118173
The number of real solutions of the equation \(x^2\) \(-|\mathbf{x}|-\mathbf{1 2}=\mathbf{0}\) is
1 2
2 3
3 1
4 4
Explanation:
A Given the equation \(\mathrm{x}^2-|\mathrm{x}|-12=0\) case : \(1 \mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(x^2-x-12=0\) \(x^2-4 x+3 x-12=0\) \(x(x-4)+3(x-4)=0\) \((x-4)(x+3)=0\) \(\mathrm{x}=-3,4, \text { but } \mathrm{x}>0 \mathrm{x}=4\) \(\text { case2: } \mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\) \(\therefore \mathrm{x}^2-(-\mathrm{x})-12=0\) \(\mathrm{x}^2+\mathrm{x}-12=0\) \(\mathrm{x}^2+4 \mathrm{x}-3 \mathrm{x}-12=0\) \(\mathrm{x}(\mathrm{x}+4)-3(\mathrm{x}+4)=0\) \((\mathrm{x}-3)(\mathrm{x}+4)=0\) \(\mathrm{x}=3,-4\) \(\text { But } \mathrm{x}=0 \Rightarrow \mathrm{x}=-4\) \(\therefore \text { Number of real solution are two }\) \(hlineright.\)
JEE Main 25.07.2021
Complex Numbers and Quadratic Equation
118174
The sum of the non-real roots of \(\left(x^2+x-2\right)\left(x^2+\right.\) \(x-3)=12\) is
1 1
2 -1
3 -6
4 6
Explanation:
B Put \(\quad x^2+x=y\) So, Equation becomes \((y-2)(y-3)=12\) \(y^2-5 y-6=0\) \((y-6)(y+1)=0\) \(y=6,-1\) when, \(y=6\), then we get \(x^2+x-6=0,\) \((x+3)(x-2)=0\) \(\mathrm{x}=-3,2\) When \(\mathrm{y}=-1\) \(\mathrm{x}^2+\mathrm{x}+1=0\) which has non real roots Sum of roots is -1
Jamia Millia Islamia-2013
Complex Numbers and Quadratic Equation
118175
The value of \(k\) for which the roots of the equation \(x^2+2(k+1) x+k^2=0\) are equal to
1 -1
2 \(-\frac{1}{2}\)
3 1
4 \(\frac{1}{2}\)
Explanation:
B Given the question. \(\mathbf{x}^2+2(\mathrm{k}+1) \mathrm{x}+\mathrm{k}^2 0\) Root are equal \(\Rightarrow \mathrm{D}=0\) \(\therefore \quad[2(\mathrm{k}+1)]^2-4 \mathrm{k}^2=0\) \(4(\mathrm{k}+1)^2-4 \mathrm{k}^2=0\) \(\Rightarrow \quad \mathrm{k}^2+1+2 \mathrm{k}-\mathrm{k}^2=0\) \(\Rightarrow \quad 1+2 \mathrm{k}=0\) \(\mathrm{k}=\frac{-1}{2}\)
Rajasthan PET-2010
Complex Numbers and Quadratic Equation
118176
By Newton-Raphson method, the positive root of the equation \(x^4-x-10=0\) is
1 1.871
2 1.868
3 1.856
4 None of these
Explanation:
A \(\mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) We have:- \(f(x)=x^4-x-10\) \(f(1)=(1)^4-1-10=-10\) \(f(2)=2^4-2-10=4\) \(f\left(x_0\right)=4\) So, \(f\left(x_0\right)=4\) \(\text { at, } \mathrm{x}_0=2\) \(f(x)=x^4-x-10\) \(f^{\prime}(x)=4 x^3-1\) \(f^{\prime}\left(x_0\right)=4\left(x_0\right)^3-1\) \(f^{\prime}(2)=4(2)^3-1\) \(=31\) \(\therefore \quad \mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) \(\mathrm{x}_1=2-\frac{4}{31}\) \(\mathrm{x}_1=\frac{62-4}{31}\) \(\mathrm{x}_1=\frac{50}{31}\) \(\mathrm{x}_1=1.871\)
118173
