JEE MAIN - Mathematics (2023 - 12th April Morning Shift - No. 8)
Let $$\alpha, \beta$$ be the roots of the quadratic equation $$x^{2}+\sqrt{6} x+3=0$$. Then $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$$ is equal to :
72
9
729
81
Explanation
Given quadratic equation: $$x^{2}+\sqrt{6} x+3=0$$
We can find the roots using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, $$a = 1$$, $$b = \sqrt{6}$$, and $$c = 3$$.
Substituting these values into the quadratic formula, we have:
$$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$$
Simplifying further :
$$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$$
$$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$$
Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $$i$$ :
$$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$$
$$ \Rightarrow $$ $$x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$$
$$ \therefore $$ $$\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$$.
The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:
$$ \begin{aligned} & =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$
Here, the exponential power of $$\sqrt{3}$$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $$(\sqrt{3})^{8} = 81$$ to simplify:
$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$
Since the cosine function has a period of $$2\pi$$, we can reduce the arguments of the cosine function in the numerator and denominator.
We have $$69\pi/4 = \pi/4 + 17\pi = \pi/4$$,
$$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$$,
$$45\pi/4 = \pi/4 + 11\pi = \pi/4$$, and
$$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$$.
Therefore, the required expression simplifies to :
$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\ & = (\sqrt{3})^{15-7} \\\\ & = (\sqrt{3})^{8} \\\\ & = 81. \end{aligned} $$
We can find the roots using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, $$a = 1$$, $$b = \sqrt{6}$$, and $$c = 3$$.
Substituting these values into the quadratic formula, we have:
$$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$$
Simplifying further :
$$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$$
$$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$$
Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $$i$$ :
$$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$$
$$ \Rightarrow $$ $$x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$$
$$ \therefore $$ $$\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$$.
The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:
$$ \begin{aligned} & =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$
Here, the exponential power of $$\sqrt{3}$$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $$(\sqrt{3})^{8} = 81$$ to simplify:
$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$
Since the cosine function has a period of $$2\pi$$, we can reduce the arguments of the cosine function in the numerator and denominator.
We have $$69\pi/4 = \pi/4 + 17\pi = \pi/4$$,
$$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$$,
$$45\pi/4 = \pi/4 + 11\pi = \pi/4$$, and
$$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$$.
Therefore, the required expression simplifies to :
$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\ & = (\sqrt{3})^{15-7} \\\\ & = (\sqrt{3})^{8} \\\\ & = 81. \end{aligned} $$
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