Answer: $\mathbf C$
Solution:
Given: $\mathrm {2f(x) + 3f(-x) = 15 - 4x }\tag{1}$
Substitute $\color {red} {\mathrm x}$ with $\color {red} {\mathrm {-x}}$, in the above question, we get:
$\mathrm {2f(-x) + 3 f(x) = 15 +4 x} \tag {2}$
On multiplying, Equation $(1)$ $\times$ $3$ and Equation $(2)$ $\times$ $2$, we get:
$\mathrm {6f(x) + 9f(-x) = 45-12x} \tag {3}$
and,
$\mathrm {4f(-x) + 6f(x) = 30 + 8x} \Rightarrow \mathrm {6f(x) + 4f(-x) = 30+8x} \tag{4}$
Subtracting equation $(3)$ and $(4)$,we get:
$\mathrm {5f(-x) = 15-20x} $
Now, substitute, $\mathrm x = -2$, we get:
$\mathrm {5f(2) = [15 - 20\times (-2)]}$
$ \Rightarrow \mathrm{5f(2)} = 55$
$\Rightarrow \mathrm{f(2) = \bf{11}}$
$\therefore \mathbf C$ is the correct option.
