2019 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the comparison of the orders of two infinitesimals by means of a Taylor expansion. Two infinitesimals are of the same order if and only if their ratio tends to a nonzero constant, so it is enough to expand $\tan x$ to a sufficiently high order.

Steps:

1. From the Maclaurin expansion $\tan x=x+\dfrac{x^3}{3}+o(x^3)$ we get $x-\tan x=-\dfrac{x^3}{3}+o(x^3)$.
2. Dividing by $x^3$: $\dfrac{x-\tan x}{x^3}\to-\dfrac13\ne0$, so $x-\tan x$ is of the same order as $x^3$.
3. If it is of the same order as $x^k$ then $k=3$, so (C) is the correct choice.

Answer:

C.

Approach:

This question tests the differentiability of a piecewise function at its breakpoint and the criterion for an extremum. The signs of the function values on the two sides of $0$ (for the extremum) and the left and right derivatives (for the differentiability) have to be examined separately.

Steps:

1. $f(0)=0$. For $x<0$ we have $f(x)=x|x|=-x^2<0$; for $0<x<1$ we have $f(x)=x\ln x<0$.
2. In a sufficiently small punctured neighbourhood of $0$ we always have $f(x)<0=f(0)$, so $x=0$ is a local maximum point.
3. The left derivative is $\lim\limits_{x\to0^-}\dfrac{f(x)-f(0)}{x}=\lim\limits_{x\to0^-}(-x)=0$.
4. The right derivative is $\lim\limits_{x\to0^+}\dfrac{f(x)-f(0)}{x}=\lim\limits_{x\to0^+}\ln x=-\infty$, so no finite right derivative exists.
5. The one-sided derivatives are not equal, so $f$ is not differentiable at $x=0$; combined with the fact that it is a local maximum point, this gives (B).

Answer:

B.

Approach:

This question tests the convergence of series of numbers. The key is to recognise that option (D) is a telescoping series whose partial sums can be written down directly; the other options need not converge under the given assumptions.

Steps:

1. For option (D): its partial sums are $S_N=\sum\limits_{n=1}^N(u_{n+1}^2-u_n^2)=u_{N+1}^2-u_1^2$, since the intermediate terms cancel.
2. Since $\{u_n\}$ is monotonically increasing and bounded, $u_{N+1}^2$ has a limit, so $S_N$ converges, that is, the series $\sum\limits_{n=1}^{\infty}(u_{n+1}^2-u_n^2)$ converges.
3. For option (A): take $u_n=n$, then the series becomes $\sum\limits_{n=1}^\infty\dfrac nn=\sum1$, which diverges.
4. For option (B): take $u_n=n$, then the series becomes $\sum\limits_{n=1}^\infty(-1)^n\dfrac1n$, which converges conditionally; while for $u_n\equiv1$ the same series converges, so (B) does not always converge.
5. For option (C): with $u_n=n$ we have $1-\dfrac{u_n}{u_{n+1}}=\dfrac{1}{n+1}$, so the series diverges, and (C) does not always converge either. In conclusion the answer is (D).

Answer:

D.

Approach:

This question tests the conditions under which a second-kind line integral in the plane is independent of the path. The condition that the integral vanishes for every closed curve is equivalent to the vector field being irrotational, that is, to $P_y=Q_x$ holding identically.

Steps:

1. The upper half-plane $y>0$ is a simply connected region, so the necessary and sufficient condition for path independence is that $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ holds identically in that region.
2. From $Q(x,y)=\dfrac{x}{y^2}$ we get $Q_x=\dfrac{1}{y^2}$, so we need $P_y=\dfrac{1}{y^2}$.
3. Integrating with respect to $y$: $P(x,y)=\displaystyle\int\dfrac{dy}{y^2}+g(x)=-\dfrac1y+g(x)$.
4. For options (A) and (B) the value of $P_y$ is not $\dfrac1{y^2}$; for option (C) the function $P=\dfrac1x-\dfrac1y$ is not defined at $x=0$, so it fails the requirement of being defined on the whole upper half-plane.
5. In option (D), $P=x-\dfrac1y$ corresponds to $g(x)=x$, which satisfies the condition, so (D) is the correct choice.

Answer:

D.

