Function MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \left( {\bf{A}} \right)f\left( { - x} \right) = \sqrt {1 + \left( { - x} \right) + {{\left( { - x} \right)}^2}} - \sqrt {1 - \left( { - x} \right) + {{\left( { - x} \right)}^2}} \cr & = \sqrt {1 - x + {x^2}} - \sqrt {1 + x + {x^2}} \cr & = - f\left( x \right) \cr & {\text{Hence, }}f\left( x \right){\text{ is odd}}{\text{.}} \cr & \left( {\bf{B}} \right)\,f\left( { - x} \right) = \log \left\{ {\frac{{1 - \left( { - x} \right)}}{{1 + \left( { - x} \right)}}} \right\} \cr & = \log \left( {\frac{{1 + x}}{{1 - x}}} \right) \cr & = - f\left( x \right) \cr & {\text{Hence, }}f\left( x \right){\text{ is odd}}{\text{.}} \cr & \left( {\bf{C}} \right)\,f\left( { - x} \right) = \log \left( { - x + \sqrt {1 + {{\left( { - x} \right)}^2}} } \right) \cr & = \log \left\{ {\frac{{\left( { - x + \sqrt {1 + {x^2}} } \right)\left( {x + \sqrt {1 + {x^2}} } \right)}}{{\left( {x + \sqrt {1 + {x^2}} } \right)}}} \right\} \cr & = \log \left\{ {\frac{1}{{\left( {x + \sqrt {1 + {x^2}} } \right)}}} \right\} \cr & = - f\left( x \right) \cr & {\text{Hence, }}f\left( x \right){\text{ is odd}}{\text{.}} \cr & \left( {\bf{D}} \right)\,f\left( { - x} \right) = \frac{{{e^{ - x}} + {e^{ - \left( { - x} \right)}}}}{2} \cr & = \frac{{{e^{ - x}} + {e^x}}}{2} \cr & = f\left( x \right) \cr & {\text{Hence, }}f\left( x \right){\text{ is even}}{\text{.}} \cr} $$

$$\eqalign{ & 3f\left( x \right) - f\left( {\frac{1}{x}} \right) = \log \,{x^4};\,\,x \equiv \frac{1}{x} \cr & 3 + \left( {\frac{1}{x}} \right) - f\left( x \right) = \log {\left( {\frac{1}{x}} \right)^4} \cr & {\text{After solving we get }}f\left( x \right) = \log \,x \cr & f\left( {{e^{ - x}}} \right) = {\log _e}{e^{ - x}} = - x \cr} $$

$$\eqalign{ & {\text{We have,}}\,\,{2^x} + {2^y} = 2 \cr & \Rightarrow {2^y} = 2 - {2^x} \cr & \Rightarrow y = {\log _2}\left( {2 - {2^x}} \right) \cr & \Rightarrow 2 - {2^x} > 0\,\,\,\,\,\left( {\therefore y{\text{ is real}}} \right) \cr & \Rightarrow {2^x} < 2 \cr & \Rightarrow {2^{x - 1}} < 1 \cr & \Rightarrow x - 1 < 0 \cr & \Rightarrow x < 1 \cr & \Rightarrow - \infty < x < 1 \cr} $$

$$\eqalign{ & {\text{Putting }}y = 1, \cr & f\left( {x + 1} \right) = f\left( x \right) + f\left( 1 \right) - x - 1 = f\left( x \right) - x \cr & \therefore f\left( {n + 1} \right) = f\left( n \right) - n < f\left( n \right) \cr & {\text{So, }}f\left( n \right) < f\left( {n - 1} \right) < f\left( {n - 2} \right) < ...... < f\left( 1 \right) = 1 \cr & \therefore f\left( n \right) = n{\text{ holds for }}n = 1\,\,{\text{only}} \cr} .$$

$$\eqalign{ & f\left( x \right) = \frac{{\alpha x}}{{x + 1}},x \ne - 1 \cr & f\left( {f\left( x \right)} \right) = x \Rightarrow \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\frac{{\alpha x}}{{x + 1}} + 1}} = x \Rightarrow \frac{{{\alpha ^2}x}}{{\left( {\alpha + 1} \right)x + 1}} = x \cr & \Rightarrow \left( {\alpha + 1} \right){x^2} + \left( {1 - {\alpha ^2}} \right)x = 0\,......\left( 1 \right) \cr & \Rightarrow \alpha + 1 = 0{\text{ and }}1 - {\alpha ^2} = 0 \Rightarrow \alpha = - 1 \cr} $$

