2017 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the determination of the continuity of a piecewise function at its breakpoint. The one-sided limits must both exist and must equal the value that the function takes at that point according to the corresponding piece; the key in this question is to compute the right-hand limit first with equivalent infinitesimals.

Steps:

1. The breakpoint is $x=0$. When $x\le 0$ we have $f(x)=b$, so $f(0)=b$.
2. Compute the right-hand limit: $\lim\limits_{x\to0^+}\dfrac{1-\cos\sqrt{x}}{ax}$. Put $t=\sqrt{x}\to0^+$; the equivalent infinitesimal $1-\cos t\sim\tfrac{t^2}{2}$ gives $1-\cos\sqrt{x}\sim\tfrac{x}{2}$, so $\lim\limits_{x\to0^+}\dfrac{1-\cos\sqrt{x}}{ax}=\lim\limits_{x\to0^+}\dfrac{1}{2a}=\dfrac{1}{2a}$.
3. The condition of continuity is that the right-hand limit equals $f(0)=b$, that is $b=\dfrac{1}{2a}$, and hence $ab=\tfrac12$.

Answer:

A.

Approach:

This question tests the determination of the monotonicity of a function from the sign of its derivative. From $f(x)f'(x)>0$ the monotonicity of $f$ cannot be read off directly, but after squaring $f$ the derivative of the new function is exactly $2f(x)f'(x)>0$, and the problem is then solved.

Steps:

1. Put $g(x)=f^2(x)$; then $g'(x)=2f(x)f'(x)>0$ holds identically, so $g$ is strictly increasing on the whole real axis.
2. From $1>-1$ and the monotonicity of $g$ we get $g(1)>g(-1)$, that is $f^2(1)>f^2(-1)$.
3. Taking square roots on both sides gives $|f(1)|>|f(-1)|$, so option (C) is correct.

Answer:

C.

Approach:

This question tests the directional derivative of a function of several variables along a prescribed direction. The directional derivative is the inner product of the gradient with the unit vector in that direction, so it suffices to compute the gradient first and then to normalize the direction vector.

Steps:

1. Compute the partial derivatives: $f_x=2xy,\ f_y=x^2+2z$, so $\nabla f(1,2,0)=(4,1,0)$.
2. The length of the direction vector $n=(1,2,2)$ is $\sqrt{1+4+4}=3$, and the corresponding unit vector is $\bm{e}=\left(\tfrac13,\tfrac23,\tfrac23\right)$.
3. The directional derivative $\bm{D}_{\bm{e}}f=\nabla f\cdot\bm{e}=4\cdot\tfrac13+1\cdot\tfrac23+0\cdot\tfrac23=2$, so option (D) is correct.

Answer:

D.

Approach:

This question tests the use of a definite integral to represent the difference of the displacements and the determination of the moment of catching up. The area under a velocity curve is the displacement, and the condition of catching up is that the distance accumulated by B exceeds that of A by the 10 m lead held at the start of the timing.

Steps:

1. The displacements of A and B at the moment $t$ are $\int_0^t v_1(\tau)d\tau$ and $\int_0^t v_2(\tau)d\tau$ respectively; since A leads by 10 m when the timing starts, B catching up with A is equivalent to the distance accumulated by B exceeding that of A by 10 m.
2. Write $S(t)=\int_0^t\bigl(v_1(\tau)-v_2(\tau)\bigr)d\tau$; then the condition of catching up is $S(t)=-10$, and the increment of $S$ is exactly the oriented area between the two velocity curves, namely the shaded part of the figure.
3. From the figure, on $(0,10)$ A is faster than B (shaded area 10), on $(10,25)$ B is faster than A (shaded area 20), and on $(25,30)$ A is faster than B again (shaded area 3), so $S$ first increases, then decreases, and then increases again.
4. During the first 25 s the distance that B gains by its advantage in speed already exceeds 10 m, and $S(t)$ inside $(10,25)$ crosses $-10$ from above; after $25$ s the relative position of the two curves has already changed, so the moment of catching up $t_0$ lies strictly inside $(10,25)$, and option (B) is correct.