Approach:

This question tests how a matrix equation determines the eigenvalues and how the canonical form of a quadratic form is obtained. The relation $A^2+A=2E$ shows that every eigenvalue satisfies the same quadratic equation, and the determinant then fixes the spectrum.

Steps:

1. Let $\lambda$ be an eigenvalue of $A$; from $A^2+A=2E$ we get $\lambda^2+\lambda-2=0$, that is $(\lambda-1)(\lambda+2)=0$.
2. Hence the eigenvalues of $A$ can only be $1$ or $-2$.
3. $A$ is a real symmetric matrix of order 3, so all eigenvalues are real and their multiplicities sum to 3; also $|A|=4>0$, and with the eigenvalues $1$ and $-2$ we get $1\cdot(-2)^2=4$, which is consistent.
4. Hence the spectrum is $\{1,-2,-2\}$, the positive index of inertia is 1 and the negative index of inertia is 2, and the canonical form is $y_1^2-y_2^2-y_3^2$, so (C) is the correct choice.

Answer:

C.

Approach:

This question tests the direction of the intersection lines of the planes and the rank of a system of linear equations. The fact that the three planes intersect pairwise and that their intersection lines are mutually parallel means that their normal vectors are linearly dependent, so the rank of the coefficient matrix is less than 3.

Steps:

1. The normal vectors of the three planes are $(a_{i1},a_{i2},a_{i3})$. If the intersection lines are mutually parallel, then all three normal vectors are perpendicular to the same direction vector, so the three normal vectors are linearly dependent.
2. Hence the rank of the coefficient matrix satisfies $r(A)\le2$; and the three normal vectors cannot be collinear (otherwise the three planes would be parallel or coincident, which is a contradiction), so $r(A)=2$.
3. The rank $r(\overline A)$ of the augmented matrix depends on the right-hand side $d_i$: if $r(\overline A)=2$, then the system has solutions and the solution space is a line (of dimension 1), that is, the three planes share a common line, in which case their intersection lines could not be mutually parallel, a contradiction.
4. Hence $r(\overline A)=3$, that is, the system has no solution (the three planes have no common point). In conclusion $r(A)=2,\ r(\overline A)=3$, so (A) is the correct choice.

Answer:

A.

Approach:

This question tests the addition formula for probabilities and the relation between difference events. Rewrite $P(A)=P(B)$ as an equality of the probabilities of two difference events and then compare with the options one by one.

Steps:

1. From $P(A)=P(A\overline B)+P(AB)$ and $P(B)=P(\overline AB)+P(AB)$, subtracting the two equations gives $P(A)-P(B)=P(A\overline B)-P(\overline AB)$.
2. Hence $P(A)=P(B)$ if and only if $P(A\overline B)=P(\overline AB)$, which is exactly the content of option (C).
3. For option (A): $P(A\cup B)=P(A)+P(B)$ is equivalent to $P(AB)=0$, which only says that $A,B$ are mutually exclusive, and that is not equivalent to $P(A)=P(B)$.
4. For options (B) and (D): they correspond respectively to the independence of $A,B$ and to a certain special relation, and neither of them implies $P(A)=P(B)$. In conclusion the answer is (C).

Answer:

C.

Approach:

This question tests the linearity property of the normal distribution. The difference of two independent identically distributed normal variables is again normally distributed, and its mean and variance can be computed directly from the original distribution.

Steps:

1. Since $X,Y$ are independent and both follow $N(\mu,\sigma^2)$, we have $X-Y\sim N(\mu-\mu,\ \sigma^2+\sigma^2)=N(0,2\sigma^2)$.
2. Hence $P\{|X-Y|<1\}=P\left\{\left|\dfrac{X-Y}{\sqrt2\,\sigma}\right|<\dfrac{1}{\sqrt2\,\sigma}\right\}$.
3. The right-hand side contains $\sigma$ but not $\mu$, so this probability is independent of $\mu$ and depends on $\sigma^2$, so (A) is the correct choice.

Answer:

A.

Approach:

This question tests the computation and simplification of partial derivatives of a composite function. Here $z$ is a function of $u=\sin y-\sin x$, and after the two partial derivatives have been obtained by the chain rule the $f'$ terms cancel exactly.

