Functions Test 41

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Functions Test 41

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Weekly Quiz Competition

Question 1
1 / -0

The function $$f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0 $$ , is continuous at $$x=0$$. Then the value of $$f(0)$$ is

A
2log 3

B
$$ (\log_{e}3)^{2} $$

C
$$ \log_{e} 6 $$

D
None of these

Solution

Given f(x) is continuous at $$x=0$$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0) $$
$$ \Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$$
$$ \Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$$ $$ (\log_e3)^{2} $$
Question 2
1 / -0

The domain of the function $$f(x) = \frac {1} {\sqrt {^{10}C_{x - 1} - 3 \times ^{10} C_x}}$$ contains the points

A
9, 10, 11

B
9, 10, 12

C
all natural numbers

D
None of these

Solution

Given function is defined if $$^{10} C_{x + 1} > 3 ^{10}C_x$$

$$\Rightarrow \dfrac{1} {11 - x} > \dfrac {3} {4} \Rightarrow 4x > 33$$

$$\Rightarrow x \geqslant 9 \ but \ x \leq 10 \Rightarrow x = 9, 10$$
Question 3
1 / -0

The domain of $$ f(x) = \cos^{-1} \left ( \frac{2 - |x|} {4} \right ) + \left [ log (3 - x) \right ]^{-1}$$ is

A
$$\left [-2, 6 \right ]$$

B
$$ \left [-6, 2 \right ) \cup \left (2, 3 \right )$$

C
$$\left [-6, 2 \right ]$$

D
$$ \left [-2, 2 \right ] \cup \left (2, 3 \right )$$

Solution

$$\cos^{-1} \left (\frac {2 - |x|} {4} \right )$$ exists if $$ -1 \leq \frac {2 - |x|} {4} \leq 1$$

$$\Rightarrow -6 \leq -|x| \leq 2$$

$$\Rightarrow -2 \leq |x| \leq 6$$

$$\Rightarrow |x| \leq 6$$

The function $$\left [ \log (3 - x) \right ]^{-1} = \frac {1} {\log(3 - x)}$$ is
defined if 3 - x > 0 and $$x \neq 2,$$

i.e. if $$x \neq 2$$ and $$x < 3$$.

Thus $$\left \{ x | - 6 \leq x \leq 6 \right \} \cap \left \{ x | x \neq 2, x <3 \right \} = \left [-6, 2 \right ) \cup \left (2, 3 \right )$$
Question 4
1 / -0

The domain of the function $$f(x) = \sqrt{ log \left (\frac{1} {|\sin x| } \right)}$$ is

A
$$ R - \left \{ -\pi, \pi \right \}$$

B
$$ R - \left \{ n\pi | n \space \epsilon \space Z\right \}$$

C
$$ R - \left \{ 2n\pi | n \space \epsilon \space Z\right \}$$

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

Solution

$$f(x)$$ is defined for $$\log \left ( \frac{1} {|sin x|} \right ) \geq 0$$
$$ \Rightarrow \frac {1} {|\sin x|} \geq 1$$ and $$|\sin x| \neq 0$$ .
$$|\sin x| \neq 0$$ $$\left [ \therefore \frac{1} {|\sin x|} \geq 1 for \space all \space x \right ]$$
$$\Rightarrow x \neq n \pi, n \epsilon Z$$
Hence, the domain of $$f(x) = R - \left \{n \pi : n \space \epsilon \space Z \right \}$$.
Question 5
1 / -0

The domain of $$f(x) = \log| \log {x}|$$ is

A
$$\left(0, \infty\right )$$

B
$$\left(1, \infty\right )$$

C
$$\left(0, 1\right )\cup \left(1, \infty \right )$$

D
$$\left(-\infty, 1\right )$$

Solution

$$f(x) = \log| \log {x}|$$, f(x) is defined if $$|log \space x| > 0$$ and x > 0, i.e.,
if x > 0 and $$ x \neq 1$$ $$(\therefore |log \space x| > 0 \space if \space x \neq 1)$$
$$ \Rightarrow x \epsilon (0, 1) \cup (1, \infty)$$
Question 6
1 / -0

Directions For Questions

$$f(x) = \begin{cases} x-1, -1 \leq x \leq 0\\x^2, 0\leq x\leq 1 \end{cases}$$ and g(x) = sin x, consider the functions.
$$h_1(x) = f(|g(x)|) \space and \space h_2(x) = |f(g(x))|$$.

...view full instructions

Which of the following is not true about $$h_1(x)$$?

