Definite IntegrationhardFree

Definite Integration: Let Continuous Function Maximum

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

Question
Let f:[0,4]→Rf:\left[0,4\right]\to\mathbb R be a continuous function such that∣f(x)∣=6∣(x−n+1)(x−n)∣\left|f(x)\right| = 6\left|\left(x-n+1\right)\left(x-n\right)\right|, where n∈Nn \in \mathbb N ∀ x∈[n−1,n]\forall\,x \in \left[n-1,n\right], and g(x)=∫0xf(t) dt−∫x4f(t) dtg(x) = \displaystyle\int_0^xf(t)\,dt-\int_x^4f(t)\,dt has its maximum at x=2x = 2. Then ∫1/24f(t) dt\displaystyle\int_{1/2}^{4}f(t)\,dt is/are
A−12-\dfrac12correct
B00
C12\dfrac12
D11
Solution
Question attachment Step 1:
g′(x)=f(x)−(−f(x))=2f(x)g'(x) = f(x)-\left(-f(x)\right) = 2f(x)
 So gg increases where f>0f>0 and decreases where f<0f<0. Step 2: On each [n−1,n]\left[n-1,n\right] the product (x−n+1)(x−n)\left(x-n+1\right)\left(x-n\right) is ≤0\le0, and ∣f∣>0\left|f\right|>0 on the open interval, so the continuous ff cannot change sign inside it:
f(x)=εn⋅(−6(x−n+1)(x−n)),εn=±1f(x) = \varepsilon_n\cdot\left(-6\left(x-n+1\right)\left(x-n\right)\right),\qquad \varepsilon_n = \pm1
 which is εn\varepsilon_n times a non-negative arch vanishing at both endpoints. Continuity at the integers is automatic, because every arch vanishes there. Step 3: Putting u=x−n+1∈[0,1]u = x-n+1 \in \left[0,1\right],
∫n−1n(−6(x−n+1)(x−n))dx=−6∫01u(u−1)du=−6(13−12)=1\int_{n-1}^{n}\left(-6\left(x-n+1\right)\left(x-n\right)\right)dx = -6\int_0^1u\left(u-1\right)du = -6\left(\frac13-\frac12\right) = 1
 So ∫n−1nf=εn\displaystyle\int_{n-1}^{n}f = \varepsilon_n and g(n)−g(n−1)=2εng(n)-g(n-1) = 2\varepsilon_n. Step 4: If ε2=−1\varepsilon_2 = -1 then g(1)=g(2)+2>g(2)g(1) = g(2)+2>g(2), and if ε3=+1\varepsilon_3 = +1 then g(3)=g(2)+2>g(2)g(3) = g(2)+2>g(2); both are impossible, so ε2=+1\varepsilon_2 = +1 and ε3=−1\varepsilon_3 = -1. With those fixed, ε1=−1\varepsilon_1 = -1 would give g(0)=g(1)+2=g(2)g(0) = g(1)+2 = g(2) and ε4=+1\varepsilon_4 = +1 would give g(4)=g(3)+2=g(2)g(4) = g(3)+2 = g(2), ties at which x=2x = 2 is no longer the maximum. Hence
ε1=ε2=+1,ε3=ε4=−1\varepsilon_1 = \varepsilon_2 = +1 ,\qquad \varepsilon_3 = \varepsilon_4 = -1
 That is, f≥0f \ge 0 on [0,2]\left[0,2\right] and f≤0f \le 0 on [2,4]\left[2,4\right]. Step 5:
∫1/21f=∫1/21(−6x(x−1))dx=−6[x33−x22]1/21=12\int_{1/2}^{1}f = \int_{1/2}^{1}\left(-6x\left(x-1\right)\right)dx = -6\left[\frac{x^3}3-\frac{x^2}2\right]_{1/2}^{1} = \frac12
 Step 6:
∫1/24f=12+ε2+ε3+ε4=12+1−1−1=−12\int_{1/2}^{4}f = \frac12+\varepsilon_2+\varepsilon_3+\varepsilon_4 = \frac12+1-1-1 = -\frac12
 Answer: (1)
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