गणित एवं संख्यात्मक विश्लेषण

Evaluation of the left-hand limit:
As $x \to 1^-$, $x \lt 1$. For values of $x$ close to 1 with $0 \lt x \lt 1$, we have $x^2 \lt 1$, which implies $x^2 - 1 \lt 0$.
By definition of absolute value, when $u \lt 0$, $|u| = -u$.

Profit maximization using differential calculus:
Cost per pair $= 50$, selling price per pair $= x$.
Profit per pair $= (x - 50)$.
Total profit function $P(x) = (x - 50) S(x) = 1000 (x - 50) e^{-0.1x}$.
Differentiating with respect to $x$ using product rule:
$P'(x) = 1000 [1 \cdot e^{-0.1x} + (x - 50)(-0.1) e^{-0.1x}]$
$= 1000 e^{-0.1x} [1 - 0.1(x - 50)] = 1000 e^{-0.1x} [1 - 0.1x + 5] = 1000 e^{-0.1x} [6 - 0.1x]$.
Setting $P'(x) = 0$ for maximum profit:
$6 - 0.1x = 0 \implies 0.1x = 6 \implies x = 60$.
Second derivative check: $P''(60) \lt 0$, confirming a local and global maximum.

Option breakdown:

• $x = 40$ is below the production cost of 50 dollars, yielding a loss.
• $x = 50$ yields zero profit.
• $x = 60$ maximizes the weekly profit function.
• $x = 70$ yields lower profit due to reduced sales volume.

Finding horizontal asymptotes of rational functions:
A horizontal asymptote is found by taking the limit of $y$ as $x \to \pm\infty$:
$\lim_{x \to \pm\infty} y = \lim_{x \to \pm\infty} \frac{-8}{x^2 - 4} = 0$.
Since the degree of the numerator (degree 0) is less than the degree of the denominator (degree 2), the line $y = 0$ (the x-axis) is the horizontal asymptote.

Option breakdown:

• $y = 0$ is the correct horizontal asymptote as $x \to \pm\infty$.
• $y = 8$ comes from the numerator constant.
• $y = 4$ comes from the denominator constant.
• $y = 2$ is incorrect; vertical asymptotes occur at $x = \pm 2$.

Step-by-step parametric differentiation:
Given $x = a\sin^3 t$ and $y = b\cos^3 t$:

• First derivatives with respect to $t$:
  $\frac{dx}{dt} = 3a\sin^2 t \cos t$, and $\frac{dy}{dt} = -3b\cos^2 t \sin t$.
• First derivative $\frac{dy}{dx}$:
  $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\frac{b}{a} \cot t$.
• Second derivative $\frac{d^2 y}{dx^2}$:
  $\frac{d^2 y}{dx^2} = \frac{d}{dt}\left(-\frac{b}{a}\cot t\right) \cdot \frac{dt}{dx} = \left(\frac{b}{a} \operatorname{cosec}^2 t\right) \cdot \frac{1}{3a\sin^2 t \cos t}$
  $= \frac{b}{3a^2} \frac{\operatorname{cosec}^2 t}{\sin^2 t \cos t} = \frac{b}{3a^2} \operatorname{cosec}^4 t \sec t$.
  
  Option breakdown:
• Inverts the trigonometric functions into numerator powers.
• $\frac{b}{3a^2}\operatorname{cosec}^4 t \sec t$ is the exact second derivative.
• Inverts the constant coefficient to $a/(3b^2)$.
• Swaps the coefficients $a$ and $b$.