118169
The number of integral values of \(m\) for which the quadratic expression, \((1+2 m) x^2-2(1+\) 3m) \(x+4(1+m), x \in R\), is always positive, is
1 6
2 8
3 7
4 3
Explanation:
C Given, the quadratic expression, \((1+2 m) x^2-2(1+3 m) x+4(1+m), x \in R\). Let, as consider the quadratic expression. \(f(x)=a x^2+b x+c\) \(f(x)=a \cdot\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a \cdot\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x+\frac{c}{a}-\left(\frac{b}{2 a}\right)^2\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2+\frac{4 a c-b^2}{4 a^2}\right]\) Now \(\mathrm{f}(\mathrm{x})\) is always positive \(\Rightarrow \mathrm{a}>0\) and \(\frac{4 a c-b^2}{4 a^2}>0\) \(b^2-4 a c\lt 0\) \(\text { Here, } a=1+2 m, b=-2(1+3 m) \text { and } c=4(1+m) \text {. }\) \(\therefore[-2(1+3 m)]^2-4(1+2 m) \cdot 4(1+m)\lt 0\) \((1+3 \mathrm{~m})^2-4[(1+2 \mathrm{~m})] 4(1+\mathrm{m})\lt 0\) \(1+9 m^2+6 m-4\left[1+2 m+m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4\left[1+3 m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4-12 m-8 m^2\lt 0\) \(\mathrm{~m}^2-6 \mathrm{~m}-3\lt 0\) \(\therefore[\mathrm{m}-(3+2 \sqrt{3})][\mathrm{m}-(3-2 \sqrt{3})]\lt 0\) \(\Rightarrow 3-2 \sqrt{3}\lt \mathrm{m}\lt 3+2 \sqrt{3}\) \(\text { And also, } 1+2 \mathrm{~m}>0\) \(\Rightarrow \quad \mathrm{m}>\frac{-1}{2}\) \(\therefore \quad-0.44\lt \mathrm{m}\lt 3+3.44\) \(\text { or } \quad-0.44\lt \mathrm{m}\lt 6.44\) \(\therefore \quad \text { m can be } 0,1,2,3,4,5,6\) Now, \(\quad 2 \sqrt{3}=2 \times(1.72)=3.44\) \(\therefore\) Total number of integral values of \(m\) is 7 .
JEE Main 12.01.2019
Complex Numbers and Quadratic Equation
118170
The number of real roots of the equation \(5+\) \(\left|2^{x^x}-1\right|=\mathbf{2}^x\left(2^x-2\right)\) is
1 1
2 3
3 4
4 2
Explanation:
A Given the equation \(5+\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\) \(\text { case } 1: 2^{\mathrm{x}}-1>0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}-1\) \(\therefore 5+2^{\mathrm{x}}-1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-3 \cdot 2^{\mathrm{x}}-4=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-4 \mathrm{t}+\mathrm{t}-4=0\) \(\text { or } \mathrm{t}(\mathrm{t}-4)+1(\mathrm{t}-4)=0\) \((\mathrm{t}+4)(\mathrm{t}-4)=0\) \(\mathrm{t}=4, \mathrm{t}=-1\) \(\therefore 2^{\mathrm{x}}=2^2 \Rightarrow \mathrm{x}=2\) \(2^{\mathrm{x}} \neq-1(\text { it is not possible) }\) \(\text { case2:- }\) \(2^{\mathrm{x}}-1\lt 0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=-\left(2^{\mathrm{x}}-1\right)\) \(\therefore \text { We get }\) \(5+\left(-\left(2^{\mathrm{x}}-1\right)=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\right.\) \(\text { or } 5-2^{\mathrm{x}}+1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-2^{\mathrm{x}}-6=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t} \text { we get }\) \(\mathrm{t}^2-\mathrm{t}-6=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}+2 \mathrm{t}-6=0\) \((\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3, \mathrm{t}=-2,2^{\mathrm{x}} \neq-2\) \(\therefore 2^{\mathrm{x}} \neq 3 \text { is the real solution }\) \(\therefore \text { Number of real roots is one. }\)
JEE Main 10.04.2019
Complex Numbers and Quadratic Equation
118171
The number of all possible positive integral values of \(\alpha\) for which the roots of the quadratic equation, \(6 x^2-11 x+\alpha=0\) are rational numbers is
1 5
2 2
3 4
4 3
Explanation:
