If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2 and f(1)=2,f(p+q)=f(p)⋅f(q),∀p,q∈R then xf(1)+yf(2)+zf(3)−(x+y+z)xf(1)+yf(2)+zf(3) is equal to

Solution:

$∵ - \frac{π}{2} \leq \sin^{- 1} x \leq \frac{π}{2}, - \frac{π}{2} \leq \sin^{- 1} y \leq \frac{π}{2}$

and $- \frac{π}{2} \leq \sin^{- 1} z \leq \frac{π}{2}$

Given that $\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{3 π}{2}$

which is possible only when

$\begin{array}{l} \sin^{- 1} x = \sin^{- 1} y = \sin^{- 1} z = \frac{π}{2} \\ \Rightarrow x = y = z = 1 \\ put p = q = 1 \\ then, f (2) = f (1) f (1) = 2 \cdot 2 = 4 \\ and put p = 1, q = 2 \\ then f (3) = f (1) f (2) = 2 \cdot 2^{2} = 8 \end{array}$

$\begin{aligned} ∴ x^{f (1)} + y^{f (2)} + z^{f (3)} - \frac{x + y + z}{x^{f (1)} + y^{f (2)} + z^{f (3)}} \\ = 1 + 1 + 1 - \frac{3}{1 + 1 + 1} = 3 - 1 = 2 \end{aligned}$

If $\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = \frac{3 π}{2}$ and $f (1) = 2, f (p + q) = f (p) \cdot f (q), \forall p, q \in R$ then $x^{f (1)} + y^{f (2)} + z^{f (3)} - \frac{(x + y + z)}{x^{f (1)} + y^{f (2)} + z^{f (3)}}$ is equal to

Solution:

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