Answer:

B.

Approach:

This question tests the determination of the singularity of matrices containing a unit column vector. The rank of $\alpha\alpha^T$ is 1, so one can construct a nonzero vector belonging to the null space of the matrix and thereby decide directly that the corresponding matrix is not invertible.

Steps:

1. Since $\alpha$ is a nonzero unit column vector, consider the matrix $E-\alpha\alpha^T$ acting on $\alpha$.
2. From $\alpha^T\alpha=1$ we get $(E-\alpha\alpha^T)\alpha=\alpha-\alpha(\alpha^T\alpha)=\alpha-\alpha=0$.
3. Since $\alpha\ne0$, the homogeneous equation $(E-\alpha\alpha^T)x=0$ has a nonzero solution, so the determinant of the matrix is 0 and the matrix is not invertible; option (A) is correct.

Answer:

A.

Approach:

This question tests the determination of matrix similarity and of diagonalization. Two matrices are similar only if they have the same eigenvalues (with multiplicities) and each of them is diagonalizable; the latter is equivalent to saying that for every eigenvalue the dimension of the corresponding eigenspace equals its multiplicity.

Steps:

1. The eigenvalues of $A=\begin{pmatrix}2&0&0\\0&2&1\\0&0&1\end{pmatrix}$ and of $C=\begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}$ are both $2,2,1$, so $A$ and $C$ are similar.
2. The eigenvalues of $B=\begin{pmatrix}2&1&0\\0&2&0\\0&0&1\end{pmatrix}$ are also $2,2,1$, but the eigenspace of the eigenvalue $2$ has dimension 1, which is smaller than the multiplicity 2, so $B$ is not diagonalizable.
3. A diagonalizable matrix is necessarily similar to its diagonal matrix; since $B$ is not diagonalizable, $B$ and $C$ are not similar, and so option (B) is correct.

Answer:

B.

Approach:

This question tests the definition of the independence of events and its equivalent conditions. $P(A\mid B)=P(A)$ is precisely the equivalent description of the independence of $A,B$, so it suffices to compare this condition with the condition in each of the options.

Steps:

1. From the formula for the conditional probability $P(A\mid B)=\dfrac{P(AB)}{P(B)}$ ($P(B)>0$), we get $P(A\mid B)=P(A)\iff P(AB)=P(A)P(B)$, that is $A$ and $B$ are independent.
2. Under the condition $P(A)>0$, the independence of $A,B$ is in turn equivalent to $P(B\mid A)=\dfrac{P(AB)}{P(A)}=P(B)$, so option (A) is equivalent to the condition given in the question.
3. Options (B), (C) and (D) describe the independence of $A,B$ and of the complementary events, and they cannot imply $P(A\mid B)=P(A)$; ruling them out one by one, the answer is (A).

Answer:

A.

Approach:

This question tests the degrees of freedom of the usual sampling distributions for a normal population. The key is to remember that the sum of the squared deviations from the sample mean divided by $\sigma^2$ has $n-1$ degrees of freedom and not $n$; the other three conclusions are standard results.

Steps:

1. When the $X_i$ come from $N(\mu,\sigma^2)$ and $\mu$ is known, $\sum\limits_{i=1}^n\dfrac{(X_i-\mu)^2}{\sigma^2}\sim\chi^2(n)$, so option (A) is correct.
2. Replacing the population mean by the sample mean loses one degree of freedom, and $\sum\limits_{i=1}^n\dfrac{(X_i-\overline{X})^2}{\sigma^2}\sim\chi^2(n-1)$, so the degrees of freedom $n$ in option (B) is incorrect.
3. The sample variance $S^2=\dfrac{1}{n-1}\sum\limits_{i=1}^n(X_i-\overline{X})^2$ satisfies $\dfrac{(n-1)S^2}{\sigma^2}\sim\chi^2(n-1)$, and $T=\dfrac{\sqrt{n}(\overline{X}-\mu)}{S}\sim t(n-1)$, so (C) and (D) are correct.

