f:R→R, $$f(x2+x+3)+2f(x2-3x+5)=6x2-10x+17$$ ∀ x∈R then the value of $$f(100)$$ is

Functions (XII)

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${f:R→R, f(x}^{2} + x + 3) + 2 f (x^{2} - 3 x + 5) = 6 x^{2} - 10 x + 17 \forall x \in R t h e n t h e v a l u e o f f (100) i s$

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Solution

$obviously f is a linear polynomial {let f(x)=ax+b hence f(x}^{2} + x + 3) + 2 f (x^{2} - 3 x + 5) = 6 x^{2} - 10 x + 17$
$[a (x^{2} + x + 3) + b] + 2 [a (x^{2} - 3 x + 5) + b] = 6 x^{2} - 10 x + 17 c o m p a r i n g t h e c o e f f i c i e n t o f x^{2} a n d c o n s t a n t s$
$\Rightarrow a + 2 a = 6 a n d 3 a + b + 10 a + 2 b = 17$
$\Rightarrow 3 a = 6 a n d 13 a + 3 b = 17 \Rightarrow a = 2 a n d 26 + 3 b = 17 b = - 3 ∴ f (x) = 2 x - 3 \Rightarrow f (100) = 197$

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Functions (XII)

Remember concepts with our Masterclasses.

f:R→R, f(x2+x+3)+2f(x2-3x+5)=6x2-10x+17 ∀ x∈R then the value of f(100) is

obviously f is a linear polynomial let f(x)=ax+b hence f(x2+x+3)+2f(x2-3x+5)=6x2-10x+17 a(x2+x+3)+b +2a(x2-3x+5)+b=6x2-10x+17 comparing the coefficient of x2 and constants ⇒a+2a=6 and 3a+b+10a+2b=17 ⇒3a=6 and 13a+3b=17 ⇒a=2 and 26+3b=17 b=-3 ∴f(x)=2x-3 ⇒f(100)=197

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${f:R→R, f(x}^{2} + x + 3) + 2 f (x^{2} - 3 x + 5) = 6 x^{2} - 10 x + 17 \forall x \in R t h e n t h e v a l u e o f f (100) i s$