NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118228
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathbf{p x}+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then
1 \(\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{pr}=0\)
2 \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
3 \(\mathrm{p}^2-\mathrm{q}^2-2 \mathrm{pr}=0\)
4 \(\mathrm{p}^2+\mathrm{q}^2+2 \mathrm{pr}=0\)
Explanation:
B Given, \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) And roots are \(\sin \alpha\) and \(\cos \alpha\) Then, \(\sin \alpha+\cos \alpha=\frac{-q}{p}\) \(\sin \alpha \cdot \cos \alpha=\frac{r}{p}\) Now, squaring the equation (i) \((\sin \alpha+\cos \alpha)^2=\left(\frac{-q}{p}\right)^2\) \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cdot \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\)\(\quad\left\{\because \sin \alpha \cdot \cos \alpha=\frac{\mathrm{r}}{\mathrm{p}}\right\}\) \(1+\frac{2 \mathrm{r}}{\mathrm{p}}=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
JCECE-2008
Complex Numbers and Quadratic Equation
118229
Let \(\alpha\) and \(\beta\) be the roots of the equation \(\mathbf{x}^2+\mathbf{x}+\mathbf{1}=\mathbf{0}\), then the equation whose roots are \(\alpha^{19}, \beta^7\), is
1 \(x^2-x-1=0\)
2 \(x^2-x+1=0\)
3 \(x^2+x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D Given, \(x^2+x+1=0\) Then, roots of given equation is \(\alpha, \beta\) \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1^2-4}}{2}\) \(x=\frac{-1 \pm i \sqrt{3}}{2}\) \(x=\alpha=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}=\beta\) and we know that cube root unity \(\alpha=\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2} \text { and } \omega^2=\beta=\frac{-1-\mathrm{i} \sqrt{3}}{2}\) Now, the equation whose root are \(\alpha^{19} \& \beta^7\) \(\alpha^{19}=\left(\omega^{19}\right)=\omega\) \(\beta^7=\left(\omega^2\right)^7=\omega^{14}=\omega^2\) So, equation is- \(x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \cdot \beta^7=0\) \(x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0\) \(\left\{\because \omega+\omega^2=-1, \omega^3=1\right\}\) \(x^2-(-1) x+1=0\) \(x^2+x+1=0\)
JCECE-2007
Complex Numbers and Quadratic Equation
118230
The number of real roots of \((6-x)^4+(8-x)^4=16\) is
1 0
2 2
3 4
4 6
Explanation:
B Given, \((6-x)^4+(8-x)^4=16\) Let, \(7-x=t\) \((t-1)^4+(t+1)^4=16\) \(t^4-4 t^3+6 t^2-4 t+1+t^4+4 t^3+6 t^2+4 t\) \(2 \mathrm{t}^4+12 \mathrm{t}^2+2=16\) \(\mathrm{t}^4+6 \mathrm{t}^2+1=8\) \(\mathrm{t}^4+6 \mathrm{t}^2-7=0\) Now, let \(\mathrm{t}^2=\mathrm{y}\) \(y^2+6 y-7=0\) \((y+7)(y-1)=0\) \(y=-7\) \(t^2 \neq-7\) \(t^2=1\) \(t= \pm 1\) So, \(x=7-t\) \(x=6,8\)Therefore, number of real roots are 2
BCECE-2017
Complex Numbers and Quadratic Equation
118231
If the roots the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}+1\)
3 \(\mathrm{p}^2=4 \mathrm{q}-1\)
4 none of these
Explanation:
B Given, \(x^2-p x+q=0\) Let roots are \(\alpha \& \beta\) Then, according to question- roots are differ by unity \(\alpha-\beta=1\) Squaring both sides- \((\alpha-\beta)^2=1^2\) \(\alpha^2+\beta^2-2 \alpha \beta=1\) \(\alpha^2+\beta^2=1+2 \alpha \beta\) Now sum of the roots of given equation \(\alpha+\beta=p\) Product of roots \(\alpha \cdot \beta=q\) Squaring equation (ii) both side \(\alpha^2+\beta^2+2 \alpha \beta=\mathrm{p}^2\) \(1+2 \alpha \beta+2 \alpha \beta=\mathrm{p}^2\) \(1+4 \alpha \beta=\mathrm{p}^2 \left\{\because \alpha^2+\beta^2=1+2 \alpha \beta\right\}\) \(\mathrm{p}^2=4 \mathrm{q}+1 \{\because \alpha \beta=\mathrm{q}\}\)
