f(x) = (3x+5/7-x) के लिए ऊर्ध्वाधर अनंतस्पर्शी (vertical... | Agrigyan

Using L'Hôpital's rule (as the limit is of the form $0/0$ when $x \to -a$ since $(-a)^9 + a^9 = -a^9 + a^9 = 0$):
$\lim_{x \to -a} \frac{\frac{d}{dx}(x^9 + a^9)}{\frac{d}{dx}(x + a)} = \lim_{x \to -a} \frac{9x^8}{1} = 9(-a)^8 = 9a^8$.
We are given that this limit equals 9:
$9a^8 = 9 \implies a^8 = 1$.
Taking real roots: $a = \pm 1$.

Detailed analysis of options:

• $a = \pm 1$ is correct ($a^8 = 1 \implies a = \pm 1$).
• $a = \pm 9$ is incorrect.
• $a = \pm 3$ is incorrect.
• $a = \pm 8$ is incorrect.

Let the price of a flopsy doll be $p$. Then the price of a mopsy doll is $2p$.
The total revenue (in hundreds of $p$) is:
$R(x) = p(x + 2y) = p\left(x + 2\cdot\frac{82-10x}{10-x}\right)$.
Note that $\frac{82-10x}{10-x} = \frac{10(10-x) - 18}{10-x} = 10 - \frac{18}{10-x}$.

First differentiate $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t$.
$\frac{dy}{dt} = a(\cos t - (\cos t - t\sin t)) = at\sin t$.
Now, the first derivative is:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t$.
Now differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at\cos t} = \frac{1}{at\cos^3 t}$.

Detailed analysis of options:

• $\frac{at}{\cos^3 t}$ inverts the position of $at$.
• $\frac{1}{at\cos^3 t}$ is the exact and correct second derivative.
• $\frac{at}{\sin^3 t}$ has incorrect trigonometric function and inverse scale.
• $\frac{1}{at\sin^3 t}$ has sine instead of cosine.