Function MCQ Questions & Answers in Calculus | Maths

$$ \Rightarrow f\left( {x - 1} \right) + f\left( {x + 1} \right) = 2f\left( x \right)....$$        (it is given that $${f\left( 0 \right) = 0}$$  )
$$\eqalign{ & {\text{Substituting }}x = 1,{\text{ we get}} \cr & \Rightarrow f\left( 0 \right) + f\left( 2 \right) = 2f\left( 1 \right) \cr & \Rightarrow f\left( 2 \right) = 2f\left( 1 \right)....\left( {f\left( 0 \right) = 0} \right) \cr & {\text{Substituting }}x = 2,{\text{ we get}} \cr & \Rightarrow f\left( 1 \right) + f\left( 3 \right) = 2f\left( 2 \right) \cr & \Rightarrow f\left( 1 \right) + f\left( 3 \right) = 4f\left( 1 \right)....\left( {f\left( 2 \right) = 2f\left( 1 \right)} \right) \cr & \Rightarrow f\left( 3 \right) = 3f\left( 1 \right) \cr & {\text{:}} \cr & {\text{:}} \cr & {\text{Therefore }}f\left( n \right) = n\left( {f\left( 1 \right)} \right) \cr} $$

$$\eqalign{ & f\left( x \right){\text{ is onto}}\,\,\therefore S = {\text{range}}\,{\text{of }}f\left( x \right) \cr & {\text{Now }}f\left( x \right) = \sin x - \sqrt 3 \cos x + 1 = 2\sin \left( {x - \frac{\pi }{3}} \right) + 1 \cr & \because - 1 \leqslant \sin \left( {x - \frac{\pi }{3}} \right) \leqslant 1 \cr & - 1 \leqslant 2\sin \left( {x - \frac{\pi }{3}} \right) + 1 \leqslant 3 \cr & \therefore f\left( x \right) \in \left[ { - 1,3} \right] = S \cr} $$

$$\eqalign{ & D\left( {f + g} \right) = D\left( f \right) \cap D\left( g \right) \cr & = \left( { - \infty ,\,0} \right] \cap \left[ {0,\,\infty } \right) \cr & = \left\{ 0 \right\} \cr} $$

\[\begin{array}{l} \left| {x - 2} \right| = \left\{ \begin{array}{l} x - 2,\,\,\,x \ge 2\\ 2 - x,\,\,\,\,x < 2 \end{array} \right.\\ \Rightarrow \left| {x - 2} \right| - \left( {x - 2} \right) = \left\{ \begin{array}{l} \,\,\,\,0,\,\,\,x \ge 2\\ 4 - 2x,\,\,\,x < 2 \end{array} \right. \end{array}\]
$$ \Rightarrow $$  given expression is defined for $$\left( { - \infty ,\,2} \right)$$

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

$$\eqalign{ & f\left( x \right) = x + \frac{1}{x} = y \Rightarrow {x^2} - yx + 1 = 0 \cr & \Rightarrow x = \frac{{y \pm \sqrt {{y^2} - 4} }}{2} \cr & \therefore x = \frac{{y + \sqrt {{y^2} - 4} }}{2}\,\,\left( {\because x \geqslant 1{\text{ and }}y \geqslant 2} \right) \cr & \therefore {f^{ - 1}}\left( x \right) = \frac{{x + \sqrt {{x^2} - 4} }}{2} \cr} $$
NOTE THIS STEP:

$$\eqalign{ & f\left( x \right) = 3x + \left| x \right| \cr & \therefore \,f\left( {2x} \right) - f\left( { - x} \right) - 6x \cr & = 6x + \left| {2x} \right| - 3( - x) - \left| { - x} \right| - 6x \cr & = 3x + 2\left| x \right| - \left| x \right| \cr & = 3x + \left| x \right| \cr & = f\left( x \right) \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right){\text{ = }}\lambda \left( x \right){\text{ where }}\lambda {\text{ is a constant}}{\text{.}} \cr & {\text{Hence, }} \cr & f\left( {x + y} \right) \cr & = \lambda \left( {x + y} \right) \cr & = \lambda \left( x \right) + \lambda \left( y \right) \cr & = f\left( x \right) + f\left( y \right) \cr & {\text{Now it has been given that }}f\left( 1 \right) = k \cr & {\text{Therefore }}\lambda = k \cr & {\text{Hence,}} \cr & f\left( x \right) = k\left( x \right) \cr & {\text{Therefore, }}f\left( n \right) = k\left( n \right) \cr} $$

$$\eqalign{ & y = x\left( {2 - x} \right) \cr & \Rightarrow {x^2} - 2x + y = 0 \cr & \Rightarrow x = \frac{{2 \pm 2\sqrt {1 - y} }}{2} = 1 \pm \sqrt {1 - y} \cr & {\text{But }}x \leqslant 1.\,\,{\text{So, }}x = 1 - \sqrt {1 - y} \cr & {\text{Therefore,}}\,\,{f^{ - 1}}\left( x \right) = 1 - \sqrt {1 - x} \cr} $$

$$\eqalign{ & f\left( x \right) = \sqrt {5 - 4x - {x^2}} + {x^2}\log \left( {x + 4} \right) \cr & \Rightarrow 5 - 4x - {x^2} \geqslant 0,\,x + 4 > 0 \cr & \Rightarrow \left( {x + 5} \right)\left( {x - 1} \right) \leqslant 0,\,x > - 4 \cr & \Rightarrow - 5 \leqslant x \leqslant 1,\,x > - 4 \cr & \Rightarrow - 4 < x \leqslant 1 \cr} $$