# find the rangef(x)=(-x2+4x-3)1/2+(sinπ/2(sinπ/2(x-1)))½Thanks

Jitender Singh IIT Delhi
9 years ago
Ans:
$f(x) = \sqrt{-x^{2}+4x-3}+\sqrt{sin(\frac{\pi }{2}(sin(\frac{\pi }{2}(x-1))))}$
1stcalculate the domain:
$-x^{2}+4x-3\geq 0$
$x^{2}-4x+3\leq 0$
$(x-1)(x-3)\leq 0$
$\Rightarrow x\in [1, 3]$
$sin(\frac{\pi }{2}(sin(\frac{\pi }{2}(x-1))))\geq 0$
$\Rightarrow 0\leq \frac{\pi }{2}(sin(\frac{\pi }{2}(x-1)))\leq \pi$
$\Rightarrow 0\leq sin(\frac{\pi }{2}(x-1))\leq 2$
$\Rightarrow 0\leq \frac{\pi }{2}(x-1)\leq \pi$
$\Rightarrow 0\leq x-1\leq 2$
$\Rightarrow 1\leq x\leq 3$
Range:
We need to find max. & min. value of f(x)
$f_{1}(x) = \sqrt{-x^{2}+4x-3}, 1\leq x\leq 3$
Inside the root, we have inverted parabola, max will be at
$x = \frac{-b}{2a} = \frac{-4}{2(-1)} = 2$
$f_{1}(2) = \sqrt{-2^{2}+4.2-3} = 1$
$f_{2}(x) = \sqrt{sin(\frac{\pi }{2}(sin(\frac{\pi }{2}(x-1))))$
$f_{2}(2) = \sqrt{sin(\frac{\pi }{2}(sin(\frac{\pi }{2}(2-1))))$
$f_{2}(2) =1$
Minima will be at x =1, 3
$f_{2}(1) =0$
$f_{2}(3) =0$
$f_{1}(1) =0$
$f_{1}(3) =0$
Range of f(x):
$[0,2]$
Thanks & Regards
Jitender Singh
IIT Delhi