Steps:

1. Put $u=\sin y-\sin x$, then $z=f(u)+xy$, $u_x=-\cos x$, $u_y=\cos y$.
2. By the chain rule $z_x=f'(u)\cdot(-\cos x)+y$ and $z_y=f'(u)\cdot\cos y+x$.
3. Hence $\dfrac{1}{\cos x}z_x+\dfrac{1}{\cos y}z_y=-f'(u)+\dfrac{y}{\cos x}+f'(u)+\dfrac{x}{\cos y}$.
4. The $f'(u)$ terms cancel each other, so the original expression $=\dfrac{y}{\cos x}+\dfrac{x}{\cos y}$.

Answer:

$\dfrac{y}{\cos x}+\dfrac{x}{\cos y}$.

Approach:

This question tests a differential equation with separable variables. Regard the left-hand side of the equation as a linear combination of the derivative of $y^2$ and of $y^2$ itself, and with the substitution $u=y^2$ the equation becomes a first-order linear one.

Steps:

1. The original equation is $2yy'-y^2-2=0$, and we note that $(y^2)'=2yy'$.
2. Put $u=y^2$, then $u'=2yy'$, and the equation becomes $u'-u-2=0$, that is $u'-u=2$.
3. The homogeneous solution is $u=Ce^x$, and the particular solution is taken to be the constant $u_0=-2$, so the general solution is $u=Ce^x-2$.
4. From $y(0)=1$ we get $u(0)=1$, and substitution gives $1=C-2$, so $C=3$, that is $u=3e^x-2$.
5. Since $y(0)>0$ we take the positive branch, giving $y=\sqrt{3e^x-2}$.

Answer:

$\sqrt{3e^x-2}$.

Approach:

This question tests the sum function of a power series. The coefficients contain $(2n)!$ while the variable appears as $x^n$, which suggests the substitution $t=\sqrt{x}$ that turns the series into an even-power series.

Steps:

1. Put $t=\sqrt{x}$ (for $x>0$ we have $t>0$), then $x^n=t^{2n}$, and the original series becomes $\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nt^{2n}}{(2n)!}$.
2. From the Taylor expansion of the cosine function $\cos t=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nt^{2n}}{(2n)!}$, the sum of the series is $\cos t$.
3. Substituting back $t=\sqrt{x}$, the sum function is $S(x)=\cos\sqrt{x}$.

Answer:

$\cos\sqrt{x}$.

Approach:

This question tests the direct computation of a second-kind surface integral on a surface. Write the upper half ellipsoid in the form $z=z(x,y)$, use the formula for the integral over a surface projected onto the $xOy$ plane to reduce it to a double integral, and then switch to polar coordinates.

Steps:

1. The upper half ellipsoid can be written as $z=\dfrac{1}{2}\sqrt{4-x^2-y^2}$, and its projection region is $D:\ x^2+y^2\le4$.
2. For the upper side (the positive orientation) $dzdx=-z_y\,dx\,dy$, where $z_y=-\dfrac{y}{2\sqrt{4-x^2-y^2}}$, so $-yz\,z_y=\dfrac{y^2}{4}$.
3. Hence $I=\iint_D\left(\dfrac{y^2}{4}+2\right)dx\,dy$.
4. Using polar coordinates $x=r\cos\theta,\ y=r\sin\theta$, $0\le r\le2$, $0\le\theta\le2\pi$: $\iint_D\dfrac{y^2}{4}dx\,dy=\dfrac14\int_0^{2\pi}\sin^2\theta\,d\theta\int_0^2r^3dr=\dfrac14\cdot\pi\cdot4=\pi$.
5. Also $\iint_D2\,dx\,dy=2\cdot\pi\cdot2^2=8\pi$, so $I=\pi+8\pi=9\pi$.

Answer:

$9\pi$.

Approach:

This question tests the solution space of the system $Ax=0$. The solution vectors of $Ax=0$ are related to the linear combinations of the column vectors that vanish, so the given linear relation immediately yields the general solution.

Steps:

1. With $A=(\alpha_1,\alpha_2,\alpha_3)$, the system $Ax=0$ is $x_1\alpha_1+x_2\alpha_2+x_3\alpha_3=0$.
2. Substituting $\alpha_3=-\alpha_1+2\alpha_2$ gives $(x_1-x_3)\alpha_1+(x_2+2x_3)\alpha_2=0$.
3. Since $\alpha_1,\alpha_2$ are linearly independent, $x_1-x_3=0,\ x_2+2x_3=0$, that is $x_1=x_3,\ x_2=-2x_3$.
4. Let $x_3=k\in\mathbb{R}$; the general solution is $x=k(1,-2,1)^T$.