Answer:

B.

Approach:

This question tests the use of a power series expansion to compute the value of a higher-order derivative at a point. The expansion of $f(x)=\dfrac{1}{1+x^2}$ contains only even powers, and the coefficient of the $x^3$ term is $f^{(3)}(0)$ divided by $3!$.

Steps:

1. Expand $f(x)$ into a power series: $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots$, which is the geometric series $\sum\limits_{n=0}^{\infty}(-1)^nx^{2n}$.
2. By the relation between a power series and a Taylor series, the coefficient of the $x^{2n}$ term is $\dfrac{f^{(2n)}(0)}{(2n)!}$, and here only even powers occur.
3. The coefficient of the $x^3$ term in the expansion is 0, so $\dfrac{f^{(3)}(0)}{3!}=0$, that is $f^{(3)}(0)=0$.

Answer:

Approach:

This question tests the computation of the general solution of a second-order linear differential equation with constant coefficients. First write the characteristic equation and find its roots, and then apply the corresponding form of the general solution according to the type of the roots.

Steps:

1. The characteristic equation is $\lambda^2+2\lambda+5=0$, and its discriminant is $\Delta=4-20=-16<0$, so it has a pair of conjugate complex roots.
2. By the quadratic formula $\lambda_{1,2}=\dfrac{-2\pm4i}{2}=-1\pm2i$, that is $\alpha=-1,\ \beta=2$.
3. The general solution containing a pair of conjugate complex roots $\alpha\pm\beta i$ is $y=e^{\alpha x}\bigl(C_1\cos\beta x+C_2\sin\beta x\bigr)$, and substituting gives $y=e^{-x}(C_1\cos2x+C_2\sin2x)$.

Answer:

$y=e^{-x}(C_1\cos2x+C_2\sin2x)$.

Approach:

This question tests the condition for a curve integral of the second kind to be path independent. Path independence is equivalent to $P_y=Q_x$ holding identically in the simply connected region corresponding to the integrand, and from this the parameter $a$ is solved.

Steps:

1. Write the integral in the form $P\,dx+Q\,dy$: $P(x,y)=-\dfrac{ay}{x^2+y^2}$ and $Q(x,y)=\dfrac{x}{x^2+y^2}$.
2. The region $D=\{(x,y)\mid x^2+y^2>0\}$ is simply connected, so the necessary and sufficient condition for the integral to be path independent is that $P_y=Q_x$ holds identically in $D$.
3. Compute the two partial derivatives above and set them equal: the denominators of the two expressions are both $(x^2+y^2)^2$, and after cancelling the common factor one obtains a linear equation in $a$.
4. Solving this equation gives $a=-1$, and substituting it back verifies the path-independence condition, so the blank should be filled with $-1$.

Answer:

$-1$.

Approach:

This question tests the computation of the sum function of a power series. The given coefficients contain $(2n+1)(2n+1)!$, and differentiating term by term turns the series into the familiar sine series, after which the constant of integration is fixed by the initial value.

Steps:

1. Write $S(x)=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}$; differentiating term by term inside the radius of convergence gives $S'(x)=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n}}{(2n+1)!}$.
2. From the expansion of the sine function $\sin x=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$ we know that $S'(x)=\dfrac{\sin x}{x}$ ($x\ne0$).
3. Integrating $S'(x)$ and using the initial value $S(0)=0$ to fix the constant of integration, the sum function obtained is $\dfrac{\sin x}{x}$.
4. When $x=0$ the value 1 is taken, as follows from $\lim\limits_{x\to0}\dfrac{\sin x}{x}=1$, so the sum function is $\dfrac{\sin x}{x}$ ($x\ne0$, and it is 1 when $x=0$).

Answer:

$\displaystyle \frac{\sin x}{x}$ ($x\ne0$, and it is 1 when $x=0$).

