KVPY SX SB Exam 05-Nov-2017 Question Paper

Question : 2 of 120

Marks: +1, -0

The number of solution pairs (x, y) of the simultaneous equations

\log_{1\/3} (x + y) + \log_3 (x -y) = 2

and

2^y^2 = 512^{x+1}

Solution:

We have,

\log _{1 \/ 3}(x+y)+\log _{3}(x-y) \;\; =2

2 y^{2} \;\; =512^{x+1}

⇒ \log _{3^{-1}}(x+y)+\log _{3}(x-y) \;\; =2

⇒-\log _{3}(x+y)+\log _{3}(x-y) \;\; =2

⇒ \;\; \;\; \;{x-y}/{x+y}=3^{2} \;\; =9

⇒ x-y=9 x+9 y

⇒ -8 x=10 y ⇒-4 x=5 y

\;\text " and "\; 2^{y^{2}}=2^{9(x+1)}

⇒ y^{2}=9(x+1)

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

(\;{-4 x}/{5})^{2}=9(x+1)

⇒ 16 x^{2}=225 x+225

⇒ 16 x^{2}-225 x-225=0

⇒ (16 x+15)(x-15)=0

⇒ x=15,-\;{15}/{16}

\;\text " When "\; x=15, y=-12

x=-\;{15}/{16}, y=\;{3}/{4}(\;\text " not possible "\;)

∴ \;\text " Only one solution "\;(15,-12)

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