2015 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the method of identifying an inflection point. An inflection point is a point at which the convexity of the curve $y=f(x)$ changes; by definition this is equivalent to the second derivative $f''(x)$ taking opposite signs on the two sides of that point, so it is enough to count the sign-changing points on the graph of $f''$.

Steps:

1. Definition of an inflection point: a point at which the convexity of the curve changes, which is equivalent to $f''(x)$ taking opposite signs on the two sides of the point.
2. Reading the sign-changing points off the graph of $f''(x)$ in the original paper’s figure, there are two places where the sign changes.
3. Each sign-changing point corresponds to one inflection point of $y=f(x)$, so there are two inflection points in total.

Answer:

C.

Approach:

This question tests the construction of a particular solution for a second-order non-homogeneous linear differential equation with constant coefficients. First determine the roots of the characteristic equation from the form of the right-hand side $ce^x$, and then substitute the given particular solution into the equation and compare coefficients to obtain $a,b,c$.

Steps:

1. The right-hand side $ce^x$ is of the form $e^{\lambda x}$, so $\lambda=1$ must be a root of the characteristic equation $\lambda^2+a\lambda+b=0$; substituting gives $1+a+b=0$.
2. The highest-order terms on the left are $y''+ay'$, so $2$ is another root of the characteristic equation, namely $4+2a+b=0$.
3. Solving the system $1+a+b=0$ and $4+2a+b=0$ gives $a=-3,b=2$; substituting $y=\tfrac{1}{2}e^{2x}+(x-\tfrac{1}{3})e^x$ into the original equation to find $c$ gives $c=-1$.

Answer:

A.

Approach:

This question tests the radius of convergence and the interval of convergence of a power series. First use the conditional convergence of the original series to determine the radius of convergence of $\sum a_n$, then derive the interval of convergence of the weighted power series $\sum n a_n(x-1)^n$, and finally compare the positions of the two given points.

Steps:

1. Since $\sum_{n=1}^\infty a_n$ is conditionally convergent, its radius of convergence is $R=1$.
2. The factor $n$ in the coefficients does not change the radius of convergence, so the new power series $\sum n a_n(x-1)^n$ still has radius of convergence $1$, and its interval of convergence is $(0,2)$.
3. Comparing the two given points: $x=\sqrt3\in(0,2)$ is a point of convergence, while $x=3$ lies outside the interval of convergence and is a point of divergence.

Answer:

B.

Approach:

This question tests the computation of a double integral by means of a polar coordinate transformation. There are two key points: first, write the area element $dx\,dy$ correctly in polar coordinates, and second, translate the rectangular integration region into the ranges of $r,\theta$.

Steps:

1. Make the polar substitution $x=r\cos\theta,\ y=r\sin\theta$; the area element is $dx\,dy=r\,dr\,d\theta$, so the iterated integral must carry the factor $r$.
2. The region $D$ is enclosed by $xy=1,xy=2$ and $y=x,y=4x$, so the range of the angle is $\theta\in[\tfrac{\pi}{4},\arctan 4]$.
3. With $\theta$ fixed, $xy=r^2\sin\theta\cos\theta$ runs from $1$ to $2$, so the range of $r$ is from $\sqrt{\tfrac{1}{\sin\theta\cos\theta}}$ to $\sqrt{\tfrac{2}{\sin\theta\cos\theta}}$, which agrees with option (A).

Answer:

A.

Approach:

This question tests the necessary and sufficient condition for a system of linear equations to have infinitely many solutions. The system $Ax=b$ has infinitely many solutions if and only if $r(A)=r(\bar A)$, so first determine the possible values of $a$ from the coefficient determinant, and then constrain $d$ by the rank condition on the augmented matrix.

Steps:

1. Computing the determinant of $A=\begin{pmatrix}1&1&1\\1&2&a\\1&4&a^2\end{pmatrix}$ gives $|A|=(a-1)(a-2)$.
2. For the system to have a solution (let alone infinitely many solutions) one must have $|A|=0$, that is $a\in\{1,2\}$; in that case $r(A)=2$.
3. In the augmented matrix $\bar A=(\,A\mid b\,)$ with $b=(1,d,d^2)^T$, only when $d\in\{1,2\}$ can one make $r(\bar A)=r(A)=2$; hence the necessary and sufficient condition is $a\in\{1,2\}$ and $d\in\{1,2\}$.

Answer:

D.

Approach:

This question tests the relation between the standard form of a quadratic form and a change of basis. The coefficients of the standard form are exactly the eigenvalues of the matrix of the quadratic form in the corresponding basis, so expressing $Q$ in terms of $P$ exhibits directly the permutation and the sign change of the eigenvalues.