Answer:

$k(1,-2,1)^T,\ k\in\mathbb{R}$.

Approach:

This question tests the properties of the distribution function and the computation of an expectation. The distribution function always takes nonnegative values, so it suffices to determine the sign of $EX-1$ in order to judge the probability of the event.

Steps:

1. Compute the expectation: $EX=\int_0^1x\cdot2x\,dx=2\int_0^1x^2\,dx=\dfrac23$, so $EX-1=-\dfrac13$.
2. The distribution function satisfies $0\le F(x)\le1$, so for the random variable $X$ we always have $F(X)\ge0$.
3. Since $0>-\dfrac13=EX-1$, the event $\{F(X)>EX-1\}$ is a sure event, and its probability is 1.

Answer:

$1$.

Approach:

This question tests the method of characteristics for a second-order partial differential equation with constant coefficients. Reducing the equation to a mixed-derivative form is equivalent to taking the two characteristic directions as the new coordinate directions, and this determines the coefficients of the transformation.

Steps:

1. For the equation $4f_{uu}+12f_{uv}+5f_{vv}=0$, find the characteristic equation: putting $f=F(u+\lambda v)$ and substituting gives $4+12\lambda+5\lambda^2=0$, whose roots are $\lambda=-2$ and $\lambda=-\dfrac25$.
2. Hence the slopes of the two characteristic directions are $dv/du=2$ and $dv/du=\dfrac25$.
3. To reduce the equation to a form involving only the mixed derivative $\dfrac{\partial^2f}{\partial x\partial y}$, the two characteristic directions must be taken as the directions of the new coordinate axes.
4. Under the form of the transformation $x=au+bv,\ y=u+v$, the correspondence between the two characteristic directions and the coordinate axes gives $a=-2,\ b=1$, that is $x=v-2u,\ y=u+v$.
5. Substituting back, the original equation reduces to a form involving only $\dfrac{\partial^2 f}{\partial x\partial y}$, that is $\dfrac{\partial^2f}{\partial x\partial y}=0$.

Answer:

$a=-2,\ b=1$, in which case the equation reduces to $\dfrac{\partial^2f}{\partial x\partial y}=0$.

Approach:

This question tests the washer method for the volume of a solid of revolution. When rotating about the line $y=1$, the radius of rotation is the distance from the curve to that line, namely $1-x^2$, and the thickness is $dx$.

Steps:

1. The curve $y=x^2$ meets the line $y=1$ at $x=\pm1$, and the bounded plane figure is $-1\le x\le1,\ x^2\le y\le1$.
2. Rotating about $y=1$, the cross-section is a disc centred on $y=1$ with radius $1-x^2$ and area $\pi(1-x^2)^2$.
3. By the washer method, $V=\pi\int_{-1}^1(1-x^2)^2\,dx$.
4. Computing: $\displaystyle\int_{-1}^1(1-x^2)^2dx=\int_{-1}^1(1-2x^2+x^4)dx=\left[x-\dfrac{2x^3}{3}+\dfrac{x^5}{5}\right]_{-1}^1=2-\dfrac43+\dfrac25=\dfrac{16}{15}$.
5. Hence $V=\dfrac{16\pi}{15}$.

Answer:

$\dfrac{16\pi}{15}$.

Approach:

This question tests the application of the mean value theorem for integrals and of Rolle’s theorem, as well as a check of the conditions under which the assertions hold. The second part requires pointing out that the conclusion does not hold under the given conditions and providing a counterexample.