$$\eqalign{ & {\text{Given}}\,f\left( x \right) = \frac{x}{{{{\left( {1 + {x^n}} \right)}^{\frac{1}{n}}}}}\,{\text{for}}\,n \geqslant 2 \cr & \therefore fof(x) = f\left[ {f\left( x \right)} \right] = f\left[ {\frac{x}{{{{\left( {1 + {x^n}} \right)}^{\frac{1}{n}}}}}} \right] \cr & = \frac{{\frac{x}{{{{\left( {1 + {x^n}} \right)}^{\frac{1}{n}}}}}}}{{\frac{x}{{{{\left[ {1 + \frac{{{x^n}}}{{1 + {x^n}}}} \right]}^{\frac{1}{n}}}}}}} = \frac{x}{{{{\left( {1 + 2{x^n}} \right)}^{\frac{1}{n}}}}} \cr & {\text{Further}}\,fofo\left( x \right) = \frac{x}{{{{\left( {1 + 3{x^n}} \right)}^{\frac{1}{n}}}}} \cr & {\text{Proceeding}}\,{\text{in}}\,{\text{the}}\,{\text{similar}}\,{\text{manner,we}}\,{\text{get}} \cr & g\left( x \right) = fofof.....of\left( x \right) = \frac{x}{{{{\left( {1 + n{x^n}} \right)}^{\frac{1}{n}}}}}\,\left( {f\,{\text{occurs}}\,n\,{\text{times}}} \right) \cr & {\text{Now}},\,\int {{x^{n - 2}}} g\left( x \right)dx = \int {\frac{{{x^{n - 1}}}}{{{{\left( {1 + n{x^n}} \right)}^{\frac{1}{n}}}}}} dx \cr & {\text{Let}}\,1 + n{x^n} = t \Rightarrow {n^2}{x^{n - 1}}dx = dt \cr & \therefore {\text{Integral becomes }} = \frac{1}{{{n^2}}}\int {{t^{ - \frac{1}{n}}}} dt = \frac{1}{{{n^2}}} \cdot \frac{{{t^{ - \frac{1}{n} + 1}}}}{{ - \frac{1}{n} + 1}} + K \cr & = \frac{1}{n} \cdot \frac{{{t^{1 - \frac{1}{n}}}}}{{n - 1}} + K = \frac{{{{\left( {1 + n{x^n}} \right)}^{1 - \frac{1}{n}}}}}{{n\left( {n - 1} \right)}} + K \cr} $$

$${\text{Clearly, }}1 - {x^2} \geqslant 0,\,1 - \sqrt {1 - {x^2}} \geqslant 0,\,1 - \sqrt {1 - \sqrt {1 - {x^2}} } \geqslant 0$$
$$1 - {x^2} \geqslant 0 \Rightarrow {x^2} \leqslant 1 \Rightarrow - 1 \leqslant x \leqslant 1.$$        For these values, the other two hold.

Given $$f\left( x \right) = \sqrt {{x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1} $$         for $$f\left( x \right)$$  to be defined, $${x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1 \geqslant 0$$
$$\eqalign{ & {\bf{Case}}\,{\bf{1}}\,{\bf{:}}\,\,x \geqslant 1 \cr & {x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1 \cr & = \left( {{x^{14}} - {x^{11}}} \right) + \left( {{x^6} - {x^3}} \right) + \left( {{x^2} + 1} \right) > 0 \cr & {\bf{Case}}\,{\bf{2}}\,{\bf{:}}\,\,0 \leqslant x \leqslant 1 \cr & {x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1 \cr & = {x^{14}} - \left\{ {\left( {{x^{11}} - {x^6}} \right) + \left( {{x^3} - {x^2}} \right)} \right\} + 1 > 0 \cr & \left\{ {\because \,{x^{11}} - {x^6} \leqslant 0,\,\,{x^3} - {x^2} \leqslant 0} \right\} \cr & {\bf{Case}}\,{\bf{3}}\,{\bf{:}}\,\,x < 0 \cr & {x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1 > 0 \cr & \left( {\because \,{x^{11}} < 0,\,{x^3} < 0,\,{x^{14}},\,{x^6},\,{x^2} < 0} \right) \cr & {\text{Thus for all area, }}x,\,{x^{14}} - {x^{11}} + {x^6} - {x^3} + {x^2} + 1 \geqslant 0 \cr & {\text{Hence, the domain of }}f\left( x \right) = R = \left( { - \infty ,\,\infty } \right) \cr} $$

\[\begin{array}{l} {\rm{We \,\,have, }}\,\,f\left( x \right) = \left\{ \begin{array}{l} \,\, - 1,\,\,\,\, - 2 \le x \le 0\\ x - 1,\,\,\,\,\,\,0 \le x \le 2 \end{array} \right.\\ f\left( {\left| x \right|} \right) = \left\{ \begin{array}{l} \,\, - 1,\,\,\,\,\,\, - 2 \le \left| x \right| \le 0\\ \left| x \right| - 1,\,\,\,\,0 \le \left| x \right| \le 2 \end{array} \right.\\ \Rightarrow f\left( {\left| x \right|} \right) = \left| x \right| - 1,\,0 \le \left| x \right| \le 2\,\left( {{\rm{as }} - 2 \le \left| x \right| \le 2{\rm{ \,\,is\,\,not\,\, possible}}} \right)\\ \Rightarrow f\left( {\left| x \right|} \right) = \left\{ \begin{array}{l} - x - 1,\,\,\,\, - 2 \le x \le 0\\ \,\,\,\,\,x - 1,\,\,\,\,\,\,\,\,0 < x \le 2 \end{array} \right. \end{array}\]

$$\eqalign{ & {\text{Let }}\,f\left( x \right) = y = {x^2} + 2x + 2 = {\left( {x + 1} \right)^2} + 1 \cr & y - 1 = {\left( {x + 1} \right)^2} = x + 1 = \sqrt {y - 1} \cr & x = \sqrt {y - 1} - 1 \cr & {\text{Since, }}y - 1 \geqslant 0 \cr & \therefore \,y \geqslant 1 \cr & \therefore \,{\text{ range is }}\left[ {1,\,\infty } \right) \cr} $$