Functions Test 15

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Functions Test 15

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle f\left ( x \right )= \log \left | 4-x^{2} \right |$$ then domain of $$x$$ is

A
$$x \in R$$ except $$x=\pm 2 $$

B
$$x \in R$$ except $$x=\pm 1 $$

C
$$x \in R$$ except $$x=\pm 4 $$

D
$$x=\pm 2 $$

Solution

Here $$\displaystyle f\left ( x \right )$$ is defined for $$|4-x^2| > 0$$

Clearly $$|4-x^2| $$ is greater than zero for every real $$x$$ except at $$x=\pm 2$$
Question 2
1 / -0

The domain of $$\displaystyle \frac{1}{\log _{10}\left ( 1-x \right )}+\sqrt{\left ( x+2 \right )}$$ is

A
$$\displaystyle \left ( -2\leq x< -1 \right )\cup \left ( 0< x< 1 \right )$$

B
$$\displaystyle \left ( -2\leq x< 2 \right )$$

C
$$\displaystyle \left ( -2\leq x< 1 \right )$$

D
$$\displaystyle \left ( -2\leq x< 0 \right )\cup \left ( 0< x< 1 \right )$$

Solution

$$f(x) = \displaystyle \frac{1}{\log _{10}\left ( 1-x \right )}+\sqrt{\left ( x+2 \right )}$$

For this function to be defined, $$(1-x) > 0,\neq 1$$ and $$(x+2) \geq 0$$

$$\Rightarrow x < 1, \neq 0 $$ and $$x\geq -2$$

Thus domain of $$f(x)$$ is $$ x\in [-2,0) \cup (0,1)$$
Question 3
1 / -0

$$\displaystyle f\left ( x \right )=\dfrac1{\left ( \left | x \right |-x \right )^{1/2}}$$. Then the domain of the function is

A
$$x>0$$

B
$$x \geq 0$$

C
$$x<0$$

D
$$x \leq 0 $$

Solution

For real values of $$y$$,

$$|x|-x>0$$ and $$|x|-x\neq 0$$

For $$x>0$$

$$|x|-x=x-x$$

$$=0$$

Thus for all $$x>0$$, $$|x|-x=0$$.

For $$x<0$$

$$|x|-x$$$$=-x-x$$$$=-2x$$

Since $$x<0$$

$$-2x>0$$

Thus domain of $$f(x)$$ is
$$x<0$$.
Question 4
1 / -0

The domain of $$\displaystyle f\left ( x \right )= \sqrt{\log \left ( \dfrac{ 5x-x^{2}}6 \right )}$$ is

A
$$2\leq x \leq 3$$

B
$$3\leq x \leq 4$$

C
$$-3\leq x \leq -2$$

D
$$1\leq x \leq 3$$

Solution

The function is defined for all values of x for which $$\displaystyle \frac{5x-x^{2}}{6}> 0$$ ...(1)

and $$\displaystyle \log \left [ \dfrac{\left( 5x-x^{2} \right )}6 \right ]\geq 0= \log 1$$ ...(2)

(1)$$\displaystyle \Rightarrow x\left ( 5-x \right )> 0$$ or $$x\left ( x-5 \right )< 0$$ ...(3)

(2)$$\displaystyle \Rightarrow \dfrac{\left ( 5x-x^{2} \right )}6\geq 1i.e. x^{2}-5x+6\leq 0$$ ...(4) or

$$\displaystyle x\left ( x-5 \right )< 0$$ and $$\displaystyle \left ( x-2 \right
)\left ( x-3 \right )\leq 0$$

$$\displaystyle \therefore D_{1}= \left [
0, 5 \right ], D_{2}= \left [ 2, 3 \right ]$$

Hence the inequalities (1)

and (2) will hold for $$\displaystyle D_{1}\cap D_{2} i.e.$$ if
$$\displaystyle 2\leq x \leq 3$$ or $$\left [ 2,3 \right ]$$

(Draw $$\displaystyle D_{1}$$ and $$D_{2}$$ on real line and take the common part of both.)
Question 5
1 / -0

Determine domain of $$\displaystyle \frac{\sqrt{4-x^{2}}}{\sin ^{-1}\left ( 2-x \right )}$$

A
(4,5)

B
[1,2]

C
both a and b

D
none of the above

Solution

$$f(x)=\dfrac{\sqrt{4-x^{2}}}{sin^{-1}(2-x)}$$
The function gives an undefined $$(\dfrac{0}{0})$$ as x approaches 2.
Therefore
$$limit_{x\rightarrow 2} f(x)$$

$$=limit_{x\rightarrow 2} \dfrac{\sqrt{4-x^{2}}}{sin^{-1}(2-x)}$$

$$=limit_{x\rightarrow2}\dfrac{\sqrt{4-x^{2}}}{2-x(\dfrac{sin^{-1}(2-x)}{2-x})}$$

$$=limit_{x\rightarrow2}\dfrac{\sqrt{2+x}}{\sqrt{2-x}\dfrac{sin^{-1}(2-x)}{2-x}}$$

$$=\dfrac{\sqrt{4}}{0\times 1}$$

$$=\infty$$...(i)
Hence as $$x\rightarrow 0$$, $$y\rightarrow \infty$$.