Steps:

1. (1) Let $F(x)=\int_0^x f(t)dt$; then $F$ is continuous on $[0,1]$, differentiable on $(0,1)$, and $F(0)=0,\ F(1)=\int_0^1f(x)dx=0$.
2. By Rolle’s theorem there exists $\xi\in(0,1)$ with $F'(\xi)=0$, and since $F'(x)=f(x)$, we get $f(\xi)=0$.
3. (2) Check whether the assertion $f''(\eta)=2f'(\eta)$ holds under the given conditions: take the counterexample $f(x)=x-\dfrac12$, for which $\int_0^1f(x)dx=\left[\dfrac{x^2}{2}-\dfrac{x}{2}\right]_0^1=0$, so the assumptions of the question are satisfied.
4. Then $f'(x)=1,\ f''(x)=0$, so $f''(\eta)=0$ while $2f'(\eta)=2$, and the two can never be equal.
5. Hence the conclusion of the second part does not hold under the given conditions, and the question needs additional assumptions before it can be proved.

Answer:

(1) There exists $\xi\in(0,1)$ such that $f(\xi)=0$;
(2) the conclusion does not hold under the given conditions: the counterexample $f(x)=x-\dfrac{1}{2}$ shows that additional assumptions are needed before it can be proved.

Approach:

This question tests the application of the divergence theorem to a closed surface. The original surface is the lateral side of the cylinder, so when the divergence theorem is used it is combined with the two discs above and below, and the caps contribute nothing.

Steps:

1. Let $\Sigma$ be the closed surface: the lateral side $x^2+y^2=1\ (-1\le z\le1)$ together with the upper and lower discs. The vector field is $F=((x-1)^3,(y-1)^3,z-1)$.
2. The divergence is $\operatorname{div}F=3(x-1)^2+3(y-1)^2+1$.
3. On the two discs the outward normal is $(0,0,\pm1)$, while the third component of $F$ is $z-1$: it is $0$ on the upper cap $z=1$ and $-2$ on the lower cap $z=-1$, so the fluxes through the two discs are $0$ and $-2\cdot\pi=-\!2\pi$ respectively.
4. By the divergence theorem, $I=\iiint_{\Omega}\left[3(x-1)^2+3(y-1)^2+1\right]dV-\text{flux of the bottom cap}$.
5. Over the unit disc, $\iint(x-1)^2dx\,dy=\iint x^2dx\,dy+\iint1dx\,dy=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}$, and similarly $\iint(y-1)^2dx\,dy=\dfrac{5\pi}{4}$, so the triple integral over the cylinder is $\left(3\cdot\dfrac{5\pi}{4}+3\cdot\dfrac{5\pi}{4}+\pi\right)\cdot2=17\pi$, and adding the contribution of the bottom cap gives $I=17\pi$.

Answer:

$17\pi$.

Approach:

This question tests proving the convergence of a recursively defined sequence by monotonicity and boundedness, and computing a power-type limit by means of equivalent infinitesimals. The first part uses $\sin x<x$ to establish monotonicity and boundedness; in the second part the power-type limit is handled by taking logarithms and using an expansion.

Steps:

1. (1) From $x_1=1>0$ and $\sin x<x\ (x>0)$, induction gives $0<x_{n+1}=\sin x_n<x_n$, so the sequence is strictly decreasing and bounded below by $0$, and hence converges.
2. Let the limit be $L\ge0$; passing to the limit ($n\to\infty$) in $x_{n+1}=\sin x_n$ gives $L=\sin L$, that is $L=0$ (for $L>0$ one has $\sin L<L$, a contradiction).
3. (2) Denote the original limit by $M$: $\ln M=\lim\limits_{n\to\infty}\dfrac{1}{x_n^2}\ln\dfrac{x_{n+1}}{x_n}$.
4. From $\sin x=x-\dfrac{x^3}{6}+o(x^3)$ we get $\dfrac{x_{n+1}}{x_n}=\dfrac{\sin x_n}{x_n}=1-\dfrac{x_n^2}{6}+o(x_n^2)$.
5. Using $\ln(1+u)=u+o(u)$ we obtain $\ln M=\lim\limits_{n\to\infty}\dfrac{-x_n^2/6+o(x_n^2)}{x_n^2}=-\dfrac16$, so $M=e^{-1/6}$.

Answer:

(1) The limit $L$ satisfies $L=\sin L$, which gives $L=0$.
(2) From $\sin x=x-\dfrac{x^3}{6}+o(x^3)$ the limit is $e^{-1/6}$.

Approach:

This question tests the eigenvalues of a matrix and diagonalization by similarity. First compute the characteristic polynomial to obtain the eigenvalues, then solve the homogeneous systems for the corresponding eigenvectors, and finally take those eigenvectors as the columns of $P$.