D Given, the roots of quadratic equation. \(6 \mathrm{x}^2-11 \mathrm{x}+\alpha=0\) is \(\mathrm{x}=\frac{11 \pm \sqrt{121-24 \alpha}}{12}\) Roots are rational if \(121-24 \alpha\) is prefect square. Let us consider integer. Put, \(\quad \alpha=1\) \(\Rightarrow \quad 121-24=97\) which is not perfect square Put, \(\quad \alpha=2\) \(\Rightarrow \quad 121-48=73\) which is not perfect square Put, \(\quad \alpha=3\) \(\Rightarrow \quad 121-72=49\) which is perfect square Put, \(\quad \alpha=4\) \(\Rightarrow \quad 121-96=25\) is again perfect square Put, \(\quad \alpha=5\) \(\Rightarrow \quad 121-120=1\) is again perfect square. So, number of integral values of \(\alpha\) is 3 .
JEE Main 09.01.2019 Shift-II
Complex Numbers and Quadratic Equation
118172
The integer ' \(k\) ' for which the inequality \(x^2-\) \(2(3 k-1) x+8 k^2-7>0\) is valid for every \(x\) in \(R\), is
1 3
2 2
3 0
4 4
Explanation:
A Given \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+8 \mathrm{x}^2-7>0 \forall \mathrm{x} \in \mathrm{R}\) \(\text { We can write it as }\) \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+(3 \mathrm{k}-1)^2+8 \mathrm{x}^2-7-(3 \mathrm{k}-1)^2>0\) \(=[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-(3 \mathrm{x}-1)^2>0\) \(\Rightarrow[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-9 \mathrm{k}^2-1+6 \mathrm{x}>0\) \(\Rightarrow(\mathrm{x}-(3 \mathrm{k}-1)]^2-\mathrm{k}^2+6 \mathrm{k}-8>0\) \(\text { It is possible when }\) \(\left(\mathrm{k}^2-6 \mathrm{k}+8\right)\lt 0\) \(\text { or }\left(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\right)\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{x}-4)\lt 0\) \(\text { or }(\mathrm{k}-4)(\mathrm{k}-2)\lt 0\) \(=2\lt \mathrm{k}\lt 4\) \(\text { But } \mathrm{k} \text { is integer } \mathrm{k}=3\)But \(\mathrm{k}\) is integer \(\mathrm{k}=3\)
118169
The number of integral values of \(m\) for which the quadratic expression, \((1+2 m) x^2-2(1+\) 3m) \(x+4(1+m), x \in R\), is always positive, is
1 6
2 8
3 7
4 3
Explanation:
C Given, the quadratic expression, \((1+2 m) x^2-2(1+3 m) x+4(1+m), x \in R\). Let, as consider the quadratic expression. \(f(x)=a x^2+b x+c\) \(f(x)=a \cdot\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a \cdot\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x+\frac{c}{a}-\left(\frac{b}{2 a}\right)^2\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2+\frac{4 a c-b^2}{4 a^2}\right]\) Now \(\mathrm{f}(\mathrm{x})\) is always positive \(\Rightarrow \mathrm{a}>0\) and \(\frac{4 a c-b^2}{4 a^2}>0\) \(b^2-4 a c\lt 0\) \(\text { Here, } a=1+2 m, b=-2(1+3 m) \text { and } c=4(1+m) \text {. }\) \(\therefore[-2(1+3 m)]^2-4(1+2 m) \cdot 4(1+m)\lt 0\) \((1+3 \mathrm{~m})^2-4[(1+2 \mathrm{~m})] 4(1+\mathrm{m})\lt 0\) \(1+9 m^2+6 m-4\left[1+2 m+m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4\left[1+3 m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4-12 m-8 m^2\lt 0\) \(\mathrm{~m}^2-6 \mathrm{~m}-3\lt 0\) \(\therefore[\mathrm{m}-(3+2 \sqrt{3})][\mathrm{m}-(3-2 \sqrt{3})]\lt 0\) \(\Rightarrow 3-2 \sqrt{3}\lt \mathrm{m}\lt 3+2 \sqrt{3}\) \(\text { And also, } 1+2 \mathrm{~m}>0\) \(\Rightarrow \quad \mathrm{m}>\frac{-1}{2}\) \(\therefore \quad-0.44\lt \mathrm{m}\lt 3+3.44\) \(\text { or } \quad-0.44\lt \mathrm{m}\lt 6.44\) \(\therefore \quad \text { m can be } 0,1,2,3,4,5,6\) Now, \(\quad 2 \sqrt{3}=2 \times(1.72)=3.44\) \(\therefore\) Total number of integral values of \(m\) is 7 .
