118260
The number of real roots of equation \(\log _e x+\) \(\mathrm{ex}=0\) is
1 0
2 1
3 2
4 3
Explanation:
B Given, \(\log _{\mathrm{e}} \mathrm{x}+\mathrm{ex}=0\) \(\because\) Both graphs intersect at only one point \(\therefore\) Number of real roots of equation will be 1 .
WB JEE-2015
Complex Numbers and Quadratic Equation
118261
The quadratic expression \((2 x+1)^2-p x+q \neq 0\) for real \(x\), if
118262
If \(p, q\) are odd integer, then the roots of the equation \(2 \mathbf{p x}^2+(2 p+q) x+q=0\) are
1 rational
2 irrational
3 non-real
4 equal
Explanation:
A Given, \(2 \mathrm{px}^2+(2 \mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{q}=0\) We know that, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\therefore \mathrm{D}=(2 \mathrm{p}+\mathrm{q})^2-4(2 \mathrm{p})(\mathrm{q})\) \(=4 p^2+q^2+4 p q-8 p q\) \(=4 \mathrm{p}^2+\mathrm{q}^2-4 \mathrm{pq}=(2 \mathrm{p}-\mathrm{q})^2\) a perfect square the roots of equation are rational. \(\therefore\) Given equation has rational roots
WB JEE-2017
Complex Numbers and Quadratic Equation
118263
Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation \(x^3-3 x+\lambda=0,(\lambda \neq 0)\). Then, \(\phi(\mathbf{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\)
D Given, \(\phi(\mathrm{x})=\frac{\mathrm{x}}{\left(\mathrm{x}^2+1\right)(\mathrm{x}+1)}\) \(\therefore \phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\) \(=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\) Given, \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are roots of cubic equation, \(x^3-3 x+\lambda=0\) \(a b+b c+c a=-3\) \(a+b+c=0\) and \(a b c=-\lambda\) Squaring equation (ii), we get, \((\mathrm{a}+\mathrm{b}+\mathrm{c})^2=0\) \(a^2+b^2+c^2+2(a b+b c+c a)=0\) \(a^2+b^2+c^2=6\) Similarly, \(a^2 b^2+b^2 c^2+c^2 a^2=9\) Put the value in \(\phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})\), we get, \(=\frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)}\) \(=\frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
118260
The number of real roots of equation \(\log _e x+\) \(\mathrm{ex}=0\) is
1 0
2 1
3 2
4 3
Explanation:
B Given, \(\log _{\mathrm{e}} \mathrm{x}+\mathrm{ex}=0\) \(\because\) Both graphs intersect at only one point \(\therefore\) Number of real roots of equation will be 1 .
WB JEE-2015
Complex Numbers and Quadratic Equation
118261
The quadratic expression \((2 x+1)^2-p x+q \neq 0\) for real \(x\), if
118262
If \(p, q\) are odd integer, then the roots of the equation \(2 \mathbf{p x}^2+(2 p+q) x+q=0\) are
1 rational
2 irrational
3 non-real
4 equal
Explanation:
A Given, \(2 \mathrm{px}^2+(2 \mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{q}=0\) We know that, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\therefore \mathrm{D}=(2 \mathrm{p}+\mathrm{q})^2-4(2 \mathrm{p})(\mathrm{q})\) \(=4 p^2+q^2+4 p q-8 p q\) \(=4 \mathrm{p}^2+\mathrm{q}^2-4 \mathrm{pq}=(2 \mathrm{p}-\mathrm{q})^2\) a perfect square the roots of equation are rational. \(\therefore\) Given equation has rational roots
WB JEE-2017
Complex Numbers and Quadratic Equation
118263
Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation \(x^3-3 x+\lambda=0,(\lambda \neq 0)\). Then, \(\phi(\mathbf{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\)
D Given, \(\phi(\mathrm{x})=\frac{\mathrm{x}}{\left(\mathrm{x}^2+1\right)(\mathrm{x}+1)}\) \(\therefore \phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\) \(=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\) Given, \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are roots of cubic equation, \(x^3-3 x+\lambda=0\) \(a b+b c+c a=-3\) \(a+b+c=0\) and \(a b c=-\lambda\) Squaring equation (ii), we get, \((\mathrm{a}+\mathrm{b}+\mathrm{c})^2=0\) \(a^2+b^2+c^2+2(a b+b c+c a)=0\) \(a^2+b^2+c^2=6\) Similarly, \(a^2 b^2+b^2 c^2+c^2 a^2=9\) Put the value in \(\phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})\), we get, \(=\frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)}\) \(=\frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
118260
The number of real roots of equation \(\log _e x+\) \(\mathrm{ex}=0\) is
1 0
2 1
3 2
4 3
Explanation:
B Given, \(\log _{\mathrm{e}} \mathrm{x}+\mathrm{ex}=0\) \(\because\) Both graphs intersect at only one point \(\therefore\) Number of real roots of equation will be 1 .