Steps:

1. The characteristic polynomial is $|\lambda E-A|=(\lambda-1)^2(\lambda+5)$, so the eigenvalues are $1$ (double) and $-5$.
2. For $\lambda=1$: solve $(A-E)x=0$; from $-2x_1+2x_2+2x_3=0$ we get $x_1=x_2+x_3$, and the eigenspace has dimension 2, equal to the algebraic multiplicity, so $A$ is diagonalizable.
3. Take two linearly independent eigenvectors for $\lambda=1$ and one eigenvector for $\lambda=-5$ (for example one obtained from $2x_1+x_2+x_3=0,\ x_1+2x_2-x_3=0,\ x_1-x_2+2x_3=0$), and use them as the columns of $P$.
4. Then $P$ is invertible and $P^{-1}AP=\operatorname{diag}(1,1,-5)$, that is, $A$ is similar to a diagonal matrix whose diagonal entries are the eigenvalues.

Answer:

The eigenvalues are $\{1,1,-5\}$; one may take $P$ built from the corresponding orthogonal eigenvectors, so that $P^{-1}AP=\operatorname{diag}(1,1,-5)$.

Approach:

This question tests the polar change of variables in a double integral and the use of symmetry. The region $D$ is the right half of the unit disc, and the cross term $xy$ in the integrand is removed by symmetry.

Steps:

1. The region $D=\{(x,y)\mid x^2+y^2\le1,\ x\ge0\}$ in polar coordinates is $0\le r\le1,\ -\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$.
2. The cross term in the integrand is $xy=r^2\cos\theta\sin\theta=\dfrac{r^2}{2}\sin2\theta$, an odd function of $\theta$ on this symmetric interval, so its integral is $0$.
3. Hence $I=\iint_D\dfrac{1}{1+x^2+y^2}dx\,dy=\int_{-\pi/2}^{\pi/2}d\theta\int_0^1\dfrac{r}{1+r^2}dr$.
4. The radial integral is $\displaystyle\int_0^1\dfrac{r}{1+r^2}dr=\dfrac12\ln(1+r^2)\Big|_0^1=\dfrac12\ln2$, and the angular integral is $\pi$.
5. Hence $I=\dfrac{\pi}{2}\ln2$.

Answer:

$\dfrac{\pi}{2}\ln 2$.

Approach:

This question tests the density, the covariance and the independence of functions of random variables. The sign of $Z=XY$ is decided by the value of $Y$, so $X$ has to be transformed separately for the two values of $Y$, and the density is then obtained with the law of total probability.

Steps:

1. (1) When $Y=1$ (probability $1-p$) we have $Z=X>0$ with density $(1-p)e^{-z}$; when $Y=-1$ (probability $p$) we have $Z=-X<0$ with density $p\,f_X(-z)=p\,e^{z}$.
2. Hence $f_Z(z)=p\,e^{z}\ (z<0)$, $(1-p)e^{-z}\ (z>0)$.
3. (2) $E(X)=1,\ E(X^2)=2$, $E(Y)=(-1)p+1\cdot(1-p)=1-2p$. Since $Z=XY$ and $X,Y$ are independent, $E(XZ)=E(X^2)E(Y)=2(1-2p)$ and $E(Z)=E(X)E(Y)=1-2p$.
4. Hence $\operatorname{Cov}(X,Z)=E(XZ)-E(X)E(Z)=2(1-2p)-(1-2p)=1-2p$, and setting this equal to zero gives $p=\dfrac12$.
5. (3) $X$ and $Z$ are not independent: the sign of $Z$ is completely determined by $Y$, so when $Y=1$ we have $Z>0$ and when $Y=-1$ we have $Z<0$; hence $\{Z>0\}$ and the range of $X$ do not satisfy the requirement of independence (for instance $P\{Z>0\}=1-p$, while $P\{X>0\}=1$, and the two are not equal).

Answer:

(1) $f_Z(z)=\begin{cases}p\,e^{z}, & z<0\\(1-p)\,e^{-z}, & z>0\end{cases}$.
(2) $X$ and $Z$ are uncorrelated when $p=\dfrac{1}{2}$.
(3) They are not independent, because the sign of $Z$ is determined by $Y$ and is related to the range of $X$.