Question

Let \(f:\mathbb{R}\to\mathbb{R}\) be the function defined by \(f(x)=3e^x\cos 2x\). If \(p(x)=ax^3+bx^2+cx+d\) is the third degree Taylor polynomial of \(f\) at \(x=0\), then \(|a|+|b|+|c|+|d|\) equals (in integer).

Hint:

For Taylor polynomial questions, expand each function only up to the required degree and ignore higher powers.

Answer 1

Correct Answer: 16

Step 1: Series for each factor.
Use $e^x=1+x+\frac{x^2}2+\frac{x^3}6+\cdots$ and $\cos2x=1-2x^2+\cdots$, keeping terms only up to degree $3$.

Step 2: Multiply them.
$(1+x+\frac{x^2}2+\frac{x^3}6)(1-2x^2)=1+x-\frac32x^2-\frac{11}6x^3$ to third order.

Step 3: Scale by $3$.
$f(x)=3+3x-\frac92x^2-\frac{11}2x^3$, so $a=-\frac{11}2$, $b=-\frac92$, $c=3$, $d=3$.

Step 4: Add the magnitudes.
$|a|+|b|+|c|+|d|=\frac{11}2+\frac92+3+3=10+6=16$.

Step 5: Conclude.
\[ \boxed{16} \]