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Functions Test 12

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Functions Test 12
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  • Question 1
    1 / -0
    The domain of the function $$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$$ is 
    Solution
    Given,
    $$\displaystyle f(x)=\log_{10}\log_{10}(1+x^{3})$$

    The value inside a log function must not be negative.
    Hence
    $$\log_{10}(1+x^3)>0$$
    $$1+x^{3}>10^{0}$$ ....(by taking anti-log).
    $$x^{3}>0$$
    $$x>0$$
    Hence the domain for $$f(x)$$ is $$x>0$$.
  • Question 2
    1 / -0
    Which of the following function is one-one?
    Solution
    Option A - Consider $$x = 2 $$ and $$ x =0 $$ , the value of $$f(x)$$ is same. Hence it is not one-one
    Option B -  If we replace $$x$$ by $$-x$$, then the value of $$g(x)$$ remains the same. Hence it is not one-one
    Option C - $$h(x)$$ is an increasing function for the given values of $$x$$. Hence it is one-one function
    Option D -$$f(x)$$ is an even function. So it is not one-one for the given values of $$x$$

  • Question 3
    1 / -0
    The domain of function satisfying $$f(x)+f(x^{-1})=\displaystyle \frac{x^{3}+1}{x}$$, is
    Solution
     $$y = f(x)+f(x^{-1})=\displaystyle \frac{x^{3}+1}{x} ..(1)$$
    Replacing $$x\to x^{-1}$$
    $$\Rightarrow y =  f(x)+f(x^{-1})=\displaystyle \frac{x^{3}+1}{x^2} ..(2)$$
    By $$(2)-(1)$$, we get
    $$\Rightarrow \displaystyle \frac{x^{3}+1}{x^2}-\displaystyle \frac{x^{3}+1}{x}=0$$
    $$\Rightarrow \displaystyle \displaystyle \frac{x^{3}+1}{x^2}(1-x)=0\Rightarrow x = \pm 1$$ which is finite set.
  • Question 4
    1 / -0
    The domain of $$f(x)=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x-2+2\sqrt{x-3}}$$, is
    Solution
    Given,
    $$f(x)=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x-2+2\sqrt{x-3}}$$

    Here $$x-3 \ge 0$$ 

    $$\Rightarrow x\ge 3$$

    Also, $$(x-2-2\sqrt{x-3}) \ge 0$$ 
    $$\Rightarrow x-2\ge 2\sqrt { x-3 } $$
    $$\Rightarrow (x-2)^{ 2 }\ge 4(x-3)$$
    $$\Rightarrow x^{ 2 }-8x+16\ge 0$$ is always true
    $$\Rightarrow x\ge3$$

    Also, $$(x-2+2\sqrt{x-3}) \ge 0$$
    $$\Rightarrow x-2\ge -2\sqrt { x-3 } $$
    This is also always true as right hand side is always negative and left hand side is always positive for $$x\geqslant3$$
    $$\Rightarrow x\ge 3 $$

    Hence, their union; $$x\ge 3$$ is the domain of $$f(x)$$
  • Question 5
    1 / -0
    The domain of $$\displaystyle f(x)=\sqrt { \log_{ x^{ 2 }-1 }(x) } $$ is
    Solution
    $$\log_{x^{2}-1}(x)\geq 0$$

    Hence,
    $$x\geq 1 \quad \dots (1)$$

    Now, $$x^{2}-1>1$$

    $$x^{2}>2$$

    $$x<-\sqrt{2}$$ and $$x>\sqrt{2} \quad \dots (2)$$

    Hence from $$(1)$$ and $$(2)$$,

    $$x\in(\sqrt{2},\infty)$$
  • Question 6
    1 / -0
    Let $$\displaystyle f:R\rightarrow A=\left \{ y: 0\leq y< \dfrac{\pi}{2} \right \}$$ be a function such that $$\displaystyle f(x)=\tan^{-1}(x^{2}+x+k),$$ where $$k$$ is a constant. The value of $$k$$ for which $$f$$ is an onto function is 
    Solution
    For $$f$$ to be an onto function, Range and co-domain of $$f$$ should be equal

