Question-153054

Question Number 153054 by daus last updated on 04/Sep/21

Answered by bobhans last updated on 04/Sep/21

{\begin{cases} g (x) = 1 - 2 x \\ f (x) = {k x}^{2} + m \end{cases} \Rightarrow (f ∙ g) (x) = x^{2} - 2 x + 5

\Rightarrow k {(1 - 2 x)}^{2} + m = x^{2} - 2 x + 5

\Rightarrow k (4 x^{2} - 4 x + 1) + m = x^{2} - 2 x + 5

\Rightarrow 4 {k x}^{2} - 4 k x + k + m = x^{2} - 2 x + 5

\to {\begin{cases} 4 k = 1 \Rightarrow k = \frac{1}{4} b u t \\ 4 k = 2 \Rightarrow k = \frac{1}{2} \end{cases}

s o m e t h i n g w r o n g t h e q u e s t i o n

Facebook Tweet Pin