Function MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( { - 2.3} \right) = 1 + \left| { - 2.3} \right| = 3.3 \cr & \therefore f\left\{ {f\left( { - 2.3} \right)} \right\} = f\left( {3.3} \right) = \left[ {3.3} \right] = 3 \cr} $$

$$\eqalign{ & {\text{Given, }}f\left( x \right) = \frac{x}{{x - 1}} \cr & {\text{Then,}}\,f\left( a \right) = \frac{a}{{a - 1}}\,{\text{and }}f\left( {a + 1} \right) = \frac{{a + 1}}{a} \cr & {\text{So, }}\frac{{f\left( a \right)}}{{f\left( {a + 1} \right)}} = \frac{a}{{a - 1}} \times \frac{a}{{a + 1}} = \frac{{{a^2}}}{{{a^2} - 1}} = f\left( {{a^2}} \right) \cr} $$

$$\eqalign{ & {\text{We have }}f\left( x \right) = {\log _e}\left\{ {\operatorname{sgn} \left( {9 - {x^2}} \right)} \right\} + \sqrt {{{\left[ x \right]}^3} - 4\left[ x \right]} \cr & {\text{We must have, }}\operatorname{sgn} \left( {9 - {x^2}} \right) > 0 \cr & \Rightarrow 9 - {x^2} > 0 \cr & \Rightarrow {x^2} - 9 < 0 \cr & \Rightarrow \left( {x - 3} \right)\left( {x + 3} \right) < 0 \cr & \Rightarrow - 3 < x < 3......({\text{i}}) \cr & {\text{Also }}{\left[ x \right]^3} - 4\left[ x \right] \geqslant 0 \cr & \Rightarrow \left[ x \right]\left( {{{\left[ x \right]}^2} - 4} \right) \geqslant 0 \cr & \Rightarrow \left[ x \right]\left( {\left[ x \right] - 2} \right)\left( {\left[ x \right] + 2} \right) \geqslant 0 \cr & \Rightarrow \left[ x \right] \geqslant - 2{\text{ or }}\left[ x \right]{\text{ lies between }} - 2{\text{ and}}\,0{\text{ i}}{\text{.e}}{\text{., }}\left[ x \right] = - 2,\, - 1\,{\text{or }}0 \cr & {\text{Now, }}\left[ x \right] \geqslant - 2 \Rightarrow x \geqslant 2......({\text{ii}}) \cr & \left[ x \right] = - 2 \Rightarrow - 2 \leqslant x < - 1; \cr & \left[ x \right] = - 1 \Rightarrow - 1 \leqslant x < 0; \cr & \left[ x \right] = 0 \Rightarrow 0 \leqslant x < 1 \cr & {\text{Hence, }}\left[ x \right] = - 2,\, - 1,\,0 \Rightarrow - 2 \leqslant x < 1 \cr & {\text{Hence,}}\,{D_f} = \left[ { - 2,\,1} \right) \cup \left[ {2,\,3} \right) \cr} $$

$$\eqalign{ & a = \sqrt {{{2011}^2} + 2012} = 2011.5\,; \cr & f\left( a \right) = \left[ {2011.5} \right] = 2011 \cr} $$

$$9 - {x^2} \geqslant 0\,\,\, \Rightarrow - 3 \leqslant x \leqslant 3$$
As $$9 - {x^2}$$   is an even function, the value of $$f\left( x \right)$$  changes in $$0 \leqslant x \leqslant 3$$
Therefore,
$$\max \,f\left( x \right) = \sqrt {9 - 0} = 3{\text{ and }}\min \,f\left( x \right) = \sqrt {9 - 9} = 0$$

