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find the range f(x)=(-x 2 +4x-3) 1/2 +(sinπ/2(sinπ/2(x-1))) ½ Thanks

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

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
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
askIITians Faculty

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