Sets and Relations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Let }}x,\,y\, \in \,R{\text{ such that }}x \geqslant - 1,\,y \geqslant - 1 \cr & {\text{Then, }}f\left( x \right) = f\left( y \right) \cr & \Rightarrow {\left( {x + 1} \right)^2} - 1 = {\left( {y + 1} \right)^2} - 1 \cr & \Rightarrow {x^2} + 2x = {y^2} + 2y \cr & \Rightarrow {x^2} - {y^2} = - 2\left( {x - y} \right) \cr & \Rightarrow \left( {x - y} \right)\left( {x + y + 2} \right) = 0 \cr & \Rightarrow x - y = 0{\text{ or }}x + y + 2 = 0 \cr & \Rightarrow x = y{\text{ or }}x = y = - 1 \cr & \therefore \,f{\text{ is one - one}}{\text{.}} \cr} $$
Also, $$f$$ is onto as for all $$y \geqslant - 1,$$   there exists $$x = - 1 + \sqrt {y + 1} \geqslant - 1$$     such that $$f\left( x \right) = y$$
$$\eqalign{ & \therefore \,f{\text{ is invertible}}{\text{.}} \cr & {\text{Let }}f\left( x \right) = y \cr & \Rightarrow {\left( {x + 1} \right)^2} - 1 = y \cr & \Rightarrow x = - 1 \pm \sqrt {y + 1} \cr & {\text{But, }}x \geqslant - 1 \cr & \therefore \,x = - 1 + \sqrt {y + 1} \cr & \Rightarrow {f^{ - 1}}\left( y \right) = - 1 + \sqrt {y + 1} \cr & {\text{Hence, }}{f^{ - 1}}\left( x \right) = - 1 + \sqrt {x + 1} \cr} $$

\[\begin{array}{l} f\left( x \right) = \left| {x - 1} \right| = \left\{ \begin{array}{l} 1 - x,\,\,\,0 < x < 1\\ x - 1,\,\,\,x \ge 1 \end{array} \right.\\ g\left( x \right) = {e^x},\,x \ge - 1\\ \left( {{\rm{fog}}} \right)\left( x \right) = \left\{ \begin{array}{l} 1 - g\left( x \right),\,\,\,0 < g\left( x \right) < 1\,\,\,{\rm{ i}}{\rm{.e}}{\rm{., }} - 1 \le x < 0\\ g\left( x \right) - 1,\,\,\,g\left( x \right) \ge 1\,\,\,{\rm{i}}{\rm{.e}}{\rm{., }}\,0 \le x \end{array} \right.\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ \begin{array}{l} 1 - {e^x},\,\,\,\,\, - 1 \le x < 0\\ {e^x} - 1,\,\,\,\,\,\,\,x \ge 0 \end{array} \right. \end{array}\]
$$\therefore {\text{ domain}} = \left[ { - 1,\,\infty } \right)$$
fog is decreasing in $$\left[ { - 1,\,0 } \right)$$  and increasing in $$\left[ {0,\,\infty } \right)$$
$$\eqalign{ & {\text{fog}}\left( { - 1} \right) = 1 - \frac{1}{e}{\text{ and fog}}\left( 0 \right) = 0 \cr & {\text{As }}x \to \infty ,\,{\text{ fog}}\left( x \right) \to \infty , \cr & \therefore {\text{ range}} = \left[ {0,\,\infty } \right) \cr} $$

$$g\left( {f\left( x \right)} \right) = \left| {\sin \,x} \right|$$    indicates that possibly $$f\left( x \right) = \sin \,x,\,g\left( x \right) = \left| x \right|$$
Assuming it correct, $$f\left( {g\left( x \right)} \right) = f\left( {\left| x \right|} \right) = \sin \left| x \right|,$$      which is not correct.
$$f\left( {g\left( x \right)} \right) = {\left( {\sin \sqrt x } \right)^2}$$    indicates that possibly
or, $$g\left( x \right) = \sin \sqrt x ,\,\,f\left( x \right) = {x^2}$$
Then $$g\left( {f\left( x \right)} \right) = g\left( {{{\sin }^2}x} \right) = \sqrt {\sin \,x} = \left| {\sin \,x} \right|$$         (for the first combination), which is given.
Hence, $$f\left( x \right) = {\sin ^2}x,\,g\left( x \right) = \sqrt x $$

