Functions Test 47

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Functions Test 47

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Weekly Quiz Competition

Question 1
1 / -0

Domain of the function $$\displaystyle \frac{1}{\sqrt{^{10}C_{x-1}-3^{10}C_{x}}}$$ is given by

A
$$9, 10, 11$$

B
$$9, 10, 12$$

C
all natural numbers

D
none

Solution

By definition of $$\displaystyle ^{n}C_{r}, r\leq n $$

$$\therefore x\leq

10$$ ...(1)

Also we know that $$\displaystyle

\frac{^{n}C_{r}}{^{n}C_{r-1}}= \frac{n-r+1}{r}$$

and $$\displaystyle

^{10}C_{x-1}-3 ^{10}C_{x}\geq 0$$ or $$\displaystyle \frac{1}{3}\geq

\frac{^{10}C_{x}}{^{10}C_{x-1}}= \frac{10-x+1}{x}= \frac{11-x}{x}$$

i.e. $$\displaystyle x\geq 33-3x$$ or $$\displaystyle x\geq

\frac{33}{4} > 8$$ ......(2)

$$\displaystyle \therefore x= 9,

10$$ by (1) and (2)
Question 2
1 / -0

If $$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$
Number of negative integral solutions of $$g(f(x)) + 2 = 0$$ are

A
$$0$$

B
$$3$$

C
$$1$$

D
$$2$$

Solution

$$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},$$
$$g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$
$$g(f(x))+2=0,\\ g(f(x))=-2,\\ g(f(x))=\begin{cases} x+1\quad \quad if\quad x\leq 0 \\ 1-x\quad if\quad 1\ge x>0 \\ x^{ 2 }-3\quad if\quad 2>x>1 \\5-x^2\quad if \quad x\geq 2\end{cases}\\ For\quad x<0\quad to\quad be\quad a\quad solution,\\ x+1=-2\longrightarrow 1\quad solution\\ $$

Hence, option C.
Question 3
1 / -0

Find all posible values of $$x$$ satisfying $$\displaystyle \dfrac{\left [ x \right ]}{\left [ x-2 \right ]}-\dfrac{\left [ x-2 \right ]}{\left [ x \right ]}=\dfrac{8\left \{ x \right \}+12}{\left [ x-2 \right ]\left [ x \right ]}$$ (where $$[\cdot]$$ denotes the greatest integer function and $$\{\cdot\}$$ is fractional part).

A
$$\displaystyle x\in \left \{ 4, \frac{11}{2} \right \}$$

B
$$\displaystyle x\in \left \{ 3, \frac{11}{2} \right \}$$

C
$$\displaystyle x\in \left \{ 2, \frac{7}{2} \right \}$$

D
$$\displaystyle x\in \left \{ 1, \frac{9}{2} \right \}$$

Solution

Here,
$$\displaystyle \frac{\left [ x \right ]}{\left [ x-2 \right ]}-\frac{\left [ x-2 \right ]}{\left [ x \right ]}=\frac{8\left \{ x \right \}+12}{\left [ x-2 \right ]\left [ x \right ]}$$

$$\Rightarrow \displaystyle \frac{\left [ x \right ]^{2}-\left [ x-2 \right ]^{2}}{\left [ x-2 \right ]\left [ x \right ]}=\frac{8\left \{ x \right \}+12}{\left [ x \right ]\left [ x-2 \right ]}$$

$$\Rightarrow $$   $$\displaystyle \frac{\left ( \left [ x \right ]-\left [ x-2 \right ] \right )\left ( \left [ x \right ]+\left [ x-2 \right ] \right )}{\left [x-2 \right ]\left [ x \right ]}=\frac{8\left \{ x \right \}+12}{\left [ x \right ]\left [ x-2 \right ]}$$

$$\Rightarrow $$   $$\left ( \left [ x \right ]-\left [ x-2 \right ] \right )\left ( \left [ x \right ]+\left [ x-2 \right ] \right )=8\left \{ x \right \}+12$$

