Function MCQ Questions & Answers in Calculus | Maths

$$\left| {\sin \frac{x}{2}} \right|$$   is a periodic function of period $$2\pi $$  and $$\left| {\cos \,x} \right|$$  is a periodic function of period $$\pi .$$

$$\eqalign{ & \left| {\cos \,x} \right| + \cos \,x > 0 \cr & \Rightarrow \cos \,x > 0 \cr & \Rightarrow 2n\pi - \frac{\pi }{2} < x < 2n\pi + \frac{\pi }{2} \cr} $$

$$\eqalign{ & f\left( x \right) = \sin x + \cos x,g\left( x \right) = {x^2} - 1, \cr & \Rightarrow g\left( {f\left( x \right)} \right) = {\left( {\sin x + \cos x} \right)^2} - 1 = \sin 2x \cr} $$
Clearly $$g\left( {f\left( x \right)} \right)$$  is invertible in $$ - \frac{\pi }{2} \leqslant 2x \leqslant \frac{\pi }{2}$$
$$\eqalign{ & \left[ {\because \sin \theta \,{\text{is}}\,{\text{invertible}}\,{\text{when}}\, - \frac{\pi }{2} \leqslant \theta \leqslant \frac{\pi }{2}} \right] \cr & \Rightarrow - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4} \cr} $$

$$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$
Consider $$f\left( x \right) = \left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|$$
$$f\left( x \right) = \left\{ {_{_{3x - 6,}^{x,}}^{_{4 - x,}^{6 - 3x,}}} \right.\,_{_{x \geqslant 3}^{2 \leqslant x < 3}}^{_{1 \leqslant x < 2}^{x < 1}}$$
NOTE THIS STEP:

Graph of $$f\left( x \right)$$ shows $$f\left( x \right)\, \geqslant 6$$   for $$x \leqslant 0$$  or $$x \geqslant 4$$

Given that $$f\left( x \right) = {x^3} + 5x + 1$$
$$\therefore f'\left( x \right) = 3{x^2} + 5 > 0,\forall x \in R$$
$$ \Rightarrow f\left( x \right)$$   is strictly increasing on $$R$$
$$ \Rightarrow f\left( x \right)$$   is one one
$$\therefore $$ Being a polynomial $$f\left( x \right)$$  is cont. and inc.
on $$R$$ with $$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = - \infty $$
and $$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \infty $$
$$\therefore $$ Range of $$f = \left( { - \infty ,\infty } \right) = R$$
Hence $$f$$ is onto also. So, $$f$$ is one one and onto $$R$$.

$$f\left( x \right) = \sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} + \sqrt { - \left( {x - 4} \right)\left( {x + 2} \right)} $$
The first part is real outside $$\left( {2,\,3} \right)$$  and the second is real in $$\left[ { - 2,\,4} \right]$$  so that the domain is $$\left[ { - 2,\,2} \right] \cup \left[ {3,\,4} \right].$$

$$\eqalign{ & f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }},{\text{define}}\,{\text{if}}\,\left| x \right| - x > 0 \cr & \Rightarrow \left| x \right| > x, \Rightarrow x < 0 \cr} $$
Hence domain of $$f\left( x \right)$$  is $$\left( { - \infty ,0} \right)$$

$$g\left\{ {f\left( x \right)} \right\} = g\left( {{x^n}} \right) = ng\left( x \right).$$      Also $$\log \,{x^n} = n\,\log \,\left| x \right|.$$    So, $$g\left( x \right) = \log \,\left| x \right|$$   is possible.

$$\eqalign{ & {\text{Given,}}\,f\left( x \right) = 4x - {x^2} \cr & \Rightarrow f\left( {a + 1} \right) - f\left( {a - 1} \right) \cr & = \left\{ {4\left( {a + 1} \right) - {{\left( {a + 1} \right)}^2}} \right\} - \left\{ {4\left( {a - 1} \right) - {{\left( {a - 1} \right)}^2}} \right\} \cr & = 8 - \left\{ {{{\left( {a + 1} \right)}^2} - {{\left( {a - 1} \right)}^2}} \right\} \cr & = 8 - 4a \cr & = 4\left( {2 - a} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = {f^{ - 1}}\left( x \right) \Rightarrow fof\left( x \right) = x \cr & {\left[ {{{\left( {x + 1} \right)}^2} - 1 + 1} \right]^2} - 1 = x \cr & \Rightarrow {\left( {x + 1} \right)^4} = x + 1 \cr & \therefore x = 0\,{\text{or}}\, - 1 \cr & \therefore {\text{Req}}{\text{.}}\,{\text{set}}\,{\text{is}}\,\left\{ {0, - 1} \right\} \cr} $$