Gujarati
8. Sequences and Series
normal

मान लीजिए कि त्रिभुज $A B C$ की भुजाएँ $a, b, c$ हैं, एवं वह $b^2=a c$ को संतुष्ट करती हैं। तब $\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}$ के सभी संभावित मानों का समुच्चय क्या होगा ?

A

$(0, \infty)$

B

$\left(0, \frac{\sqrt{5}+1}{2}\right)$

C

$\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$

D

$\left(\frac{\sqrt{5}-1}{2}, \infty\right)$

(KVPY-2021)

Solution

(c)

$\frac{\sin A \frac{\cos C}{\sin C}+\cos A}{\sin B \frac{\cos C}{\sin C}+\cos B}=\frac{\frac{\sin A \cos C+\sin C \cos A}{\sin C}}{\frac{\sin B \operatorname{Cos} C+\cos B \sin C}{\sin C}}$

$=\frac{\sin (A+C)}{\sin (B+C)}=\frac{b}{a}$

a, b, c $\rightarrow$ G.P.

$b =a r ; c =a r ^2$

$C-I$ If $r > 1$ then

$a+a r > a r^2 \Rightarrow r^2-r-1 < 0$

$r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ but $r > 1$

Hence $r \in\left(1, \frac{1+\sqrt{5}}{2}\right) \quad \ldots .( I ) \ldots$

$C-II$ $0 < r < 1$

$a r+a r^2 > a \Rightarrow r^2+r-1 > 0$

$r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$

but $r \in(0,1)$

$r \in\left(\frac{\sqrt{5}-1}{2}, 1\right) \quad \ldots (II) ….$

$C-III$ when $r=1$ then $\Delta$ is equilateral

Hence $r \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$

Standard 11
Mathematics

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