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Let $C$ denote a green pear on the first draw and $D$ a red apple on the second. The two categories are disjoint, so after $C$ there are still three red apples among seventeen remaining fruits. Thus $P(C\cap D)=(8/18)(3/17)$. Using $P(D)=1/6$ from (c), conditional probability givesEquivalently, conditioning on the second fruit being one particular category removes a red apple and leaves the other seventeen fruits symmetrically eligible for the first position.

Condition on whether the first selection was a red apple. Its probability is $3/18$. If it was, two red apples remain among seventeen fruits; otherwise three remain. The law of total probability givesThe same result follows from exchangeability: before either selection is observed, each ordered draw position has the original red-apple fraction $3/18$, despite dependence between the draws.

Insert the values from (a) into Bayes' theorem:Thus the identity is satisfied. Equivalently, both $P(A\mid B)P(B)$ and $P(B\mid A)P(A)$ equal the joint probability $4/9$.

There are eighteen equally likely first choices. Counting the relevant subsets givesFor conditional probabilities, restrict the denominator to the conditioning subset:The five counts come from the same finite sample space, so no independence assumption is involved.

Factor $1+e^x=2e^{x/2}\cosh (x/2)$. This givesThe last function is even. Its Taylor series begins $\ln \cosh (x/2)=x^2/8+O(x^4)$, since $\cosh (x/2)=1+x^2/8+O(x^4)$ and $\ln (1+w)=w+O(w^2)$. ThereforeIn particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.

The required derivatives are $f^{\prime }(x)=1/(2\sqrt{x})$, $f^{\prime \prime }(x)=-1/(4x^{3/2})$, $f^{(3)}(x)=3/(8x^{5/2})$, and $f^{(4)}(x)=-15/(16x^{7/2})$. The four-term Taylor polynomial about one, evaluated at two, isThe Taylor remainder is $R_3(2)=-5/(128{\xi }^{7/2})$ for some $1<\xi <2$. ThusIts negative sign also shows that this approximation is an overestimate. Bounding the fourth derivative throughout the interval establishes the requested error estimate without using a decimal value of $\sqrt{2}$.

A cubic version of Taylor's theorem requires the corresponding higher differentiability, rather than only one derivative. For example, if $f$ has four continuous derivatives on the interval between $a$ and $x$, thenwhere the Lagrange form of the Taylor remainder isfor some $\xi$ strictly between $a$ and $x$ when they differ. At $x=a$, the remainder is zero. These hypotheses apply to both functions used below.

With $a=2$, integrate the $dx$ coefficient with respect to $x$:Its other partial derivative is ${\Phi }_y=x^2/2-2xy+g^{\prime }(y)$. Matching the $dy$ coefficient gives $g^{\prime }=0$. Therefore the differential equation is $d\Phi =0$, and its implicit general solution isThese are level curves of the potential of a conservative vector field. Where $N\ne 0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.

Write the differential one-form as $Mdx+Ndy$. An exact differential form must satisfy equality of mixed partial derivatives, $M_y=N_x$. HereEquality at every $(x,y)$ holds precisely when $\boxed{a=2}$. Since these polynomials are smooth on the simply connected plane, the closed-form condition is also sufficient for exactness; the explicit potential of a conservative vector field constructed in (c) verifies that sufficiency directly.

The multivariable chain rule givesSubstituting the three parametrizing functions also writes the complete derivative asAt $t=\pi /2$, the point is $(0,3,\pi )$, its velocity is $(-3,0,2)$, and $\nabla F=(3,0,1)$. Hence

The condition at infinity gives $C_{\mathrm{out}}=0$. Matching the electric potential at $r=a$ givesConsequentlyThis function is finite and differentiable at the origin. At the joining radius, both $V$ and $V^{\prime }$ match, giving $V(a)=-k/a$ and $V^{\prime }(a)=k/a^2$. Thus the electric field is continuous there as well as being conservative on each side. This is the potential of a uniform spherical source, with source coefficient $3k/a^3$.

For a radial electric potential, $\mathbf{E}=-V^{\prime }(r)\mathbf{r}/r$. Comparing radial coefficients givesIntegrating these one-variable derivatives gives the general local expressionsThe two constants are initially independent because these integrations were performed in separate radial regions.

Inside the sphere, the electric field has the form $\mathbf{E}=\nabla [-k{r}^2/(2a^3)]$, including at the origin. Thus it is a conservative vector field. Outside, use the gradient of a radial function to obtainThis is also an explicit potential of a conservative vector field, single-valued throughout the exterior region. Equivalently, the curl vanishes in both regions: for a radial coefficient $h(r)$, $\nabla \times (h\mathbf{r})=\nabla h\times \mathbf{r}+h\nabla \times \mathbf{r}=0$ because $\nabla h$ is radial. The interior formula avoids any apparent $1/r$ issue at the origin. The vector-field potentials here have the opposite sign from the electric potential convention used in (b).

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.

The characteristic equation factors as $(r+1)(r+2)=0$, giving $y_h=C_1{e}^{-x}+C_2{e}^{-2x}$. The superposition principle lets us find particular solutions for the two forcing terms separately. A trial $Ax+B$ gives $2A=1$ and $3A+2B=0$, so it is $x/2-3/4$. A trial $C{e}^{3x}$ gives $(9+9+2)C=1$, so $C=1/20$. Therefore

The characteristic equation has the repeated root $r=-3$. Its homogeneous solution is therefore $(C_1+C_{2}x)e^{-3x}$. For a particular solution, try $y_p=Ax+B$. Substitution gives $9Ax+6A+9B=3x+1$, so $A=1/3$ and $B=-1/9$. Thus the general solution isThe extra factor $x$ in the second homogeneous term is required by the repeated root.

The characteristic equation is $r^2+r-2=(r-1)(r+2)=0$, giving $y=A{e}^x+B{e}^{-2x}$. The initial conditions give $A+B=1$ and $A-2B=-1$. Solving yields $A=1/3$, $B=2/3$, henceThe two distinct characteristic roots supply all solutions of this second-order linear ordinary differential equation.