$$\eqalign{ & \frac{{1 - \left| x \right|}}{2} \leqslant - 1\,\,\,{\text{or}}\,\,\,\frac{{1 - \left| x \right|}}{2} \geqslant 1 \cr & \frac{{1 - \left| x \right|}}{2} \leqslant - 1\,\,\, \Rightarrow \left| x \right| \geqslant 3\,\, \Rightarrow x\, \in \left( { - \infty ,\, - 3} \right] \cup \left[ {3,\, + \infty } \right) \cr & \frac{{1 - \left| x \right|}}{2} \geqslant 1\,\, \Rightarrow \left| x \right| \leqslant - 1\left( {{\text{absurd}}} \right). \cr & {\text{Also 0}} \leqslant {\sec ^{ - 1}}\left\{ {\frac{{1 - \left| x \right|}}{2}} \right\} \leqslant \pi \cr} $$

$$\eqalign{ & 16 - x > 0,\,2x - 1 \geqslant 0,\,16 - x \geqslant 2x - 1,\,20 - 3x > 0,\,4x - 5 \geqslant 0,\,20 - 3x \geqslant 4x - 5 \cr & \Rightarrow x < 16,\,x \geqslant \frac{1}{2},\,x \leqslant \frac{{17}}{3},\,x < \frac{{20}}{3},\,x \geqslant \frac{5}{4},\,x \leqslant \frac{{25}}{7} \cr & \Rightarrow \frac{5}{4} \leqslant x \leqslant \frac{{25}}{7} \cr} $$
But $$16 - x\, \in \,N.$$   So, $$x$$ must be integer. Hence, $$x=2,\,3$$

$$\eqalign{ & y = \frac{1}{{{{\log }_{10}}\left( {1 - x} \right)}} + \sqrt {x + 2} \cr & y = f\left( x \right) + g\left( x \right) \cr} $$
NOTE THIS STEP: Then domain of given function is $${D_f} \cap {D_g}$$
Now, for domain of $$f\left( x \right) = \frac{1}{{{{\log }_{10}}\left( {1 - x} \right)}}$$
We know it is defined only when $$1 - x > 0$$   and $$1 - x \ne 1 \Rightarrow x < 1\,{\text{and}}\,x \ne 0$$
$$\eqalign{ & \therefore {D_f} = \left( { - \infty ,1} \right) - \left\{ 0 \right\} \cr & {\text{For}}\,{\text{domain}}\,{\text{of}}\,g\left( x \right) = \sqrt {x + 2} \cr & x + 2 \geqslant 0 \cr & \Rightarrow x \geqslant - 2 \cr & \therefore {D_g} = \left[ { - 2,\infty } \right) \cr} $$



$$\therefore $$ Common domain is $$\left[ { - 2,1} \right) - \left\{ 0 \right\}$$

$$\eqalign{ & f\left( x \right) = f\left( {x + k} \right) \cr & \Rightarrow f\left( {kx + a} \right) = f\left( {kx + a + k} \right) = f\left\{ {k\left( {x + 1} \right) + a} \right\} \cr} $$
So, the period is 1.

$${x^2} - 1 > 0,\,{x^2} - 1 \ne 1,\,x > 0\,{\text{ and }}{\log _{{x^2} - 1}}\left( x \right) \geqslant 0$$
The first three imply $$x > 1{\text{ but }}x \ne \sqrt 2 $$
$$\eqalign{ & {\log _{{x^2} - 1}}\left( x \right) \geqslant 0 \Rightarrow x \geqslant 1{\text{ if }}{x^2} - 1 > 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \leqslant 1{\text{ if }}{x^2} - 1 < 1 \cr} $$
Now, $${x^2} - 1 > 1$$   or $$x > \sqrt 2 \,\,\left( {\because x > 1} \right)$$    then $$x \geqslant 1$$  which is true for all $$x > \sqrt 2 $$
$${x^2} - 1 < 1$$   or $$1 < x < \sqrt 2 \,\,\left( {\because x > 1} \right)$$     then $$x \leqslant 1$$  which is not true for any $$x$$ in $$\left( {1,\,\sqrt 2 } \right)$$
This gives no value of $$x.$$  Thus $$x > \sqrt 2 $$