Self Studies

Functions Test 23

Result Self Studies

Functions Test 23
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    Given that, $$x$$ is a real number satisfying $$\dfrac{5x^2-26x+5}{3x^2-10x+3}<0$$, then
    Solution
    We have, 
    $$\dfrac{5x^2-26x+5}{3x^2-10x+3}< 0$$

    $$\Rightarrow \dfrac{5x^2-25x-x+5}{3x^2-9x-x+3}< 0$$

    $$\Rightarrow \dfrac{5x(x-5)-1(x-5)}{(3x(x-3)-1(x-3)}< 0$$

    $$\Rightarrow \dfrac{(5x-1)(x-5)}{((3x-1)(x-3)}< 0$$

    $$\therefore x \in \left(\dfrac{1}{5},\dfrac{1}{3}\right)\cup (3, 5)$$
  • Question 2
    1 / -0
    If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are defined by $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$ for $$x\in R$$, then
    $$g\left( f\left( x \right)  \right) :\left\{ -\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 }  \right\} $$ is equal to
    [.] is Greatest integer function
    Solution
    Given that, $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$
    For $$-\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } , 0\le f\left( x \right) < \dfrac { 8 }{ 5 } $$
    Now, for $$0 < f \left( x \right) < 1$$,
             $$g\left( f\left( x \right)  \right) = \left[ f\left( x \right) -3 \right] $$
                      $$=-3\quad \because -3\le f\left( x \right)  -3 < -2$$
    Again, for $$1 < f\left( x \right) < 1.6$$
               $$g\left( f\left( x \right)  \right) =-2$$
                                        $$\because -2\le f\left( x \right)  -3 < -1.4$$
    Hence, required set is $$\left\{ -3,-2 \right\} $$
  • Question 3
    1 / -0
    Let $$R$$ be the set of real numbers and the functions $$f: R \rightarrow R$$ and $$g: R\rightarrow R$$ be defined by $$f(x) = x^{2} + 2x - 3$$ and $$g(x) = x + 1$$. Then the value of $$x$$ for which $$f(g(x)) = g(f(x))$$ is
    Solution
    According to the question,

    $$f(g(x)) = g(f(x))$$
    $$\Rightarrow f(x + 1) = g(x^{2} + 2x - 3)$$

    $$\Rightarrow (x + 1)^{2} + 2 (x + 1) - 3 = x^{2} + 2x - 3 + 1$$

    $$\Rightarrow x^{2} + 1 + 2x + 2x + 2 - 3 = x^{2} + 2x - 2$$

    $$\Rightarrow x^{2} + 4x = x^{2} + 2x - 2$$
    $$\Rightarrow x^{2} + 4x - x^{2} - 2x + 2 = 0$$
    $$\Rightarrow 2x + 2 = 0$$

    $$\Rightarrow 2x = -2$$

    $$\Rightarrow x = -1$$
  • Question 4
    1 / -0
    Let $$f:R\rightarrow R$$ be such that $$f$$ is injective and $$f(x)f(y)=f(x+y)$$ for all $$x,y\in R$$, if $$f(x), f(y)$$ and $$f(z)$$ are in GP, then $$x,y$$ and $$z$$ are in
    Solution
    Let the funtion $$f(x)={a}^{kx}$$
    which define in $$f:R\rightarrow R$$ and injective also.