WB JEE-2015
Complex Numbers and Quadratic Equation
118261
The quadratic expression \((2 x+1)^2-p x+q \neq 0\) for real \(x\), if
118262
If \(p, q\) are odd integer, then the roots of the equation \(2 \mathbf{p x}^2+(2 p+q) x+q=0\) are
1 rational
2 irrational
3 non-real
4 equal
Explanation:
A Given, \(2 \mathrm{px}^2+(2 \mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{q}=0\) We know that, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\therefore \mathrm{D}=(2 \mathrm{p}+\mathrm{q})^2-4(2 \mathrm{p})(\mathrm{q})\) \(=4 p^2+q^2+4 p q-8 p q\) \(=4 \mathrm{p}^2+\mathrm{q}^2-4 \mathrm{pq}=(2 \mathrm{p}-\mathrm{q})^2\) a perfect square the roots of equation are rational. \(\therefore\) Given equation has rational roots
WB JEE-2017
Complex Numbers and Quadratic Equation
118263
Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation \(x^3-3 x+\lambda=0,(\lambda \neq 0)\). Then, \(\phi(\mathbf{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\)
D Given, \(\phi(\mathrm{x})=\frac{\mathrm{x}}{\left(\mathrm{x}^2+1\right)(\mathrm{x}+1)}\) \(\therefore \phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\) \(=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\) Given, \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are roots of cubic equation, \(x^3-3 x+\lambda=0\) \(a b+b c+c a=-3\) \(a+b+c=0\) and \(a b c=-\lambda\) Squaring equation (ii), we get, \((\mathrm{a}+\mathrm{b}+\mathrm{c})^2=0\) \(a^2+b^2+c^2+2(a b+b c+c a)=0\) \(a^2+b^2+c^2=6\) Similarly, \(a^2 b^2+b^2 c^2+c^2 a^2=9\) Put the value in \(\phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})\), we get, \(=\frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)}\) \(=\frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
118260
The number of real roots of equation \(\log _e x+\) \(\mathrm{ex}=0\) is
1 0
2 1
3 2
4 3
Explanation:
B Given, \(\log _{\mathrm{e}} \mathrm{x}+\mathrm{ex}=0\) \(\because\) Both graphs intersect at only one point \(\therefore\) Number of real roots of equation will be 1 .
WB JEE-2015
Complex Numbers and Quadratic Equation
118261
The quadratic expression \((2 x+1)^2-p x+q \neq 0\) for real \(x\), if
118262
If \(p, q\) are odd integer, then the roots of the equation \(2 \mathbf{p x}^2+(2 p+q) x+q=0\) are
1 rational
2 irrational
3 non-real
4 equal
Explanation:
A Given, \(2 \mathrm{px}^2+(2 \mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{q}=0\) We know that, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\therefore \mathrm{D}=(2 \mathrm{p}+\mathrm{q})^2-4(2 \mathrm{p})(\mathrm{q})\) \(=4 p^2+q^2+4 p q-8 p q\) \(=4 \mathrm{p}^2+\mathrm{q}^2-4 \mathrm{pq}=(2 \mathrm{p}-\mathrm{q})^2\) a perfect square the roots of equation are rational. \(\therefore\) Given equation has rational roots
WB JEE-2017
Complex Numbers and Quadratic Equation
118263
Let \(\phi(x)=\frac{x}{\left(x^2+1\right)(x+1)}\). If \(a, b\) and \(c\) are the roots of the equation \(x^3-3 x+\lambda=0,(\lambda \neq 0)\). Then, \(\phi(\mathbf{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\)
D Given, \(\phi(\mathrm{x})=\frac{\mathrm{x}}{\left(\mathrm{x}^2+1\right)(\mathrm{x}+1)}\) \(\therefore \phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})=\) \(=\frac{a b c}{(1+a)(1+b)(1+c)\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\) Given, \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are roots of cubic equation, \(x^3-3 x+\lambda=0\) \(a b+b c+c a=-3\) \(a+b+c=0\) and \(a b c=-\lambda\) Squaring equation (ii), we get, \((\mathrm{a}+\mathrm{b}+\mathrm{c})^2=0\) \(a^2+b^2+c^2+2(a b+b c+c a)=0\) \(a^2+b^2+c^2=6\) Similarly, \(a^2 b^2+b^2 c^2+c^2 a^2=9\) Put the value in \(\phi(\mathrm{a}) \phi(\mathrm{b}) \phi(\mathrm{c})\), we get, \(=\frac{-\lambda}{(1+0-3-\lambda)\left(1+6+9+\lambda^2\right)}\) \(=\frac{-\lambda}{(-2-\lambda)\left(\lambda^2+16\right)}=\frac{\lambda}{(\lambda+2)\left(\lambda^2+16\right)}\)