Also, for real values of f(x)

$$4-x^{2}\geq0$$

$$x^{2}\leq 4$$
$$|x|\leq 2$$
$$-2\leq x\leq 2$$...(ii)
And
$$\dfrac{-\pi}{2}\leq sin^{-1}(2-x)\leq \dfrac{\pi}{2}$$

$$-1\leq 2-x\leq 1$$

Therefore
$$2-x\leq 1$$
$$x\geq1$$

And

$$-1\leq 2-x$$
$$x\leq 3$$

$$x\epsilon [1,3]$$...(iii)
From i, ii, and iii

The domain of x is
$$x\epsilon [1,2]$$
Question 6
1 / -0

The domain of $$\displaystyle \log \left [ 1-\log _{10}\left ( x^{2}-5x+16 \right ) \right ]$$ is

A
$$\displaystyle 2< x< 3$$

B
$$\displaystyle 1< x< 3 $$

C
$$\displaystyle -2< x< 3 $$

D
$$\displaystyle -2< x< 2 $$

Solution

Let $$f(x) = \displaystyle \log \left [ 1-\log _{10}\left ( x^{2}-5x+16 \right ) \right ]$$

For $$f$$ to be defined, we must have $$\displaystyle x^{2}-5x+16> 0$$ which is true as

$$\displaystyle \Delta = $$ -ive and the sign of first term is +ive. Again by definition of $$\displaystyle \log x$$ we must have

$$\displaystyle 1-\log _{10}\left ( x^{2}-5x+16 \right )> 0$$

$$\Rightarrow \displaystyle \log _{10}\left ( x^{2}-5x+16 \right )< 1= \log _{10}10$$

$$\Rightarrow \displaystyle

x^{2}-5x+16< 10 $$

or $$\displaystyle

x^{2}-5x+6< 0$$ or $$\Rightarrow \displaystyle \left ( x-2 \right )\left ( x-3

\right )< 0$$

or $$\displaystyle 2< x< 3$$

$$\therefore x \in \left ( 2, 3 \right )$$
Question 7
1 / -0

If $$\displaystyle f\left ( x \right )=\frac{ax+b}{cx+d}$$ and $$\displaystyle \left ( fof \right )x=x,$$ then d=?

A
$$a$$

B
$$-a$$

C
$$b$$

D
$$-b$$

Solution

$$\displaystyle \left ( fof \right )x=x\Rightarrow f\left [ f\left ( x \right ) \right ]=x$$ or $$\displaystyle f\left [ \frac{ax+b}{cx+d} \right ]=x$$ or $$\displaystyle\frac{a\left [ \frac{ax+b}{cx+d}\right ]+b}{c\left [\frac{ax+b}{cx+d} \right ]+d}=x$$ $$\displaystyle \therefore \frac{x\left ( a^{2}+bc \right )+b\left ( a+d \right )}{cx\left ( a+d \right )+\left ( bc+d^{2} \right )}=x$$ Clearly if $$\displaystyle a+d=0$$ or $$\displaystyle d=-a$$ then in that case $$\displaystyle L.H.S.=\frac{x\left ( a^{2}+bc \right )+0}{0+\left ( bc+a^{2} \right )}=x$$ $$\displaystyle \because d=-a$$
Question 8
1 / -0

The domain of the function $$\displaystyle \sqrt{\left \{ \frac{\log _{0.3}\left | x-2 \right |}{\left | x \right |} \right \}}$$ is

A
$$\displaystyle x \in \left [ \left ( 1, 2 \right )\cup \left ( 2, 3 \right ) \right ]$$

B
$$\displaystyle x \in \left [ \left ( 1, 3 \right )\cup \left ( 3, 5 \right ) \right ]$$

C
$$\displaystyle x \in \left [ \left ( 0, 2 \right )\cup \left ( 2, 3 \right ) \right ]$$

D
$$\displaystyle x \in ( 1, 3)$$

Solution

The given function is defined if $$\displaystyle x\neq 0, x\neq 2$$ and

$$\displaystyle \log _{0.3}\left | x-2 \right |> 0$$

Now $$\displaystyle

\log _{0.3}\left | x-2 \right |> 0$$

$$\displaystyle

\Rightarrow \left | x-2 \right |< 1$$ , as the base $$0.3$$ is less than $$1$$, the inequality is reversed.

$$\displaystyle \therefore -1\leq x-2\leq 1$$

$$\displaystyle 1\leq x\leq 3.$$

Out of the above, we have to exclude $$0$$ and $$2$$.