The number of real solutions of the equation \(x^2\) \(-|\mathbf{x}|-\mathbf{1 2}=\mathbf{0}\) is
1 2
2 3
3 1
4 4
Explanation:
A Given the equation \(\mathrm{x}^2-|\mathrm{x}|-12=0\) case : \(1 \mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(x^2-x-12=0\) \(x^2-4 x+3 x-12=0\) \(x(x-4)+3(x-4)=0\) \((x-4)(x+3)=0\) \(\mathrm{x}=-3,4, \text { but } \mathrm{x}>0 \mathrm{x}=4\) \(\text { case2: } \mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\) \(\therefore \mathrm{x}^2-(-\mathrm{x})-12=0\) \(\mathrm{x}^2+\mathrm{x}-12=0\) \(\mathrm{x}^2+4 \mathrm{x}-3 \mathrm{x}-12=0\) \(\mathrm{x}(\mathrm{x}+4)-3(\mathrm{x}+4)=0\) \((\mathrm{x}-3)(\mathrm{x}+4)=0\) \(\mathrm{x}=3,-4\) \(\text { But } \mathrm{x}=0 \Rightarrow \mathrm{x}=-4\) \(\therefore \text { Number of real solution are two }\) \(hlineright.\)
JEE Main 25.07.2021
Complex Numbers and Quadratic Equation
118174
The sum of the non-real roots of \(\left(x^2+x-2\right)\left(x^2+\right.\) \(x-3)=12\) is
1 1
2 -1
3 -6
4 6
Explanation:
B Put \(\quad x^2+x=y\) So, Equation becomes \((y-2)(y-3)=12\) \(y^2-5 y-6=0\) \((y-6)(y+1)=0\) \(y=6,-1\) when, \(y=6\), then we get \(x^2+x-6=0,\) \((x+3)(x-2)=0\) \(\mathrm{x}=-3,2\) When \(\mathrm{y}=-1\) \(\mathrm{x}^2+\mathrm{x}+1=0\) which has non real roots Sum of roots is -1
Jamia Millia Islamia-2013
Complex Numbers and Quadratic Equation
118175
The value of \(k\) for which the roots of the equation \(x^2+2(k+1) x+k^2=0\) are equal to
1 -1
2 \(-\frac{1}{2}\)
3 1
4 \(\frac{1}{2}\)
Explanation:
B Given the question. \(\mathbf{x}^2+2(\mathrm{k}+1) \mathrm{x}+\mathrm{k}^2 0\) Root are equal \(\Rightarrow \mathrm{D}=0\) \(\therefore \quad[2(\mathrm{k}+1)]^2-4 \mathrm{k}^2=0\) \(4(\mathrm{k}+1)^2-4 \mathrm{k}^2=0\) \(\Rightarrow \quad \mathrm{k}^2+1+2 \mathrm{k}-\mathrm{k}^2=0\) \(\Rightarrow \quad 1+2 \mathrm{k}=0\) \(\mathrm{k}=\frac{-1}{2}\)
Rajasthan PET-2010
Complex Numbers and Quadratic Equation
118176
By Newton-Raphson method, the positive root of the equation \(x^4-x-10=0\) is
1 1.871
2 1.868
3 1.856
4 None of these
Explanation:
A \(\mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) We have:- \(f(x)=x^4-x-10\) \(f(1)=(1)^4-1-10=-10\) \(f(2)=2^4-2-10=4\) \(f\left(x_0\right)=4\) So, \(f\left(x_0\right)=4\) \(\text { at, } \mathrm{x}_0=2\) \(f(x)=x^4-x-10\) \(f^{\prime}(x)=4 x^3-1\) \(f^{\prime}\left(x_0\right)=4\left(x_0\right)^3-1\) \(f^{\prime}(2)=4(2)^3-1\) \(=31\) \(\therefore \quad \mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) \(\mathrm{x}_1=2-\frac{4}{31}\) \(\mathrm{x}_1=\frac{62-4}{31}\) \(\mathrm{x}_1=\frac{50}{31}\) \(\mathrm{x}_1=1.871\)
118173
