Functions Test 5

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Functions Test 5

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Weekly Quiz Competition

Question 1
1 / -0

The number of one-one functions that can be defined from $$A=\{4,8,12,16\}$$ to $$B$$ is $$5040,$$ then $$n(B)=$$

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

The first element $$4$$ of $$A$$ can be mapped to any of the $$n$$ elements in $$B$$
Similarly second element $$8$$ of $$A$$ can be mapped to any of $$(n-1)$$ elements in $$B$$
$$\Rightarrow$$ Total no of one - one functions is
$$n(n-1)(n-2)(n-3)=5040$$
$$\Rightarrow n=10$$
Question 2
1 / -0

If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=x^{2}, g(x)=\cos x$$ then $$(gof)(x)=$$

A
$$\cos 2x$$

B
$$x^{2}\cos x$$

C
$$\cos x^{2}$$

D
$$\cos^{2} x^{2}$$

Solution

$$gof\left ( x \right )=g\left ( f\left ( x \right ) \right )=g\left ( x^{2} \right )$$
$$=\cos x^{2}$$
Question 3
1 / -0

Let $$f(x)=\dfrac{Kx}{x+1}(x\neq -1)$$ then the value of $$K$$ for which $$(fof)(x)=x$$ is

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$\sqrt{2}$$

Solution

$$f\left ( f\left ( x \right ) \right )=f\left ( \dfrac{kx}{x+1} \right )$$
$$=\dfrac{k\left ( \dfrac{kx}{x+1} \right )}{\dfrac{kx}{x+1}=1}=\dfrac{k^{2}x}{\left ( k+1 \right )x+1}$$
$$\therefore$$ for $$f\left ( f\left ( x \right ) \right )=x$$
$$\Rightarrow k+1=0$$
$$k^{2}=1$$
$$\therefore k=-1$$
Question 4
1 / -0

If $$n (A) = 4$$ and $$n(B) = 6$$, then the number of surjections from $$A$$ to $$B$$ is

A
$$4^{6}$$

B
$$6^{4}$$

C
$$0$$

D
$$24$$

Solution

$$n(A)<n(B)$$
$$\therefore$$ Co-domain can never be equal to range.
$$\therefore$$ No of surjections is zero
Question 5
1 / -0

$$f:\left ( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right )\rightarrow \left ( -\infty ,\infty \right )$$ defined by $$f(x)=1+3x$$ is

A
one-one but not onto

B
onto but not one-one

C
neither one - one nor onto

D
bijective

Solution

$$f(x_{1})=f(x_{2}) \Rightarrow 1+3x_{1}=1+3x$$
$$x_{1}=x_{2}$$
$$\therefore f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}$$
$$\therefore$$ f is one-one.
$$f$$ lies b/w $$(1-\dfrac{3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\therefore$$ co-domain $$(-\infty ,\infty )$$ is not equal to range $$(\dfrac{1-3\pi }{2},\dfrac{1+3\pi }{2})$$
$$\Rightarrow$$ f is not onto.
$$\therefore$$ f is one-one but not onto.
Question 6
1 / -0

Let $$A=\{1,2,3\}, B =\{a, b, c\}$$ and If $$f=\{(1,a),(2,b),(3,c)\}, g=\{(1,b),(2,a),(3,b)\}, h=\{(1,b)(2,c),(3,a)\}$$ then

A
$$g$$ and $$h$$ are injections

B
$$f$$ and $$h$$ are injections

C
$$f$$ and $$g$$ injections

D
$$f,g$$ and $$h$$ are injections

Solution

$$g(1)=g(3)=b \Rightarrow$$ g is not one-one.
domain of $$f =$$ domain of $$h = A.$$
range of $$f =$$ range of $$h = B$$
$$n(A)=n(B)=3$$
$$\therefore f$$ and $$h$$ are injections.
Question 7
1 / -0

