Function MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & g\left( x \right) = f\left( {x + 1} \right) = \left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \cr & {\text{If }}x < 1,\,\,\,g\left( x \right) = - x + 1 - x + 2 - x + 3 = - 3x + 6 \cr & {\text{If 1}} \leqslant x < 2,\,\,\,g\left( x \right) = x - 1 - x + 2 - x + 3 = - x + 4 \cr & {\text{If 2}} \leqslant x < 3,\,\,\,g\left( x \right) = x - 1 + x - 2 - x + 3 = x \cr & {\text{If }}x \geqslant 3,\,\,\,g\left( x \right) = x - 1 + x - 2 + x - 3 = 3x - 6 \cr} $$

The domain of $$\sqrt {\left( {x + 2} \right)\left( {5 - x} \right)} $$    is the set of values of $$x$$ for which $$\left( {x + 2} \right)\left( {5 - x} \right) \geqslant 0,\,\,{\text{i}}{\text{.e}}{\text{., }}x\, \in \left[ { - 2,\,5} \right].$$
The domain of $$\frac{1}{{\sqrt {{x^2} - 4} }}$$   is the set of values of $$x$$ for which $${x^2} - 4 > 0,\,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,x\, \in \left( { - \infty ,\, - 2} \right) \cup \left( {2,\, + \infty } \right)$$
The common values are in $$\left( {2,\,5} \right]$$

$$\eqalign{ & {\text{Given }}f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x{\text{ for }}x \in \left( { - 1,1} \right) \cr & {\text{If }}x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right) \cr & \Rightarrow 2{\tan ^{ - 1}}x \in \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right) \cr} $$
Clearly, range of $$f\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
For $$f$$ to be onto, co-domain = range
Co-domain of function $$ = B = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

$$\eqalign{ & {\text{We have }} - 1 \leqslant x \leqslant 0 \Rightarrow 0 \leqslant {x^2} \leqslant 1......({\text{i}}) \cr & {\text{and }}0 \leqslant x \leqslant 1 \Rightarrow 0 \leqslant {x^2} \leqslant 1......({\text{ii}}) \cr & \therefore \,E = \left\{ {x\, \in \,{\bf{R}}: - 1 \leqslant x \leqslant 0} \right\} \cr & \Rightarrow f\left( E \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\}{\text{ from (i)}} \cr & {\text{Also, }}F = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\} \cr & \Rightarrow f\left( F \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\}{\text{ from (ii)}} \cr & {\text{Hence, }}f\left( E \right) = f\left( F \right) \cr & {\text{Again }}E \cap F = \left\{ 0 \right\} \subset f\left( E \right) \cap f\left( F \right) \cr & \left[ {{\text{Since }}f\left( E \right) = f\left( F \right)\,\therefore \,f\left( E \right) \cap f\left( F \right) = f\left( E \right) = f\left( F \right)} \right]{\text{ }} \cr & {\text{Also }}E \cap F = \left\{ 0 \right\} \Rightarrow f\left( {E \cap F} \right) = \left\{ 0 \right\} \cr & {\text{Next }}E \cup F = \left\{ {x\, \in \,{\bf{R}}: - 1 \leqslant x \leqslant 1} \right\} \cr & {\text{and }}f\left( E \right) \cup f\left( F \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\} \cr & \therefore \,E \cup F \subset f\left( E \right) \cup f\left( F \right) \cr} $$

$$\eqalign{ & {\text{Given that }}f\left( x \right) = {x^2}\,{\text{and }}g\left( x \right) = \sin x,\forall x \in R \cr & {\text{Then }}\left( {gof} \right)\left( x \right) = \sin {x^2} \Rightarrow \left( {gogof} \right)\left( x \right) = \sin \left( {\sin {x^2}} \right) \cr & \Rightarrow \left( {fogogof} \right)\left( x \right) = {\sin ^2}\left( {\sin {x^2}} \right) \cr & {\text{As given that }}\left( {fogogof} \right)\left( x \right) = \left( {gogof} \right)\left( x \right) \Rightarrow {\sin ^2}\left( {\sin {x^2}} \right) = \sin \left( {\sin {x^2}} \right) \cr & \Rightarrow \sin \left( {\sin {x^2}} \right) = 0,1 \cr & \Rightarrow \sin {x^2} = n\pi \,{\text{or }}\left( {\left( {4n + 1} \right)\frac{\pi }{2}} \right.\,{\text{where }}n \in Z \cr & \Rightarrow \sin {x^2} = 0\,\,\because \sin {x^2} \in \left[ { - 1,1} \right] \Rightarrow {x^2} = n\pi \cr & \Rightarrow x = \pm \sqrt {n\pi } {\text{ where }}n \in W \cr} $$