The number of real solutions of the equation \(x^2\) \(-|\mathbf{x}|-\mathbf{1 2}=\mathbf{0}\) is
1 2
2 3
3 1
4 4
Explanation:
A Given the equation \(\mathrm{x}^2-|\mathrm{x}|-12=0\) case : \(1 \mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(x^2-x-12=0\) \(x^2-4 x+3 x-12=0\) \(x(x-4)+3(x-4)=0\) \((x-4)(x+3)=0\) \(\mathrm{x}=-3,4, \text { but } \mathrm{x}>0 \mathrm{x}=4\) \(\text { case2: } \mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\) \(\therefore \mathrm{x}^2-(-\mathrm{x})-12=0\) \(\mathrm{x}^2+\mathrm{x}-12=0\) \(\mathrm{x}^2+4 \mathrm{x}-3 \mathrm{x}-12=0\) \(\mathrm{x}(\mathrm{x}+4)-3(\mathrm{x}+4)=0\) \((\mathrm{x}-3)(\mathrm{x}+4)=0\) \(\mathrm{x}=3,-4\) \(\text { But } \mathrm{x}=0 \Rightarrow \mathrm{x}=-4\) \(\therefore \text { Number of real solution are two }\) \(hlineright.\)
JEE Main 25.07.2021
Complex Numbers and Quadratic Equation
118174
The sum of the non-real roots of \(\left(x^2+x-2\right)\left(x^2+\right.\) \(x-3)=12\) is
1 1
2 -1
3 -6
4 6
Explanation:
B Put \(\quad x^2+x=y\) So, Equation becomes \((y-2)(y-3)=12\) \(y^2-5 y-6=0\) \((y-6)(y+1)=0\) \(y=6,-1\) when, \(y=6\), then we get \(x^2+x-6=0,\) \((x+3)(x-2)=0\) \(\mathrm{x}=-3,2\) When \(\mathrm{y}=-1\) \(\mathrm{x}^2+\mathrm{x}+1=0\) which has non real roots Sum of roots is -1
Jamia Millia Islamia-2013
Complex Numbers and Quadratic Equation
118175
The value of \(k\) for which the roots of the equation \(x^2+2(k+1) x+k^2=0\) are equal to
1 -1
2 \(-\frac{1}{2}\)
3 1
4 \(\frac{1}{2}\)
Explanation:
B Given the question. \(\mathbf{x}^2+2(\mathrm{k}+1) \mathrm{x}+\mathrm{k}^2 0\) Root are equal \(\Rightarrow \mathrm{D}=0\) \(\therefore \quad[2(\mathrm{k}+1)]^2-4 \mathrm{k}^2=0\) \(4(\mathrm{k}+1)^2-4 \mathrm{k}^2=0\) \(\Rightarrow \quad \mathrm{k}^2+1+2 \mathrm{k}-\mathrm{k}^2=0\) \(\Rightarrow \quad 1+2 \mathrm{k}=0\) \(\mathrm{k}=\frac{-1}{2}\)
Rajasthan PET-2010
Complex Numbers and Quadratic Equation
118176
By Newton-Raphson method, the positive root of the equation \(x^4-x-10=0\) is
1 1.871
2 1.868
3 1.856
4 None of these
Explanation:
A \(\mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) We have:- \(f(x)=x^4-x-10\) \(f(1)=(1)^4-1-10=-10\) \(f(2)=2^4-2-10=4\) \(f\left(x_0\right)=4\) So, \(f\left(x_0\right)=4\) \(\text { at, } \mathrm{x}_0=2\) \(f(x)=x^4-x-10\) \(f^{\prime}(x)=4 x^3-1\) \(f^{\prime}\left(x_0\right)=4\left(x_0\right)^3-1\) \(f^{\prime}(2)=4(2)^3-1\) \(=31\) \(\therefore \quad \mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) \(\mathrm{x}_1=2-\frac{4}{31}\) \(\mathrm{x}_1=\frac{62-4}{31}\) \(\mathrm{x}_1=\frac{50}{31}\) \(\mathrm{x}_1=1.871\)
118173