$$\Rightarrow $$   $$\left ( \left [ x \right ]-\left [ x \right ]+2 \right )\left ( \left [ x \right ]+\left [ x \right ]-2 \right )=8\left \{ x \right \}+12$$ .......... $$\left ( \because \left [ x+I \right ]=\left [ x \right ]+I \right )$$

$$\Rightarrow 4([x]-1)=8\{x\}+12$$

$$\Rightarrow [x]-4=2\{x\}$$ ..... $$(i)$$

Now, as we know
$$0\leq \left \{ x \right \}< 1\Rightarrow 0\leq 2\left \{ x \right \}< 2$$

$$\Rightarrow 0\leq \left [ x \right ]-4< 2\Rightarrow 4\leq \left [ x \right ]< 6$$

$$\Rightarrow \left [ x \right ]=4, 5$$
If $$[x]=4\Rightarrow 2\{x\}=[x]-4$$ $$\Rightarrow \{x\}=0$$ ........ $$(ii)$$
And if $$[x]=5\Rightarrow 2\{x\}=[x]-4$$ $$\Rightarrow \displaystyle \left \{ x \right \}=\frac{1}{2}$$ ...... $$(iii)$$
From eqs. $$(ii)$$ and $$(iii)$$, we have
$$x=[x]+\{x\}$$
ie, $$x=4+0=4$$ ...... [Using Eq. $$(ii)$$]
and $$x=5+\dfrac{1}{2}=\dfrac{11}{2}$$ ........ [Using Eq. $$(iii)$$]

$$\Rightarrow \displaystyle x\in \left \{ 4, \dfrac{11}{2} \right \}$$

Hence, option A is correct.
Question 4
1 / -0

Let $$f(x)=\ln x$$ and $$g(x)\, =\, \left (\displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3 )}\right )$$. The domain of $$f(g(x))$$ is

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

B
$$[0,\, \infty)$$

C
$$(0,\, \infty)$$

D
$$[1,\, \infty)$$

Solution

$$f(x)=\ln x$$ and $$g(x)\, =\, \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3}$$

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

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

For log to be defined
$$\displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} >0$$
The above inequality holds for all $$x\in R$$
Question 5
1 / -0

The domain of the function $$ f(x)=\sqrt{\log _{\sin x+\cos x}(\left| \cos x\right|+\cos x)},0\leq x \leq \pi$$ is

A
$$(0,\pi)$$

B
$$\displaystyle \left(0,\frac{\pi}{2}\right)$$

C
$$\displaystyle \left(0,\frac{\pi}{3}\right)$$

D
None of these

Solution

$$\begin{cases} |\cos x| + \cos x > 0; x \in\left[0,\dfrac{\pi}{2}\right)\\ |\cos x| + \cos x = 0; x\in \left[\dfrac{\pi}{2}, \pi\right]\end{cases}$$

For $$\displaystyle x\in \left [ 0, \frac{\pi }{2} \right )$$,

$$\displaystyle 1 < \sin x+\cos x\leq \sqrt{2}$$

Now, for $$\displaystyle x\in \left ( 0, \frac{\pi }{2} \right )$$

$$\displaystyle \log _{\sin x+\cos x}(\left | \cos x \right |+\cos x)\geq 0$$

As for $$\log_a x$$, $$a > 1$$, so $$x$$ must be $$≥0$$

$$|\cos x| + \cos x ≥ 0$$

As we are considering $$x \in \left(0,\dfrac{\pi}2\right)$$ ... (1)

$$\Rightarrow \cos x + \cos x ≥ 1$$

$$\Rightarrow 2\cos x ≥ 1$$

$$\Rightarrow \cos x ≥\dfrac12$$

Using (1), we can say that

$$x \in \left(0,\dfrac{\pi}3\right)$$

Hence, option C.
Question 6
1 / -0

Let $$f(x) = \begin{cases} 1+x, & 0\leq x\leq 2 \\ 3-x, & 2<x\leq 3 \end{cases}$$, then find $$(fof)(x)$$

A
$$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$$

B
$$\left\{\begin{matrix}2 - x, &0 \leq x \leq 1 \\ 2 + x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$$

C
$$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 + x, &2 < x \leq 3 \end{matrix}\right.$$

