Function MCQ Questions & Answers in Calculus | Maths

If $$f\left( x \right) = f\left( { - x} \right),$$   i.e., $$f\left( {0 + x} \right) = f\left( {0 - x} \right),$$     the graph is symmetrical about $$x=0.$$  Therefore, the graph is symmetrical about $$x=2$$  if $$f\left( {2 + x} \right) = f\left( {2 - x} \right).$$

$$\eqalign{ & {\text{We know that the range of }}{\sin ^{ - 1}}{\text{ is }}\left[ {\frac{{ - \pi }}{2},\,\frac{\pi }{2}} \right] \cr & {\text{and domain is }}\left[ { - 1,\,1} \right] \cr & {\text{Now in }}f\left( x \right) = \sin \left( {x + \left[ x \right]} \right) \cr & {\text{The input should be in the domain }}\left[ { - 1,\,1} \right] \cr & x + \left[ x \right] \geqslant - 1 \cr & \Rightarrow x \geqslant 0 \cr & {\text{and }}x + \left[ x \right] \leqslant 1 \cr & \Rightarrow x \leqslant 1 \cr & {\text{However, }}\left[ x \right]{\text{ is discontinuous at }}x = 1 \cr & {\text{Hence domain of }}f\left( x \right) = {\sin ^{ - 1}}\left( {x + \left[ x \right]} \right){\text{ is }}\left[ {0,\,1} \right) \cr} $$

We have $$f:N \to I$$
If $$x$$ and $$y$$ are two even natural numbers, then $$f\left( x \right) = f\left( y \right) \Rightarrow \frac{{ - x}}{2} = \frac{{ - y}}{2} \Rightarrow x = y$$
Again if $$x$$ and $$y$$ are two odd natural numbers then $$f\left( x \right) = f\left( y \right) \Rightarrow \frac{{x - 1}}{2} = \frac{{y - 1}}{2} \Rightarrow x = y$$
$$\therefore f$$  is onto.
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
$$\therefore f$$  is onto.
Hence $$f$$ is one one and onto both.

$$\frac{{f\left( { - a} \right)}}{{f\left( b \right)}} = \frac{{{e^a}}}{{{e^{ - b}}}} = {e^a}{e^b} = {e^{a + b}} = {e^{ - \left( { - a - b} \right)}} = f\left( { - a - b} \right)$$

$$\eqalign{ & f\left( x \right) \ne b \Rightarrow \cr & f\left( x \right) = a,\,f\left( y \right) = c,\,f\left( z \right) = b\,\,\left( {{\text{uses (iii) also}}} \right) \cr & f\left( x \right) = a,\,f\left( y \right) = b,\,f\left( z \right) = c\,\,\left( {{\text{uses (ii) also}}} \right) \cr & f\left( x \right) = c,\,f\left( y \right) = a,\,f\left( z \right) = b\,\,\left( {{\text{uses (iii) also}}} \right) \cr & f\left( x \right) = c,\,f\left( y \right) = b,\,f\left( z \right) = a\,\,\left( {{\text{uses (ii) also}}} \right) \cr & f\left( y \right) = b \Rightarrow \cr & f\left( x \right) = a,\,f\left( y \right) = b,\,f\left( z \right) = c\,\,\left( {{\text{uses (i) also}}} \right) \cr & f\left( x \right) = c,\,f\left( y \right) = b,\,f\left( z \right) = a\,\,\left( {{\text{uses (i) also}}} \right) \cr & f\left( z \right) \ne a \Rightarrow \cr & f\left( x \right) = a,\,f\left( y \right) = b,\,f\left( z \right) = c\,\,\left( {{\text{uses (i) also}}} \right) \cr & f\left( x \right) = a,\,f\left( y \right) = c,\,f\left( z \right) = b\,\,\left( {{\text{uses (i) also}}} \right) \cr & f\left( x \right) = b,\,f\left( y \right) = a,\,f\left( z \right) = c\,\,\left( {{\text{possible}}} \right) \cr & f\left( x \right) = c,\,f\left( y \right) = a,\,f\left( z \right) = b\,\,\left( {{\text{uses (i) also}}} \right) \cr} $$

On simplification, $$f\left( x \right) = \frac{5}{4} = $$   constant. So, $$g\left\{ {f\left( x \right)} \right\} = g\left( {\frac{5}{4}} \right) = 1$$     (from the question).

$$\eqalign{ & f\left( x \right){\text{ is defined if }} - {\log _{\frac{1}{2}}}\left( {1 + \frac{1}{{{x^{\frac{1}{4}}}}}} \right) - 1 > 0 \cr & \Rightarrow {\log _{\frac{1}{2}}}\left( {1 + \frac{1}{{{x^{\frac{1}{4}}}}}} \right) < - 1 \cr & \Rightarrow 1 + \frac{1}{{{x^{\frac{1}{4}}}}} > {\left( {\frac{1}{2}} \right)^{ - 1}} \cr & \Rightarrow \frac{1}{{{x^{\frac{1}{4}}}}} > 1 \cr & \Rightarrow 0 < x < 1 \cr} $$

$$f\left( x \right) = x\left\{ {{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{{11}}{4}} \right\} + \sin \,x.$$
Clearly $$x\left\{ {{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{{11}}{4}} \right\}$$     increases with $$x$$ and its value changes over $$R.$$ Also, $$ - 1 \leqslant \sin \,x \leqslant 1$$
So, the range of $$f=R$$
Hence, $$f$$ is onto and one-one.

$$\eqalign{ & {\text{Here, }}x - \sqrt {1 - {x^2}} \geqslant 0\,\,\,\,\therefore x \geqslant \sqrt {1 - {x^2}} > 0 \cr & \therefore {x^2} \geqslant 1 - {x^2}{\text{ or }}{x^2} \geqslant \frac{1}{2} \cr & {\text{Also }}1 - {x^2} \geqslant 0{\text{ or }}{x^2} \leqslant 1 \cr & {x^2} \geqslant \frac{1}{2} \Rightarrow x < - \frac{1}{{\sqrt 2 }}{\text{ or }}x \geqslant \frac{1}{{\sqrt 2 }} \cr & {x^2} \leqslant 1 \Rightarrow - 1 \leqslant x \leqslant 1 \cr & {\text{Thus }}x > 0,\,{x^2} \geqslant \frac{1}{2}{\text{ and }}\,{x^2} \leqslant 1 \Rightarrow x\, \in \left[ {\frac{1}{{\sqrt 2 }},\,1} \right] \cr} $$

$$\eqalign{ & f\left( x \right) = \cos \left( {\log x} \right) \cr & \therefore f\left( x \right)f\left( y \right) - \frac{1}{2}\left[ {f\left( {\frac{x}{y}} \right) + f\left( {xy} \right)} \right] \cr & = \cos \left( {\log x} \right)\cos \left( {\log y} \right) - \frac{1}{2}\left[ {\cos \left( {\log x - \log y} \right) + \cos \left( {\log x + \log y} \right)} \right] \cr & = \cos \left( {\log x} \right)\cos \left( {\log y} \right) - \frac{1}{2}\left[ {2\cos \left( {\log x} \right)\cos \left( {\log y} \right)} \right] \cr & = 0 \cr} $$