For the linear programming problem (LPP): Maximize Z = x + 1.5y, subject to constraints, x + 2y ≤ 40, 2x + y ≤ 40, x + y ≤ 25, x ≥ 0, y ≥ 0. Which of…

For $f(x)$f(x) to be continuous at $x = \pi/2$x=π/2, the left-hand limit must equal the right-hand limit.

From the left $(x \leq \pi/2)$(x≤π/2):

$$\lim_{x \to \pi/2^-} f(x) = \sin\left(\frac{\pi}{2}\right) + n$$x→π/2−lim​f(x)=sin(2π​)+n

$$= 1 + n$$=1+n

From the right $(x \geq \pi/2)$(x≥π/2):

$$\lim_{x \to \pi/2^+} f(x) = m \cdot \frac{\pi}{2} + 1$$x→π/2+lim​f(x)=m⋅2π​+1

$$= \frac{m\pi}{2} + 1$$=2mπ​+1

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Setting the two limits equal for continuity:

$$1 + n = \frac{m\pi}{2} + 1$$1+n=2mπ​+1

$$n = \frac{m\pi}{2}$$n=2mπ​

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Substituting into $\sin 2n$sin2n:

$$\sin(2n) = \sin\left(2 \cdot \frac{m\pi}{2}\right)$$sin(2n)=sin(2⋅2mπ​)

$$= \sin(m\pi)$$=sin(mπ)

Since $m \in \mathbb{Z}$m∈Z, we have $\sin(m\pi) = 0$sin(mπ)=0 for any integer $m$m.

Therefore, $\sin 2n = 0$sin2n=0