Function MCQ Questions & Answers in Calculus | Maths

$$f\left( x \right)$$  is to be defined when $${x^2} - 1 > 0$$   and $$3 + x > 0$$   and $$3 + x \ne 1.$$   i.e., $${x^2} > 1$$  and $$x > - 3$$   and $$x \ne - 2,$$   i.e., $$x < - 1$$  or $$x > 1$$  and $$x > - 3$$   and $$x \ne - 2$$
$$\therefore \,{D_f} = \left( { - 3,\, - 2} \right) \cup \left( { - 2,\, - 1} \right) \cup \left( {1,\,\infty } \right)$$

For $$f\left( x \right)$$  to be defined, we must have
$$\eqalign{ & x - \sqrt {1 - {x^2}} \geqslant 0 \cr & {\text{or }}x \geqslant \sqrt {1 - {x^2}} > 0 \cr & {\text{or }}{x^2} \geqslant 1 - {x^2} \cr & {\text{or }}{x^2} \geqslant \frac{1}{2} \cr & {\text{Also, }}1 - {x^2} \geqslant 0{\text{ }}\,{\text{or}}\,{\text{ }}{x^2} \leqslant 1 \cr & {\text{Now, }}{x^2} \geqslant \frac{1}{2} \cr & \Rightarrow \left( {x - \frac{1}{{\sqrt 2 }}} \right)\left( {x + \frac{1}{{\sqrt 2 }}} \right) \geqslant 0 \cr & \Rightarrow x \leqslant - \frac{1}{{\sqrt 2 }}{\text{ }}\,{\text{or}}\,{\text{ }}x \geqslant \frac{1}{{\sqrt 2 }} \cr & {\text{Also, }}{x^2} \leqslant 1 \cr & \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) \leqslant 0 \cr & \Rightarrow - 1 \leqslant x \leqslant 1 \cr & {\text{Thus, }}x > 0,\,{x^2} \geqslant \frac{1}{2}{\text{ and }}{x^2} \leqslant 1\, \Rightarrow x\, \in \left[ {\frac{1}{{\sqrt 2 }},\,1} \right]\, \cr} $$

$$\eqalign{ & {\text{We have, }}F\left( {x + 2} \right) = 2F\left( x \right) - F\left( {x + 1} \right).....\left( {\text{i}} \right) \cr & x = 0{\text{ in }}\left( 1 \right) \Rightarrow F\left( 2 \right) = 2F\left( 0 \right) - F\left( 1 \right) = 2\left( 2 \right) - 3 = 1 \cr & x = 1{\text{ in }}\left( 1 \right) \Rightarrow F\left( 3 \right) = 2F\left( 1 \right) - F\left( 2 \right) = 2\left( 3 \right) - 1 = 5 \cr & x = 2{\text{ in }}\left( 1 \right) \Rightarrow F\left( 4 \right) = 2F\left( 2 \right) - F\left( 3 \right) = 2\left( 1 \right) - 5 = - 3 \cr & x = 3{\text{ in }}\left( 1 \right) \Rightarrow F\left( 5 \right) = 2F\left( 3 \right) - F\left( 4 \right) = 2\left( 5 \right) - \left( { - 3} \right) = 13 \cr} $$
$$\therefore $$  The correct answer is (D).

$$f\left( x \right) = x + \left| x \right|.$$    Clearly, $$f\left( { - 1} \right) = f\left( { - 2} \right) = ...... = 0$$
So, $$f$$ is many-one. Also, $$f\left( x \right) \geqslant 0$$   for all $$x.$$ So, it is not surjective.

$$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$$
Function should be $$f\left( x \right) = mx$$
$$\eqalign{ & f\left( 1 \right) = 7;\,\,\therefore m = 7,f\left( x \right) = 7x \cr & \sum\limits_{r = 1}^n f \left( r \right) = 7\sum\limits_1^n r = \frac{{7n\left( {n + 1} \right)}}{2} \cr} $$

$$f\left( x \right) = \frac{1}{{\sqrt {\left( {x + 1} \right)\left( {{e^x} - 1} \right)\left( {x - 4} \right)\left( {x + 5} \right)\left( {x - 6} \right)} }}$$
$$f\left( x \right)$$  is defined is $$\left( {x + 1} \right)\left( {{e^x} - 1} \right)\left( {x - 4} \right)\left( {x + 5} \right)\left( {x - 6} \right) > 0$$
Hence, $$x\, \in \left( { - 5,\, - 1} \right) \cup \left( {0,\,4} \right) \cup \left( {6,\,\infty } \right)$$

$$\eqalign{ & D\left( f \right) = R,\,D\left( g \right) = R - \left\{ 0 \right\} \cr & \therefore \,D\left( h \right) = R - \left\{ 0 \right\}{\text{ and }}h\left( x \right) = f\left( x \right)g\left( x \right) = x \times \frac{1}{x} = 1 \cr & \therefore \,h\left( x \right) = 1{\text{ if and only if }}x\, \in \,R - \left\{ 0 \right\} \cr} $$

$$f\left( x \right) = {e^{{x^2}}} + {e^{ - {x^2}}} \Rightarrow f'\left( x \right) = 2x\left( {{e^{{x^2}}} - {e^{ - {x^2}}}} \right) \geqslant 0,\forall {\text{x}} \in \left[ {0,1} \right]$$
$$\therefore f\left( x \right)$$  is an increasing function on $$\left[ {0,1} \right]$$
$$\eqalign{ & {\text{Hence}}\,{f_{\max }} = f\left( 1 \right) = e + \frac{l}{e} = a \cr & g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}} \cr & \Rightarrow g'\left( x \right) = \left( {2{x^2} + 1} \right){e^{{x^2}}} - 2x{e^{ - {x^2}}} \geqslant 0,\forall x \in \left[ {0,1} \right] \cr} $$
$$\therefore g\left( x \right)$$  is an increasing function on $$\left[ {0,1} \right]$$
$$\eqalign{ & \therefore {g_{\max }} = g\left( 1 \right) = e + \frac{1}{e} = b \cr & h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}} \cr & h'\left( x \right) = 2x\left[ {{e^{{x^2}}}\left( {1 + {x^2}} \right) - {e^{ - {x^2}}}} \right] \geqslant 0,\forall x \in \left[ {0,1} \right] \cr} $$
$$\therefore h\left( x \right)$$  is an increasing function on $$\left[ {0,1} \right]$$
$$\therefore {h_{\max }} = h\left( 1 \right) = e + \frac{1}{e} = c$$
Hence $$a = b = c$$

$$x - \left[ x \right]$$  has the period 1 and $$\cos \,x$$  has the period $$2\pi .$$  Clearly, 1 and $$2\pi $$ do not have a common multiple.

$$\eqalign{ & f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {\frac{x}{2} - 1} \right) + \log (\cos x) \cr & f\left( x \right)\,{\text{is}}\,{\text{defined}}\,{\text{if}}\, - 1 \leqslant \left( {\frac{x}{2} - 1} \right) \leqslant 1\,{\text{and}}\,\cos x > 0 \cr & {\text{or}}\,0 \leqslant \frac{x}{2} \leqslant 2\,{\text{and}}\, - \frac{\pi }{2} < x < \frac{\pi }{2} \cr & {\text{or}}\,0 \leqslant x \leqslant 4\,{\text{and}}\, - \frac{\pi }{2} < x < \frac{\pi }{2} \cr & \therefore x \in \left[ {0,\frac{\pi }{2}} \right) \cr} $$