Hence the domain is $$\displaystyle 1\leq x\leq 3$$ but

$$\displaystyle x\neq 0, x\neq 2$$ or $$\displaystyle x \in \left [

\left ( 1, 2 \right )\cup \left ( 2, 3 \right ) \right ]$$
Question 9
1 / -0

The total number of injective mappings from a set with $$m$$ elements to a set with $$n$$ elements,$$\displaystyle m\leq n,$$ is

A
$$\displaystyle m^{n}$$

B
$$\displaystyle n^{m}$$

C
$$\displaystyle \frac{n!}{\left ( n-m \right )!}$$

D
$$\displaystyle n!$$

Solution

Let $$A=\{a_1,a_2, a_3.....a_m\}$$
and $$B=\{b_1,b_2, b_3.....b_n\}$$ where $$m \le n$$
Given $$f:A\rightarrow B$$ be an injective mapping.
So, for $$a_1 \in A$$, there are $$n$$ possible choices for $$f(a_1)\in B$$.
For $$a_2 \in A$$, there are $$(n-1)$$ possible choices for $$f(a_2)\in B$$.
Similarly for $$a_m \in A$$, there are $$(n-m-1)$$ choices for $$f(a_m)\in B$$
So, there are $$n(n-1)(n-2).....(n-m-1)=\dfrac{n!}{(n-m)!}$$ injective mapping from $$A$$ to $$B.$$
Question 10
1 / -0

Are the following sets of ordered pairs functions? If so, examine whether the mapping is surjective or injective :
{(x, y): x is a person, y is the mother of x}

A
injective (one- one ) and surjective (into)

B
injective (one- one ) and not surjective (into)

C
not injective (one- one ) and surjective (into)

D
not injective (one- one ) and not surjective (into)

Solution

We have {(x, y) : x is a person, y is the mother of x}. Clearly each person 'x' has only one biological mother. So above set of ordered pair is a function. Now more than one person may have same mother. So function is many-one and surjective.

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Re-Attempt Weekly Quiz Competition

Functions Test 15

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Functions Test 15

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Question 1

$$x \in R$$ except $$x=\pm 2 $$

$$x \in R$$ except $$x=\pm 1 $$

$$x \in R$$ except $$x=\pm 4 $$

$$x=\pm 2 $$

Solution

Question 2

$$\displaystyle \left ( -2\leq x< -1 \right )\cup \left ( 0< x< 1 \right )$$

$$\displaystyle \left ( -2\leq x< 2 \right )$$

$$\displaystyle \left ( -2\leq x< 1 \right )$$

$$\displaystyle \left ( -2\leq x< 0 \right )\cup \left ( 0< x< 1 \right )$$

Solution

Question 3

$$x>0$$

$$x \geq 0$$

$$x<0$$

$$x \leq 0 $$

Solution

Question 4

$$2\leq x \leq 3$$

$$3\leq x \leq 4$$

$$-3\leq x \leq -2$$

$$1\leq x \leq 3$$

Solution

Question 5

(4,5)

[1,2]

both a and b

none of the above

Solution

Question 6

$$\displaystyle 2< x< 3$$

$$\displaystyle 1< x< 3 $$

$$\displaystyle -2< x< 3 $$

$$\displaystyle -2< x< 2 $$

Solution

Question 7

$$a$$

$$-a$$

$$b$$

$$-b$$

Solution

Question 8

$$\displaystyle x \in \left [ \left ( 1, 2 \right )\cup \left ( 2, 3 \right ) \right ]$$

$$\displaystyle x \in \left [ \left ( 1, 3 \right )\cup \left ( 3, 5 \right ) \right ]$$

$$\displaystyle x \in \left [ \left ( 0, 2 \right )\cup \left ( 2, 3 \right ) \right ]$$

$$\displaystyle x \in ( 1, 3)$$

Solution

Question 9

$$\displaystyle m^{n}$$

$$\displaystyle n^{m}$$

$$\displaystyle \frac{n!}{\left ( n-m \right )!}$$

$$\displaystyle n!$$

Solution

Question 10

injective (one- one ) and surjective (into)

injective (one- one ) and not surjective (into)

not injective (one- one ) and surjective (into)

not injective (one- one ) and not surjective (into)

Solution

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