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Differential Calculus> find the range f(x)=(-x 2 +4x-3) 1/2 +(si...

question mark

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

vineet nimesh , 11 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 11 Years ago

Ans:

$f(x) = \sqrt{-x^{2}+4x-3}+\sqrt{sin(\frac{\pi }{2}(sin(\frac{\pi }{2}(x-1))))}$

1^stcalculate 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

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