$\quad = (4 - 2{\rm{x}} - \sin {\rm{y}})\overrightarrow {\rm{b}} + \left( {{{\rm{x}}^2} - 1} \right)\vec c$
$ \Rightarrow \overrightarrow a \cdot \overrightarrow {\rm{c}} + \overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = 4 - 2{\rm{x}} - \sin {\rm{y}},\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = 1 - {{\rm{x}}^2}$
Also, $\quad(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}) \overline{\mathrm{a}}=\overrightarrow{\mathrm{c}}$
$\Rightarrow \quad (\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}} $
$|\overrightarrow{\mathrm{c}}|^{2} \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|^{2} $
$ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=1$
Now ${1 + \vec a \cdot \vec b = 4 - 2x - \sin y}$
${ \Rightarrow \quad {x^2} - 2x + 1 = \sin y - 1 \le 0}$
${ \Rightarrow \quad x = 1,y = \pi /2}$
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$f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2$
$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$
આપેલ છે કે જ્યાં $P(x)$ એ બહુપદી છે કે જેથી $P^{\prime \prime}(x)$ એ હંમેશા અચળ થાય છે અને $P(3)=9$ છે જો વિધેય $f(x)$ એ $x=2$ આગળ સતત હોય તો $P(5)$ ની કિમંત મેળવો.