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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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
askIITians Faculty

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