JEE Main 12.01.2019
Complex Numbers and Quadratic Equation
118170
The number of real roots of the equation \(5+\) \(\left|2^{x^x}-1\right|=\mathbf{2}^x\left(2^x-2\right)\) is
1 1
2 3
3 4
4 2
Explanation:
A Given the equation \(5+\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\) \(\text { case } 1: 2^{\mathrm{x}}-1>0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}-1\) \(\therefore 5+2^{\mathrm{x}}-1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-3 \cdot 2^{\mathrm{x}}-4=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-4 \mathrm{t}+\mathrm{t}-4=0\) \(\text { or } \mathrm{t}(\mathrm{t}-4)+1(\mathrm{t}-4)=0\) \((\mathrm{t}+4)(\mathrm{t}-4)=0\) \(\mathrm{t}=4, \mathrm{t}=-1\) \(\therefore 2^{\mathrm{x}}=2^2 \Rightarrow \mathrm{x}=2\) \(2^{\mathrm{x}} \neq-1(\text { it is not possible) }\) \(\text { case2:- }\) \(2^{\mathrm{x}}-1\lt 0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=-\left(2^{\mathrm{x}}-1\right)\) \(\therefore \text { We get }\) \(5+\left(-\left(2^{\mathrm{x}}-1\right)=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\right.\) \(\text { or } 5-2^{\mathrm{x}}+1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-2^{\mathrm{x}}-6=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t} \text { we get }\) \(\mathrm{t}^2-\mathrm{t}-6=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}+2 \mathrm{t}-6=0\) \((\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3, \mathrm{t}=-2,2^{\mathrm{x}} \neq-2\) \(\therefore 2^{\mathrm{x}} \neq 3 \text { is the real solution }\) \(\therefore \text { Number of real roots is one. }\)
JEE Main 10.04.2019
Complex Numbers and Quadratic Equation
118171
The number of all possible positive integral values of \(\alpha\) for which the roots of the quadratic equation, \(6 x^2-11 x+\alpha=0\) are rational numbers is
1 5
2 2
3 4
4 3
Explanation:
D Given, the roots of quadratic equation. \(6 \mathrm{x}^2-11 \mathrm{x}+\alpha=0\) is \(\mathrm{x}=\frac{11 \pm \sqrt{121-24 \alpha}}{12}\) Roots are rational if \(121-24 \alpha\) is prefect square. Let us consider integer. Put, \(\quad \alpha=1\) \(\Rightarrow \quad 121-24=97\) which is not perfect square Put, \(\quad \alpha=2\) \(\Rightarrow \quad 121-48=73\) which is not perfect square Put, \(\quad \alpha=3\) \(\Rightarrow \quad 121-72=49\) which is perfect square Put, \(\quad \alpha=4\) \(\Rightarrow \quad 121-96=25\) is again perfect square Put, \(\quad \alpha=5\) \(\Rightarrow \quad 121-120=1\) is again perfect square. So, number of integral values of \(\alpha\) is 3 .
