x=1\) \(\text { When } t=-2\) \(x^{1 / 3}=-2\) \(x=-8\) \(\text { Therefore root are } 1,-8\) \(hline right.\)Therefore root are \(1,-8\)
118082
The equation \(4^{\left(x^2+2\right)}-9 \cdot 2^{\left(x^2+2\right)}+8=0\) has the solution
1 \(x=1\)
2 \(x=0\)
3 \(\mathrm{x}=\sqrt{2}\)
4 \(x=-\sqrt{2}\)
Explanation:
A Given equation is- \(4^{\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(2^{2\left(\mathrm{x}^2+2\right)}-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) \(\left\{2^{\left(\mathrm{x}^2+2\right)}\right\}^2-9.2^{\left(\mathrm{x}^2+2\right)}+8=0\) Let \(2^{\left(\mathrm{x}^2+2\right)}=\mathrm{t}\) Then, \(\mathrm{t}^2-9 \mathrm{t}+8=0\) \(\mathrm{t}^2-8 \mathrm{t}-\mathrm{t}+8=0\) \(\mathrm{t}(\mathrm{t}-8)-1(\mathrm{t}-8)=0\) \((\mathrm{t}-1)(\mathrm{t}-8)=0\) \(t=1\), and 8 Now, When \(t=1\) \(2^{\left(\mathrm{x}^2+2\right)}=1\) \(2^{\left(\mathrm{x}^2+2\right)}=2^0\) \(\mathrm{x}^2+2=0\) \(x^2 \neq-2\) Which is not possible? When \(\mathrm{t}=8\) \(2^{\left(x^2+2\right)}=8\) \(2^{\left(x^2+2\right)}=2^3\) \(x^2+2=3\) \(x^2=1\) \(x= \pm 1\)Therefore, \(\mathrm{x}=1\) and \(\mathrm{x}=-1\)
UPSEE-2014
Complex Numbers and Quadratic Equation
118084
If \(\alpha, \beta\) are the roots of the equation \(6 x^2-5 x+1\) \(=0\), then the value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) is:
1 0
2 \(\pi / 4\)
3 1
4 \(\pi / 2\)
Explanation:
B Given equation is- \(6 x^2-5 x+1=0\) and \(\alpha, \beta\) are roots of the equation- then \(\alpha+\beta=\frac{5}{6}\) \(\alpha \cdot \beta=\frac{1}{6}\) Now, value of \(\tan ^{-1} \alpha+\tan ^{-1} \beta\) We know that \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{(\alpha+\beta)}{1-\alpha \cdot \beta}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}\left[\frac{\frac{5}{6}}{1-\frac{1}{6}}\right]\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1}(1)=\tan ^{-1}(\tan \pi / 4)\) \(\tan ^{-1} \alpha+\tan ^{-1} \beta=\pi / 4\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118085
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3+x+1=0\), then the value of \(\alpha^3+\beta^3+\gamma^3\) is:
1 0
2 3
3 -3
4 -1
Explanation:
C Given \( \alpha, \beta, \gamma \text { are the roots of the equation- }\) \(\mathrm{x}^3+\mathrm{x}+1=0\) \(\text { So, } \alpha+\beta+\gamma=0\) \(\alpha \beta+\beta \gamma+\gamma \alpha=1\) \(\alpha \cdot \beta \cdot \gamma=-1\) \(\text { We, know that, }\) \(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)+\) \(3 \mathrm{abc}\) \(\text { Now }\) \(\alpha^2+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)\) \(+3 \alpha \beta \gamma\) \(\alpha^3+\beta^3+\gamma^3=0+3 \alpha \beta \gamma\) \(=3(-1)\) \(\alpha^3+\beta^3+\gamma^3=-3\)
UPSEE-2006
Complex Numbers and Quadratic Equation
118086
The number of solutions of the equation \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)=\mathbf{5}^{\mathrm{x}}+\mathbf{5}^{-\mathbf{x}}\), are:
1 no solution
2 one solution
3 two solutions
4 infinitely many solutions
Explanation:
A Let \(5^{\mathrm{x}}=\mathrm{t}\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}=\mathrm{t}+\frac{1}{\mathrm{t}}\) \(\left(\sqrt{\mathrm{t}}-\frac{1}{\sqrt{\mathrm{t}}}\right)^2+2\) \(\Rightarrow 5^{\mathrm{x}}+5^{-\mathrm{x}}>2\) So, \(2 \cos \left(\mathrm{e}^{\mathrm{x}}\right)>2\) \(\cos \left(\mathrm{e}^{\mathrm{x}}\right)>1\) \(\because-1\lt \cos \left(\mathrm{e}^{\mathrm{x}}\right)\lt 1\) \(\therefore\) equation (i) is not possible \(\therefore\) There will be no solution for given equation.
UPSEE-2004
Complex Numbers and Quadratic Equation
118087
The real roots of the equation \(x^{2 / 3}+x^{1 / 3}-2=0\) are:
1 1,8
2 \(-1,-8\)
3 \(-1,8\)
4 \(1,-8\)
Explanation:
D Given equation- \(x^{2 / 3}+x^{1 / 3}-2=0\) \(Let x^{1 / 3}=t\) \(Then\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) \((t-1)=0\) \(t=1\) \((t+2)=0\) \(t=-2\) \(\text { when } t=1\) \(x^{1 / 3}=1\) \(x=1\) \(\text { When } t=-2\) \(x^{1 / 3}=-2\) \(x=-8\) \(\text { Therefore root are } 1,-8\) \(hline right.\)Therefore root are \(1,-8\)