If $$f:R\rightarrow R, g:R\rightarrow R$$ are defined by $$f(x)=4x-1,g(x)=x^{3}+2,$$ then $$(gof)\left(\dfrac{a+1}{4}\right)=$$

A
$$43$$

B
$$4a^3-1$$

C
$$a^{3}+2$$

D
$$64a^3 - 8a^{2}-1$$

Solution

$$g\left ( f\left ( \dfrac{a+1}{4} \right ) \right )=g\left ( 4\left ( \dfrac{a+1}{4} \right )-1 \right )$$
$$=g\left ( a \right )$$
$$=a^{3}+2$$
Question 8
1 / -0

If $$f(x)=2x+1$$ and $$g(x)=x^{2}+1$$ then $$ (go(fof))(2)=$$

A
$$112$$

B
$$122$$

C
$$12$$

D
$$124$$

Solution

$$(go(fof))\left( x \right) =g\left( f\left( f\left( x \right) \right) \right) $$
$$=g\left ( f\left ( 2x+1 \right ) \right )$$
$$=g\left ( 2\left ( 2x+1 \right )+1 \right )$$
$$=g\left ( 4x+3 \right )$$
$$=\left ( 4x+3 \right )^{2}+1$$
$$(go(fof))\left ( 2 \right )=\left ( 11 \right )^{2}+1=122$$
Question 9
1 / -0

If $$f:R\rightarrow R,f(x)=3x-2$$ then $$ (fof)(x)+2=$$

A
$$f(x)$$

B
$$2f(x)$$

C
$$3f(x)$$

D
$$-f(x)$$

Solution

$$fof (x)=f(f(x))$$
$$=f(3x-2)$$
$$=3(3x-2)-2$$
$$=9x-8$$
$$fof(x)+2=9x-6=3(3x-2)$$
$$=3f(x)$$
Question 10
1 / -0

If $$f(x)=\dfrac{1}{x}, g(x)=\sqrt{x}$$ and $$ (go\sqrt{f})(16)=$$

A
$$2$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$4$$

Solution

$$\sqrt{f\left ( x \right )}=\dfrac{1}{\sqrt{x}}$$
$$g\left ( \sqrt{f\left ( x \right )} \right )=g\left ( \dfrac{1}{\sqrt{x}} \right )=\dfrac{1}{\sqrt{\sqrt{x}}}=\dfrac{1}{x^{\frac{1}{4}}}$$
$$\left ( go\sqrt{f} \right )(16)=\dfrac{1}{\left ( 16 \right )^{\dfrac{1}{4}}}=\dfrac{1}{2}$$

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Re-Attempt Weekly Quiz Competition

Functions Test 5

Result

Functions Test 5

Score

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Rank

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SHARING IS CARING

Question 1

$$7$$

$$8$$

$$9$$

$$10$$

Solution

Question 2

$$\cos 2x$$

$$x^{2}\cos x$$

$$\cos x^{2}$$

$$\cos^{2} x^{2}$$

Solution

Question 3

$$1$$

$$-1$$

$$2$$

$$\sqrt{2}$$

Solution

Question 4

$$4^{6}$$

$$6^{4}$$

$$0$$

$$24$$

Solution

Question 5

one-one but not onto

onto but not one-one

neither one - one nor onto

bijective

Solution

Question 6

$$g$$ and $$h$$ are injections

$$f$$ and $$h$$ are injections

$$f$$ and $$g$$ injections

$$f,g$$ and $$h$$ are injections

Solution

Question 7

$$43$$

$$4a^3-1$$

$$a^{3}+2$$

$$64a^3 - 8a^{2}-1$$

Solution

Question 8

$$112$$

$$122$$

$$12$$

$$124$$

Solution

Question 9

$$f(x)$$

$$2f(x)$$

$$3f(x)$$

$$-f(x)$$

Solution

Question 10

$$2$$

$$1$$

$$\dfrac{1}{2}$$

$$4$$

Solution

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