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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x - 16 = 0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$.
If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then
$\dfrac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \dfrac{Q_{25} - Q_{23}}{Q_{24}}$
is equal to
Solution
Given
$ x^2 + \sqrt{3}x - 16 = 0 $
$ \alpha + \beta = -\sqrt{3}, ;; \alpha\beta = -16 $
Recurrence:
$ P_n = -\sqrt{3}P_{n-1} + 16P_{n-2} $
Divide by $P_{n-1}$:
Let $ r_n = \dfrac{P_n}{P_{n-1}} $
$ r_n = -\sqrt{3} + \dfrac{16}{r_{n-1}} $
For large $n$, $ r_n \to \alpha $ (root)
So,
$ \dfrac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \dfrac{P_{24}(r_{25}+\sqrt{3})}{2P_{23}} = \dfrac{r_{24}(r_{25}+\sqrt{3})}{2} $
Using root relation:
$ \alpha^2 + \sqrt{3}\alpha -16 = 0 \Rightarrow \alpha + \sqrt{3} = \dfrac{16}{\alpha} $
So expression becomes:
$ \dfrac{\alpha \cdot \frac{16}{\alpha}}{2} = 8 $
Now second equation
$ x^2 + 3x - 1 = 0 $
$ \gamma + \delta = -3, ;; \gamma\delta = -1 $
Recurrence:
$ Q_n = -3Q_{n-1} + Q_{n-2} $
Similarly,
$ \dfrac{Q_{25} - Q_{23}}{Q_{24}} = \dfrac{Q_{24}r_{25} - Q_{24}/r_{24}}{Q_{24}} = r_{25} - \dfrac{1}{r_{24}} $
Using root equation:
$ \gamma^2 + 3\gamma -1=0 \Rightarrow \gamma - \dfrac{1}{\gamma} = -3 $
So value = $ -3 $
Final Answer:
$ 8 - 3 = \boxed{5} $
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