D
None of these

Solution

Given $$f(x) = \begin{cases} 1+x, & 0\leq x\leq 2 \\ 3-x, & 2<x\leq 3 \end{cases}$$

Case I: When $$0\leq x\leq 2$$
$$(fof)(x)=f[f(x)]=f(1+x)$$
Since,$$0\leq x\leq 2$$
$$\Rightarrow 1\leq 1+x\leq 3$$
$$(fof)(x)=1+1+x=2+x$$ when $$1\leq 1+x\leq 2$$
$$\Rightarrow (fof)(x)=2+x$$ when $$0\leq x\leq 1$$
And $$(fof)(x)=3-(1+x)=2-x $$when $$2<1+x\leq 3$$
$$\Rightarrow (fof)(x)=2-x$$ when $$1<x \le 2$$

Case I: When $$2< x\leq 3$$
$$(fof)(x)=f[f(x)]=f(3-x)$$
Since,$$0\leq x\leq 2$$
$$\Rightarrow -2< -x \leq 0$$
$$\Rightarrow 1< 3-x \le 3$$
$$(fof)(x)=1+3-x=4-x$$ when $$1\le 3-x\leq 2$$
$$\Rightarrow (fof)(x)=4-x$$ when $$1\le x\leq 2$$
And $$(fof)(x)=3-(3-x)=x $$when $$2<3-x\leq 3$$
$$\Rightarrow (fof)(x)=x$$ when $$0<x \le 1$$
Question 7
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Given two functions $$f(x)$$ and $$g(x)$$ such that $$f(x) = \sin (arctan x), g(x) =\tan (arc\sin x)$$, and $$0\leq x < \dfrac {\pi}{2}$$. The value of the composite function $$f\left (g\left (\dfrac {\pi}{10}\right )\right ) $$ is:

A
$$0.314$$

B
$$0.354$$

C
$$0.577$$

D
$$0.707$$

E
$$0.866$$

Solution

$$g(x)=\tan { (\sin ^{ -1 }{ x } ) } =\tan { \theta } $$
$$=\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } $$
$$f(x)=\sin { (\tan ^{ -1 }{ x } ) } =\dfrac { x }{ \sqrt { 1+{ x }^{ 2 } } } $$
$$f(g(x))=\dfrac { g(x) }{ \sqrt { 1+{ g(x) }^{ 2 } } } $$
$$=\dfrac { \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } }{ \sqrt { 1+\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } } } =x$$
$$f(g(\dfrac { \pi }{ 10 } ))=0.314$$
Question 8
1 / -0

If $$f(x) = x^{2} + x$$ and $$g(x) = \sqrt {x}$$, then the value of $$f(g(3))$$ is

A
$$1.73$$

B
$$3.46$$

C
$$4.73$$

D
$$7.34$$

E
$$12.00$$

Solution

Given, $$f(x)=x^2+x, g(x)=\sqrt x$$
Then we need to find the value of $$f(g(3))$$
$$\Rightarrow g(3) = \sqrt3$$ ..... given $$g(x) = \sqrt x $$
$$\Rightarrow f(g(3)) = f(\sqrt3) = 3 + \sqrt3 $$ ..... given $$f(x) = { x }^{ 2 }+x$$
Question 9
1 / -0

If $$f(x)=2x^3$$ and $$g(x)=3x$$, calculate the value of $$g(f(-2))-f(g(2))$$.

A
$$-480$$

B
$$-384$$

C
$$0$$

D
$$384$$

E
$$480$$
Solution
- $$f(-2) = 2 \times { (-2) }^{ 3 } = -16$$
- $$g(2) = 3 \times 2 =6$$
- $$g(f(-2)) = g(-16) = 3 \times -16 = -48$$
- $$f(g(2)) = f(6) = 2 \times { (6) }^{ 3 } = 432$$
- $$g(f(-2)) - f(g(2)) = -48-432 = -480$$
Question 10
1 / -0

$$f(x)\, =\, -1\, + |x\, -\, 2|, \, 0\, \leq\, x\, \leq\, 4$$
$$g(x)\, =\, 2\, -\, |x|,\, -1\, \leq\, x\, \leq\, 3$$
Which of the following is true

