Answer
$L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$
Work Step by Step
We have $f(x) = \cosh x$
and $f'(x)= \sinh x ; \\f''(x)= \cosh x$
$\implies f(0)=1; f'(0)=1; f''(0)=1$
Therefore, $L(x)=f(0)+xf'(0)=1$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=1+\dfrac{x^2}{2!}$
So, $L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$
