Sets and Relations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = \frac{{ax + d}}{{cx + b}} \cr & f\left( {f\left( x \right)} \right) = \frac{{a\left( {\frac{{ax + d}}{{cx + b}}} \right) + b}}{{c\left( {\frac{{ax + d}}{{cx + b}}} \right) + b}} = \frac{{{a^2}x + ad + cdx + bd}}{{cax + cd + bcx + {b^2}}} \cr & f\left( {f\left( x \right)} \right) = x \cr & \Rightarrow \frac{{{a^2}x + ad + cdx + bd}}{{cax + cd + bcx + {b^2}}} = x \cr & \Rightarrow c\left( {a + b} \right){x^2} - \left( {{a^2} - {b^2}} \right)x - \left( {a + b} \right)d = 0 \cr & \Rightarrow \left( {a + b} \right)\left( {c{x^2} - \left( {a - b} \right)x - d} \right) = 0 \cr & \Rightarrow a + b = 0 \cr & {\text{As }}c{x^2} - \left( {a - b} \right)x - d \ne 0{\text{ for all }}x \cr} $$

$$\eqalign{ & f\left( x \right){\text{ is onto}}\,\,\,\therefore \,S = {\text{range of }}f\left( x \right) \cr & {\text{Now }}f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1 = 2\,\sin \,\left( {x - \frac{\pi }{3}} \right) + 1 \cr & \because \, - 1 \leqslant \sin \left( {x - \frac{\pi }{3}} \right) \leqslant 1 \cr & - 1 \leqslant 2\,\sin \left( {x - \frac{\pi }{3}} \right) + 1 \leqslant 3 \cr & \therefore \,f\left( x \right)\, \in \left[ { - 1,\,3} \right] = S\, \cr} $$

We know that, if $$X$$ and $$Y$$ are any two finite sets having $$m$$ and $$n$$ elements respectively, where $$1 \leqslant n \leqslant m,$$   then the number of onto functions from $$X$$ to $$Y$$ is given by $$\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{n - r}}\,{}^n{C_r}{r^m}.r = 1} $$
Thus, the number of surjective mapping is $$\sum\limits_{r = 1}^2 {{{\left( { - 1} \right)}^{2 - r}}\,{C_r}{r^n} = \left( {{2^n} - 2} \right)} $$

$$\eqalign{ & n\left( {A \times B \times C \times .....} \right) = n\left( A \right) \times n\left( B \right) \times n\left( C \right)...... \cr & \therefore \,n\left( {A \times A \times B} \right) \cr & = n\left( A \right) \times n\left( A \right) \times n\left( B \right) \cr & = 3 \times 3 \times 5 = 45\,\,\,\,\,\left[ {\because \,n\left( A \right) = 3,\,\,n\left( B \right) = 5} \right] \cr} $$

$$\eqalign{ & {\text{We have }}\left( {a,\,b} \right)R\left( {a,\,b} \right)\,{\text{for all }}\left( {a,\,b} \right)\, \in \,N \times N \cr & {\text{As}}\,a + b = b + a{\text{. Hence }}R\,{\text{is reflexive}} \cr & R\,{\text{is symmetric for we have }}\left( {a,\,b} \right)R\left( {c,\,d} \right) \cr & \Rightarrow a + d = b + c\,\,\,\,\,\, \Rightarrow d + a = c + b \cr & \Rightarrow c + b = d + a\,\,\,\,\,\, \Rightarrow \left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & {\text{Then, by defination of }}R,\,{\text{we have}} \cr & a + d = b + c{\text{ and }}c + f = d + e \cr & {\text{So, by addition, we get}} \cr & a + d + c + f = b + c + d + c{\text{ or }}a + f = b + e \cr & {\text{Hence, }}\left( {a,\,b} \right)R\left( {e,\,f} \right) \cr & {\text{Thuse, }}\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & \Rightarrow \,\left( {a,\,b} \right)R\left( {e,\,f} \right) \cr} $$

$$\eqalign{ & P = \left\{ {\theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta } \right\} \cr & \sin \theta = \left( {\sqrt 2 + 1} \right)\cos \theta ,\tan \theta = \sqrt 2 + 1 \cr & Q = \left\{ {\theta :\sin \theta + \cos \theta = \sqrt 2 \cos \theta } \right\} \cr & \cos \theta = \left( {\sqrt 2 - 1} \right)\sin \theta \,\,{\text{or}}\tan \theta = \sqrt 2 + 1 \cr & \therefore \,\,P = Q \cr} $$

$$\eqalign{ & y = \frac{1}{2}\left[ {\cos \left( {2x + 2} \right) + \cos \,2 - \left\{ {1 + \cos \left( {2x + 2} \right)} \right\}} \right] \cr & {\text{or }}y = - \frac{1}{2}\left( {1 - \cos \,2} \right) = - {\sin ^2}1{\text{ i}}{\text{.e}}{\text{., constant}} \cr} $$
$$\therefore $$  Graph is a line parallel to $$x$$-axis. Also when $$x = \frac{\pi }{2},\,y = - {\cos ^2}\left( {\frac{\pi }{2} + 1} \right) = - {\sin ^2}1$$        and hence it passes through the point $$\left( {\frac{\pi }{2},\, - {{\sin }^2}1} \right)$$

Replacing $$x$$ by $${g^{ - 1}}\left( x \right),$$  we get $$x = 2f\left( {{g^{ - 1}}\left( x \right)} \right) + 5$$
$$\eqalign{ & \therefore \,f\left( {{g^{ - 1}}\left( x \right)} \right) = \frac{{x - 5}}{2} \cr & \therefore \,{g^{ - 1}}\left( x \right) = {f^{ - 1}}\left( {\frac{{x - 5}}{2}} \right) \cr} $$

$$\eqalign{ & {\text{Let }}y = 1 - {2^{ - x}} \cr & {\text{or }}{2^{ - x}} = 1 - y \cr & {\text{or }} - x = {\log _2}\left( {1 - y} \right) \cr & {\text{or }}{f^{ - 1}}\left( x \right) = g\left( x \right) = - {\log _2}\left( {1 - x} \right) \cr} $$

Given, $$n\left( A \right) = 4,\,n\left( B \right) = 3$$
Since, the sets $$A$$ and $$B$$ are not known, then cardinality of the set $$A\Delta B$$  cannot be determined.