    Now, we have
    $$f(x)f(y)=f(x+y)$$

    $$\Rightarrow$$ $${a}^{kx}.{a}^{ky}={a}^{k(x+y)}$$

    $$\Rightarrow$$ $${a}^{k(x+y)}={a}^{k(x+y)}$$

    $$\because$$ $$f(x), f(y)$$ and $$f(z)$$ are in GP

    $$\therefore$$ $$f({y}^{2})=f(x).f(z)$$

    $$\Rightarrow$$ $${a}^{2ky}={a}^{kx}.{a}^{kz}$$

    $$\Rightarrow$$ $${e}^{2ky}={e}^{k(x+z)}$$

    On comparing, we get
    $$2ky=k(x+z)$$ $$\Rightarrow$$ $$2y=x+z$$

    $$\Rightarrow$$ $$x,y$$ and $$z$$ are in AP
  • Question 5
    1 / -0
    Let Q be the set of all rational numbers in [0, 1] and $$f : [0, 1]\rightarrow [0, 1]$$ be defined by $$f(x)=\begin{cases}x&for&x\in Q\\ 1-x&for&x\notin Q\end{cases}$$
    Then the set $$S=\{x\in [0, 1]: (f\, o \, f)(x)=x\}$$ is equal to
    Solution
    Let $$x\in Q$$, then,
    $$f(x)=x$$ where $$x\in Q$$
    So, $$fof(x)=f(f(x))=f(x)=x$$   as  $$x\in Q$$
    $$\therefore fof(x)=x$$   when $$x \in Q$$
    Now, 
    Let $$x\notin Q$$ then
    $$f(x) =1-x$$
    $$\therefore fof(x)=1-(1-x) = x$$
    as $$1-x\notin Q$$ as $$x\notin Q$$
    where $$x\notin Q$$
    $$fof(x)=\begin{cases}x \,where\, x\in Q \,\&\, x\in [0, 1] \\ x\, where x\notin Q\, \&\, x\in [0, 1] \end{cases}$$
    $$\therefore$$ the set $$S = [0, 1]$$
  • Question 6
    1 / -0
    If $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$, then the set of real numbers $$x$$ such that $$f\left\{ g(x) \right\} =x$$ is
    Solution
    Given $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$

    Now $$fo{g(x)}=x$$

    $$\Rightarrow$$ $$f({3}^{100}.x+1)=x\\$$
    $$\Rightarrow$$ $${2}^{100}({3}^{100}.x+1)+1=x\\$$
    $$\Rightarrow$$ $${6}^{100}.x+{2}^{100}+1=x\\$$

    $$\Rightarrow$$ $$x(1-{6}^{100})=(1+{2}^{100})$$

    $$\Rightarrow$$ $$x=\cfrac{1+{2}^{100}}{1-{6}^{100}}$$

    Hence $$fog(x)=x$$ represent a singleton set.
  • Question 7
    1 / -0
    If $$f(x) =$$ $$\sqrt{x}$$ and $$g(x) =$$ $$\sqrt{x^2+4}$$, calculate the value of $$f(g(2))$$.
    Solution
    Given functions are $$f(x)=\sqrt {x}$$ and $$g(x)=\sqrt {x^2+4}$$.
    Now $$g(2)=\sqrt{2^2+4}=2\sqrt2$$
    Then $$f(g(2))=\sqrt { g(2) } = \sqrt { 2\sqrt { 2 }  } ={ 2 }^{ 3/4 }=1.681$$
  • Question 8
    1 / -0
    If $$f: R\rightarrow R^{+}$$ and $$g: R^{+} \rightarrow R$$ are such that $$g(f(x)) = |\sin x|$$ and $$f(g(x)) = (\sin \sqrt {x})^{2}$$, then a possible choice for f and g is
    Solution
    As f returns only positive values and takes any real number it can be mod or square hence option 2 is eliminated 
    For g only positive values has to be given and it will return any real number
    so on trying the option only option $$A$$ and $$C$$ results in the given equation ie $$g(f(x)) = |sin x|$$

    Now function g should take positive values and return real values which is satisfied by only option $$A$$ as in option $$C$$ square root of positive number will always result in positive number while $$sin\sqrt{x}$$ will give -ve real numbers as well.
  • Question 9
    1 / -0
    The domain of the function $$f(x) = \log (1 - x) + \sqrt {x^{2} - 1}$$
    Solution
    $$\log(1 - x)$$ is defined if $$1 - x > 0\Rightarrow x < 1 \Rightarrow x\epsilon (-\infty, 1) ..... (1)$$
    $$\sqrt {x^{2} - 1}$$ is defined if $$x^{2} - 1 \geq 0\Rightarrow x^{2} \geq 1$$
    $$\Rightarrow x \geq 1$$ or $$ x \leq -1 ..... (2)$$
    $$\therefore$$ Required domain is the instruction of $$(1)$$ and $$(2)$$
    i.e. $$(-\infty, -1]$$
  • Question 10
    1 / -0
    If $$h(x)={x}^{3}+x$$ and $$g(x)=2x+3$$, then calculate $$g(h(2))$$.
    Solution
    Given, $$h(x)=x^3+x$$
    $$\therefore h(2)=2^3+2$$
    $$\Rightarrow h(2)=8+2=10$$

    Also given, $$g(x)=2x+3$$
    $$\Rightarrow g(h(2))=g(10)$$
    $$\because h(2)=10$$
    Then the value of $$g(10)=2\times 10+3$$
    $$\Rightarrow g(10)=20+3=23$$


Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now