118228
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathbf{p x}+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then
1 \(\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{pr}=0\)
2 \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
3 \(\mathrm{p}^2-\mathrm{q}^2-2 \mathrm{pr}=0\)
4 \(\mathrm{p}^2+\mathrm{q}^2+2 \mathrm{pr}=0\)
Explanation:
B Given, \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) And roots are \(\sin \alpha\) and \(\cos \alpha\) Then, \(\sin \alpha+\cos \alpha=\frac{-q}{p}\) \(\sin \alpha \cdot \cos \alpha=\frac{r}{p}\) Now, squaring the equation (i) \((\sin \alpha+\cos \alpha)^2=\left(\frac{-q}{p}\right)^2\) \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cdot \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\)\(\quad\left\{\because \sin \alpha \cdot \cos \alpha=\frac{\mathrm{r}}{\mathrm{p}}\right\}\) \(1+\frac{2 \mathrm{r}}{\mathrm{p}}=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
JCECE-2008
Complex Numbers and Quadratic Equation
118229
Let \(\alpha\) and \(\beta\) be the roots of the equation \(\mathbf{x}^2+\mathbf{x}+\mathbf{1}=\mathbf{0}\), then the equation whose roots are \(\alpha^{19}, \beta^7\), is
1 \(x^2-x-1=0\)
2 \(x^2-x+1=0\)
3 \(x^2+x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D Given, \(x^2+x+1=0\) Then, roots of given equation is \(\alpha, \beta\) \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1^2-4}}{2}\) \(x=\frac{-1 \pm i \sqrt{3}}{2}\) \(x=\alpha=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}=\beta\) and we know that cube root unity \(\alpha=\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2} \text { and } \omega^2=\beta=\frac{-1-\mathrm{i} \sqrt{3}}{2}\) Now, the equation whose root are \(\alpha^{19} \& \beta^7\) \(\alpha^{19}=\left(\omega^{19}\right)=\omega\) \(\beta^7=\left(\omega^2\right)^7=\omega^{14}=\omega^2\) So, equation is- \(x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \cdot \beta^7=0\) \(x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0\) \(\left\{\because \omega+\omega^2=-1, \omega^3=1\right\}\) \(x^2-(-1) x+1=0\) \(x^2+x+1=0\)
JCECE-2007
Complex Numbers and Quadratic Equation
118230
The number of real roots of \((6-x)^4+(8-x)^4=16\) is
1 0
2 2
3 4
4 6
Explanation:
B Given, \((6-x)^4+(8-x)^4=16\) Let, \(7-x=t\) \((t-1)^4+(t+1)^4=16\) \(t^4-4 t^3+6 t^2-4 t+1+t^4+4 t^3+6 t^2+4 t\) \(2 \mathrm{t}^4+12 \mathrm{t}^2+2=16\) \(\mathrm{t}^4+6 \mathrm{t}^2+1=8\) \(\mathrm{t}^4+6 \mathrm{t}^2-7=0\) Now, let \(\mathrm{t}^2=\mathrm{y}\) \(y^2+6 y-7=0\) \((y+7)(y-1)=0\) \(y=-7\) \(t^2 \neq-7\) \(t^2=1\) \(t= \pm 1\) So, \(x=7-t\) \(x=6,8\)Therefore, number of real roots are 2
BCECE-2017
Complex Numbers and Quadratic Equation
118231
If the roots the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}+1\)
3 \(\mathrm{p}^2=4 \mathrm{q}-1\)
4 none of these
Explanation:
B Given, \(x^2-p x+q=0\) Let roots are \(\alpha \& \beta\) Then, according to question- roots are differ by unity \(\alpha-\beta=1\) Squaring both sides- \((\alpha-\beta)^2=1^2\) \(\alpha^2+\beta^2-2 \alpha \beta=1\) \(\alpha^2+\beta^2=1+2 \alpha \beta\) Now sum of the roots of given equation \(\alpha+\beta=p\) Product of roots \(\alpha \cdot \beta=q\) Squaring equation (ii) both side \(\alpha^2+\beta^2+2 \alpha \beta=\mathrm{p}^2\) \(1+2 \alpha \beta+2 \alpha \beta=\mathrm{p}^2\) \(1+4 \alpha \beta=\mathrm{p}^2 \left\{\because \alpha^2+\beta^2=1+2 \alpha \beta\right\}\) \(\mathrm{p}^2=4 \mathrm{q}+1 \{\because \alpha \beta=\mathrm{q}\}\)