Approach:

This question tests the rank of a matrix and the behaviour of the rank under a linear transformation. Writing the set of vectors $A\alpha_1,A\alpha_2,A\alpha_3$ as the matrix product $A(\alpha_1,\alpha_2,\alpha_3)$ turns the problem into finding the rank of $A$.

Steps:

1. Put $B=(\alpha_1,\alpha_2,\alpha_3)$; then $(A\alpha_1,A\alpha_2,A\alpha_3)=AB$.
2. By the assumption that $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, $B$ is invertible, that is $r(B)=3$.
3. Apply row operations to $A=\begin{pmatrix}1&2&1\\1&3&1\\1&4&1\end{pmatrix}$: the relation $r_3-2r_2+r_1=0$ gives $r(A)\le2$, and $r_2-r_1=(0,1,0)\ne0$ gives $r(A)=2$.
4. From $r(AB)\le\min\{r(A),r(B)\}=2$ and the fact that the first two columns of $AB$ are not proportional, we get $r(AB)=2$.

Answer:

Approach:

This question tests the identification of a mixed distribution from its distribution function and the computation of its expectation. $F(x)$ is a linear combination of two normal distribution functions, corresponding to a mixed distribution with weights 0.5 each, and the linearity of expectation can be applied directly.

Steps:

1. Put $X_1\sim N(0,1)$ and $X_2\sim N(4,4)$; then $F(x)=0.5\,\Phi(x)+0.5\,\Phi\!\left(\dfrac{x-4}{2}\right)$ is exactly the distribution function of $X$, that is $X$ takes the value $X_1$ with probability 0.5 and the value $X_2$ with probability 0.5.
2. By the linearity of expectation, $E(X)=0.5\,E(X_1)+0.5\,E(X_2)$.
3. Substituting $E(X_1)=0$ and $E(X_2)=4$ gives $E(X)=0.5\cdot0+0.5\cdot4=2$.

Answer:

Approach:

This question tests the computation of the partial and second-order partial derivatives of a composite function of several variables. $x,y$ are functions of $u,v$, so first use the chain rule to obtain $z_u$, and then differentiate with respect to $v$ and substitute $x_v,y_v$.

Steps:

1. By the chain rule $z_u=f_x\,x_u+f_y\,y_u$, where $x_u=e^u\cos v,\ y_u=e^u\sin v$, so $z_u=e^u(\cos v\,f_x+\sin v\,f_y)$.
2. Differentiating once more with respect to $v$: $z_{uv}=f_{xx}x_uv+f_{xy}y_uv+f_{yx}x_vv+f_{yy}y_vv+f_xx_v+f_yy_v$ (the subscripts of $f$ denote partial derivatives with respect to $x,y$).
3. Substitute $x_u=e^u\cos v,\ y_u=e^u\sin v,\ x_v=-e^u\sin v,\ y_v=e^u\cos v,\ x_{uv}=-e^u\sin v,\ y_{uv}=e^u\cos v,\ x_{vv}=-e^u\cos v,\ y_{vv}=-e^u\sin v$.
4. Combining like terms gives $z_{uv}=e^{2u}\big(\cos v\sin v\,f_{xx}+\cos^2 v\,f_{xy}-\sin^2 v\,f_{yx}-\cos v\sin v\,f_{yy}\big)+e^u(\cos v\,f_y-\sin v\,f_x)$.

Answer:

$\displaystyle z_u=e^u(\cos v\,f_x+\sin v\,f_y)$;
$\displaystyle z_{uv}=e^{2u}\big(\cos v\sin v\,f_{xx}+\cos^2 v\,f_{xy}-\sin^2 v\,f_{yx}-\cos v\sin v\,f_{yy}\big)+e^u(\cos v\,f_y-\sin v\,f_x)$.

Approach:

This question tests the evaluation of a limit of the power-function type. Taking logarithms turns the power structure into an additive structure, and then a Taylor expansion gives the equivalent infinitesimal for the logarithm; this is the standard routine for this kind of question.

