Function MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = \sqrt {\left( {\frac{2}{{{x^2} - x + 1}} - \frac{1}{{x + 1}} - \frac{{2x - 1}}{{{x^3} + 1}}} \right)} \cr & {\text{We must have }}\frac{2}{{{x^2} - x + 1}} - \frac{1}{{x + 1}} - \frac{{2x - 1}}{{{x^3} + 1}} \geqslant 0 \cr & {\text{or }}\frac{{2\left( {x + 1} \right) - \left( {{x^2} - x + 1} \right) - \left( {2x - 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \geqslant 0 \cr & {\text{or }}\frac{{ - \left( {{x^2} - x - 2} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \geqslant 0 \cr & {\text{or }}\frac{{ - \left( {x - 2} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} \geqslant 0 \cr & {\text{or }}\frac{{2 - x}}{{{x^2} - x + 1}} \geqslant 0,{\text{ where}}\,x \ne - 1 \cr & {\text{or }}2 - x \geqslant 0,\,x \ne - 1\,\,\,\,\left( {{\text{as }}{x^2} - x + 1 > 0\,\forall \,x\, \in \,R} \right) \cr & {\text{or }}x \leqslant 2,\,x \ne - 1 \cr} $$
Hence, domain of the function is $$\left( { - \infty ,\, - 1} \right) \cup \left( { - 1,\,2} \right].$$
$${\text{or }}\left( { - \infty ,\,2} \right] - \left\{ { - 1} \right\}$$

$$\eqalign{ & f\left( {{x^2}} \right) + 2 = {x^2} + \frac{1}{{{x^2}}} + 2 \cr & = {\left( {x + \frac{1}{x}} \right)^2} \cr & = {\left\{ {f\left( x \right)} \right\}^2}\,{\text{and}} \cr & {\text{ }}f\left( {{x^3}} \right) + 3\,f\left( x \right) = {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right) \cr & = {\left( {x + \frac{1}{x}} \right)^3} \cr & = {\left\{ {f\left( x \right)} \right\}^3} \cr} $$
Thus, both 1 and 2 are correct.

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