Functions Test 46

$$^{ n }C_{ r }$$ is defined when $$n,r \in N, \; n \geq r$$ ($$r$$ can be zero)
$$ \displaystyle ^{ 16-x }C_{ 2x-1 }$$ is defined for 
$$16-x>0.....(1)\\ 2x-1\ge 0.....(2)\\ 16-x\ge 2x-1.....(3)$$

Combining $$(1), (2)$$ and $$(3)$$ we get
.
$$x=\{1,2,3,4,5 \}$$.........$$(i)$$
.
Similarly, $$^{ 20-3x }P_{ 4x-5 }$$ is defined for 
$$20-3x>0....(4)\\ 4x-5\ge 0....(5)\\ 20-3x\ge 4x-5.....(6)$$
.
Combining $$(4), (5)$$ and $$(6)$$ we get
.
$$x= \{2,3 \}$$.........$$(ii)$$
.
From $$(i)$$ and $$(ii)$$, the common values of $$x$$ are $$2$$ and $$3$$.

$$\sqrt { \left( x+2 \right) \left( 5-x \right)  } $$ is defined for 

$$a^{x+y}=a^{x}+a^{y}$$

$$a_{1} \in $$A can have $$n$$ images in $$B$$, but the element $$a_{2}$$  will have only $$(n-1)$$ images as the mappings are to be one-one (injective). 

$$g\left( f\left( x \right)  \right) \le f\left( g\left( x \right)  \right) \Rightarrow g\left( \left| x \right|  \right) \le f\left( \left[ x \right]  \right) \Rightarrow \left[ \left| x \right|  \right] \le \left[ \left| x \right|  \right] $$
This is true for each $$x\in R$$

Let $$\displaystyle F\left ( x \right )=h\left ( x \right )-h\left ( 1 \right )=f\left ( g\left ( x \right ) \right )-h\left ( 1 \right )$$ $$\displaystyle F'\left ( x \right )=f'\left ( g\left ( x \right ) \right ).g'\left ( x \right )=\left ( + \right )\left ( - \right )=-ive.$$ (As f is increasing function $$\displaystyle f'\left ( g\left ( x \right ) \right )$$ is +ive and as g is decreasing function $$\displaystyle g'\left ( x \right )$$ is-ive.) 
Since F'(x) is-ive therefore F(x)i.e.h(x)-h(1) is decreasing function.
Now split the interval $$\displaystyle I=\left [ 0,\infty  \right ]$$ into two intervals $$\displaystyle I_{1},0\leq x< 1and I_{2}, 1\leq x< \infty ,$$ 
Apply the definition of decreasing function on $$\displaystyle h\left ( x \right )-h\left ( 1 \right ):$$ on $$\displaystyle I_{1},0\leq x< 1,\underset{\left ( Big \right )}{h\left ( x \right )}-\underset{\left ( Less \right )}{h\left ( 1 \right )}=+ive$$ On $$\displaystyle I_{2},1\leq x< \infty ,\underset{\left ( Less \right )}{h\left ( x \right )}-\underset{\left ( Big \right )}{h\left ( 1 \right )}=-ive$$ Hence for $$\displaystyle I,h\left ( x \right )-h\left ( 1 \right )$$ is neither always zero nor always +ive nor always-ive,nor strictly increasing throughout.
Hence (v) is the correct answer.

There are two components, Let's split them.
Component 1:
$$\displaystyle { \cos }^{ -1 }\left(\frac { 2-\left| x \right|  }{ 4 }\right)$$
We know that domain of $$\displaystyle {\cos}^{-1}(x)$$ is $$[-1, 1]$$,
$$\displaystyle \therefore -1 \le \frac { 2-\left| x \right|  }{ 4 } \le 1$$
Multiply both sides by $$4$$ and then subtract $$2$$ from both sides, inequality reduces to
$$\displaystyle -6\le -\left|x\right| \le 2$$, if this is multiplied by $$-1$$ then the equality will be reversed, $$\displaystyle -2\le\left|x\right|\le6$$, which gives the range of $$\displaystyle x$$ for component $$1$$, $$\displaystyle x\in[-6,6]$$ ..... $$(i)$$

Component 2:
$$\displaystyle \left\{ \log(3-x) \right\}^{ -1 }$$
Domain of generalized $$\displaystyle { \log(x) }^{ -1 }$$ is $$\displaystyle (0,1)\cup(1,\infty)$$,
Here $$1$$ is excluded because $$\displaystyle \log(x)$$ is in the denominator.
$$\displaystyle \therefore (3-x)>0$$, which gives $$\displaystyle x<3$$
Domain of component $$2$$ is $$\displaystyle (\infty,2)\cup(2,3)$$ ..... $$(ii)$$ 

Final domain of equation $$\displaystyle { \cos }^{ -1 }\left(\frac { 2-\left| x \right|  }{ 4 }\right)  + { \log(3-x) }^{ -1 }$$ is the intersection of $$(i)$$ and $$(ii)$$

Given $$\displaystyle f\left ( x \right )=\frac{ax^{2}+2x+1}{2x^{2}-2x+1}$$

$$\displaystyle (i) \log_{a}b$$ hold good if $$\displaystyle a> 0,a\neq 1,a> 1$$