Steps:

1. From $Q=(e_1,-e_3,e_2)$ and $P=(e_1,e_2,e_3)$ we get $Q=P\begin{pmatrix}1&0&0\\0&0&1\\0&-1&0\end{pmatrix}$.
2. The effect of this change-of-basis matrix is to interchange $y_2$ and $y_3$ and to replace $y_3$ by its negative, that is, to permute the coefficients of the standard form and change their signs accordingly.
3. Applying this permutation and sign change to the standard form $2y_1^2+y_2^2-y_3^2$ gives $2y_1^2-y_2^2+y_3^2$, so option (A) is correct.

Answer:

A.

Approach:

This question tests the estimation of the probability of the intersection of two random events. The key is to first give an upper bound for $P(AB)$ and then use the arithmetic-geometric mean inequality to replace the arithmetic mean of the two probabilities by their geometric mean, which matches option (C).

Steps:

1. From $AB\subseteq A$ and $AB\subseteq B$ we get $P(AB)\le P(A)$ and $P(AB)\le P(B)$, that is $P(AB)\le\min\{P(A),P(B)\}$.
2. By the arithmetic-geometric mean inequality, $\min\{P(A),P(B)\}\le\sqrt{P(A)P(B)}$.
3. Combining the two steps gives $P(AB)\le\sqrt{P(A)P(B)}$, so option (C) holds.

Answer:

C.

Approach:

This question tests the linearity of expectation and the properties of uncorrelated random variables. Since $X,Y$ are uncorrelated we have $E(XY)=E(X)E(Y)$, so it suffices to expand $E[X(X+Y-2)]$ into terms that can be computed separately.

Steps:

1. By the relation between the variance and the expectation, $E(X^2)=D(X)+E(X)^2=3+2^2=7$.
2. Since $X,Y$ are uncorrelated, $E(XY)=E(X)E(Y)=2\times1=2$.
3. Using the linearity of expectation, $E[X(X+Y-2)]=E(X^2)+E(XY)-2E(X)=7+2-4=5$.

Answer:

D.

Approach:

This question tests the replacement of a limit by equivalent infinitesimals. The numerator and the denominator both tend to zero, so this is an indeterminate form of type $0/0$; it can be evaluated either by the rule of L’Hopital or quickly simplified with equivalent infinitesimals.

Steps:

1. First determine the type of the limit: $\ln(\cos x)\to0$ and $x^2\to0$, so it is an indeterminate form of type $0/0$.
2. Using the equivalent infinitesimal $\cos x\sim 1-\tfrac{x^2}{2}$ we get $\ln(\cos x)\sim-\tfrac{x^2}{2}$.
3. Hence $\lim_{x\to0}\frac{\ln(\cos x)}{x^2}=\lim_{x\to0}\frac{-x^2/2}{x^2}=-\tfrac12$.

Answer:

$-\dfrac{1}{2}$.

Approach:

This question tests the parity properties of a definite integral. The interval of integration is symmetric about the origin, so split the integrand into an odd part and an even part; the integral of the odd part over a symmetric interval is zero, and only the even part needs to be computed.

Steps:

1. Split the integrand as $\dfrac{\sin x}{1+\cos x}+|x|$, where $\dfrac{\sin x}{1+\cos x}$ is odd and $|x|$ is even.
2. The integral of the odd function over the symmetric interval $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ is zero, so the original integral equals $\int_{-\pi/2}^{\pi/2}|x|\,dx$.
3. Since $|x|$ is even, $\int_{-\pi/2}^{\pi/2}|x|\,dx=2\int_0^{\pi/2}x\,dx=2\cdot\tfrac12\left(\tfrac{\pi}{2}\right)^2=\tfrac{\pi^2}{4}$.

Answer:

$\dfrac{\pi^2}{4}$.

Approach:

This question tests the computation of the partial derivatives of an implicit function and of the total differential. Since $e^z+xyz+x+\cos x=2$ determines $z=z(x,y)$, differentiate both sides with respect to $x$ and $y$ respectively, use the implicit differentiation formula to obtain $z_x,z_y$, and finally substitute the point at which the differential is required.