The number of real solutions of the equation \(x^2\) \(-|\mathbf{x}|-\mathbf{1 2}=\mathbf{0}\) is
1 2
2 3
3 1
4 4
Explanation:
A Given the equation \(\mathrm{x}^2-|\mathrm{x}|-12=0\) case : \(1 \mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(x^2-x-12=0\) \(x^2-4 x+3 x-12=0\) \(x(x-4)+3(x-4)=0\) \((x-4)(x+3)=0\) \(\mathrm{x}=-3,4, \text { but } \mathrm{x}>0 \mathrm{x}=4\) \(\text { case2: } \mathrm{x}\lt 0 \Rightarrow|\mathrm{x}|=-\mathrm{x}\) \(\therefore \mathrm{x}^2-(-\mathrm{x})-12=0\) \(\mathrm{x}^2+\mathrm{x}-12=0\) \(\mathrm{x}^2+4 \mathrm{x}-3 \mathrm{x}-12=0\) \(\mathrm{x}(\mathrm{x}+4)-3(\mathrm{x}+4)=0\) \((\mathrm{x}-3)(\mathrm{x}+4)=0\) \(\mathrm{x}=3,-4\) \(\text { But } \mathrm{x}=0 \Rightarrow \mathrm{x}=-4\) \(\therefore \text { Number of real solution are two }\) \(hlineright.\)
JEE Main 25.07.2021
Complex Numbers and Quadratic Equation
118174
The sum of the non-real roots of \(\left(x^2+x-2\right)\left(x^2+\right.\) \(x-3)=12\) is
1 1
2 -1
3 -6
4 6
Explanation:
B Put \(\quad x^2+x=y\) So, Equation becomes \((y-2)(y-3)=12\) \(y^2-5 y-6=0\) \((y-6)(y+1)=0\) \(y=6,-1\) when, \(y=6\), then we get \(x^2+x-6=0,\) \((x+3)(x-2)=0\) \(\mathrm{x}=-3,2\) When \(\mathrm{y}=-1\) \(\mathrm{x}^2+\mathrm{x}+1=0\) which has non real roots Sum of roots is -1
Jamia Millia Islamia-2013
Complex Numbers and Quadratic Equation
118175
The value of \(k\) for which the roots of the equation \(x^2+2(k+1) x+k^2=0\) are equal to
1 -1
2 \(-\frac{1}{2}\)
3 1
4 \(\frac{1}{2}\)
Explanation:
B Given the question. \(\mathbf{x}^2+2(\mathrm{k}+1) \mathrm{x}+\mathrm{k}^2 0\) Root are equal \(\Rightarrow \mathrm{D}=0\) \(\therefore \quad[2(\mathrm{k}+1)]^2-4 \mathrm{k}^2=0\) \(4(\mathrm{k}+1)^2-4 \mathrm{k}^2=0\) \(\Rightarrow \quad \mathrm{k}^2+1+2 \mathrm{k}-\mathrm{k}^2=0\) \(\Rightarrow \quad 1+2 \mathrm{k}=0\) \(\mathrm{k}=\frac{-1}{2}\)
Rajasthan PET-2010
Complex Numbers and Quadratic Equation
118176
By Newton-Raphson method, the positive root of the equation \(x^4-x-10=0\) is
1 1.871
2 1.868
3 1.856
4 None of these
Explanation:
A \(\mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) We have:- \(f(x)=x^4-x-10\) \(f(1)=(1)^4-1-10=-10\) \(f(2)=2^4-2-10=4\) \(f\left(x_0\right)=4\) So, \(f\left(x_0\right)=4\) \(\text { at, } \mathrm{x}_0=2\) \(f(x)=x^4-x-10\) \(f^{\prime}(x)=4 x^3-1\) \(f^{\prime}\left(x_0\right)=4\left(x_0\right)^3-1\) \(f^{\prime}(2)=4(2)^3-1\) \(=31\) \(\therefore \quad \mathrm{x}_1=\mathrm{x}_0-\frac{\mathrm{f}\left(\mathrm{x}_0\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_0\right)}\) \(\mathrm{x}_1=2-\frac{4}{31}\) \(\mathrm{x}_1=\frac{62-4}{31}\) \(\mathrm{x}_1=\frac{50}{31}\) \(\mathrm{x}_1=1.871\)