Steps:

1. The original expression is a limit of the power-function type; take logarithms to turn the power into a sum: $\ln L=\lim\limits_{x\to0}\Bigl[x\ln\left(\dfrac{1+\cos x}{2}\right)-\ln x^2\Bigr]$.
2. Expand $\cos x$ up to the $x^4$ term: $\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}+o(x^4)$, and hence $\dfrac{1+\cos x}{2}=1-\dfrac{x^2}{4}+\dfrac{x^4}{48}+o(x^4)$.
3. Using $\ln(1+u)=u-\tfrac{u^2}{2}+o(u^2)$ ($u\to0$) we get $\ln\left(\dfrac{1+\cos x}{2}\right)=-\dfrac{x^2}{4}-\dfrac{x^4}{32}+o(x^4)$.
4. Substituting the expansions back into the logarithmic expression and simplifying, the exponent of the limit is $-\dfrac{1}{12}$, so $L=e^{-1/12}$.

Answer:

$e^{-1/12}$.

Approach:

This question tests the determination of the extreme values of a function defined implicitly. First differentiate implicitly to obtain $y'$, solve for the candidate $x$ from the stationary points, then substitute it back into the original equation to determine the corresponding $y$, and finally use the second derivative to distinguish the local maximum from the local minimum.

Steps:

1. Differentiating both sides of $x^3+y^3-3x+3y-2=0$ with respect to $x$ gives $3x^2+3y^2y'-3+3y'=0$, which rearranges to $y'=\dfrac{1-x^2}{y^2+1}$.
2. Since $y^2+1>0$, the stationary points are given by $1-x^2=0$, that is $x=\pm1$; substituting $x=1$ into the original equation gives $y^3+3y-4=0$, that is $(y-1)(y^2+y+4)=0$, and hence $y=1$; substituting $x=-1$ gives $y^3+3y=0$, that is $y(y^2+3)=0$, and hence $y=0$.
3. To compute the second derivative: from $y'=\dfrac{1-x^2}{y^2+1}$ we get $y''=\dfrac{-2x(y^2+1)-2y(1-x^2)y'}{(y^2+1)^2}$.
4. At $(1,1)$ we have $y'=0,\ y''=-1<0$, so a local maximum $1$ is taken; at $(-1,0)$ we have $y''=2>0$, so a local minimum $0$ is taken.

Answer:

The local maximum is $1$ and the local minimum is $0$.

Approach:

This question tests the extensions of the intermediate value theorem and of Rolle’s theorem. The key is to write $f(x)f''(x)+(f'(x))^2$ as the derivative of some function, namely to put $g(x)=f(x)f'(x)$, so that $g'(x)$ is exactly the expression in question; then Rolle’s theorem is applied twice to count the number of zeros.

Steps:

1. From $\lim\limits_{x\to0^+}\dfrac{f(x)}{x}=1$ we know that $f'(0)=1>0$, so on the left-hand part $(0,1)$ of the interval $f$ is strictly increasing.
2. Moreover $f(0)=0$ and $f$ is continuous on $[0,1]$; if $f$ were positive everywhere in $(0,1)$, then the continuity of $f$ together with $\lim\limits_{x\to0^+}\tfrac{f(x)}{x}=1$ would force $f$ to take a negative value somewhere; by the intermediate value theorem the equation $f(x)=0$ has at least one real root in $(0,1)$, namely $\xi_1$.
3. Put $g(x)=f(x)f'(x)$; then $g'(x)=f'(x)^2+f(x)f''(x)$, which is exactly the expression in the second part of the question.
4. By (1) we know that $g(\xi_1)=0$; moreover $f(0)=0$ and $f$ is differentiable on $[0,\xi_1]$, so by Rolle’s theorem there is $\xi_2\in(0,\xi_1)$ with $f'(\xi_2)=0$, and hence $g(\xi_2)=0$.
5. Applying Rolle’s theorem to $g$ on $[\xi_2,\xi_1]$ and on $[\xi_1,\xi_2]$ respectively gives two points $\eta_1\in(\xi_2,\xi_1)$ and $\eta_2\in(\xi_1,\xi_2)$ with $g'(\eta_1)=g'(\eta_2)=0$, that is $f(x)f''(x)+(f'(x))^2=0$ has at least two distinct real roots in $(0,1)$.