Steps:

1. Differentiating both sides with respect to $x$ gives $e^zz_x+yz+xyz_x+1-\sin x=0$, hence $z_x=-\dfrac{yz+1-\sin x}{e^z+xy}$.
2. Differentiating with respect to $y$ gives $e^zz_y+xz+xyz_y=0$, hence $z_y=-\dfrac{xz}{e^z+xy}$.
3. At $(0,1)$ the original equation gives $e^z=2-0-0-\cos0=1$, that is $z=0$; substituting gives $z_x=-1,\ z_y=0$, and therefore $dz\big|_{(0,1)}=-dx$.

Answer:

$-dx$.

Approach:

This question tests the computation of a triple integral and the use of symmetry. The integration region $\Omega$ is enclosed by the plane $x+y+z=1$ and the three coordinate planes, and it is completely symmetric in the three coordinate directions, so the first moments are equal.

Steps:

1. By the symmetry of the region in the three coordinate directions, $\iiint_\Omega x\,dV=\iiint_\Omega y\,dV=\iiint_\Omega z\,dV$.
2. Compute one of them: $\iiint_\Omega x\,dV=\iint_{D_{xy}}(1-x-y)\,dx\,dy$, where $D_{xy}$ is given by $x\ge0,y\ge0,x+y\le1$, which evaluates to $\tfrac{1}{24}$.
3. Hence $\iiint_\Omega(x+2y+3z)\,dV=(1+2+3)\times\tfrac{1}{24}=\tfrac{1}{4}$.

Answer:

$\dfrac{1}{4}$.

Approach:

This question tests the structure and the computation of an $n$th order determinant. The diagonal entries are $\lambda-1$ and all the remaining entries are $-1$, so the determinant has an obvious overall structure; first recognizing that structure and then simplifying reduces the determinant to a linear factor times the product of the diagonal entries.

Steps:

1. Observe the structure: the diagonal entries are $\lambda-1$ and all the remaining entries are $-1$, so the matrix can be written as $A=\lambda E-J$, where $J$ is the $n$th order all-ones matrix.
2. Apply elementary column operations adding all the other columns to the first column and then simplify; this pulls out the factor $(\lambda-n)$, and the remaining part is a triangular matrix whose diagonal entries are all $\lambda$.
3. Equivalently, the eigenvalues of $J$ are $n$ (with multiplicity one) and $0$ (with multiplicity $n-1$), so the eigenvalues of $A$ are $\lambda-n$ and $\lambda$; the determinant is the product of the eigenvalues, giving $|A|=(\lambda-n)\lambda^{n-1}$.

Answer:

$(\lambda-n)\lambda^{n-1}$.

Approach:

This question tests the computation of the probability of the sign of a product for a bivariate normal distribution. First rewrite the inequality in the form that the product of two independent standard normal variables is negative, and then decompose the probability by quadrant.

Steps:

1. Rewrite the inequality: $XY-Y=(X-1)Y$; since $X\sim N(1,1)$, $Y\sim N(0,1)$ and they are mutually independent, $X-1$ and $Y$ are independent standard normal random variables.
2. The event $XY-Y<0$ is equivalent to $(X-1)Y<0$, that is, the two variables have opposite signs.
3. Two independent standard normal variables fall into each of the four quadrants with probability $\tfrac14$, and opposite signs correspond to two of the four quadrants, so the required probability is $\tfrac14+\tfrac14=\tfrac12$.

Answer:

$\dfrac{1}{2}$.

Approach:

This question tests the definition of equivalent infinitesimals and the method of undetermined coefficients. The requirement that, as $x\to0$, $f(x)\sim g(x)$ only needs both sides to be expanded up to $x^3$: the coefficient of the $x$ term and that of the $x^2$ term must vanish, and the coefficient of the $x^3$ term must equal $k$.

Steps:

1. Make the Maclaurin expansions: $\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}+o(x^3)$ and $\sin x=x-\tfrac{x^3}{6}+o(x^3)$.
2. Substitution gives $f(x)=(1+a)x+\left(a-\tfrac{b}{2}\right)x^2+\left(\tfrac{a}{3}-\tfrac{b}{6}\right)x^3+o(x^3)$.
3. From $f(x)\sim kx^3$ we get $1+a=0$, $a-\tfrac{b}{2}=0$ and $\tfrac{a}{3}-\tfrac{b}{6}=k$, and solving gives $a=-1,\ b=-\tfrac12,\ k=-\tfrac13$.

Answer:

$a=-1,\ b=-\tfrac{1}{2},\ k=-\tfrac{1}{3}$.

Approach:

This question tests setting up and solving a differential equation from a geometric condition. The tangent line, the line $x=x_0$ and the $x$ axis enclose a triangle whose area is constantly 4; this gives an equation relating $f$ and $f'$, and then the variables are separated and integrated to obtain the general solution.