JEE Main 09.01.2019 Shift-II
Complex Numbers and Quadratic Equation
118172
The integer ' \(k\) ' for which the inequality \(x^2-\) \(2(3 k-1) x+8 k^2-7>0\) is valid for every \(x\) in \(R\), is
1 3
2 2
3 0
4 4
Explanation:
A Given \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+8 \mathrm{x}^2-7>0 \forall \mathrm{x} \in \mathrm{R}\) \(\text { We can write it as }\) \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+(3 \mathrm{k}-1)^2+8 \mathrm{x}^2-7-(3 \mathrm{k}-1)^2>0\) \(=[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-(3 \mathrm{x}-1)^2>0\) \(\Rightarrow[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-9 \mathrm{k}^2-1+6 \mathrm{x}>0\) \(\Rightarrow(\mathrm{x}-(3 \mathrm{k}-1)]^2-\mathrm{k}^2+6 \mathrm{k}-8>0\) \(\text { It is possible when }\) \(\left(\mathrm{k}^2-6 \mathrm{k}+8\right)\lt 0\) \(\text { or }\left(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\right)\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{x}-4)\lt 0\) \(\text { or }(\mathrm{k}-4)(\mathrm{k}-2)\lt 0\) \(=2\lt \mathrm{k}\lt 4\) \(\text { But } \mathrm{k} \text { is integer } \mathrm{k}=3\)But \(\mathrm{k}\) is integer \(\mathrm{k}=3\)
118169
The number of integral values of \(m\) for which the quadratic expression, \((1+2 m) x^2-2(1+\) 3m) \(x+4(1+m), x \in R\), is always positive, is
1 6
2 8
3 7
4 3
Explanation:
C Given, the quadratic expression, \((1+2 m) x^2-2(1+3 m) x+4(1+m), x \in R\). Let, as consider the quadratic expression. \(f(x)=a x^2+b x+c\) \(f(x)=a \cdot\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a \cdot\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x+\frac{c}{a}-\left(\frac{b}{2 a}\right)^2\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2+\frac{4 a c-b^2}{4 a^2}\right]\) Now \(\mathrm{f}(\mathrm{x})\) is always positive \(\Rightarrow \mathrm{a}>0\) and \(\frac{4 a c-b^2}{4 a^2}>0\) \(b^2-4 a c\lt 0\) \(\text { Here, } a=1+2 m, b=-2(1+3 m) \text { and } c=4(1+m) \text {. }\) \(\therefore[-2(1+3 m)]^2-4(1+2 m) \cdot 4(1+m)\lt 0\) \((1+3 \mathrm{~m})^2-4[(1+2 \mathrm{~m})] 4(1+\mathrm{m})\lt 0\) \(1+9 m^2+6 m-4\left[1+2 m+m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4\left[1+3 m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4-12 m-8 m^2\lt 0\) \(\mathrm{~m}^2-6 \mathrm{~m}-3\lt 0\) \(\therefore[\mathrm{m}-(3+2 \sqrt{3})][\mathrm{m}-(3-2 \sqrt{3})]\lt 0\) \(\Rightarrow 3-2 \sqrt{3}\lt \mathrm{m}\lt 3+2 \sqrt{3}\) \(\text { And also, } 1+2 \mathrm{~m}>0\) \(\Rightarrow \quad \mathrm{m}>\frac{-1}{2}\) \(\therefore \quad-0.44\lt \mathrm{m}\lt 3+3.44\) \(\text { or } \quad-0.44\lt \mathrm{m}\lt 6.44\) \(\therefore \quad \text { m can be } 0,1,2,3,4,5,6\) Now, \(\quad 2 \sqrt{3}=2 \times(1.72)=3.44\) \(\therefore\) Total number of integral values of \(m\) is 7 .