118228
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathbf{p x}+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then
1 \(\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{pr}=0\)
2 \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
3 \(\mathrm{p}^2-\mathrm{q}^2-2 \mathrm{pr}=0\)
4 \(\mathrm{p}^2+\mathrm{q}^2+2 \mathrm{pr}=0\)
Explanation:
B Given, \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) And roots are \(\sin \alpha\) and \(\cos \alpha\) Then, \(\sin \alpha+\cos \alpha=\frac{-q}{p}\) \(\sin \alpha \cdot \cos \alpha=\frac{r}{p}\) Now, squaring the equation (i) \((\sin \alpha+\cos \alpha)^2=\left(\frac{-q}{p}\right)^2\) \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cdot \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\)\(\quad\left\{\because \sin \alpha \cdot \cos \alpha=\frac{\mathrm{r}}{\mathrm{p}}\right\}\) \(1+\frac{2 \mathrm{r}}{\mathrm{p}}=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
JCECE-2008
Complex Numbers and Quadratic Equation
118229
Let \(\alpha\) and \(\beta\) be the roots of the equation \(\mathbf{x}^2+\mathbf{x}+\mathbf{1}=\mathbf{0}\), then the equation whose roots are \(\alpha^{19}, \beta^7\), is
1 \(x^2-x-1=0\)
2 \(x^2-x+1=0\)
3 \(x^2+x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D Given, \(x^2+x+1=0\) Then, roots of given equation is \(\alpha, \beta\) \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1^2-4}}{2}\) \(x=\frac{-1 \pm i \sqrt{3}}{2}\) \(x=\alpha=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}=\beta\) and we know that cube root unity \(\alpha=\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2} \text { and } \omega^2=\beta=\frac{-1-\mathrm{i} \sqrt{3}}{2}\) Now, the equation whose root are \(\alpha^{19} \& \beta^7\) \(\alpha^{19}=\left(\omega^{19}\right)=\omega\) \(\beta^7=\left(\omega^2\right)^7=\omega^{14}=\omega^2\) So, equation is- \(x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \cdot \beta^7=0\) \(x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0\) \(\left\{\because \omega+\omega^2=-1, \omega^3=1\right\}\) \(x^2-(-1) x+1=0\) \(x^2+x+1=0\)
JCECE-2007
Complex Numbers and Quadratic Equation
118230
The number of real roots of \((6-x)^4+(8-x)^4=16\) is
1 0
2 2
3 4
4 6
Explanation:
B Given, \((6-x)^4+(8-x)^4=16\) Let, \(7-x=t\) \((t-1)^4+(t+1)^4=16\) \(t^4-4 t^3+6 t^2-4 t+1+t^4+4 t^3+6 t^2+4 t\) \(2 \mathrm{t}^4+12 \mathrm{t}^2+2=16\) \(\mathrm{t}^4+6 \mathrm{t}^2+1=8\) \(\mathrm{t}^4+6 \mathrm{t}^2-7=0\) Now, let \(\mathrm{t}^2=\mathrm{y}\) \(y^2+6 y-7=0\) \((y+7)(y-1)=0\) \(y=-7\) \(t^2 \neq-7\) \(t^2=1\) \(t= \pm 1\) So, \(x=7-t\) \(x=6,8\)Therefore, number of real roots are 2
BCECE-2017
Complex Numbers and Quadratic Equation
118231
If the roots the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}+1\)
3 \(\mathrm{p}^2=4 \mathrm{q}-1\)
4 none of these
Explanation:
B Given, \(x^2-p x+q=0\) Let roots are \(\alpha \& \beta\) Then, according to question- roots are differ by unity \(\alpha-\beta=1\) Squaring both sides- \((\alpha-\beta)^2=1^2\) \(\alpha^2+\beta^2-2 \alpha \beta=1\) \(\alpha^2+\beta^2=1+2 \alpha \beta\) Now sum of the roots of given equation \(\alpha+\beta=p\) Product of roots \(\alpha \cdot \beta=q\) Squaring equation (ii) both side \(\alpha^2+\beta^2+2 \alpha \beta=\mathrm{p}^2\) \(1+2 \alpha \beta+2 \alpha \beta=\mathrm{p}^2\) \(1+4 \alpha \beta=\mathrm{p}^2 \left\{\because \alpha^2+\beta^2=1+2 \alpha \beta\right\}\) \(\mathrm{p}^2=4 \mathrm{q}+1 \{\because \alpha \beta=\mathrm{q}\}\)
