Function MCQ Questions & Answers in Calculus | Maths

$$\cos \,3x$$  has the period $$\frac{{2\pi }}{3}$$ and $$\sin \,\sqrt 3 x$$  has the period $$\frac{{2\pi }}{{\sqrt 3 }}.$$
As $$\frac{{2\pi }}{3}$$ and $$\frac{{2\pi }}{{\sqrt 3 }}$$ do not have a common multiple, $$f\left( x \right)$$  is not periodic.

$$f\left( x \right) = \log \,x,$$   is not periodic.
$$f\left( x \right) = {e^x},$$   is not periodic.
$$f\left( x \right) = x - \left[ x \right] = \left\{ x \right\},$$     has period 1
$$f\left( x \right) = x + \left[ x \right],$$    is not periodic.

\[f\left( x \right) = \frac{{{e^x} - {e^{\left| x \right|}}}}{{{e^x} + {e^{\left| x \right|}}}} = \left\{ \begin{array}{l} \,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ge 0\\ \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}},\,\,\,\,\,\,\,\,\,\,x < 0 \end{array} \right.\]
Clearly, $$f\left( x \right)$$  is identically zero if $$x \geqslant 0.....(1)$$
$$\eqalign{ & {\text{If }}x < 0,\,{\text{let }}y = f\left( x \right) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}{\text{ or }}{e^{2x}} = \frac{{1 + y}}{{1 - y}} \cr & \because \,x < 0;\,{e^{2x}} < 1{\text{ or }}0 < {e^{2x}} < 1 \cr & \therefore \,\,0 < \frac{{1 + y}}{{1 - y}} < 1 \cr & {\text{or }}\frac{{1 + y}}{{1 - y}} > 0{\text{ and }}\frac{{1 + y}}{{1 - y}} < 1 \cr & {\text{or }}\left( {y + 1} \right)\left( {y - 1} \right) < 0{\text{ and }}\frac{{2y}}{{1 - y}} < 0 \cr & {\text{i}}{\text{.e}}{\text{., }} - 1 < y < 1{\text{ and }}y < 0{\text{ or }}y > 1 \cr & {\text{or }} - 1 < y < 0.....(2) \cr} $$

The period of $$f\left( x \right)$$  is 7. So, the period of $$f\left( {\frac{x}{3}} \right)$$   is $$\frac{7}{{\frac{1}{3}}} = 21$$
The period of $$g\left( x \right)$$  is 11. So, the period of $$g\left( {\frac{x}{5}} \right)$$   is $$\frac{{11}}{{\frac{1}{5}}} = 55$$
Hence, $${T_1} = $$  period of $$f\left( x \right)g\left( {\frac{x}{5}} \right) = 7 \times 55 = 385$$
and $${T_2} = $$  period of $$g\left( x \right)f\left( {\frac{x}{3}} \right) = 11 \times 21 = 231$$
$$\eqalign{ & \therefore {\text{ Period of }}F\left( x \right) = {\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\left\{ {{T_1},\,{T_2}} \right\} \cr & = {\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\left\{ {385,\,231} \right\} \cr & = 7 \times 11 \times 3 \times 5 \cr & = 1155 \cr} $$

An even function is symmetrical about the $$y$$-axis.
Clearly from the graph, the definition given in option (A) is correct.

\[{\rm{Here,\, }}f\left( x \right) = \left\{ \begin{array}{l} x\left| x \right|,\,x \le - 1\\ \,{x^2}\sin \left( {\frac{{\pi x}}{2}} \right),\, - 1 < x < 1\\ x\left| x \right|,\,x \ge 1 \end{array} \right.\]
$$\eqalign{ & {\text{Let }}k \geqslant 0.{\text{ Then }}f\left( { - 1 - k} \right) = \left( { - 1 - k} \right)\left| { - 1 - k} \right| = - {\left( {1 + k} \right)^2} \cr & {\text{and}}\,\,f\left( {1 + k} \right) = \left( {1 + k} \right)\left| {1 + k} \right| = {\left( {1 + k} \right)^2} \cr & {\text{Therefore, }}f\left( {1 + k} \right) = - f\left( { - \overline {1 + k} } \right) \cr & f\left( { - 1 + k} \right) = {\left( { - 1 + k} \right)^2}\sin \frac{\pi }{2}\left( { - 1 + k} \right) \cr & = - {\left( {1 - k} \right)^2}\sin \frac{\pi }{2}\left( {1 - k} \right) \cr & = - f\left( {1 - k} \right) \cr & = - f\left\{ { - \left( { - 1 + k} \right)} \right\} \cr} $$
$$\therefore \,f\left( x \right) = - f\left( { - x} \right)$$    for all $$x.$$ Also, none of the pieces of definition are periodic.

$$\eqalign{ & {\text{For}}\,{\text{domain}}\,{\text{of}}\,f\left( x \right) = \frac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}} \cr & {x^2} + 3x + 2 \ne 0\,{\text{and}}\,x + 3 > 0 \cr & \Rightarrow x \ne - 1, - 2\,{\text{and}}\,x > - 3 \cr & \therefore {D_f} = \left( { - 3,\infty } \right) - \left\{ { - 1, - 2} \right\} \cr} $$

The period of $$f\left( x \right) = \frac{{2\pi }}{{\sqrt p }} = \pi $$    (from the question).
$$\eqalign{ & \therefore \sqrt p = 2{\text{ or }}p = 4 \cr & \therefore \left[ a \right] = 4 \cr & \therefore \,\,4 \leqslant a < 5 \cr} $$

$$\eqalign{ & {\text{Let }}2x + 3y = A{\text{ and }}2x - 7y = B \cr & {\text{Then, }}7A + 3B = 20x \cr & \therefore \,f\left( {A,\,B} \right) = 7A + 3B \cr & \therefore \,f\left( {x,\,y} \right) = 7x + 3y \cr} $$

$$^{7 - x}{P_{x - 3}}$$   is defined if
$$\eqalign{ & 7 - x \geqslant 0,x - 3 \geqslant 0{\text{ and }}7 - x \geqslant x - 3 \cr & \Rightarrow 3 \leqslant x \leqslant 5\,{\text{and}}\,x \in I \cr & \therefore x = 3,4,5 \cr & \therefore f(3){ = ^{7 - 3}}{P_{3 - 3}}{ = ^4}{P_0} = 1 \cr & \therefore f(4){ = ^{7 - 4}}{P_{4 - 3}}{ = ^3}{P_1} = 3 \cr & \therefore f(5){ = ^{7 - 5}}{P_{5 - 3}}{ = ^2}{P_2} = 2 \cr} $$
Hence range = {I, 2, 3}