$$\eqalign{ & {\text{Given, }}f\left( x \right) = \ln |x| + b{x^2} + ax \cr & \therefore f'\left( x \right) = \frac{1}{x} + 2bx + a \cr & {\text{At }}x = - 1,f'\left( x \right) = - 1 - 2{\text{b}} + a = 0 \cr & \Rightarrow a - 2b = 1\,......\left( {\text{i}} \right) \cr & {\text{At }}x = 2,\,f'\left( x \right) = \frac{1}{2} + 4b + a = 0 \cr & \Rightarrow a + 4b = - \frac{1}{2}\,......\left( {{\text{ii}}} \right) \cr & {\text{On solving }}\left( {\text{i}} \right){\text{ and}}\,\left( {{\text{ii}}} \right){\text{ we get }}a = \frac{1}{2},b = - \frac{1}{4} \cr & {\text{Thus, }}f'\left( x \right) = \frac{1}{x} - \frac{x}{2} + \frac{1}{2} = \frac{{2 - {x^2} + x}}{{2x}} \cr & = \frac{{ - {x^2} + x + 2}}{{2x}} = \frac{{ - \left( {{x^2} - x - 2} \right)}}{{2x}} = \frac{{ - \left( {x + 1} \right)\left( {x - 2} \right)}}{{2x}} \cr} $$

So maxima at $$x = - 1,2$$
Hence both the statements are true and statement 2 is a correct explanation for 1.

$$\eqalign{ & f\left( 0 \right) = \sin \,0 = 0,\,f\left( {{0^ + }} \right) \to {0^ + } \cr & f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \sin \left( {{x^2} - 3x} \right) = \mathop {\lim }\limits_{h \to 0} \sin \left( {{h^2} + 3h} \right) \to {0^ + } \cr & {\text{Thus, }}f\left( {{0^ + }} \right) > f\left( 0 \right){\text{ and }}f\left( {{0^ - }} \right) > f\left( 0 \right) \cr} $$
Hence, $$x = 0$$  is a point of minima.

$$\eqalign{ & \frac{{dy}}{{dx}} = 3{x^2} - 2ax + 1 \cr & {\text{From the question,}}\,\,\,\,\frac{{dy}}{{dx}} \geqslant 0 \cr & \Rightarrow 3{x^2} - 2ax + 1 \geqslant 0\,\,{\text{for all }}x \cr & \therefore D \leqslant 0\,\,\,\,\,{\text{or,}}\,\,4{a^2} - 12 \leqslant 0\,\,\,\,\,\,\,\, \Rightarrow - \sqrt 3 \leqslant a \leqslant \sqrt 3 \cr} $$

We have, $$a = \frac{{{d^2}x}}{{d{t^2}}} = - 9.8$$
The initial conditions are $$x\left( 0 \right) = 19.6{\text{ and }}v\left( 0 \right) = 0$$
$$\eqalign{ & {\text{So, }}v = \frac{{dx}}{{dt}} = - 9.8t + v\left( 0 \right) = - 9.8t \cr & \therefore \,x = - 4.9{t^2} + x\left( 0 \right) = - 4.9{t^2} + 19.6 \cr} $$
Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set $$x = 0$$  and solve for $$t;\,\,0 = - 4.9{t^2} + 19.6 \Rightarrow t = 2$$

$$\eqalign{ & {\text{Given, }}\,f\left( x \right) = {e^x}\sin \,x \cr & \Rightarrow f'\left( x \right) = {e^x}\cos \,x + {e^x}\sin \,x \cr & \Rightarrow {\text{ slope}} = {e^x}\left( {\cos \,x + \sin \,x} \right) \cr & {\text{Now, }}\frac{d}{{dx}}\left( {\cos \,x + \sin \,x} \right) = 0 \cr & \Rightarrow - \sin \,x + \cos \,x = 0 \cr & \Rightarrow \sin \,x = \cos \,x \cr & \Rightarrow \tan \,x = 1 \cr & \Rightarrow x = \frac{\pi }{4} \cr} $$