JEE Main 12.01.2019
Complex Numbers and Quadratic Equation
118170
The number of real roots of the equation \(5+\) \(\left|2^{x^x}-1\right|=\mathbf{2}^x\left(2^x-2\right)\) is
1 1
2 3
3 4
4 2
Explanation:
A Given the equation \(5+\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\) \(\text { case } 1: 2^{\mathrm{x}}-1>0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}-1\) \(\therefore 5+2^{\mathrm{x}}-1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-3 \cdot 2^{\mathrm{x}}-4=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-4 \mathrm{t}+\mathrm{t}-4=0\) \(\text { or } \mathrm{t}(\mathrm{t}-4)+1(\mathrm{t}-4)=0\) \((\mathrm{t}+4)(\mathrm{t}-4)=0\) \(\mathrm{t}=4, \mathrm{t}=-1\) \(\therefore 2^{\mathrm{x}}=2^2 \Rightarrow \mathrm{x}=2\) \(2^{\mathrm{x}} \neq-1(\text { it is not possible) }\) \(\text { case2:- }\) \(2^{\mathrm{x}}-1\lt 0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=-\left(2^{\mathrm{x}}-1\right)\) \(\therefore \text { We get }\) \(5+\left(-\left(2^{\mathrm{x}}-1\right)=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\right.\) \(\text { or } 5-2^{\mathrm{x}}+1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-2^{\mathrm{x}}-6=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t} \text { we get }\) \(\mathrm{t}^2-\mathrm{t}-6=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}+2 \mathrm{t}-6=0\) \((\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3, \mathrm{t}=-2,2^{\mathrm{x}} \neq-2\) \(\therefore 2^{\mathrm{x}} \neq 3 \text { is the real solution }\) \(\therefore \text { Number of real roots is one. }\)
JEE Main 10.04.2019
Complex Numbers and Quadratic Equation
118171
The number of all possible positive integral values of \(\alpha\) for which the roots of the quadratic equation, \(6 x^2-11 x+\alpha=0\) are rational numbers is
1 5
2 2
3 4
4 3
Explanation:
D Given, the roots of quadratic equation. \(6 \mathrm{x}^2-11 \mathrm{x}+\alpha=0\) is \(\mathrm{x}=\frac{11 \pm \sqrt{121-24 \alpha}}{12}\) Roots are rational if \(121-24 \alpha\) is prefect square. Let us consider integer. Put, \(\quad \alpha=1\) \(\Rightarrow \quad 121-24=97\) which is not perfect square Put, \(\quad \alpha=2\) \(\Rightarrow \quad 121-48=73\) which is not perfect square Put, \(\quad \alpha=3\) \(\Rightarrow \quad 121-72=49\) which is perfect square Put, \(\quad \alpha=4\) \(\Rightarrow \quad 121-96=25\) is again perfect square Put, \(\quad \alpha=5\) \(\Rightarrow \quad 121-120=1\) is again perfect square. So, number of integral values of \(\alpha\) is 3 .
JEE Main 09.01.2019 Shift-II
Complex Numbers and Quadratic Equation
118172
The integer ' \(k\) ' for which the inequality \(x^2-\) \(2(3 k-1) x+8 k^2-7>0\) is valid for every \(x\) in \(R\), is
1 3
2 2
3 0
4 4
Explanation:
A Given \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+8 \mathrm{x}^2-7>0 \forall \mathrm{x} \in \mathrm{R}\) \(\text { We can write it as }\) \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+(3 \mathrm{k}-1)^2+8 \mathrm{x}^2-7-(3 \mathrm{k}-1)^2>0\) \(=[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-(3 \mathrm{x}-1)^2>0\) \(\Rightarrow[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-9 \mathrm{k}^2-1+6 \mathrm{x}>0\) \(\Rightarrow(\mathrm{x}-(3 \mathrm{k}-1)]^2-\mathrm{k}^2+6 \mathrm{k}-8>0\) \(\text { It is possible when }\) \(\left(\mathrm{k}^2-6 \mathrm{k}+8\right)\lt 0\) \(\text { or }\left(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\right)\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{x}-4)\lt 0\) \(\text { or }(\mathrm{k}-4)(\mathrm{k}-2)\lt 0\) \(=2\lt \mathrm{k}\lt 4\) \(\text { But } \mathrm{k} \text { is integer } \mathrm{k}=3\)But \(\mathrm{k}\) is integer \(\mathrm{k}=3\)
118169
The number of integral values of \(m\) for which the quadratic expression, \((1+2 m) x^2-2(1+\) 3m) \(x+4(1+m), x \in R\), is always positive, is
1 6
2 8
3 7
4 3
Explanation:
C Given, the quadratic expression, \((1+2 m) x^2-2(1+3 m) x+4(1+m), x \in R\). Let, as consider the quadratic expression. \(f(x)=a x^2+b x+c\) \(f(x)=a \cdot\left[x^2+\frac{b}{a} x+\frac{c}{a}\right]\) \(=a \cdot\left[x^2+\left(\frac{b}{2 a}\right)^2+\frac{b}{a} x+\frac{c}{a}-\left(\frac{b}{2 a}\right)^2\right]\) \(=a\left[\left(x+\frac{b}{2 a}\right)^2+\frac{4 a c-b^2}{4 a^2}\right]\) Now \(\mathrm{f}(\mathrm{x})\) is always positive \(\Rightarrow \mathrm{a}>0\) and \(\frac{4 a c-b^2}{4 a^2}>0\) \(b^2-4 a c\lt 0\) \(\text { Here, } a=1+2 m, b=-2(1+3 m) \text { and } c=4(1+m) \text {. }\) \(\therefore[-2(1+3 m)]^2-4(1+2 m) \cdot 4(1+m)\lt 0\) \((1+3 \mathrm{~m})^2-4[(1+2 \mathrm{~m})] 4(1+\mathrm{m})\lt 0\) \(1+9 m^2+6 m-4\left[1+2 m+m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4\left[1+3 m+2 m^2\right]\lt 0\) \(1+9 m^2+6 m-4-12 m-8 m^2\lt 0\) \(\mathrm{~m}^2-6 \mathrm{~m}-3\lt 0\) \(\therefore[\mathrm{m}-(3+2 \sqrt{3})][\mathrm{m}-(3-2 \sqrt{3})]\lt 0\) \(\Rightarrow 3-2 \sqrt{3}\lt \mathrm{m}\lt 3+2 \sqrt{3}\) \(\text { And also, } 1+2 \mathrm{~m}>0\) \(\Rightarrow \quad \mathrm{m}>\frac{-1}{2}\) \(\therefore \quad-0.44\lt \mathrm{m}\lt 3+3.44\) \(\text { or } \quad-0.44\lt \mathrm{m}\lt 6.44\) \(\therefore \quad \text { m can be } 0,1,2,3,4,5,6\) Now, \(\quad 2 \sqrt{3}=2 \times(1.72)=3.44\) \(\therefore\) Total number of integral values of \(m\) is 7 .
JEE Main 12.01.2019
Complex Numbers and Quadratic Equation
118170
The number of real roots of the equation \(5+\) \(\left|2^{x^x}-1\right|=\mathbf{2}^x\left(2^x-2\right)\) is
1 1
2 3
3 4
4 2
Explanation:
A Given the equation \(5+\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\) \(\text { case } 1: 2^{\mathrm{x}}-1>0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=2^{\mathrm{x}}-1\) \(\therefore 5+2^{\mathrm{x}}-1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-3 \cdot 2^{\mathrm{x}}-4=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}-4=0\) \(\text { or } \mathrm{t}^2-4 \mathrm{t}+\mathrm{t}-4=0\) \(\text { or } \mathrm{t}(\mathrm{t}-4)+1(\mathrm{t}-4)=0\) \((\mathrm{t}+4)(\mathrm{t}-4)=0\) \(\mathrm{t}=4, \mathrm{t}=-1\) \(\therefore 2^{\mathrm{x}}=2^2 \Rightarrow \mathrm{x}=2\) \(2^{\mathrm{x}} \neq-1(\text { it is not possible) }\) \(\text { case2:- }\) \(2^{\mathrm{x}}-1\lt 0 \Rightarrow\left|2^{\mathrm{x}}-1\right|=-\left(2^{\mathrm{x}}-1\right)\) \(\therefore \text { We get }\) \(5+\left(-\left(2^{\mathrm{x}}-1\right)=2^{\mathrm{x}}\left(2^{\mathrm{x}}-2\right)\right.