Let R be the region bounded by the curve y = 4-|x| and the x-axis. Wh…

The correct answer is Option (2) → (A), (B) and (D) only

The curve meets the x-axis where y = 0:

    \(4-|x| = 0,\ \ x = \pm4\)

Since \(|-x| = |x|\), the curve is symmetric about the y-axis. For \(x \ge 0\) it is the line y = 4 − x, so the area is twice the right half:

    \( \displaystyle A = 2\int_{0}^{4}(4-x)\,dx \)

    \( \displaystyle = 2\left[4x-\frac{x^{2}}{2}\right]_{0}^{4} \)

    \(= 2(16-8) = 16\)

The integral in (C) uses y = 4 − x on the left half too, where the curve is really y = 4 + x:

    \( \displaystyle \int_{-4}^{4}(4-x)\,dx = \left[4x-\frac{x^{2}}{2}\right]_{-4}^{4} \)

    \(= 8-(-24) = 32\)

So the curve meets the x-axis at x = ±4, the area of R is 16, and the integral in (C) equals 32.

Checking the statements

(A) The curve meets the x-axis at x = −4 and x = 4 — Correct. y = 0 gives |x| = 4.

(B) The area of R is \(\displaystyle 2\int_{0}^{4}(4-x)\,dx\) — Correct. The region is symmetric and its right half lies under y = 4 − x.

(C) The area of R is \(\displaystyle \int_{-4}^{4}(4-x)\,dx\) — Incorrect. This integral equals 32, not the area 16.

(D) The area of R is 16 sq. units — Correct. \(2\times8 = 16\).