Functions Test 21

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Functions Test 21

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f(x) =\frac {ax+b} {cx+d}$$. Then fof(x) = x provided that.

A
d =- a

B
d = a

C
a = b = c = d = 1

D
a = b = 1

Solution

$$f(f(x))=\dfrac{af(x) + b}{cf(x)+d}$$

$$f(f(x))=\dfrac{a\dfrac{ax+b}{cx+d} + b}{c\dfrac{ax+b}{cx+d}+d}$$

$$f(f(x))=\dfrac{a^2x + ab+bcx+bd}{acx+bc+d^2+dcx}=x$$

$$\Rightarrow a^2x+ab+bcx+bd=acx^2+bcx+d^2x+dcx^2$$

Given $$a=-d$$, the above equation can also be verified
Question 2
1 / -0

If $$f(x) =\dfrac {1}{1-x}, x \neq 0, 1$$ then the graph of the function $$y = f[f\{f(x)\}]$$ for $$x > 1 $$ is

A
a straight line

B
a circle

C
an ellipse

D
a pair of straight lines

Solution

Given $$f(x) = \cfrac{1}{1-x}$$
$$\displaystyle \Rightarrow f\{f(x)\} = \frac{1}{1-f(x)} = \frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}=1-\frac{1}{x}$$
$$\displaystyle \therefore f[f\{f(x)\}] = \frac{1}{1- f\{f(x)\}}=\frac{1}{1-1+\frac{1}{x}}=x$$, which is a straight line.
Question 3
1 / -0

The domain of the function f (x) = $$\displaystyle \frac{1}{\sqrt{x-3}}$$ is given by :

A
$$x < 3$$

B
$$x > 3$$

C
$$x$$ $$\geq$$ $$-3$$

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

Solution

The function will not be defined when the denominator is 0 or when the square root has a negative number.

Hence combination of the conditions

$$x\neq 3$$ and $$x\geq 3$$ gives:

$$x>3$$
Question 4
1 / -0

Let $$\displaystyle f\left ( x \right )=\frac{3}{2}+\sqrt{x-\frac{3}{4}}$$ be a function and $$g\left ( x \right )$$ be another function such that $$g\left ( f\left ( x \right ) \right )=x,$$ then the value of $$g\left ( 20 \right )$$ will be

A
$$333$$

B
$$335$$

C
$$338$$

D
$$343$$

Solution

$$g\left ( x \right )$$ is inverse of $$f\left ( x \right )$$

$$\Rightarrow $$ $$\displaystyle x=\frac{3}{2}+\sqrt{y-\frac{3}{4}}$$

$$\Rightarrow $$ $$\displaystyle y=\left ( x-\frac{3}{2} \right )^2+\frac{3}{4}=f^{-1}(x)=g(x)$$

$$\therefore $$ $$\displaystyle g\left ( 20 \right )=\left ( 20-\frac{3}{2} \right )^{2}+\frac{3}{4}=343$$
Question 5
1 / -0

Let $$f : R \rightarrow R, g : R \rightarrow R$$ be two function such that
$$f(x) = 2x- 3, g(x) = x^3 + 5$$
The function $$(fog)^{1}(x)$$ is equal to.

A
$$\left ( \dfrac {x+7}{2} \right )^{1/3}$$

B
$$\left (x- \dfrac {7}{2} \right )^{1/3}$$

C
$$\left ( \dfrac {x-2}{7} \right )^{1/3}$$

D
$$\left ( \dfrac {x-7}{2} \right )^{1/3}$$

Solution

Given $$f(x) = 2x-3$$ and $$g(x) = x^3+ 5$$
$$(fog)(x) = fg(x) = f(x^3+5) = 2(x^3+5) -3 = 2x^3 +7$$
Let $$(fog)(x) = y = 2x^3+7$$
$$y-7 = 2x^3$$
$$x = (\dfrac{y-7}{2})^{\dfrac{1}{3}}$$
$$\therefore (fog)^{-1}(x) = (\dfrac{x-7}{2})^{\dfrac{1}{3}}$$
Question 6
1 / -0

Number of integers in the domain of the function $$f\left ( x \right )=\log _{\left ( 6-x \right )}\left ( 7x-x^{2} \right )$$ is