    $$\Rightarrow f(x)\ge 0 \forall x\in R$$

    $$\Rightarrow \tan^{-1}(x^2+x+k)\ge 0$$

    For the above equation to be valid for all $$x$$

    We must have, the discriminant of $$x^2+x+k=0$$, is zero

    $$ b^2-4ac=0$$

    $$\Rightarrow 1-4k=0\Rightarrow k =\dfrac{1}{4}$$
  • Question 7
    1 / -0
    The domain of the function $$\displaystyle f(x)=\sqrt{\sec^{-1}\left \{ \frac{1-|x|}{2} \right \}}$$ is
    Solution
    $${\sec^{-1}\left \{ \frac{1-|x|}{2} \right \}}>0$$

    $$\dfrac{1-|x|}{2}\leq -1$$ and $$\dfrac{1-|x|}{2}\geq 1$$

    $$3\leq |x|$$ and $$-1\geq |x|$$

    Hence $$3\leq |x|$$

    $$x\in(-\infty,-3]\cup[3,\infty)$$
  • Question 8
    1 / -0
    The domain of $$\displaystyle f(x)=\sin^{-1} \left ( \frac{1+x^{2}}{2x} \right )+\sqrt{1-x^{2}}$$ is
    Solution
    Let $$g\left( x

    \right)={\sin ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{2x}}} \right)$$

    and $$h\left( x \right)=\sqrt {1 - {x^2}}$$.  

    Hence $$f\left( x \right) = g\left( x \right) + h\left( x \right)$$ and the domain set of $$f\left( x \right)$$ is the intersection of the domain sets of 

    $$g\left( x \right)$$ and $$h\left(x \right)$$.

    Now, the domain of $$h\left( x \right)$$ is $$\left[ { - 1,1} \right]$$.

    Since, $$1 + {x^2} \ge 2x$$ for all real values of $$x$$, 

    we have  $$\dfrac{{1 + {x^2}}}{{2x}} \ge 1$$ for $$x > 0$$ and $$\dfrac{{1 + {x^2}}}{{2x}} \le -1$$ for $$x < 0$$, 

    Also the domain of $${sin ^{ - 1}}\left( x \right)$$ is $$\left[ { - 1,1} \right]$$, only possible values for which $${\sin ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{2x}}} \right)$$ is defined is $$\left\{ { - 1,1} \right\}$$. 

    That is the domain of  $$g\left( x \right)$$ is $$\left\{ { - 1,1} \right\}$$.

    Hence, the domain set of $$f\left( x \right)$$ is $$\left\{ { - 1,1} \right\}$$.
  • Question 9
    1 / -0
    Let $$\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \}$$ be a one-one function and only one of the conditions $$(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a$$ is true then the function $$f$$  is given by the set 
    Solution
    $$f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\} $$  is  a  one-one  function 

    $$ \Rightarrow$$   each  element  in $$ \left\{ x,y,z \right\} $$  will  have  exactly  one  image  in  $$\left\{ a,b,c \right\}$$

    and  no  two  elements  of $$ \left\{ x,y,z \right\}$$   will  have  same  image  in  $$\left\{ a,b,c \right\} $$

     Coming to the given  3  conditions, only one is true.

     1)  if  $$f\left( x \right) \neq b$$  is  true  then  $$f\left( y \right) =b$$  is  false  which  makes  $$f\left( z \right) \neq a$$  true  $$\Longrightarrow   f\left( x \right) \neq b$$  is  false.

     2)  if  $$f\left( y \right) =b$$  is  true  then  $$f\left( x \right) \neq b$$  will  also  be  true  $$\Longrightarrow   f\left( y \right) =b$$  is  false

    $$ \therefore   f\left( z \right) \neq a$$  is  the  true  condition  and  remainig two  are  false  conditions.

    $$\therefore   f\left( x \right) =b,  f\left( y \right) =a,  f\left( z \right) =c$$

     hence  $$f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right)  \right\} $$
  • Question 10
    1 / -0
    If the real-valued function $$\displaystyle f(x)=px+\sin\:x$$ is a bijective function then the set of possible values of $$p\:\in\:R$$ is 
    Solution
    $$\displaystyle f(x)=px+\sin\:x$$
    $$\displaystyle f'(x) = p +\cos x$$
    For given function to be bijective,
    $$f'(x)> 0 \forall  x \in R \Rightarrow |p| > 1$$
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