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BITSAT 2017 Slot 2 Solved Paper

Section: Mathematics

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Question : 145 of 150

Marks: +1, -0

If

\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}

f(1)=2

f(p+q)=f(p) \cdot f(q), \forall p, q \in R

x^{f(1)}+y^{f(2)}+z^{f(3)}-\frac{x+y+z}{x^{f(1)}+y^{f(2)}+z^{f(3)}}

is equal to

0
1
2
3

Solution:

\because \frac{-\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}, \frac{-\pi}{2} \le \sin^{-1} y \le \frac{\pi}{2}

and

\frac{-\pi}{2} \le \sin^{-1} z \le \frac{\pi}{2}

Given that,

\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}

Which is possible only when

\sin^{-1} x = \sin^{-1} y = \sin^{-1} z = \frac{\pi}{2}

\Rightarrow x = y = z = 1

Put

p=q=1

Then,

f(2)=f(1) f(1)=2 \times 2 = 4

and put

p=1, q=2

then,

f(3)=f(1) f(2)

=2 \cdot 2^{2}=8

\therefore x^{t(1)}+y^{t(2)}+z^{t(3)}-\frac{x+y+z}{x^{t(1)}+y^{t(2)}+z^{t(3)}}

=1+1+1-\frac{3}{1+1+1}=3-1=2

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