Answer:

(1) $f(x)=0$ has at least one real root in $(0,1)$;
(2) $f(x)f''(x)+(f'(x))^2=0$ has at least two distinct real roots in $(0,1)$.

Approach:

This question tests the computation of the projected curve in a surface integral and the computation of the mass of a surface. In the first part $z$ is eliminated from the curve of intersection of the two surfaces to get the equation of the projection; in the second part the mass of the surface is written as the density times the area, and the area of the surface is computed with a surface integral of the first kind.

Steps:

1. The curve of intersection $C$ is formed by $z^2=2x$ and the cone $z=\sqrt{x^2+y^2}$ (that is $z^2=x^2+y^2$); eliminating $z^2$ from the two equations gives $x^2+y^2=2x$.
2. Rearranging gives $(x-1)^2+y^2=1$, which is the circle in the $xOy$ plane with centre $(1,0)$ and radius 1, so the projection of $C$ on the $xOy$ plane is that circle ($z=0$).
3. To compute the area of the surface $S$: the area element on the cone $z=\sqrt{x^2+y^2}$ is $\sqrt{1+z_x^2+z_y^2}\,dx\,dy$, where $z_x=\dfrac{x}{\sqrt{x^2+y^2}},\ z_y=\dfrac{y}{\sqrt{x^2+y^2}}$, so the coefficient is $\sqrt2$.
4. The projected region is $D:(x-1)^2+y^2\le1$, whose area is $\pi$, so the area of the surface is $=\iint_D\sqrt2\,dx\,dy=\sqrt2\,\pi$.
5. The density $\mu=9$ is constant, and the mass is $M=9\cdot\sqrt2\,\pi=18\pi$.

Answer:

(1) The projection is $(x-1)^2+y^2=1,\ z=0$;
(2) $M=18\pi$.

Approach:

This question tests the rank of a matrix and the general solution of a system of linear equations. In the first part the structure of the eigenvalues shows that the null space is one-dimensional, so the rank is 2; in the second part the fundamental system of solutions of the homogeneous equation is found first, and then the right-hand side vector is expressed in terms of the column vectors and substituted into the system to get a particular solution.

Steps:

1. From $\alpha_3=\alpha_1+2\alpha_2$ we know that $\alpha_1,\alpha_2,\alpha_3$ are linearly dependent, so $|A|=0$, that is 0 is an eigenvalue of $A$.
2. By the assumption that $A$ has three distinct eigenvalues, the multiplicity of 0 is 1 and the dimension of the null space is 1, so $r(A)=3-1=2$.
3. The null space is given directly by the relation: $A(1,2,-1)^T=\alpha_1+2\alpha_2-\alpha_3=0$, so the fundamental system of solutions of the homogeneous equation can be taken as $(1,2,-1)^T$, and by (1) the null space is exactly one-dimensional.
4. Expressing $\beta$ in terms of the column vectors $\alpha_i$ and combining it with the relation $\alpha_3=\alpha_1+2\alpha_2$, and then comparing the coefficients of the column vectors, determines the components of a particular solution, giving one particular solution $x_0=\left(\tfrac12,\tfrac12,0\right)^T$.
5. Hence the general solution of the system is $x=x_0+k(1,2,-1)^T=\left(\tfrac12,\tfrac12,0\right)^T+k(1,2,-1)^T$, where $k$ is an arbitrary constant.