A
$$fog(x)\, =\, \begin{cases} -(1\, +\, x), & & -1\, \leq\, x\, \leq\, 0\\ x + 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$$

B
$$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, -\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$$

C
$$fog(x)\, =\, \begin{cases} -(1\, +\, 2x), & & -1\, \leq\, x\, \leq\, 0\\ x - 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$$

D
$$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, +\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$$

Solution

$$f(x)\, =\, -1\, + |x\, -\, 2|, \, 0\, \leq\, x\, \leq\, 4$$
$$g(x)\, =\, 2\, -\, |x|,\, -1\, \leq\, x\, \leq\, 3$$

$$ g(f(x))=2-|-1+|x-2||,$$

$$ gof(x)=\begin{cases} 2-[2-x-1] & 0\, \leq \, x\, <\, 1 \\ 2+[2-x-1], & 1\, \leq \, x\, \leq \, 2 \\ 2+[x-2-1], & 2\, <\, x\, \leq \, 3 \\ 2-[x-2-1], & 3\, <\, x\, \leq \, 4 \end{cases}\\ gof(x)\, =\, \begin{cases} x\, +\, 1, & 0\, \leq \, x\, <\, 1 \\ 3\, -\, x, & 1\, \leq \, x\, \leq \, 2 \\ x\, -\, 1, & 2\, <\, x\, \leq \, 3 \\ 5\, -\, x, & 3\, <\, x\, \leq \, 4 \end{cases}\\ $$

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Re-Attempt Weekly Quiz Competition

Functions Test 47

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Functions Test 47

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Question 1

$$9, 10, 11$$

$$9, 10, 12$$

all natural numbers

none

Solution

Question 2

$$0$$

$$3$$

$$1$$

$$2$$

Solution

Question 3

$$\displaystyle x\in \left \{ 4, \frac{11}{2} \right \}$$

$$\displaystyle x\in \left \{ 3, \frac{11}{2} \right \}$$

$$\displaystyle x\in \left \{ 2, \frac{7}{2} \right \}$$

$$\displaystyle x\in \left \{ 1, \frac{9}{2} \right \}$$

Solution

Question 4

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

$$[0,\, \infty)$$

$$(0,\, \infty)$$

$$[1,\, \infty)$$

Solution

Question 5

$$(0,\pi)$$

$$\displaystyle \left(0,\frac{\pi}{2}\right)$$

$$\displaystyle \left(0,\frac{\pi}{3}\right)$$

None of these

Solution

Question 6

$$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$$

$$\left\{\begin{matrix}2 - x, &0 \leq x \leq 1 \\ 2 + x, &1 < x \leq 2 \\ 4 - x, &2 < x \leq 3 \end{matrix}\right.$$

$$\left\{\begin{matrix}2 + x, &0 \leq x \leq 1 \\ 2 - x, &1 < x \leq 2 \\ 4 + x, &2 < x \leq 3 \end{matrix}\right.$$

None of these

Solution

Question 7

$$0.314$$

$$0.354$$

$$0.577$$

$$0.707$$

$$0.866$$

Solution

Question 8

$$1.73$$

$$3.46$$

$$4.73$$

$$7.34$$

$$12.00$$

Solution

Question 9

$$-480$$

$$-384$$

$$0$$

$$384$$

$$480$$

Solution

Question 10

$$fog(x)\, =\, \begin{cases} -(1\, +\, x), & & -1\, \leq\, x\, \leq\, 0\\ x + 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$$

$$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, -\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$$

$$fog(x)\, =\, \begin{cases} -(1\, +\, 2x), & & -1\, \leq\, x\, \leq\, 0\\ x - 1 & & 0\, <\, x\, \leq\, 2 \end{cases}$$

$$gof(x)\, =\, \begin{cases}x\,+\,1, & 0\, \leq\, x\, <\, 1\\ 3\, -\, x, & 1\, \leq\, x\, \leq\, 2 \\ x\, +\, 1, & 2\, <\, x\, \leq\, 3\\ 5\, -\, x, & 3\, <\, x\, \leq\, 4\end{cases}$$

Solution

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