Skip to content
dronacharya.ai
← JEE Advanced 2024 · Paper 2
MathematicsInteger+4Question 8 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 8

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a function such that f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y) for all x,y∈Rx, y \in \mathbb{R}, and g:R→(0,∞)g : \mathbb{R} \to (0, \infty) be a function such that g(x+y)=g(x)g(y)g(x+y) = g(x)g(y) for all x,y∈Rx, y \in \mathbb{R}. If f(−35)=12f\left(\dfrac{-3}{5}\right) = 12 and g(−13)=2g\left(\dfrac{-1}{3}\right) = 2, then the value of (f(14)+g(−2)−8)g(0)\left(f\left(\dfrac14\right) + g(-2) - 8\right)g(0) is _____.

Answer and worked solution

Official answer: 51

f. ff is additive, so f(rx)=rf(x)f(rx)=rf(x) for rational rr and f(x)=cxf(x)=cx on the rationals. From f(−35)=12f(-\frac35)=12, c=−20c=-20, so f(14)=−5f(\frac14)=-5.

g. g(x+y)=g(x)g(y)g(x+y)=g(x)g(y) with g>0g>0 gives g(r)=brg(r)=b^r on the rationals.

  • From g(−13)=2g(-\frac13)=2: b−1/3=2b^{-1/3}=2, so b=18b=\frac18.
  • Then g(−2)=82=64g(-2)=8^{2}=64 and g(0)=1g(0)=1.

(f(14)+g(−2)−8)g(0)=(−5+64−8)⋅1=51.\left(f(\tfrac14)+g(-2)-8\right)g(0)=(-5+64-8)\cdot1=51.