JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 19)
For some a, b, c $$\in\mathbb{N}$$, let $$f(x) = ax - 3$$ and $$\mathrm{g(x)=x^b+c,x\in\mathbb{R}}$$. If $${(fog)^{ - 1}}(x) = {\left( {{{x - 7} \over 2}} \right)^{1/3}}$$, then $$(fog)(ac) + (gof)(b)$$ is equal to ____________.
Answer
2039
Explanation
$f(x)=a x-3$
$g(x)=x^{b}+c$
$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$\Rightarrow a=2, b=3, c=5$
$fog(a c)+gof(b)$
$\because f(x)=2 x-3$
$g(x)=x^{3}+5$
$fog(10)+g o f(3)$
$=2007+32$
$=2039$
$g(x)=x^{b}+c$
$(fog)^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$(fog)^{-1}(x)=\left(\frac{x+3-c a}{a}\right)^{\frac{1}{b}}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
$\Rightarrow a=2, b=3, c=5$
$fog(a c)+gof(b)$
$\because f(x)=2 x-3$
$g(x)=x^{3}+5$
$fog(10)+g o f(3)$
$=2007+32$
$=2039$
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