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: Use Taylor expansions.
We know that
\[ e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots \] and
\[ \cos 2x=1-\frac{(2x)^2}{2!}+\cdots \] \[ \cos 2x=1-2x^2+\cdots \]

Step 2: Multiply the expansions up to degree \(3\).
\[ e^x\cos 2x= \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)(1-2x^2) \] \[ =1+x+\frac{x^2}{2}+\frac{x^3}{6}-2x^2-2x^3 \] \[ =1+x-\frac{3}{2}x^2-\frac{11}{6}x^3 \]

Step 3: Multiply by \(3\).
\[ f(x)=3e^x\cos 2x \] \[ =3+3x-\frac{9}{2}x^2-\frac{11}{2}x^3 \] Thus,
\[ p(x)=-\frac{11}{2}x^3-\frac{9}{2}x^2+3x+3 \] So,
\[ a=-\frac{11}{2},\quad b=-\frac{9}{2},\quad c=3,\quad d=3 \]

Step 4: Find the required value.
\[ |a|+|b|+|c|+|d| = \frac{11}{2}+\frac{9}{2}+3+3 \] \[ =10+6 \] \[ =16 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{16} \]