Test Index

TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

© examsnet.com

Question : 9 of 160

Marks: +1, -0

a + \frac{\cos\theta}{2} + \frac{\cos 2\theta}{4} + \frac{\cos 3\theta}{8} + \dots + \frac{a - 2\cos\theta}{5 + b\cos\theta}

a,b \in R,

(a-b)^2=

0
64
36
125

Solution:

We have,

1+\; \frac{\cos\theta}{2} +\; \frac{\cos 2\theta}{4} +\; \frac{\cos 3\theta}{8} +\dots = \; \frac{a-2\cos\theta}{5+b\cos\theta}

Put,

\theta = 0

\therefore \;\; 1+\;\frac{1}{2}+\;\frac{1}{4}+\;\frac{1}{8}\dots =\;\frac{a-2}{5+b}

\Rightarrow \;\; \;\frac{1}{1-\frac{1}{2}} = \;\frac{a-2}{5+b} \Rightarrow 2 = \;\frac{a-2}{5+b}

\Rightarrow \;\; 10+2b = a-2 \Rightarrow a-2b = 12 \;\; \dots (i)

Put,

\theta = \pi

1-\;\frac{1}{2}+\;\frac{1}{4}-\;\frac{1}{8}+\dots =\;\frac{a+2}{5-b}

\Rightarrow \;\; \;\frac{1}{1+\frac{1}{2}} = \;\frac{a+2}{5-b} \Rightarrow \;\frac{2}{3} = \;\frac{a+2}{5-b}

\Rightarrow \;\; 10-2b = 3a+6 \Rightarrow 3a+2b = 4

... (ii)

From Eq. (i) and (ii), we get

a=4, b=-4

\therefore \;\; (a-b)^2 = (4+4)^2 = 8^2 = 64

© examsnet.com

Go to Question:

Prev Question

Next Question