A
6

B
5

C
4

D
infinite

Solution

$$f(x)=log_{(6-x)}(7x-x^{2})$$

$$=\dfrac{log(7x-x^{2})}{log(6-x)}$$

Now
$$7x-x^{2}>0$$
$$x(7-x)>0$$

$$x>0$$ and $$x<7$$ ...(i)

And $$6-x>0$$
Or
$$x<6$$ ...(ii)
Hence
$$x\epsilon (0,6)$$
The integral points will be $$\{1,2,3,4,5\}$$
However
as $$x\rightarrow 5$$

$$log(6-x)\rightarrow 0$$.
Hence
$$f(x)\rightarrow \infty$$.
Thus the integral points will be $$\{1,2,3,4\}$$
Hence 4 integral points.
Question 7
1 / -0

The domain of the function f(x)=$$\displaystyle \frac{\sqrt{-\log_{0.3}(x-1)}} {\sqrt{-x^2+2x+8}}$$ is

A
$$(1, 4)$$

B
$$(2, 4)$$

C
$$[2, 4)$$

D
none of these

Solution

Given,
f(x)=$$\displaystyle \frac{\sqrt{-\log_{0.3}(x-1)}} {\sqrt{-x^2+2x+8}}$$

For given function to be defined,
$$-\log_{0.3}(x-1) \geq 0\Rightarrow \log_{10/3}(x-1) \geq 0\Rightarrow x-1\geq 1\Rightarrow x\geq 2$$

and,

$$-x^2+2x+8> 0\Rightarrow x^2-2x-8< 0\Rightarrow (x+2)(x-4) < 0\Rightarrow -2 < x< 4$$

Combining above two we get required domain $$[2,4)$$
Question 8
1 / -0

If $$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$$ then (gofogofogogog) (x) is.

A
an odd function

B
an even function

C
a polynomial function

D
an identity function

Solution

If $$f(x) = -x^2+1$$......

$$f(x) = -x^2+1, g(x) = -\sqrt[3]{x}$$........

$$\Rightarrow$$ $$f$$ is even and $$g$$ is odd

$$\therefore$$ $$gogog$$ is even

$$\Rightarrow$$ $$fogogog$$ is even

$$\Rightarrow$$ $$gofogofogogog$$ is even.
Question 9
1 / -0

The domain of the function $$f(x)=\sqrt {1-{\sqrt{1-\sqrt{1-x^2}}}}$$ is .

A
$$(\infty , 1) $$

B
$$(1, \infty) $$

C
$$[0, 1]$$

D
$$[-1, 1]$$

Solution

For given function to be defined,

$$(1-x^2) > 0\Rightarrow x^2-1\leq 0 \Rightarrow x \in [-1,1]$$
Question 10
1 / -0

If f(x)=2x-1 and g(x)=3x+2 then find (fog) (x)

A
2(3x+1)

B
2(3x+2)

C
3(2x+1)

D
3(3x+1)

Solution

$$(fog)(x) = f(g(x)) = 2(3x+2) -1 = 6x+4-1 = 6x+3 = 3(2x+1) $$

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Functions Test 21

Result

Functions Test 21

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SHARING IS CARING

Question 1

d =- a

d = a

a = b = c = d = 1

a = b = 1

Solution

Question 2

a straight line

a circle

an ellipse

a pair of straight lines

Solution

Question 3

$$x < 3$$

$$x > 3$$

$$x$$ $$\geq$$ $$-3$$

$$x$$ $$\leq$$ $$3$$

Solution

Question 4

$$333$$

$$335$$

$$338$$

$$343$$

Solution

Question 5

$$\left ( \dfrac {x+7}{2} \right )^{1/3}$$

$$\left (x- \dfrac {7}{2} \right )^{1/3}$$

$$\left ( \dfrac {x-2}{7} \right )^{1/3}$$

$$\left ( \dfrac {x-7}{2} \right )^{1/3}$$

Solution

Question 6

6

5

4

infinite

Solution

Question 7

$$(1, 4)$$

$$(2, 4)$$

$$[2, 4)$$

none of these

Solution

Question 8

an odd function

an even function

a polynomial function

an identity function

Solution

Question 9

$$(\infty , 1) $$

$$(1, \infty) $$

$$[0, 1]$$

$$[-1, 1]$$

Solution

Question 10

2(3x+1)

2(3x+2)

3(2x+1)

3(3x+1)

Solution

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