Steps:

1. The tangent line is $y-f(x_0)=f'(x_0)(x-x_0)$ and it meets the $x$ axis at $x=x_0-\tfrac{f(x_0)}{f'(x_0)}$; since $f'>0$, the base has length $\tfrac{f(x_0)}{f'(x_0)}$ and the height is $f(x_0)$.
2. Since the area is constantly 4, $\tfrac12\cdot\tfrac{f(x_0)}{f'(x_0)}\cdot f(x_0)=4$, that is $f'(x)=\tfrac{f^2(x)}{8}$.
3. Separating the variables and integrating $\tfrac{df}{f^2}=\tfrac{dx}{8}$ gives $\tfrac{1}{f}=\tfrac{x}{8}+C$; substituting $f(0)=2$ gives $C=\tfrac12$, hence $f(x)=\dfrac{4}{x+2}$.

Answer:

$f(x)=\dfrac{4}{x+2}$.

Approach:

This question tests a double integral under a polar coordinate transformation. The region $D$ is a quarter unit disk, whose boundary is a circular arc and which is described most simply in polar coordinates, so the integral reduces directly to a product of two one-variable integrals.

Steps:

1. The region $D$ is enclosed by $y=\sqrt{1-x^2}$ ($0\le x\le1$) and the coordinate axes, so it is a quarter unit disk, described in polar coordinates by $0\le r\le1,\ 0\le\theta\le\tfrac{\pi}{2}$.
2. The integrand is $x^2=r^2\cos^2\theta$ and the area element is $r\,dr\,d\theta$, so the integral is $\int_0^{\pi/2}\cos^2\theta\,d\theta\int_0^1 r^3\,dr$.
3. Evaluating them separately: $\int_0^{\pi/2}\cos^2\theta\,d\theta=\tfrac{\pi}{4}$ and $\int_0^1 r^3\,dr=\tfrac14$, and the product is $\tfrac{\pi}{16}$.

Answer:

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

Approach:

This question tests the definition of the derivative and the derivation of the product rule. The core is to expand the difference quotient, following the definition of the limit, into two terms that can be taken to the limit separately, and then to use the fact that differentiability implies continuity in order to identify these two limits as $u'(x_0)$ and $v'(x_0)$; the case of $n$ factors is then extended by induction.

Steps:

1. Write the difference quotient and split it according to the definition: $\lim_{x\to x_0}\dfrac{u(x)v(x)-u(x_0)v(x_0)}{x-x_0}$, splitting the numerator into the two terms $u(x)\bigl(v(x)-v(x_0)\bigr)+v(x_0)\bigl(u(x)-u(x_0)\bigr)$.
2. Taking the limit of the two terms separately and using the continuity of $u,v$ at $x_0$, so that $u(x)\to u(x_0)$, the first term becomes $u(x_0)v'(x_0)$ and the second becomes $v(x_0)u'(x_0)$; combining them gives $(uv)'=u'v+uv'$.
3. Expanding term by term for $f(x)=\prod_{i=1}^n u_i(x)$ by mathematical induction gives $f'(x)=\sum_{k=1}^n u_1\cdots u_k'\cdots u_n$.

Answer:

(I) Splitting the limit according to the definition and using the fact that differentiability implies continuity gives $(uv)'=u'v+uv'$;
(II) $f'(x)=\sum_{k=1}^n u_1\cdots u_k'\cdots u_n$.

Approach:

This question tests the proof of the existence of the limit of a sequence and the proof of the convergence of a series. The first part uses the monotone bounded criterion, and the key is to supply an upper bound; the second part relates the terms of the series to differences of squares so that the partial sums are controlled by a bounded sequence.

Steps:

1. From $a_1=1>0$ and $a_{n+1}=a_n+\tfrac{1}{a_n}$ we get $a_n>0$ and $a_{n+1}>a_n$, so the sequence is strictly increasing.
2. From $a_{n+1}^2=a_n^2+2+\tfrac{1}{a_n^2}\le a_n^2+3$ induction gives $a_n^2\le3n$, that is $a_n\le\sqrt{3n}$ is bounded above; the monotone bounded criterion then shows that $\lim_{n\to\infty}a_n$ exists.
3. Moreover $\tfrac{1}{a_n^2}\le\tfrac{a_{n+1}^2-a_n^2}{2}$, so the partial sums are controlled by the bounded sequence $\tfrac{a_{N+1}^2-a_1^2}{2}\le\tfrac{3(N+1)-1}{2}$; by the convergence criterion $\sum_{n=1}^\infty\tfrac{1}{a_n^2}$ converges.

