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Range of f(x)=∫₁ˣ|t|dt on [−1/2, 1/2] | JEE

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Question
The range of the function f(x)=∫1x∣t∣ dtf(x)=\displaystyle\int_{1}^{x}|t|\,dt, x∈[−12,12]x\in\left[-\dfrac12,\dfrac12\right], is
A[38,58]\left[\dfrac38,\dfrac58\right]
B[−58,38]\left[-\dfrac58,\dfrac38\right]
C[−38,58]\left[-\dfrac38,\dfrac58\right]
D[−58,−38]\left[-\dfrac58,-\dfrac38\right]correct
Solution
Step 1: Differentiate using the Fundamental Theorem of Calculus: f′(x)=∣x∣f'(x)=|x|. On [−12,12]\left[-\dfrac12,\dfrac12\right], ∣x∣≥0|x|\ge 0, so ff is non-decreasing (strictly increasing away from 00). Therefore the range is [f(−12), f(12)]\big[f(-\tfrac12),\,f(\tfrac12)\big]. Step 2: Compute f ⁣(12)=∫11/2∣t∣ dt=−∫1/21t dt=−[t22]1/21=−(12−18)=−38.f\!\left(\dfrac12\right)=\displaystyle\int_{1}^{1/2}|t|\,dt=-\int_{1/2}^{1}t\,dt=-\left[\dfrac{t^{2}}{2}\right]_{1/2}^{1}=-\left(\dfrac12-\dfrac18\right)=-\dfrac38. Step 3: Compute f ⁣(−12)=∫1−1/2∣t∣ dt=−∫−1/21∣t∣ dtf\!\left(-\dfrac12\right)=\displaystyle\int_{1}^{-1/2}|t|\,dt=-\int_{-1/2}^{1}|t|\,dt. Split at 00:
∫−1/21∣t∣ dt=∫−1/20(−t) dt+∫01t dt=18+12=58,\int_{-1/2}^{1}|t|\,dt=\int_{-1/2}^{0}(-t)\,dt+\int_{0}^{1}t\,dt=\dfrac18+\dfrac12=\dfrac58,
 so f ⁣(−12)=−58.f\!\left(-\dfrac12\right)=-\dfrac58. Step 4: Since ff increases, the range is [f(−12),f(12)]=[−58,−38].\left[f(-\tfrac12),f(\tfrac12)\right]=\left[-\dfrac58,-\dfrac38\right]. Correct answer: (4)
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