Functions Test 7

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Functions Test 7

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(n+1)=f(n)$$ for all $$n\in N, f(7)=5$$ then $$f(35)=$$

A
$$25$$

B
$$49$$

C
$$35$$

D
$$5$$
Question 2
1 / -0

If $$f(x)=\displaystyle \dfrac{x}{\sqrt{1-x^{2}}},g(x)=\displaystyle \dfrac{x}{\sqrt{1+x^{2}}} $$, then $$(fog)(x)=$$

A
$$x$$

B
$$\dfrac{x}{\sqrt{1+x^{2}}}$$

C
$$\sqrt{1+x^{2}}$$

D
$$2x$$

Solution

$$fog\left(x\right)=f\left(g\left(x\right)\right)$$
$$\displaystyle =f\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
$$\displaystyle =\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1-\dfrac{x^{2}}{1+x^{2}}}}=x$$
Question 3
1 / -0

The domain of $$f(x)= \displaystyle \frac{1}{\sqrt{(x-2)(7-x)}} is$$

A
$$(2,\infty )$$

B
$$(7,\infty )$$

C
$$(2,7)$$

D
$$(-\infty ,2)$$

Solution

For $$f(x)$$ to be defined,
$$(x-2)(7-x)> 0\Rightarrow (x-2)(x-7) < 0\Rightarrow 2< x<7$$
Hence domain of $$f(x)$$ is $$(2,7)$$
Question 4
1 / -0

If the function is $$f:R\rightarrow R, g:R\rightarrow R$$ are defined as $$f(x)=2x+3, g(x)=x^{2}+7$$ and $$f[g(x)]=25$$ then $$x=$$

A
$$f(x)$$

B
$$\pm 2$$

C
$$\pm 3$$

D
$$\pm 4$$

Solution

$$f(g(x))=f(x^{2}+7) =25$$
$$=2(x^{2}+7)+3$$
$$=2x^{2}+17$$
$$\Rightarrow 2x^{2}=8$$
$$x^{2}=4 \Rightarrow x=\pm 2$$
Question 5
1 / -0

If $$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$$ then $$fofofof(x)=$$

A
$$f(x)$$

B
$$2\left ( \dfrac{x+1}{x-1} \right )$$

C
$$\dfrac{x-1}{x+1}$$

D
$$x$$

Solution

$$fofofof\left(x\right)=fofof\left(\dfrac{x+1}{x-1}\right)$$
$$=fof\left(\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\right)=fof\left(\dfrac{2x}{2}\right)$$
$$=fof\left(x\right)$$
$$=f\left(\dfrac{x+1}{x-1}\right)=x$$
Question 6
1 / -0

Let $$Z$$ be the set of integers and $$f:Z\rightarrow Z$$ is a bijective function then

A
$$f(x)=x^{2}+4$$

B
$$f(x)=2x+3$$

C
$$f(x)=x$$

D
$$f(x)=x^{3}+3$$

Solution

Let us check all the options one by one:
Option A:
$$f(x)=x^2+4$$
Here, $$f(1) = 1^2 +4 = 1+4 = 5$$
$$f(-1) = (-1)^2 +4 = 1+4 = 5 $$
Thus, $$f(1) = f(-1)$$but $$1 \neq -1 $$
So, it is not one - one and hence not a bijective function.

Option B:
$$f(x)=2x+3$$
It is one - one but it is not onto, because for $$y=0, x=\frac{-3}{2}\notin Z$$
Hence, not a bijective function.

Option 3:
$$f(x)=x$$
It is one - one & onto. So it is a bijective function.

$$f(x)=x^3+3$$
it is one - one but not onto, because for $$y=0, x\notin Z$$
Hence, not a bijective function.
Question 7
1 / -0

If $$F(n)=(-1)^{k-1}(n-1), G(n)=n-F(n)$$ then $$ (GoG)(n)=$$ (where $$k$$ is odd)

A
$$1$$

B
$$n$$

C
$$2$$

D
$$n-1$$

Solution

$$G(n)=n-(-1)^{k-1}(n-1)$$
$$GoG(n)=G(n-(-1)^{k-1}(n-1))$$
$$=n-(-1)^{k-1}(n-1)-(-1)^{k-1}((n-1)-(-1)^{k-1}(n-1))$$
$$=n-(n-1)$$
$$=1$$
Question 8
1 / -0

