Function MCQ Questions & Answers in Calculus | Maths

Clearly $$f$$ is one one and onto, so invertible
$$\eqalign{ & {\text{Also}}\,f\left( x \right) = 4x + 3 = y \Rightarrow x = \frac{{y - 3}}{4} \cr & \therefore g\left( y \right) = \frac{{y - 3}}{4} \cr} $$

$$\eqalign{ & {x_0} = a,\,{x_1} = f\left( x \right) = \frac{{{x_0}}}{{1 - {x_0}}} = \frac{a}{{1 - a}}; \cr & {x_2} = f\left( {{x_1}} \right) = \frac{{{x_1}}}{{1 - {x_1}}} = \frac{{\frac{a}{{1 - a}}}}{{1 - \frac{a}{{1 - a}}}} = \frac{a}{{1 - 2a}} \cr & \therefore\, {x_{2009}} = \frac{a}{{1 - 2009\,a}} = 1 \cr & \Rightarrow 1 - 2009\,a = a \cr & \Rightarrow a = \frac{1}{{2010}} \cr} $$

$$\eqalign{ & {\text{Given}}\,{\text{that}}\,f\left( x \right) = 2x + \sin x,\,\,x \in R \Rightarrow f'\left( x \right) = 2 + \cos x \cr & {\text{But}}\, - 1 \leqslant \cos x \leqslant 1 \Rightarrow \,1 \leqslant 2 + \cos x \leqslant 3 \Rightarrow 1 \leqslant 2 + \cos x \leqslant 3 \cr & \therefore f'\left( x \right) > 0,\forall x \in R \cr} $$
$$ \Rightarrow f\left( x \right)$$  is strictly increasing and hence one-one
Also as $$x \to \infty ,f\left( x \right) \to \infty \,{\text{and}}\,x \to - \infty ,f\left( x \right) \to - \infty $$
$$\therefore $$ Range of $$f\left( x \right) = R = $$   domain of $$f\left( x \right) \Rightarrow f\left( x \right)$$   is onto.
Thus, $$f\left( x \right)$$  is one-one and onto.

$$\eqalign{ & f\left( x \right) = \frac{3}{{4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right) \cr & 4 - {x^2} \ne 0;{x^3} - x > 0 \cr & x \ne \pm \sqrt 4 {\text{ and }} - 1 < x < 0{\text{ or }}1 < x < \infty \cr} $$

$$\eqalign{ & \therefore D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\} \cr & D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right) \cr} $$

$$\eqalign{ & {\text{We}}\,{\text{have}}\,f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1 \cr & \Rightarrow f'\left( x \right) = 6{x^2} - 30x + 36 \cr & = 6\left( {{x^2} - 5x + 6} \right) \cr & = 6\left( {x - 2} \right)\left( {x - 3} \right) \cr & \because f'\left( x \right) > 0\forall x \in \left[ {0,2} \right]\,{\text{and}}\,f'\left( x \right) < 0\,\forall \,x \in \left[ {2,3} \right] \cr} $$
$$\therefore f\left( x \right)$$  is increasing on [0, 2] and decreasing on [2,3]
$$\therefore f\left( x \right)$$  is many one on [0, 3]
$${\text{Also}}\,f\left( 0 \right) = 1,f\left( 2 \right) = 29,f\left( 3 \right) = 28$$
$$\therefore $$ Global min = 1 and Global max = 29
i.e., Range of $$f$$ = [1, 29] = codomain
$$\therefore f$$  is onto.

Given that $$f:\left[ {0,\infty } \right) \to \left[ {0,\infty } \right)$$
Such that $$\,f\left( x \right) = \frac{x}{{1 + x}}$$
Then $$f'\left( x \right) = \frac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}} > 0\forall x$$
$$\therefore f$$  is an increasing function $$ \Rightarrow f$$  is one-one.
Also, $${D_f} = \left[ {0,\infty } \right)$$
And for range let $$\frac{x}{{1 + x}} = y \Rightarrow x = \frac{y}{{1 - y}}$$
$$x \geqslant 0 \Rightarrow 0 \leqslant y < 1$$
$$\therefore {R_f} = \left[ {0,1} \right) \ne \,{\text{Co - domain}}$$
$$\therefore f$$  is not onto.

$$\eqalign{ & f\left( x \right) = {x^2} + \frac{1}{{{x^2} + 1}} - 1 + 1 = 1 + {x^2} - \frac{{{x^2}}}{{1 + {x^2}}} \cr & = 1 + {x^2}\left( {1 - \frac{1}{{1 + {x^2}}}} \right) \geqslant 1{\text{ for all }}x\, \in \,R \cr} $$
The domain $$f=R.$$   Clearly, as $$x$$ increases $$f\left( x \right)$$  increases.

Let us consider a graph symm. with respect to line $$x = 2$$  as shown in the figure.

$$\eqalign{ & {\text{From the figure }}f\left( {{x_1}} \right) = f\left( {{x_2}} \right),\,{\text{where}}\,{x_1} = 2 - x\,{\text{and}}\,{x_2} = 2 + x \cr & \therefore f\left( {2 - x} \right) = f\left( {2 + x} \right) \cr} $$

$$\eqalign{ & f\left( {x + y} \right).f\left( {x - y} \right) \cr & = \frac{{{2^{x + y}} + {2^{ - x - y}}}}{2}.\frac{{{2^{x - y}} + {2^{ - x + y}}}}{2} \cr & = \frac{{{2^{2x}} + {2^{2y}} + {2^{ - 2x}} + {2^{ - 2y}}}}{{2 \times 2}} \cr & = \frac{1}{2}\left[ {\frac{{{2^{2x}} + {2^{ - 2x}}}}{2} + \frac{{{2^{2y}} + {2^{ - 2y}}}}{2}} \right] \cr & = \frac{1}{2}\left[ {f\left( {2x} \right) + f\left( {2y} \right)} \right] \cr} $$

$$\eqalign{ & {\text{Given,}}\,\,f\left( x \right) = \frac{1}{{\sqrt {2x - 1} }} - \sqrt {1 - {x^2}} = p\left( x \right) - q\left( x \right) \cr & {\text{where }}p\left( x \right) = \frac{1}{{\sqrt {2x - 1} }}{\text{ and }}q\left( x \right){\text{ = }}\sqrt {1 - {x^2}} \cr & {\text{Now, domain of }}p\left( x \right)\,{\text{exist when}}\,2x - 1 \ne 0 \cr & \Rightarrow x = \frac{1}{2}{\text{ and }}2x - 1 > 0 \cr & \Rightarrow x = \frac{1}{2}{\text{ and }}x > \frac{1}{2} \cr & \therefore \,x\, \in \left( {\frac{1}{2},\,\infty } \right) \cr & {\text{and domain of }}q\left( x \right)\,{\text{exists when }}1 - {x^2} \geqslant 0 \cr & \Rightarrow {x^2} \leqslant 1 \cr & \Rightarrow \left| x \right| \leqslant {\text{1}} \cr & \therefore \, - 1 \leqslant x \leqslant 1 \cr & \therefore \,{\text{Common domain is }}\left] {\frac{1}{2},\,1} \right[ \cr} $$