$$\eqalign{ & {\left\{ {\left( {x - 1} \right)} \right\}^2} = {\left\{ {\left( x \right) - 1} \right\}^2} = {\left( x \right)^2} - 2\left( x \right) + 1 \cr & {\left\{ {\left[ {x + 1} \right]} \right\}^2} = {\left\{ {\left[ x \right] + 1} \right\}^2} = {\left[ x \right]^2} + 2\left[ x \right] + 1 \cr & \therefore \,\,{\text{equation gives }}\left[ x \right] - \left( x \right) + 1 = 0. \cr & {\text{If }}x = n \in Z,n - n + 1 = 0\left( {{\text{absurd}}} \right). \cr & {\text{If }}x = n + k,n \in Z,0 < k < 1\,\,{\text{then }}n - \left\{ {n + 1} \right\} + 1 = 0. \cr} $$
This is true for all $$n$$ and $$k.$$

$$x - y$$  is a factor
⇒ putting $$y = x,$$  expression = 0 which gives $$1 - 3 + \lambda + \mu = 0.$$
$$y - 2x$$  is a factor
⇒ putting $$y = 2x,$$  expression = 0 which gives $$1 - 6 + 4\lambda + 8\mu = 0.$$

Given that
$$\eqalign{ & \alpha {\text{ + }}\beta = - p\,\,{\text{and }}{\alpha ^3} + {\beta ^3} = q \cr & \Rightarrow \,\,{\left( {\alpha + \beta } \right)^3} - 3\alpha \beta \left( {\alpha + \beta } \right) = q \cr & \Rightarrow \,\, - {p^3} - 3\alpha \beta \left( { - p} \right) = q \cr & \Rightarrow \,\,\alpha \beta = \frac{{{p^3} + q}}{{3p}} \cr} $$
Now for required quadratic equation,
sum of roots $$ = \frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$$
$$\eqalign{ & = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr & = \frac{{{p^2} - 2\left( {\frac{{{p^3} + q}}{{3p}}} \right)}}{{\frac{{{p^3} + q}}{{3p}}}} \cr & = \,\frac{{3{p^3} - 2{p^3} - 2q}}{{{p^3} + q}} \cr & = \frac{{{p^3} - 2q}}{{{p^3} + q}} \cr} $$
and product of roots $$ = \frac{\alpha }{\beta }.\frac{\beta }{\alpha } = 1$$
∴ Required equation is $${x^2} - \left( {\frac{{{p^3} - 2q}}{{{p^3} + q}}} \right)x + 1 = 0$$
or $$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} - 2q} \right)x + \left( {{p^3} + q} \right) = 0$$

$$\eqalign{ & {x^2} - 2kx + {k^2} - 4 = 0 \cr & \Rightarrow {\left( {x - k} \right)^2} - {2^2} = 0 \cr & \Rightarrow \left( {x - k - 2} \right)\left( {x - k + 2} \right) = 0 \cr & \Rightarrow x = k + 2,k - 2. \cr & \Rightarrow k + 2 < 5\,\& \,k - 2 > - 3 \cr & \Rightarrow k < 3\,\& \,k > - 1 \cr & \Rightarrow - 1 < k < 3 \cr} $$

Let $$\alpha $$ be the common root of given equations, then
$$\eqalign{ & {\alpha ^2} + b \alpha - 1 = 0\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}\,{\alpha ^2} + \alpha + b = 0\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Subtracting (2) from (1), we get
$$\eqalign{ & \left( {b - 1} \right)\alpha - \left( {b + 1} \right) = 0 \cr & {\text{or }}\alpha = \frac{{b + 1}}{{b - 1}} \cr} $$
Substituting this value of $$\alpha $$ in equation (1), we get
$$\eqalign{ & {\left( {\frac{{b + 1}}{{b - 1}}} \right)^2} + b\left( {\frac{{b + 1}}{{b - 1}}} \right) - 1 = 0\,\,\,{\text{or}}\,\,{b^3} + 3b = 0 \cr & \Rightarrow \,\,b = 0,i\sqrt 3 , - i\sqrt 3 \cr} $$

Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
put $${e^{\sin x}} = t$$   in the given equation, we get
$$\eqalign{ & {t^2} - 4t - 1 = 0 \cr & \Rightarrow \,\,t = \frac{{4 \pm \sqrt {16 + 4} }}{2} \cr & = \frac{{4 \pm \sqrt {20} }}{2} \cr & = \frac{{4 \pm 2\sqrt 5 }}{2} \cr & = 2 \pm \sqrt 5 \cr & \Rightarrow \,\,{e^{\sin x}} = 2 \pm \sqrt 5 \left( {\because \,t = {e^{\sin x}}} \right) \cr & \Rightarrow \,\,{e^{\sin x}} = 2 - \sqrt 5 \,\,{\text{and }}{e^{\sin x}} = 2 + \sqrt 5 \cr & \Rightarrow \,\,{e^{\sin x}} = 2 - \sqrt 5 \, < 0\,{\text{and }}\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1 \cr & \,{\text{So rejected }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{So rejected}} \cr} $$
Hence given equation has no solution.
∴ The equation has no real roots.

$$y = x + \frac{1}{x}\,\,{\text{or}}\,\,\,\frac{{dy}}{{dx}} = 1 - \frac{1}{{{x^2}}}$$
For max. or min., $$1 - \frac{1}{{{x^2}}} = 0$$
$$\eqalign{ & \Rightarrow x = \pm 1 \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{2}{{{x^3}}} \cr & \Rightarrow \,\,{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 2}} = 2\left( { + ve\,\,{\text{minima}}} \right) \cr & \therefore \,\,x = 1 \cr} $$

$$\eqalign{ & 2\cos \left( {a + bx} \right) = x - \frac{{{x^2} + 5}}{2} = \frac{{ - {x^2} + 2x - 5}}{2} \cr & 2\cos \left( {a + bx} \right) = \frac{1}{2}\left[ { - 4 - \left( {{x^2} - 2x + 1} \right)} \right] \cr & {\text{or, }}2\cos \left( {a + bx} \right) = \frac{1}{2}\left\{ { - 4 - {{\left( {x - 1} \right)}^2}} \right\} \cr & 2\cos \left( {a + bx} \right) = - 2 - \frac{1}{2}{\left( {x - 1} \right)^2} \leqslant - 2 \cr} $$
$$ \Rightarrow \,\,\cos \left( {a + bx} \right) = - 1$$     and it takes place when $$a + bx = 2n\pi - \pi $$    and $$x = 1.$$
$$\therefore \,\,a + b = \pi ,3\pi ,.....;$$     but $$a + b$$  is less than $$10.$$

$$\eqalign{ & {\sin ^2}\phi = \cos \theta \cdot \sin \theta \cr & \therefore \,\,\cos 2\phi = 1 - \sin 2\theta = {\left( {\cos \theta - \sin \theta } \right)^2}. \cr & D = 4{\cot ^2}\phi - 4 = 4 \cdot \frac{{\cos 2\phi }}{{{{\sin }^2}\phi }} = 4 \cdot {\left( {\frac{{\cos \theta - \sin \theta }}{{\sin \phi }}} \right)^2} > 0\,\,{\text{because}}\,\cos \theta = \sin \theta \cr} $$
⇒ three numbers are equal which is a special case.

We have
$$\eqalign{ & {\left( {2x} \right)^{\ell n2}} = {\left( {3y} \right)^{\ell n3}} \cr & \Rightarrow \,\,\ell n2.\,\,\ell n2x = \ell n3.\,\,\ell n3y \cr & \Rightarrow \,\,\ell n2.\,\,\ell n2x = \ell n3.\left( {\ell n3 + \ell ny} \right)\,\,\,\,.....\left( 1 \right) \cr & {\text{Also given }}{3^{\ell nx}} = {2^{\ell ny}} \cr & \Rightarrow \,\,\ell nx.\,\,\ell n3 = \ell ny.\,\,\ell n2 \cr & \Rightarrow \,\,\ell ny = \frac{{\ell nx.\,\,\ell n3}}{{\ell n2}} \cr} $$
Substituting this value of $${\ell ny}$$  in equation (1), we get
$$\eqalign{ & \ell n2.\,\,\ell n2x = \ell n3\left[ {\ell n3 + \frac{{\ell nx.\,\,\ell n3}}{{\ell n2}}} \right] \cr & \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x = {\left( {\ell n3} \right)^2}\ell n2 + {\left( {\ell n3} \right)^2}\ell nx \cr & \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x = {\left( {\ell n3} \right)^2}\left( {\ell n2 + \ell nx} \right) \cr & \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x - {\left( {\ell n3} \right)^2}\ell n2x = 0 \cr & \Rightarrow \,\,\left[ {{{\left( {\ell n2} \right)}^2} - {{\left( {\ell n3} \right)}^2}} \right]\ell n2x = 0 \cr & \Rightarrow \,\,\ell n2x = 0 \cr & \Rightarrow \,\,2x = 1\,\,{\text{or }}x = \frac{1}{2} \cr} $$