Domain - Functions

1) f(x) = √x² - |x| - 2
We impose various restrictions(set theory ke mutaabik) as per definition of functions involved. Then we take intersection of those restrictions to get the domain.

R1 : Square root function. For √x, x ≥ 0.
So x² - |x| - 2 ≥ 0
=> |x|² - |x| - 2 ≥ 0 (As x² is the same as |x|²)
=> (|x| + 1)(|x| - 2) ≥ 0
Using the wavy curve method, we can easily see that |x| ≤ -1 U |x| ≥ 2 for the inequality to be satisfied(where U is union).
As |x| ≤ -1 is not possible, we discard it from the union(As any set U Null set = the set itself).
So |x| ≥ 2 or x ≤ -2 U x ≥ 2.
Hence R1 -> x ≤ -2 U x ≥ 2

R2 : Modulus function. For |x|, x ≥ 0.
Here simply x ≥ 0.

Now taking intersection, R1 ∩ R2 => x ≥ 2 is the domain of the function...

2) Dunno bout this one...it never came right for me when I tried it.

3) You have to be careful of the base of the logarithms here.
As per def of log function,
log₂log_2/Ï€(Y) ≥ 0 where Y is (tan^-1x)^-1 for simplicity.
=> log₂log_2/Ï€(Y) ≥ log₂1
=> log_2/π(Y) ≥ 1
=> log_2/π(Y) ≥ log_2/π(2/π)
=> Y ≤ 2/π [Note the inequality change, this is because logarithm base was smaller than 1].
=> 1/tan^-1x ≤ 2/π
I think we can cross multiply. This is because 1/tan^-1x will have to be positive anyway, for the rest of the logs to work. Sooo...
=> tan^-1x ≥ π/2...
Feeling sleepy...lol...I'll be back for the rest later [13]

2)
either question or answer is wrong because putting x =1 we get i (iota)

5)

here cases arise as we get -1=< [2x²-3]>=1 in sin^-1[2x²-3] ...so.....

0 =< 2x²-3 < 1..........eq.1 we get sin^-10

-1 =< 2x²-3 <0...........eq.2 we get sin^-1-1

2x²-3= 1...........eq.3 we get sin^-11

2x²-3= -1..........eq.4 we get sin^-1-1

2x²-3= 0..........eq.5 we get sin^-10

now find out x using these equations.....

for 2nd one may be u did some mistake while typing the sumas x = 1 is clearly not in domain

pritish for 5th one u made a mistake

-1 ≤[2x² - 3 ] ≤1

so -1 ≤ 2x² -3 < 2

1 ≤ x² < 5/2

so x≡ (-√52 , -1] U [1 , √52)

3rd one
y= log_{10}log_{2}log_{2}log_{2/\pi}(\frac{1}{tan^{-1}x})

\Rightarrow log_{2}log_{2}log_{2/\pi}(\frac{1}{tan^{-1}x})>0

\Rightarrow log_{2}log_{2/\pi}(\frac{1}{tan^{-1}x})>1

\Rightarrow log_{2/\pi}(\frac{1}{tan^{-1}x})>2
\Rightarrow 0<(\frac{1}{tan^{-1}x})< (\frac{2}{\pi })^{2}

i.e \Rightarrow \infty>tan^{-1}x >(\frac{\pi}{2})^{2}

wich isnt possible ,
hence domain is an empty set ( if i didnt miss anything)

Thanks a lot Pritish for the detailed explanation.[1]

@Mavarick, thanks a lot. You din make any mistake while solving the questions [1], nice explanation and solution[1]
You did a small mistake in the 5th question while solving
[2x²-3] < 2

here x will lie between -√(5/2) and √(5/2), Qwerty has given the solution too.

Thanks a lot Nihal and Qwerty [1]

I got all the question now , yay yay except the 2nd one of course.

Thanks a lot lot friends [1][1]

for 2nd one

case (i) | x | > 1

thus x⁸ + 1 + 1 > x⁸ + 1

thus x⁴(x⁸ + 1) + 1 > x (x⁸ + 1)

i.e x¹² + x⁴ + 1 > x⁹ + x

hence all x satisfying |x| > 1 are in domain

now case (ii)

|x| ≤ 1
i.e x is a fraction
hence x⁴ > x⁹

1 > x

x¹² > 0

hence

x¹² + x⁴ + 1 > x + x⁹

hence all x satisfying |x| ≤1 also lie in the domain

hence all reals lie in the domain

Thanks a lot Qwerty [1]