$$\eqalign{ & y = {\left( {\frac{1}{x}} \right)^{2{x^2}}} \Rightarrow \log \,y = 2{x^2}\log \frac{1}{x} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,\,\,\frac{1}{y}.\frac{{dy}}{{dx}} = 4x\log \frac{1}{x} + 2{x^2}.\left( { - \frac{1}{x}} \right) \cr & {\text{Then }}\frac{{dy}}{{dx}} = 0{\text{ if }} - 4x\log \,x - 2x = 0 \cr & {\text{or,}}\,\,\,2\log \,x + 1 = 0 \cr & \therefore \,\,x = {e^{ - \,\frac{1}{2}}} \cr & {\text{Verify that }}\frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ at }}x = {e^{ - \,\frac{1}{2}}} \cr} $$

$$\eqalign{ & {\text{Let base }} = b \cr & {\text{Altitude }}\left( {{\text{or perpendicular}}} \right) = \sqrt {{h^2} - {b^2}} \cr} $$

$$\eqalign{ & {\text{Area, }}A = \frac{1}{2} \times {\text{base}} \times {\text{altitude}} \cr & {\text{ = }}\frac{1}{2} \times b \times \sqrt {{h^2} - {b^2}} \cr & \Rightarrow \frac{{dA}}{{db}}{\text{ = }}\frac{1}{2}\left[ {\sqrt {{h^2} - {b^2}} + b.\frac{{ - 2b}}{{2\sqrt {{h^2} - {b^2}} }}} \right] \cr & \Rightarrow \frac{{dA}}{{db}} = \frac{1}{2}\left[ {\frac{{{h^2} - 2{b^2}}}{{\sqrt {{h^2} - {b^2}} }}} \right] \cr & {\text{Put }}\frac{{dA}}{{db}} = 0,b = \frac{h}{{\sqrt 2 }} \cr & {\text{Maximum area}} = \frac{1}{2} \times \frac{h}{{\sqrt 2 }} \times \sqrt {{h^2} - \frac{{{h^2}}}{2}} = \frac{{{h^2}}}{4} \cr} $$

$$\eqalign{ & {\text{Given rule is :}} \cr & {\text{Distance, }}s = 2 - 3t + 4{t^3} \cr & \Rightarrow {\text{Velocity}} = \frac{{ds}}{{dt}} = - 3 + 12{t^2} \cr & \Rightarrow {\text{Acceleration}} = \frac{{{d^2}s}}{{d{t^2}}} = 16t \cr & {\text{Since, velocity is zero}} \cr & \therefore \,\frac{{ds}}{{dt}} = 0\,{\text{;}} \cr & \Rightarrow 0 = - 3 + 12{t^2} \cr & \Rightarrow t = \frac{{\sqrt 3 }}{{12}} = \frac{1}{2} \cr & \Rightarrow \frac{{{d^2}s}}{{d{t^2}}} = 16t \cr & {\text{Acceleration }}\left( {{\text{when velocity is zero}}} \right) = 16 \times \frac{1}{2} = 8{\text{ units}} \cr} $$

$$\eqalign{ & {y^2} = 18x \Rightarrow 2y\frac{{dy}}{{dx}} = 18 \Rightarrow \frac{{dy}}{{dx}} = \frac{9}{y} \cr & {\text{Given }}\frac{{dy}}{{dx}} = 2 \Rightarrow \frac{9}{y} = 2 \Rightarrow y = \frac{9}{2} \cr & {\text{Putting in }}{y^2} = 18x \Rightarrow x = \frac{9}{8} \cr & \therefore {\text{Required point is }}\left( {\frac{9}{8},\frac{9}{2}} \right) \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{f}}\left( x \right) = \,{x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0 \cr & \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\forall \,x \in R \cr & \Rightarrow f\,{\text{is}}\,{\text{an}}\,{\text{increasing}}\,{\text{function}}\,{\text{on}}\,R \cr & {\text{Also}}\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \infty \,{\text{and}}\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = - \infty \cr & \Rightarrow \,{\text{The}}\,{\text{curve}}\,y = f\left( x \right)\,{\text{crosses}}\,x\,{\text{ - axis}}\,{\text{only}}\,{\text{once}}. \cr & \therefore f\left( x \right) = 0\,{\text{has}}\,{\text{exactly}}\,{\text{one}}\,{\text{real}}\,{\text{root}}. \cr} $$