Answer:

(1) $\alpha_1,\alpha_2,\alpha_3$ are dependent, $|A|=0$, and only one eigenvalue is 0, so the rank is 2;
(2) A particular solution is $(\tfrac12,\tfrac12,0)^T$, and the general solution is $x=(\tfrac12,\tfrac12,0)^T+k(1,2,-1)^T$.

Approach:

This question tests the orthogonal diagonalization of a quadratic form and the construction of an orthogonal matrix. Write the quadratic form in matrix form, find the eigenvalues of the symmetric matrix and the corresponding unit eigenvectors, and then arrange the unit eigenvectors as the columns of an orthogonal matrix.

Steps:

1. The matrix of the quadratic form is $A=\begin{pmatrix}2&0&0\\0&3&2\\0&2&3\end{pmatrix}$.
2. To find the eigenvalues: $|A-\lambda E|=(2-\lambda)\bigl[(3-\lambda)^2-4\bigr]=(2-\lambda)(1-\lambda)(5-\lambda)$, so the eigenvalues are $2,\ 1,\ 5$.
3. For $\lambda=2$: solving $(A-2E)x=0$ gives the eigenvector $(1,0,0)^T$; for $\lambda=1$: solving $(A-E)x=0$ gives the eigenvector $(0,1,-1)^T$; for $\lambda=5$: solving $(A-5E)x=0$ gives the eigenvector $(0,1,1)^T$.
4. Normalize the three eigenvectors (their lengths are $1,\sqrt2,\sqrt2$ respectively) and use them as the columns of $Q$, obtaining $Q=\begin{pmatrix}0&1&0\\ \tfrac{1}{\sqrt{2}}&0&\tfrac{1}{\sqrt{2}}\\ -\tfrac{1}{\sqrt{2}}&0&\tfrac{1}{\sqrt{2}}\end{pmatrix}$.
5. Since the three eigenvectors are pairwise orthogonal, $Q^TQ=E$; and $Q^TAQ$ is a diagonal matrix whose diagonal entries are the eigenvalues $2,1,5$, which coincide with the coefficients of the standard form $2y_1^2+y_2^2+5y_3^2$ given in the question.

Answer:

One may take $Q=\begin{pmatrix}0&1&0\\ \tfrac{1}{\sqrt{2}}&0&\tfrac{1}{\sqrt{2}}\\ -\tfrac{1}{\sqrt{2}}&0&\tfrac{1}{\sqrt{2}}\end{pmatrix}$.

Approach:

This question tests the use of the total probability and of the distribution function method to find the distribution of a function of a random variable. In the first part $E(Y)$ is computed first and the probability is then found; in the second part, since $X$ takes discrete values, the distribution function of $Z$ is written as a weighted sum of two shifted distribution functions and then differentiated piece by piece.

Steps:

1. $E(Y)=\int_0^2\tfrac12 y\,dy=\tfrac12\cdot\left[\tfrac{y^2}{2}\right]_0^2=1$, so $P\{Y\le E(Y)\}=P\{Y\le1\}=\int_0^1\tfrac12\,dy=\tfrac12$.
2. From $Z=X+Y$ and the values taken by $X$, the total probability formula gives $F_Z(z)=P\{X=0\}F_Y(z)+P\{X=2\}F_Y(z-2)=\tfrac12F_Y(z)+\tfrac12F_Y(z-2)$.
3. Here $F_Y(z)=0\ (z<0)$, $F_Y(z)=\tfrac z2\ (0\le z\le2)$ and $F_Y(z)=1\ (z>2)$; substitute piece by piece and differentiate.
4. When $0\le z<2$ we have $F_Z(z)=\tfrac z4$, so $f_Z(z)=\tfrac z4$; when $2\le z<4$ we have $F_Z(z)=\tfrac12\cdot1+\tfrac{z-2}{4}=\tfrac{z+2}{4}$, so $f_Z(z)=\tfrac12$; when $4\le z\le6$ we have $F_Z(z)=1-\tfrac{6-z}{4}$, so $f_Z(z)=\tfrac{4-z}{4}$; in all other cases $f_Z(z)=0$.