If _ x → 0 e ^ ax - ( bx ) - cx e ^ - cx 2 1 - ( 2x ) = 17 , then 5...

Given that

$⇒ \lim_{x \to 0}\dfrac{{e}^{ax} - \cos ( bx ) - \dfrac{cx}{2}{e}^{- cx}}{1 - \cos 2x} = 17$

We know that ${e}^{ax} = 1 + ax + \dfrac{( ax{)}^{2}}{2 !} + \ldots$ and $\cos \left(bx\right) = 1 - \dfrac{( bx{)}^{2}}{2 !} + \ldots$

$⇒ \lim_{x \to 0}\dfrac{\left(1 + ax + \dfrac{{\left(ax\right)}^{2}}{2 !} + \ldots\right) - \left(1 - \dfrac{{\left(bx\right)}^{2}}{2 !} + \ldots\right) - \dfrac{cx}{2}\left(1 - \left(cx\right) + \dfrac{{\left(cx\right)}^{2}}{2 !} - \ldots\right)}{\left(\dfrac{1 - \cos 2x}{{\left(2x\right)}^{2}}\right) \times 4{x}^{2}} = 17$

$⇒ \lim_{x \to 0}\dfrac{\left(a - \dfrac{c}{2}\right)x + \left(\dfrac{{a}^{2} + {b}^{2} + {c}^{2}}{2}\right){x}^{2} + \ldots}{\dfrac{1}{2} \times 4{x}^{2}} = 17$

Now for limit to exist, $a - \dfrac{c}{2} = 0$$⇒ c = 2a$

$⇒ \lim_{x \to 0}\dfrac{\left(\dfrac{{a}^{2} + {b}^{2} + {c}^{2}}{2}\right){x}^{2} + \ldots}{\dfrac{1}{2} \times 4{x}^{2}} = 17$

$⇒ \dfrac{{a}^{2} + {b}^{2} + {c}^{2}}{4} = 17$

$⇒ {a}^{2} + {b}^{2} + 4{a}^{2} = 68$

$⇒ 5{a}^{2} + {b}^{2} = 68$

Hence this is the correct option.