Maths Olympiad Prep

Track / Stage 3 / 53 of 260 #53 of 1964

Problem 53

AMC 10/12, early questions
Algebra Difficulty 3.1 Find the answer

Find the value of the complex number 2−i3+4i\frac{2-i}{3+4i}; Given f(x)=x2+3xf′(2)f(x)=x^2+3xf'(2), find the value of 1+f′(1)1+f'(1).

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Next problem →

Official solution

(1) Analysis
The problem involves the division operation of complex numbers. We can solve it according to the rules of operation.

Solution
2−i3+4i=(2−i)(3−4i)(3+4i)(3−4i)=2−11i25=225−1125i\frac{2-i}{3+4i} = \frac{(2-i)(3-4i)}{(3+4i)(3-4i)} = \frac{2-11i}{25} = \frac{2}{25} - \frac{11}{25}i,

Hence, the answer is 225−1125i\boxed{\frac{2}{25} - \frac{11}{25}i}.

(2) Analysis
This problem involves the operation of derivatives. First, find the derivative function, then substitute x=2x=2 to find the expression, and then solve the problem.

Solution
f′(x)=2x+3f′(2)f'(x)=2x+3f'(2), substituting x=2x=2 gives f′(2)=−2f'(2)=-2, hence f′(x)=2x−6f'(x)=2x-6,

1+f′(1)=1+2−6=−31+f'(1)=1+2-6=-3.

Hence, the answer is −3\boxed{-3}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.