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By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

Apply the chain rule twice: once to the square and once to the argument $x^3$ of the sine. The inner derivative is $y^{\prime }=3x^2\cos (x^3)-2\sin (2x)$. ThereforeThe original PDF has $\sin (x^3)$, rather than the cube of $\sin x$.

The first term is $ar$, so the finite geometric series is $S=a(r+r^2+\cdots +r^{100})$. Subtracting $rS$ cancels the interior terms and gives $(1-r)S=ar(1-r^{100})$. ThereforeThe rational expression holds for every real $a$ and every real $r\ne 1$, including $r=0$; the same algebra works for complex numbers. No restriction $|r|<1$ is needed for a finite geometric series. The exceptional case follows by adding one hundred equal terms, or by taking the limit as $r\to 1$.

Differentiate the two powers to locate the stationary point:For $x>0$, this is negative before $2$ and positive afterwards, so the global minimum isThe curve crosses the horizontal axis at $x=1$. Its limits are $+\infty$ as $x\to 0^{+}$ and $0$ from below as $x\to \infty$, giving the vertical asymptote $x=0$ and horizontal asymptote $y=0$. These features determine the required sketch.

The two directed segments have vectors $\mathbf{u}=(1,0,-1)$ and $\mathbf{v}=(0,1,1)$. Their dot product is $-1$ and both norms equal $\sqrt{2}$. HenceThe common starting point fixes the obtuse angle between these segments.

Use integration by substitution with $u=3-2x$, so $dx=-du/2$ and $x=(3-u)/2$. The indefinite integral becomesThus, with an arbitrary constant of integration,This real antiderivative applies for $x<3/2$ and extends continuously to the stated endpoint. Its left derivative there agrees with the limiting integrand, zero.

Set $X=x^2$ and $Y=y^2$ in the two conic sections. A real intersection requires $X,Y\ge 0$. Solving the resulting linear equations givesFor $a>0$ the denominator and $X$ are positive, so intersections exist exactly when $a\ge 1$. At $a=1$ they are $(\pm 1,0)$; for $a>1$ both choices of sign for each square root give intersections. Consequently

Apply the product rule to $x\sin (1/x)$ and the chain rule to the sine. Since $d(1/x)/dx=-1/x^2$, the derivative is

The Pythagorean trigonometric identity makes the second factor $1+{\cos }^2\theta$, which is strictly positive. It is therefore enough to solve $\sin \theta -\cos \theta \le 0$. The sine addition formula givesThe shifted angle lies between $\pi /4$ and $5\pi /4$. On this interval the sine is nonpositive precisely from $\pi$ onwards. Restoring the original angle and retaining its strict upper bound gives