$$\eqalign{ & {x^2} - x - 2 \ne 0{\text{ and }}\left[ {x + \frac{1}{2}} \right] > 0,\,\left[ {x + \frac{1}{2}} \right] \ne 1 \cr & {x^2} - x - 2 \ne 0\,\, \Rightarrow x \ne - 1,\,2{\text{ and }}\left[ {x + \frac{1}{2}} \right] > 0\,\,\,\, \Rightarrow x \geqslant \frac{1}{2} \cr & \left[ {x + \frac{1}{2}} \right] \ne 1\,\, \Rightarrow x\, \notin \left[ {\frac{1}{2},\,\frac{3}{2}} \right) \cr & {\text{Thus }}x\, \in \left[ {\frac{3}{2},\,2} \right) \cup \left( {2,\, + \infty } \right) \cr} $$

$$\left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| \leqslant 1.$$   This is true for all $$x\, \in \,R.$$   So, the domain $$=R$$
$$\eqalign{ & {\text{Now }}\frac{{{x^2}}}{{1 + {x^2}}} = 1 - \frac{1}{{1 + {x^2}}} \cr & \therefore \,\,0 \leqslant \frac{{{x^2}}}{{1 + {x^2}}} < 1 \cr & \therefore \,\,{\cos ^{ - 1}}0 \geqslant {\cos ^{ - 1}}\frac{{{x^2}}}{{1 + {x^2}}} > {\cos ^{ - 1}}1 \cr & \Rightarrow \frac{\pi }{2} \geqslant {\cos ^{ - 1}}\frac{{{x^2}}}{{1 + {x^2}}} > 0 \cr & \therefore {\text{The range }} = \left( {0,\,\frac{\pi }{2}} \right] \cr} $$

$$\eqalign{ & {\text{Putting }}x = 1,\,y = 1\,\,{\text{we get,}} \cr & f\left( 2 \right) = {\left\{ {f\left( 1 \right)} \right\}^2} = {3^2} \cr & {\text{Putting }}x = 2,\,y = 1\,\,{\text{we get}} \cr & f\left( 3 \right) = f\left( 2 \right) \cdot f\left( 1 \right) = {\left\{ {f\left( 1 \right)} \right\}^3} = {3^3},{\text{ etc}}{\text{.}} \cr & \therefore \,\sum\limits_{r = 1}^n {f\left( r \right)} = \sum\limits_{r = 1}^n {{3^r}} = \frac{{3\left( {{3^n} - 1} \right)}}{{3 - 1}} \cr} $$

Given that f $$f\left( x \right) = {\left( {x + 1} \right)^2},x \geqslant - 1$$
Now if $$g\left( x \right)$$  is the reflection of $$f\left( x \right)$$  in the line $$y = x$$   then it can be obtained by interchanging $$x$$ and $$y$$ in $$f\left( x \right)$$
i.e., $$y = {\left( {x + 1} \right)^2}$$   changes to $$x = {\left( {y + 1} \right)^2}$$
$$\eqalign{ & \Rightarrow y + 1 = \sqrt x \cr & \left[ {y + 1 \ne - \sqrt x ,\,{\text{since}}\,y \geqslant - 1\,{\text{as}}\,{\text{in}}\,{\text{figure}}.} \right. \cr & \Rightarrow y = \sqrt x - 1\,{\text{defined}}\,\forall x \geqslant 0 \cr} $$

$$\therefore g\left( x \right) = \sqrt x - 1\,\forall x \geqslant 0$$

$$f\left( x \right) = \frac{3}{{4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$$

$$\eqalign{ & 4 - {x^2} \ne 0;\,{x^3} - x > 0;\,x \ne \pm \sqrt 4 {\text{ and }} - 1 < x < 0\,{\text{or}}\,1 < x < \infty \cr & \therefore \,D = \left( { - 1,\,0} \right) \cup \left( {1,\,\infty } \right) - \left\{ {\sqrt 4 } \right\} \cr & D = \left( { - 1,\,0} \right) \cup \left( {1,\,2} \right) \cup \left( {2,\,\infty } \right) \cr} $$