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Differential Calculus> Domain Image is attached...

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Drake , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$f(x) = \sqrt{cos(sinx)} + \sqrt{log_{x}(x)}$

1st Part:

$cos(sinx)\geq 0$

$\Rightarrow x\in R$

Explaination:

$cos(x)\geq 0$

when

$\frac{-\pi }{2}\leq x\leq \frac{\pi }{2}$

$-1\leq sinx\leq 1$

[-1, 1] lie in the interval [-pi/2, pi/2].

2nd Part:

$x > 0$

$log_{x}(x)\geq 0$

Suppose for ths ‘x’ in the base

$0<x<1$

$\Rightarrow x\leq 1$

Suppose for ths ‘x’ in the base

$x>1$

$\Rightarrow x\geq 1$

Final Solution:

$x\in (0,1)\cup (1,\infty )$

Thanks & Regards

Jitender Singh

IIT Delhi

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