Question

The Taylor series for f(x)=7x^2-6x+1 at x=2 is given by a+b(x-2)+c(x-2)^2. Find the value of a+b+c.

(a) -1

(b) 0

(c) 17

(d) 46

I have been asked this question in quiz.

The above asked question is from Expansion of Trigonometric Functions in division Complex Numbers of Engineering Mathematics

Answer 1

Right option is (d) 46

To explain I would say: We know

\(f(x)=7x^2-6x+1\)

\(f'(x)=14x-6\)

\(f”(x)=14\)

\(f”'(x)=0\)

Thus for n>=3, the derivative of the function is 0.

As per the Taylor Series,

\(7x^2-6x+1=\sum_{n=0}^{\infty} \frac{f^n (2)(x-2)^n}{n!}\)

\(7x^2-6x+1=f(2)+f'(2)(x-2)+\frac{1}{2} f”(2) (x-2)^2+0\)

\(7x^2-6x+1=17+22(x-2)+7(x-2)^2\)

Thus, a=17, b=22, c=7

a+b+c=46

Thus the answer is 46.