$$\eqalign{ & \frac{{{f^{ - 1}}\left( {{x_1}} \right) + {f^{ - 1}}\left( {{x_2}} \right) + ..... + {f^{ - 1}}\left( {{x_n}} \right)}}{n} = f\left( {\frac{{{x_1} + {x_2} + ..... + {x_n}}}{n}} \right) \cr & {\text{and }}\frac{{{f^{ - 1}}\left( {{x_1}} \right) + {f^{ - 1}}\left( {{x_2}} \right) + ..... + {f^{ - 1}}\left( {{x_n}} \right)}}{n} = f\left( {\frac{{{x_1} + {x_2} + ..... + {x_n}}}{n}} \right) \cr & \Rightarrow f\left( {\overline x } \right) = \overline x ,{\text{ where }}\overline x = \frac{{{x_1} + {x_2} + ..... + {x_n}}}{n} \cr & \Rightarrow {\overline x ^2} + 3\overline x - 3 = \overline x \,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\overline x ^2} + 2\overline x - 3 = 0 \cr & \Rightarrow \overline x = - 3,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \overline x = 1{\text{ as }}\overline x > 0 \cr} $$

$$R$$ be a relation on $$N$$ defined by $$x + 2y = 8$$
$$\therefore \,R = \left\{ {\left( {2,\,3} \right);\left( {4,\,2} \right);\left( {6,\,1} \right)} \right\}$$
Hence, Domain of $$R = \left\{ {2,\,4,\,6} \right\}$$

$$\eqalign{ & a + e + f + g = 285,\,\,b + d + f + g = 195 \cr & c + d + e + f = 115,\,\,e + g = 45,\,\,f + g = 70,\,\,d + g = 50 \cr & a + b + c + d + e + f + g = 500 - 50 = 450 \cr & {\text{We obtain,}} \cr & a + f = 240,\,\,b + d = 125,\,\,c + e = 65 \cr & a + e = 215,\,\,b + f = 145,\,\,b + c + d = 165 \cr & a + c + e = 255;\,\,a + b + f = 335 \cr & {\text{Solving we get}} \cr & b = 95,\,\,c = 40,\,\,a = 190,\,\,d = 30,\,\,e = 25,\,\,f = 50\,{\text{and}}\,g = 20 \cr & {\text{Desired quantity}} = a + b + c = 325 \cr} $$

Given function is defined if
$$\eqalign{ & {}^{10}{C_{x - 1}} > 3{}^{10}{C_x} \cr & {\text{or }}\frac{1}{{11 - x}} > \frac{3}{x} \cr & {\text{or }}4x > 33 \cr & {\text{or }}x \geqslant 9 \cr & {\text{But }}x \leqslant 10 \cr & \therefore \,x = 9,\,10 \cr} $$

Since, $$A$$ and $$B$$ are two disjoints therefore $$A \cap B = \phi $$
$$\therefore \,A - B = A - \left( {A \cap B} \right)$$

In the given Venn diagram shaded region is $$C \cup \left( {C \cap A} \right) \cup \left( {C \cap B} \right).$$

$$\eqalign{ & \because \,f\left( x \right) = \frac{x}{{x - 1}} \cr & \therefore \,\left( {fof} \right)\left( x \right) = f\left\{ {f\left( x \right)} \right\} \cr & = f\left( {\frac{x}{{x - 1}}} \right) \cr & = \frac{{\frac{x}{{x - 1}}}}{{\frac{x}{{x - 1}} - 1}} \cr & = \frac{{\frac{x}{{x - 1}}}}{{\frac{{x - x + 1}}{{x - 1}}}} \cr & = \frac{{\frac{x}{{x - 1}}}}{{\frac{1}{{x - 1}}}} \cr & = x \cr & \Rightarrow \left( {fofof} \right)\left( x \right) = f\left( {fof} \right)\left( x \right) = f\left( x \right) = \frac{x}{{x - 1}} \cr & \Rightarrow \underbrace {\left( {fofof.....of} \right)}_{19{\text{ times}}}\left( x \right) = f\left( {fof} \right)\left( x \right) = f\left( x \right) = \frac{x}{{x - 1}} \cr} $$