Function MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{For }}f\left( x \right) = g\left( x \right) \cr & \Rightarrow 2{x^2} - 1 = 1 - 3x \cr & \Rightarrow 2{x^2} + 3x - 2 = 0 \cr & \Rightarrow 2{x^2} + 4x - x - 2 = 0 \cr & \Rightarrow 2x\left( {x + 2} \right) - 1\left( {x + 2} \right) = 0 \cr & \Rightarrow \left( {x + 2} \right)\left( {2x - 1} \right) = 0 \cr & \Rightarrow x = - 2,\,\frac{1}{2} \cr} $$
$$\therefore $$  The domain for which the function $$f\left( x \right) = g\left( x \right){\text{ is }}\left\{ { - 2,\,\frac{1}{2}} \right\}$$

$$\eqalign{ & f\left\{ {f\left( x \right)} \right\} = \frac{1}{{1 - f\left( x \right)}} = \frac{1}{{1 - \frac{1}{{1 - x}}}} = \frac{{1 - x}}{{ - x}} = \frac{{x - 1}}{x} \cr & \therefore f\left\{ {f\left( {f\left( x \right)} \right)} \right\} = \frac{1}{{1 - f\left\{ {f\left( x \right)} \right\}}} = \frac{1}{{1 - \frac{{x - 1}}{x}}} = x \cr & \therefore {\text{ the graph has the equation }}y = x \cr} $$

For the function to be defined the argument of the logarithm must be greater than zero.
$$\eqalign{ & \therefore \,x - \left[ x \right] > 0 \cr & x - \left[ x \right] = \left\{ x \right\}{\text{ where}}\left\{ \cdot \right\}{\text{ denotes the fractional part}} \cr & \therefore \,\left\{ x \right\} > 0 \cr} $$
This is always true unless $$x$$ is an integer.
Therefore, answer is option 'B'.

$$\eqalign{ & f\left( x \right) = {\log _{\frac{1}{2}}}\left( {{x^2} - 5x + 7} \right) > 0 \cr & \Rightarrow {x^2} - 5x + 7 > 0,\,{x^2} - 5x + 7 < 1,\,x\, \in \,R \cr & \Rightarrow {x^2} - 5x + 6 < 0 \cr & \Rightarrow x\, \in \left( {2,\,3} \right) \cr} $$

$$\eqalign{ & y = {2^{x\left( {x - 1} \right)}} \cr & \Rightarrow x\left( {x - 1} \right) = {\log _2}y \cr & \Rightarrow {x^2} - x - {\log _2}y = 0 \cr & \therefore \,\,x = \frac{{1 \pm \sqrt {1 + 4\,{{\log }_2}y} }}{2}\, \cr & {\text{But}}\,\,x \geqslant 1.\,\,\,\,\,\,\,\,\therefore \,\,x = \frac{{1 + \sqrt {1 + 4\,{{\log }_2}y} }}{2}\, \cr & \therefore \,\,{f^{ - 1}}\left( x \right) = \frac{1}{2}\left\{ {1 + \sqrt {1 + 4\,{{\log }_2}y} } \right\} \cr} $$

$$f\left( x \right) = {x^2}$$   is many one as $$f\left( 1 \right) = f\left( { - 1} \right) = 1$$
Also $$f$$ is into as $$- ve$$  real number have no pre-image.
$$\therefore F$$  is neither injective nor surjective.

$$f(x) = \left| {px - q} \right| + r\left| x \right|$$
\[ = \left\{ {\begin{array}{*{20}{c}} { - px + q - rx,}\\ { - px + q + rx,}\\ {px - q + rx,} \end{array}\,\,\begin{array}{*{20}{c}} {x \le 0}\\ {0 < x \le \frac{q}{p}}\\ {\frac{q}{p} < x} \end{array}} \right.\]
$$\eqalign{ & {\text{For}}\,r = p,f'\left( x \right) < 0\,{\text{if}}\,x < 0 \cr & = 0\,{\text{if}}\,0 < x < \frac{q}{p} \cr & > 0\,{\text{if}}\,x > \frac{q}{p} \cr} $$

From graph (i) infinite many points for min value of $$f\left( x \right)$$
$$\eqalign{ & {\text{for}}\,r < p,f'\left( x \right) < 0\,{\text{if}}\,x \leqslant 0 \cr & < 0\,{\text{if}}\,{\text{0 < }}\,x \leqslant \frac{q}{p} \cr & > 0\,{\text{if}}\,x < \frac{q}{p} \cr} $$

From graph (ii) only pt. of min of $$f\left( x \right)$$ at $$x = \frac{q}{p}$$
$$\eqalign{ & {\text{For}}\,r > p,f'\left( x \right) < 0\,{\text{if}}\,x \leqslant 0 \cr & > 0\,{\text{if}}\,0 < x \cr} $$

From graph (iii) only one pt. of min of $$f\left( x \right)$$  at $$x = 0$$

\[f\left( x \right) = x - \left[ x \right] = \left\{ {\begin{array}{*{20}{c}} { \ldots .}\\ {\begin{array}{*{20}{c}} {x - 1,}\\ {\begin{array}{*{20}{c}} {x - 2,}\\ {x - 3,} \end{array}} \end{array}}\\ { \ldots .} \end{array}\,\,\begin{array}{*{20}{c}} {1 \le x < 2}\\ {2 \le x < 3}\\ {3 \le x < 4} \end{array}} \right.\]
graph of function is

Clearly it is a periodic function with period 1.
$$\therefore \left( a \right)$$  is the correct alternative.

$$g\left( x \right) = 1 + x - \left[ x \right];$$
\[f\left( x \right)\left\{ {\begin{array}{*{20}{c}} { - 1,}\\ {0,}\\ {1,} \end{array}} \right.\,\begin{array}{*{20}{c}} {x < 0}\\ {x = 0}\\ {x > 0} \end{array}\]
For integral values of $$x;g\left( x \right) = 1$$
For $$x < 0;$$  (but not integral value) $$x - \left[ x \right] > 0 \Rightarrow g\left( x \right) > 1$$
For $$x > 0;$$  (but not integral value) $$x - \left[ x \right] > 0 \Rightarrow g\left( x \right) > 1$$
$$\therefore g\left( x \right) \geqslant 1,\forall x\,\therefore f\left( {g\left( x \right)} \right) = 1\forall x$$

$$\eqalign{ & \because f\left( {x + y} \right) = f\left( x \right).f\left( y \right) \cr & \Rightarrow {\text{ Let }}f\left( x \right) = {t^x} \cr & \because f\left( a \right) = 2\therefore t = 2 \cr & \Rightarrow f\left( x \right) = {2^x} \cr & {\text{Since, }}\sum\limits_{k = 1}^{10} f \left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right) \cr & {\text{Then }},\sum\limits_{k = 1}^{10} {{2^{a + k}}} = 16\left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2^a}\sum\limits_{k = 1}^{10} {{2^k}} = 16\left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2^a} \times \frac{{\left( {\left( {{2^{10}}} \right) - 1} \right) \times 2}}{{(2 - 1)}} = 16 \times \left( {{2^{10}} - 1} \right) \cr & \Rightarrow {2.2^a} = 16 \Rightarrow {\text{a}} = 3 \cr} $$