Functions Test 32

Result

Functions Test 32

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Question 1
1 / -0

The domain of the function $$\phi \left( x \right) = {\log _7}\left( { - {{\log }_{\frac{1}{2}}}\left( {1 + \frac{1}{{\sqrt[4]{x}}}} \right) - 1} \right)$$ is

A
$$[x:1<x<2]$$

B
$$[x:0<x<1]$$

C
$$[x:0<x<2]$$

D
$$[x:1<x<3]$$

Solution

$$\phi(x)=log_7\left(-log_{1/2}\left(1+\cfrac{1}{\sqrt[4]{x}}\right)-1\right)$$ is
$$\Rightarrow log_{1/2}\left(1+\cfrac{1}{\sqrt[4]{x}}\right)<-1$$
$$\Rightarrow 1-\cfrac{1}{\sqrt[4]{x}}>(2^{-1})^{-1}=2$$
$$\Rightarrow \cfrac{1}{4\sqrt{x}}>1\quad; 0<x<1$$
Hence Domain is $$[x:0<x<1]$$
Question 2
1 / -0

If f is even function and g is an odd function, then $$f_og$$ is ............function.

A
Even

B
Odd

C
Neither even nor odd

D
Either even

Solution

$$fog$$ function is an even function

Let $$f\left(x \right)$$ is a even function and $$g \left( - x \right)$$ is odd function.
So, $$f\left( {g\left( { - x} \right)} \right) = f\left( { - g\left( x \right)} \right) = even$$
Question 3
1 / -0

If $$f\left( x \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$$ then $$f\left( f\left( x \right) \right) $$ is given by

A
$$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 4-x,\quad x<0 \end{cases}$$

B
$$f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}$$

C
$$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x< 0 \\ x,\quad x\ge 0 \end{cases}$$

D
$$f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x\ge 0 \\ x,\quad x<0 \end{cases}$$

Solution
Question 4
1 / -0

The domain of the function $$y(x)$$ given by $${2^x} + {2^r} = 2$$ for all $$r \in ( - \infty ,1)$$ is:

A
$$(0,-1]$$

B
$$[0,1]$$

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

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

Solution

For domain of $$x$$ let's assume $$r$$ as a function of $$x:$$

Now,
equation
$$2^x+2^r=2$$

$$=>r=log_2(2-2^x)$$

So,

$$=>2-2^x>0$$

$$=>2^x<2$$

$$=>x<1$$

Therefore,solution is set form we have
$$=>x\epsilon(-\infty,1)$$

Foe domain of $$r$$ let's assume $$x$$ as a function of $$r$$

$$=>x=log_2(2-2^r)$$--------------------$$[2^x=log_2 x]$$

So,

$$=>2-2^y>0$$

$$=>2^y<2$$

$$=>y<1$$

Therefore solution is set form we have $$r\epsilon(-\infty,1)$$

Hence,

$$x\epsilon(-\infty,1)$$ for $$r\epsilon(-\infty,1)$$
Question 5
1 / -0

Let $$f:X \to \left[ {1,\,27} \right]$$ be a function by $$f\left( x \right) = 5\sin x + 12\cos x + 14$$. The set $$X$$ so that $$f$$ is one-one and onto is

A
$$\left[ { - \pi /2,\pi /2\,} \right]$$

B
$$\left[ {0,\,\pi } \right]$$

C
$$\left[ {0,\,\pi /2} \right]$$

D
non of these

Solution
Question 6
1 / -0

If : $$f(x) = 5 {x}^{2}$$, $$g(x) = 3x^{4}$$, then : $$(fog) (-1) =$$

A
$$45$$

B
$$-54$$

C
$$-32$$

D
$$-64$$

Solution
Question 7
1 / -0

Let $$g\left( x \right) =1+x-\left[ x \right] $$ and $$f\left( x \right) =\begin{cases} -1,x<0 \\ 0,x=0 \\ 1,x>0 \end{cases}$$ Then for all $$x,f\left( g\left( x \right) \right) $$ is equal to (where $$\left[ . \right] $$ represents the greatest integer function)

A
$$x$$

B
$$1$$

C
$$f\left( x \right)$$

D
$$g\left( x \right)$$

Solution
Question 8
1 / -0

The distinct linear functions which maps from $$[-1,1]$$ onto $$[0,2]$$ are

A
$$x+1$$, $$-x+1$$

B
$$x-1$$, $$x+1$$

C
$$-x+1$$

D
$$-x+2$$

Solution

$$[-1,1]\rightarrow [0,2]$$ onto are
$$x1!=f(x)$$
$$f(-1)=0$$
$$f(1)=2$$
$$-1+2=f(x)$$
$$f(-1)=2$$
$$f(1)=0$$
Question 9
1 / -0

For $$a,\ b\ \in \ R-\left\{ 0 \right\}$$, let $$f(x)=ax^{2}+bx+a$$ satisfies $$f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \in\ R$$.
Also the equation $$f(x)=7x+a$$ has only one real distinct solution. The minimum value of $$f(x)$$ in $$\left[0,\dfrac{3}{2}\right]$$ is equal to

A
$$\dfrac{-33}{8}$$

B
$$0$$

C
$$4$$

D
$$-2$$

Solution
Question 10
1 / -0

If function $$f\left( x \right) = \frac{1}{2} - \tan \left( {\frac{{\pi x}}{2}} \right);\left( { - 1 < x < 1} \right)$$ and $$g\left( x \right) = \sqrt {3 + 4x - 4{x^2}} $$, then the domain of $$gof$$ is

A
$$(-1,1)$$

B
$$\left[ { - \frac{1}{2},\frac{1}{2}} \right]$$

C
$$\left[ { - 1,\frac{1}{2}} \right]$$

D
$$\left[ { - \frac{1}{2}, - 1} \right]$$

Solution