118228
If \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathbf{p x}+\mathbf{q x}+\mathbf{r}=\mathbf{0}\), then
1 \(\mathrm{p}^2+\mathrm{q}^2-2 \mathrm{pr}=0\)
2 \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
3 \(\mathrm{p}^2-\mathrm{q}^2-2 \mathrm{pr}=0\)
4 \(\mathrm{p}^2+\mathrm{q}^2+2 \mathrm{pr}=0\)
Explanation:
B Given, \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) And roots are \(\sin \alpha\) and \(\cos \alpha\) Then, \(\sin \alpha+\cos \alpha=\frac{-q}{p}\) \(\sin \alpha \cdot \cos \alpha=\frac{r}{p}\) Now, squaring the equation (i) \((\sin \alpha+\cos \alpha)^2=\left(\frac{-q}{p}\right)^2\) \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cdot \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\)\(\quad\left\{\because \sin \alpha \cdot \cos \alpha=\frac{\mathrm{r}}{\mathrm{p}}\right\}\) \(1+\frac{2 \mathrm{r}}{\mathrm{p}}=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(\mathrm{p}^2-\mathrm{q}^2+2 \mathrm{pr}=0\)
JCECE-2008
Complex Numbers and Quadratic Equation
118229
Let \(\alpha\) and \(\beta\) be the roots of the equation \(\mathbf{x}^2+\mathbf{x}+\mathbf{1}=\mathbf{0}\), then the equation whose roots are \(\alpha^{19}, \beta^7\), is
1 \(x^2-x-1=0\)
2 \(x^2-x+1=0\)
3 \(x^2+x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D Given, \(x^2+x+1=0\) Then, roots of given equation is \(\alpha, \beta\) \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1^2-4}}{2}\) \(x=\frac{-1 \pm i \sqrt{3}}{2}\) \(x=\alpha=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}=\beta\) and we know that cube root unity \(\alpha=\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2} \text { and } \omega^2=\beta=\frac{-1-\mathrm{i} \sqrt{3}}{2}\) Now, the equation whose root are \(\alpha^{19} \& \beta^7\) \(\alpha^{19}=\left(\omega^{19}\right)=\omega\) \(\beta^7=\left(\omega^2\right)^7=\omega^{14}=\omega^2\) So, equation is- \(x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \cdot \beta^7=0\) \(x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0\) \(\left\{\because \omega+\omega^2=-1, \omega^3=1\right\}\) \(x^2-(-1) x+1=0\) \(x^2+x+1=0\)
JCECE-2007
Complex Numbers and Quadratic Equation
118230
The number of real roots of \((6-x)^4+(8-x)^4=16\) is
1 0
2 2
3 4
4 6
Explanation:
B Given, \((6-x)^4+(8-x)^4=16\) Let, \(7-x=t\) \((t-1)^4+(t+1)^4=16\) \(t^4-4 t^3+6 t^2-4 t+1+t^4+4 t^3+6 t^2+4 t\) \(2 \mathrm{t}^4+12 \mathrm{t}^2+2=16\) \(\mathrm{t}^4+6 \mathrm{t}^2+1=8\) \(\mathrm{t}^4+6 \mathrm{t}^2-7=0\) Now, let \(\mathrm{t}^2=\mathrm{y}\) \(y^2+6 y-7=0\) \((y+7)(y-1)=0\) \(y=-7\) \(t^2 \neq-7\) \(t^2=1\) \(t= \pm 1\) So, \(x=7-t\) \(x=6,8\)Therefore, number of real roots are 2
BCECE-2017
Complex Numbers and Quadratic Equation
118231
If the roots the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}+1\)
3 \(\mathrm{p}^2=4 \mathrm{q}-1\)
4 none of these
Explanation:
B Given, \(x^2-p x+q=0\) Let roots are \(\alpha \& \beta\) Then, according to question- roots are differ by unity \(\alpha-\beta=1\) Squaring both sides- \((\alpha-\beta)^2=1^2\) \(\alpha^2+\beta^2-2 \alpha \beta=1\) \(\alpha^2+\beta^2=1+2 \alpha \beta\) Now sum of the roots of given equation \(\alpha+\beta=p\) Product of roots \(\alpha \cdot \beta=q\) Squaring equation (ii) both side \(\alpha^2+\beta^2+2 \alpha \beta=\mathrm{p}^2\) \(1+2 \alpha \beta+2 \alpha \beta=\mathrm{p}^2\) \(1+4 \alpha \beta=\mathrm{p}^2 \left\{\because \alpha^2+\beta^2=1+2 \alpha \beta\right\}\) \(\mathrm{p}^2=4 \mathrm{q}+1 \{\because \alpha \beta=\mathrm{q}\}\)
