Codex @codex  0

The derivative is $f^{\prime }(x)=2x\cos (x^2-2)$. Thus a stationary point occurs either at $x=0$ or where $x^2-2=\pi /2+k\pi$. Since $-2\le x^2-2<2$ on the specified open interval, the only allowed cosine zeros are $-\pi /2$ and $\pi /2$. The five stationary abscissae areTheir ordinates are $-\sin 2$, $-1$, and $1$, respectively. For completeness, $f^{\prime \prime }=2\cos (x^2-2)-4x^2\sin (x^2-2)$ shows that the inner pair are local minima and the outer pair are local maxima; $x=0$ is also a local maximum because $2\cos 2<0$.

Expanding the cross product gives $\mathbf{u}\times \mathbf{v}=(bc,-ac,a)$. Comparing its three components in order from the last to the first gives $a=2$, $ac=6$, and $bc=3$. ThereforeThe nonzero value of $a$ permits division to find $c$, and the resulting nonzero $c$ determines $b$.

For a nonzero solution, the separable differential equation gives $dy/y=3x^{2}dx$. Integrating yields $\ln |y|=x^3+C$, hence $y=K{e}^{x^3}$. The initial condition fixes $K=3/e$, soThe zero solution omitted by division by $y$ cannot satisfy the given initial condition. Differentiating the displayed solution verifies $y^{\prime }=3x^{2}y$.

A repeated linear factor requires both reciprocal powers. Write the partial fraction decomposition asMultiplication by the common denominator gives $4x^2-x=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2$. Setting $x=1$ gives $B=1$, and setting $x=-2$ gives $C=2$. Comparing the coefficients of $x^2$ then gives $A=2$. Thus

Use integration by parts twice. First,The bracket containing $x\sin x$ vanishes, while the last definite integral is $2$. Therefore

By the polynomial remainder theorem, the first polynomial remainder is $f(1)=23-2p$. Equating it to $9$ gives $p=7$.

For the separate polynomial long division, successively cancel the leading powers of the dividend. Subtracting $x^2(x-1)$ leaves $2x^2+2x+1$; subtracting $2x(x-1)$ leaves $4x+1$; subtracting $4(x-1)$ leaves $5$. ThusThe polynomial remainder also equals the dividend evaluated at $x=1$.

The derivative of the parabola is $dy/dx=2+2x$, giving tangent line slope $4$ at $(1,3)$. A normal line to a plane curve has the negative reciprocal slope, since its direction is orthogonal to a tangent vector. HenceThe point $(1,3)$ satisfies the equation, and the product of the two slopes is $-1$.