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Definite Integration: Continuous Increasing Function Satisfying Choose Correct Opt

JEE Maths question with a full step-by-step solution.

Question
f(x)f(x) is a continuous and increasing function satisfying f(x)=f(x−3)+4f(x) = f\left(x-3\right)+4 and ∫06f(x) dx=0\displaystyle\int_0^6f(x)\,dx = 0. Choose the CORRECT option(s)
AArea bounded by y=f(x)y = f(x), the xx-axis and the lines x=6x = 6 and x=9x = 9 is 99
BArea bounded by y=f(x)y = f(x), the xx-axis and the lines x=6x = 6 and x=9x = 9 is 1818correct
Cy=f(x)y = f(x) is a periodic function
Dif y=f(x)+h(x)y = f(x)+h(x) is a periodic function (where h(x)h(x) is a continuous function), then
lim⁡x→∞(h(x)−x)(h(x)+x)x2=79\lim_{x\to\infty}\frac{\left(h(x)-x\right)\left(h(x)+x\right)}{x^2} = \frac79
correct
Solution
Step 1: f(x)−f(x−3)=4f(x)-f\left(x-3\right) = 4 says ff rises by 44 over every step of 33, i.e. at the average rate 43\tfrac43, so put
g(x)=f(x)−43xg(x) = f(x)-\frac43x
 Then
g(x)−g(x−3)=[f(x)−f(x−3)]−43⋅3=4−4=0g(x)-g\left(x-3\right) = \left[f(x)-f\left(x-3\right)\right]-\frac43\cdot3 = 4-4 = 0
 so gg is periodic with period 33, and
f(x)=g(x)+43xf(x) = g(x)+\frac43x
 Step 2:
∫06f=∫06g+43∫06x dx=2∫03g+43⋅362=2∫03g+24=0\int_0^6f = \int_0^6g+\frac43\int_0^6x\,dx = 2\int_0^3g+\frac43\cdot\frac{36}2 = 2\int_0^3g+24 = 0
∫03g=−12\int_0^3g = -12
 Step 3:
∫69f=∫69g+43∫69x dx=∫03g+43⋅81−362=−12+30=18\int_6^9f = \int_6^9g+\frac43\int_6^9x\,dx = \int_0^3g+\frac43\cdot\frac{81-36}2 = -12+30 = 18
 Step 4: ff is increasing, so f≤f(6)f \le f(6) on [0,6]\left[0,6\right]. If f(6)≤0f(6) \le 0 then f≤0f \le 0 there and ∫06f=0\int_0^6f = 0 would implies that f≡0f \equiv 0 on [0,6]\left[0,6\right], contradicting f(6)=f(0)+8f(6) = f(0)+8. So f(6)>0f(6)>0, hence f>0f>0 throughout [6,9]\left[6,9\right] and the integral is the area:
area=∫69f=18\text{area} = \int_6^9f = 18
 (2) is TRUE, (1) is FALSE. Step 5: f(x)=g(x)+43xf(x) = g(x)+\tfrac43x with gg bounded (continuous and periodic), so f(x)→∞f(x)\to\infty. A periodic function is bounded, so ff is not periodic. (3) is FALSE. Also ff is increasing and non-constant, which already rules out periodicity. Step 6: If f+h=pf+h = p is periodic, then
h(x)=p(x)−f(x)=p(x)−g(x)⏟periodic, hence bounded−43xh(x) = p(x)-f(x) = \underbrace{p(x)-g(x)}_{\text{periodic, hence bounded}}-\frac43x
 So h(x)x→−43\dfrac{h(x)}x \to -\dfrac43 as x→∞x\to\infty, and
(h−x)(h+x)x2=h2−x2x2=(hx)2−1 ⟶ 169−1=79.\frac{\left(h-x\right)\left(h+x\right)}{x^2} = \frac{h^2-x^2}{x^2} = \left(\frac hx\right)^2-1 \ \longrightarrow\ \frac{16}9-1 = \frac79 .
 (4) is TRUE. Answer: (2), (4)
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