If $$f(x)=\dfrac{x}{\sqrt{1-x^{2}}}$$, then $$ (fof)(x)=$$

A
$$\dfrac{x}{\sqrt{1-x^{2}}}$$

B
$$\dfrac{x}{\sqrt{1-2x^{2}}}$$

C
$$\dfrac{x}{\sqrt{1-3x^{2}}}$$

D
$$x$$

Solution

$$fof\left ( x \right )=f\left ( f\left ( x \right ) \right )=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-\left ( x^{2}/\left ( 1-x^{2} \right ) \right )} \right )$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\sqrt{\left ( 1-x^{2}-x^{2} \right )/\left ( 1-x^{2} \right )}$$
$$=\left ( x/\sqrt{1-x^{2}} \right )/\left ( \sqrt{1-2 x^{2}} /\sqrt{1-x^{2}}\right)$$
$$=x/\sqrt{1-2x^{2}}$$
Question 9
1 / -0

If $$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}-10x+21 $$ then $$ f^{-1}(-3)$$ is

A
$$\left \{ -4,6 \right \}$$

B
$$\left \{ 4,6 \right \}$$

C
$$\left \{ -4, 4, 6 \right \}$$

D
Not Invertible

Solution

Let $$f^{-1}(-3)=t$$
$$\Rightarrow f(t)=-3$$
$$t^{2}-10t+21=-3$$
$$t^{2}-10t+24=0$$
$$t^{2}-6t-4t+24=0$$
$$t(t-6)-4(t-6)=0$$
$$\Rightarrow t=6,4 = f^{-1}(-3)$$
Question 10
1 / -0

If $$f:[1,\infty )\rightarrow B$$ defined by the function $$ f(x)=x^{2}-2x+6$$ is a surjection, then $$B$$ is equals to

A
$$[1,\infty )$$

B
$$[5,\infty )$$

C
$$[6,\infty )$$

D
$$[2,\infty )$$

Solution

$$f(x)=x^2-2x+6$$ is a surjection.
So the range of $$f(x)$$ will be equal to its codomain.
$$f(x)=x^2-2x+6$$
$$f^1(x)=2x-2$$
$$=2(x-1)$$
$$f(x)$$ will be increasing when $$x\geqslant 1$$.
$$\therefore f(1)=1-2+6$$
$$=5$$
$$\therefore B=[5, \infty)$$

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Re-Attempt Weekly Quiz Competition

Functions Test 7

Result

Functions Test 7

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SHARING IS CARING

Question 1

$$25$$

$$49$$

$$35$$

$$5$$

Question 2

$$x$$

$$\dfrac{x}{\sqrt{1+x^{2}}}$$

$$\sqrt{1+x^{2}}$$

$$2x$$

Solution

Question 3

$$(2,\infty )$$

$$(7,\infty )$$

$$(2,7)$$

$$(-\infty ,2)$$

Solution

Question 4

$$f(x)$$

$$\pm 2$$

$$\pm 3$$

$$\pm 4$$

Solution

Question 5

$$f(x)$$

$$2\left ( \dfrac{x+1}{x-1} \right )$$

$$\dfrac{x-1}{x+1}$$

$$x$$

Solution

Question 6

$$f(x)=x^{2}+4$$

$$f(x)=2x+3$$

$$f(x)=x$$

$$f(x)=x^{3}+3$$

Solution

Question 7

$$1$$

$$n$$

$$2$$

$$n-1$$

Solution

Question 8

$$\dfrac{x}{\sqrt{1-x^{2}}}$$

$$\dfrac{x}{\sqrt{1-2x^{2}}}$$

$$\dfrac{x}{\sqrt{1-3x^{2}}}$$

$$x$$

Solution

Question 9

$$\left \{ -4,6 \right \}$$

$$\left \{ 4,6 \right \}$$

$$\left \{ -4, 4, 6 \right \}$$

Not Invertible

Solution

Question 10

$$[1,\infty )$$

$$[5,\infty )$$

$$[6,\infty )$$

$$[2,\infty )$$

Solution

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