The given eq. is $$\sin {\left( e \right)^x} = {5^x} + {5^{ - x}}$$
We know $${5^x}$$ and $${5^{ - x}}$$ both are $$+ve$$  real numbers using
$$\eqalign{ & {\text{AM}} \geqslant {\text{GM}} \cr & \therefore \,\,{{\text{5}}^x} + \frac{1}{{{5^x}}} \geqslant 2 \cr & \Rightarrow \,\,{5^x} + {5^{ - x}} \geqslant 2 \cr & \therefore \,\,{\text{R}}{\text{.H}}{\text{.S of given eq}}{\text{. }} \geqslant {\text{2}} \cr & {\text{While }}\sin {e^x} \in \left[ { - 1,1} \right] \cr & {\text{i}}{\text{.e}}{\text{. L}}{\text{.H}}{\text{.S }} \in \left[ { - 1,1} \right] \cr & \therefore \,\,{\text{The equation is not possible for any real value of }}\,x. \cr & {\text{Hence}}\,\,{\text{(A)}}\,\,{\text{is the correct answer}}{\text{.}} \cr} $$

$$\eqalign{ & {5^x} + {\left( {2\sqrt 3 } \right)^{2x}} > {13^2} \cr & \Rightarrow \,\,{5^x} + {12^x} > {13^2}. \cr & {\text{We know }}{5^2} + {12^2} = {13^2}.\,{\text{So, for }}x > 2,{5^x} + {12^x} > {5^2} + {12^2} = {13^2}. \cr} $$

The equation is
$$ab{c^2}{x^2} + \left( {3{a^2}c + {b^2}c} \right)x - 6{a^2} - ab + 2{b^2} = 0$$
Discriminant
$$\eqalign{ & D = {\left( {3{a^2} + {b^2}} \right)^2}{c^2} - 4ab{c^2}\left( { - 6{a^2} - ab + 2{b^2}} \right) \cr & = 9{a^4}{c^2} + {b^4}{c^2} + 6{a^2}{b^2}{c^2} + 24{a^3}b{c^2} + 4{a^2}{b^2}{c^2} - 8a{b^3}{c^2} \cr & = 9{a^4}{c^2} + 16{a^2}{b^2}{c^2} + {b^4}{c^2} + 24{a^3}b{c^2} - 8a{b^3}{c^2} - 6{a^2}{b^2}{c^2} \cr & = {\left( {3{a^2}c + 4abc - {b^2}c} \right)^2} \cr} $$
Since, the discriminant is a prefect square, therefore the roots are rational provided $$a, b, c$$  are rational.

4 is the root of $${x^2} + px + 12 = 0$$
$$\eqalign{ & \Rightarrow \,\,16 + 4p + 12 = 0 \cr & \Rightarrow \,\,p = - 7 \cr} $$
Now, the equation $${x^2} + px + q = 0$$    has equal roots.
$$\eqalign{ & \therefore \,\,{p^2} - 4q = 0 \cr & \Rightarrow \,\,q = \frac{{{p^2}}}{4} = \frac{{49}}{4} \cr} $$

If $$k\left( { > 0} \right)$$  be the common difference then the equation is
$$\eqalign{ & 3{x^2} - \left( {6a + 10k} \right)x + a\left( {a + 2k} \right) + 2\left( {a + k} \right)\left( {a + 3k} \right) = 0 \cr & \therefore \,\,D = {\left( {6a + 10k} \right)^2} - 4 \cdot 3 \cdot \left\{ {{a^2} + 2ak + 2{a^2} + 8ak + 6{k^2}} \right\} = 28{k^2} > 0. \cr} $$

Solve the inequations $${x^2} - 3x + 2 \leqslant 0$$    and $$2{x^2} - 3x - 5 \geqslant 0.$$    If $${S_1},{S_2}$$  are the respective solution sets then $${S_1} \cap {S_2}$$  is the required solution set.

Clearly, $${e^x} > x$$  for all $$x.$$

If roots are $$\alpha - k,\alpha ,\alpha + k$$    then
$$\eqalign{ & \alpha - k + \alpha + \alpha + k = 12 \cr & \Rightarrow \,\,\alpha = 4, \cr & \left( {\alpha - k} \right)\alpha \left( {\alpha + k} \right) = 28 \cr & \Rightarrow \,\,\left( {{4^2} - {k^2}} \right)4 = 28 \cr & \Rightarrow \,\,{k^2} = 9. \cr} $$

From the given system, we have
$$\eqalign{ & \frac{{k + 1}}{k} = \frac{8}{{k + 3}} \ne \frac{{4k}}{{3k - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \,\,{\text{System has no solution}}} \right) \cr & \Rightarrow \,\,{k^2} + 4k + 3 = 8k \cr & \Rightarrow \,\,k = 1,3 \cr & {\text{If }}k = 1\,\,\,\,{\text{then }}\frac{8}{{1 + 3}} \ne \frac{{4.1}}{2}\,\,{\text{which is false}} \cr & {\text{And if }}k = 3 \cr} $$
$${\text{then }}\frac{8}{6} \ne \frac{{4.3}}{{9 - 1}}$$   which is true, therefore $$k = 3$$
Hence for only one value of $$k$$ System has no solution.

Let $$\sin \alpha $$  and $$\cos \alpha $$  be the roots of $$a{x^2} + bx + c = 0$$
Now, $$\sin \alpha + \cos \alpha = \frac{{ - b}}{a}\,$$   and $$\sin \alpha \,\,\cos \alpha = \frac{{c}}{a}$$
Consider $$\sin \alpha + \cos \alpha = \frac{{ - b}}{a}$$
Squaring both side, $${\left( {\sin \alpha + \cos \alpha } \right)^2} = \frac{{{b^2}}}{{{a^2}}}$$
$$\eqalign{ & \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha + 2\sin \alpha \cos \alpha = \frac{{{b^2}}}{{{a^2}}} \cr & \Rightarrow 1 + \frac{{2c}}{a} = \frac{{{b^2}}}{{{a^2}}} \cr & \Rightarrow \frac{{a + 2c}}{a} = \frac{{{b^2}}}{{{a^2}}} \cr & \Rightarrow a + 2c = \frac{{{b^2}}}{a} \cr & \Rightarrow {a^2} + 2ac = {b^2} \cr & \Rightarrow {b^2} - {a^2} = 2ac \cr} $$