\) \(\text { or } 5-2^{\mathrm{x}}+1=\left(2^{\mathrm{x}}\right)^2-2 \cdot 2^{\mathrm{x}}\) \(\text { or }\left(2^{\mathrm{x}}\right)^2-2^{\mathrm{x}}-6=0\) \(\text { Let } 2^{\mathrm{x}}=\mathrm{t} \text { we get }\) \(\mathrm{t}^2-\mathrm{t}-6=0\) \(\text { or } \mathrm{t}^2-3 \mathrm{t}+2 \mathrm{t}-6=0\) \((\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3, \mathrm{t}=-2,2^{\mathrm{x}} \neq-2\) \(\therefore 2^{\mathrm{x}} \neq 3 \text { is the real solution }\) \(\therefore \text { Number of real roots is one. }\)
JEE Main 10.04.2019
Complex Numbers and Quadratic Equation
118171
The number of all possible positive integral values of \(\alpha\) for which the roots of the quadratic equation, \(6 x^2-11 x+\alpha=0\) are rational numbers is
1 5
2 2
3 4
4 3
Explanation:
D Given, the roots of quadratic equation. \(6 \mathrm{x}^2-11 \mathrm{x}+\alpha=0\) is \(\mathrm{x}=\frac{11 \pm \sqrt{121-24 \alpha}}{12}\) Roots are rational if \(121-24 \alpha\) is prefect square. Let us consider integer. Put, \(\quad \alpha=1\) \(\Rightarrow \quad 121-24=97\) which is not perfect square Put, \(\quad \alpha=2\) \(\Rightarrow \quad 121-48=73\) which is not perfect square Put, \(\quad \alpha=3\) \(\Rightarrow \quad 121-72=49\) which is perfect square Put, \(\quad \alpha=4\) \(\Rightarrow \quad 121-96=25\) is again perfect square Put, \(\quad \alpha=5\) \(\Rightarrow \quad 121-120=1\) is again perfect square. So, number of integral values of \(\alpha\) is 3 .
JEE Main 09.01.2019 Shift-II
Complex Numbers and Quadratic Equation
118172
The integer ' \(k\) ' for which the inequality \(x^2-\) \(2(3 k-1) x+8 k^2-7>0\) is valid for every \(x\) in \(R\), is
1 3
2 2
3 0
4 4
Explanation:
A Given \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+8 \mathrm{x}^2-7>0 \forall \mathrm{x} \in \mathrm{R}\) \(\text { We can write it as }\) \(\mathrm{x}^2-2 \mathrm{x}(3 \mathrm{k}-1)+(3 \mathrm{k}-1)^2+8 \mathrm{x}^2-7-(3 \mathrm{k}-1)^2>0\) \(=[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-(3 \mathrm{x}-1)^2>0\) \(\Rightarrow[\mathrm{x}-(3 \mathrm{k}-1)]^2+8 \mathrm{k}^2-7-9 \mathrm{k}^2-1+6 \mathrm{x}>0\) \(\Rightarrow(\mathrm{x}-(3 \mathrm{k}-1)]^2-\mathrm{k}^2+6 \mathrm{k}-8>0\) \(\text { It is possible when }\) \(\left(\mathrm{k}^2-6 \mathrm{k}+8\right)\lt 0\) \(\text { or }\left(\mathrm{k}^2-4 \mathrm{k}-2 \mathrm{k}+8\right)\lt 0\) \(\mathrm{k}(\mathrm{k}-4)-2(\mathrm{x}-4)\lt 0\) \(\text { or }(\mathrm{k}-4)(\mathrm{k}-2)\lt 0\) \(=2\lt \mathrm{k}\lt 4\) \(\text { But } \mathrm{k} \text { is integer } \mathrm{k}=3\)But \(\mathrm{k}\) is integer \(\mathrm{k}=3\)
