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Olympiad workbook
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Olympiad workbook
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IOQM
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Functions

Olympiad workbook · IOQM · Functions

Let f: ℝ arrow ℝ be a function satisfying the relation 4 f(3-x)+3...

Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function satisfying the relation 4f(3−x)+3f(x)=x24 f(3-x)+3 f(x)=x^2 for any real xx. Find the value of f(27)−f(25)f(27)-f(25) to the nearest integer. (Here R\mathbb{R} denotes the set of real numbers.)

Correct answer

8

Step-by-step solution

4f(3−x)+3f(x)=x2∀x∈R4f(3−(3−x))+3f(3−x)=(3−x)2=4f(x)+3f(3−x)=(x−3)2⇒12f(3−x)+9f(x)=3x212f(3−x)+16f(x)=4(x−3)2\begin{aligned} & 4 f(3-x)+3 f(x)=x^2 \quad \forall x \in R \\ & 4 f(3-(3-x))+3 f(3-x)=(3-x)^2 \\ & =4 f(x)+3 f(3-x)=(x-3) 2 \\ & \Rightarrow \quad 12 f(3-x)+9 f(x)=3 x^2 \\ & 12 f(3-x)+16 f(x)=4(x-3)^2\end{aligned} ⇒7f(x)=4(x−3)2=3x2=(4(24)2−3.272)−(4(22)2−3.252)=4(242−222)+3(252−272)=4(46)(2)+3(52)(−2)=8×46×6×52⇒f(27)−f(25)=17(46×8−6×52)=(567)=8\begin{aligned} \Rightarrow \quad & 7 f(x)=4(x-3)^2=3 x^2 \\ & =\left(4(24)^2-3.27^2\right)-\left(4(22)^2-3.25^2\right) \\ & =4\left(24^2-22^2\right)+3\left(25^2-27^2\right) \\ & =4(46)(2)+3(52)(-2) \\ & =8 \times 46 \times 6 \times 52 \\ \Rightarrow & f(27)-f(25)=\frac{1}{7}(46 \times 8-6 \times 52)=\left(\frac{56}{7}\right)=8\end{aligned}
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