A
It is periodic function with period $$\pi$$

B
Range is [0,1]

C
Domain if R

D
None of these

Solution
Question 7
1 / -0

Consider the function
f(x) = $$\begin{cases} (x + 1) , x \leq 1\\ 2x + 1, 1< x \leq 2 \end{cases} and \space g(x) = \begin{cases} x^2, -1 \leq x < 2\\ x+1, 2 \leq x \leq 3 \end{cases}$$

The domain of the function f(g(x)) is

A
$$[0, \sqrt{2}]$$

B
$$[-1, 2]$$

C
$$[-1, \sqrt{2}]$$

D
None of these

Solution

$$f(x) = \begin{cases} x + 1, x \leq 1\\ 2x + 1, 1 < x \leq 2 \end{cases}$$

$$g(x) = \begin{cases} x^2, -1 \leq x < 2\\ x+2, 2 \leq x \leq 3 \end{cases}$$

$$\Rightarrow f(x) = \begin{cases} g(x) + 1, g(x) \leq 1\\ 2g(x) +1, 1 < g(x) \leq 2 \end{cases}$$

$$\Rightarrow \space f(g(x)) = \begin{cases} x^2 + 1, x^2 \leq 1, -1 \leq x < 2\\ \space x + 2 + 1, x + 2 \leq 1, 2 \leq x \leq 3\\ \space 2x^2 + 1, 1 < x^2 \leq 2, -1 \leq x < 2\\ \space2(x + 2) + 1, 1 < x + 2 \leq 2, 2 \leq x \leq 3 \end{cases}$$

$$\Rightarrow f(g(x)) = \begin{cases} x^2 + 1, -1 \leq x \leq 1 \\ 2x^2 + 1, 1< x < \sqrt{2} \end{cases}$$

Hence, the domain of f(x) is [$$-1, \sqrt{2}$$]
Question 8
1 / -0

Directions For Questions

If $$a_{o} = x, a_{n+1} = f(a_n)$$, where n = 0, 1, 2,.....then answer the following question

...view full instructions

If f: $$R\rightarrow R$$ be given by $$f(x) = 3 + 4x$$ and $$a_n = A + Bx$$, then which of the following is not true?

A
A + B + 1 = $$2^{2n + 1}$$

B
| A - B| = 1`

C
$$lim_{n \to \infty} \dfrac{A}{B} = -1$$

D
None of these

Solution
Question 9
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Consider two functions $$f(x) = \begin{cases} [x], -2 \leq x \leq -1\\ |x| + 1, -1 < x \leq 2 \end{cases}$$ and g(x) = \begin{cases} [x], -\pi \leq x < 0 \\ sin x, 0 \leq x \leq \pi \end{cases} where [.] denotes thegreatest integer function

The exhaustive domain of g(f(x)) is

A
[0,2]

B
[-2, 0]

C
[-2,2]

D
[-1,2]

Solution
Question 10
1 / -0

Let $$g(x) = f(x) - 1$$. If $$f(x) + f(1 - x) = 2 \space \forall \space x \space \epsilon \space R$$, then $$g(x)$$ is symmetrical about

A
Origin

B
The line $$x = \frac{1} {2}$$

C
The point $$\left (1, 0 \right )$$

D
The point $$\left (\frac {1} {2}, 0 \right )$$

Solution

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Re-Attempt Weekly Quiz Competition

Functions Test 41

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Functions Test 41

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SHARING IS CARING

Question 1

2log 3

$$ (\log_{e}3)^{2} $$

$$ \log_{e} 6 $$

None of these

Solution

Question 2

9, 10, 11

9, 10, 12

all natural numbers

None of these

Solution

Question 3

$$\left [-2, 6 \right ]$$

$$ \left [-6, 2 \right ) \cup \left (2, 3 \right )$$

$$\left [-6, 2 \right ]$$

$$ \left [-2, 2 \right ] \cup \left (2, 3 \right )$$

Solution

Question 4

$$ R - \left \{ -\pi, \pi \right \}$$

$$ R - \left \{ n\pi | n \space \epsilon \space Z\right \}$$

$$ R - \left \{ 2n\pi | n \space \epsilon \space Z\right \}$$

$$( - \infty, \infty )$$

Solution

Question 5

$$\left(0, \infty\right )$$

$$\left(1, \infty\right )$$

$$\left(0, 1\right )\cup \left(1, \infty \right )$$

$$\left(-\infty, 1\right )$$

Solution

Question 6

It is periodic function with period $$\pi$$

Range is [0,1]

Domain if R

None of these

Solution

Question 7

$$[0, \sqrt{2}]$$

$$[-1, 2]$$

$$[-1, \sqrt{2}]$$

None of these

Solution

Question 8

A + B + 1 = $$2^{2n + 1}$$

| A - B| = 1`

$$lim_{n \to \infty} \dfrac{A}{B} = -1$$

None of these

Solution

Question 9

[0,2]

[-2, 0]

[-2,2]

[-1,2]

Solution

Question 10

Origin

The line $$x = \frac{1} {2}$$

The point $$\left (1, 0 \right )$$

The point $$\left (\frac {1} {2}, 0 \right )$$

Solution

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