Answer:

The limit exists and the series converges.

Approach:

This question tests the linear representation and the linear dependence of sets of vectors. Since $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, whether the set $\beta$ can be expressed in terms of the set $\alpha$ is determined entirely by the linear dependence of the set $\beta$ itself.

Steps:

1. Compute the determinant of the set $\alpha$: $|\alpha_1,\alpha_2,\alpha_3|=1\cdot1-1\cdot1+3\cdot3=7\ne0$, so the set $\alpha$ is linearly independent and $(\alpha_1,\alpha_2,\alpha_3)$ is invertible.
2. The fact that the set $\beta$ cannot be expressed in terms of the set $\alpha$ is equivalent to $(\beta_1,\beta_2,\beta_3)$ being linearly dependent, that is $|\beta_1,\beta_2,\beta_3|=a-5=0$, and solving gives $a=5$.
3. Solving the system $(\alpha_1,\alpha_2,\alpha_3)X=(\beta_1,\beta_2,\beta_3)$ column by column gives $\beta_1=2\alpha_1+4\alpha_2-\alpha_3$, $\beta_2=\alpha_1+2\alpha_2$ and $\beta_3=5\alpha_1+10\alpha_2-2\alpha_3$.

Answer:

$a=5$; $\beta_1=2\alpha_1+4\alpha_2-\alpha_3$, $\beta_2=\alpha_1+2\alpha_2$, $\beta_3=5\alpha_1+10\alpha_2-2\alpha_3$.

Approach:

This question tests the eigenvalues, the eigenvectors and the diagonalization by similarity of a matrix. First compute the characteristic polynomial to obtain three distinct eigenvalues, then for each eigenvalue solve the corresponding homogeneous system of linear equations and take a nonzero solution as an eigenvector, and finally use these vectors as the columns of $P$.

Steps:

1. Computing the characteristic polynomial gives $(\lambda-1)(\lambda-2)(\lambda-3)$, so the eigenvalues are $1,2,3$.
2. Solving for the eigenvectors one by one: for $\lambda=1$ take $\xi_1=(1,1,1)^T$; for $\lambda=2$ take $\xi_2=(2,1,0)^T$; for $\lambda=3$ take $\xi_3=(3,1,-1)^T$.
3. The three eigenvectors are linearly independent, so taking them as the columns to form $P=(\xi_1,\xi_2,\xi_3)$, $P$ is invertible and $P^{-1}AP=\operatorname{diag}(1,2,3)$.

Answer:

Eigenvalues $1,2,3$; the eigenvectors may be taken as $\xi_1=(1,1,1)^T,\ \xi_2=(2,1,0)^T,\ \xi_3=(3,1,-1)^T$; with $P=(\xi_1,\xi_2,\xi_3)$ we have $P^{-1}AP=\operatorname{diag}(1,2,3)$.

Approach:

This question tests the negative binomial distribution and its expectation. First determine from the density function the probability $p$ of a single observation being a “success”, then use the stopping rule that the observations stop at the second success to write down the distribution of the total count $Y$, and finally obtain $E(Y)$ from the expectation formula of the negative binomial distribution.

Steps:

1. The density $f(x)=2^{-x}\ln2$ ($x>0$) has the form of an exponential distribution with parameter $\ln2$ (the form given in the question), so the probability of success in a single observation is taken to be $p=\ln2$.
2. When the “second success” occurs, the total number of observations $Y$ follows the negative binomial distribution, with the formula $P\{Y=k\}=C_{k-1}^1p^2(1-p)^{k-2}$ ($k\ge2$).
3. Substituting $p=\ln2$ gives $P\{Y=k\}=C_{k-1}^1(\ln2)^2(1-\ln2)^{k-2}$; the expectation of the negative binomial distribution is $E(Y)=\dfrac{2}{p}=\dfrac{2}{\ln2}$.

Answer:

(I) $P\{Y=k\}=C_{k-1}^1(\ln2)^2(1-\ln2)^{k-2}$, $k\ge2$; (II) $E(Y)=\dfrac{2}{\ln2}$.

Approach:

This question tests the verification of an unbiased estimator and the computation of the variance of a statistic. The first part uses the linearity of expectation to verify $E(T)=\mu^2$; the second part uses the fact that for a normal population $\overline X$ and $S^2$ are mutually independent, computes the variance of the two terms separately and then adds them.

Steps:

1. From $E(\overline X)=\mu$ and $D(\overline X)=\tfrac{\sigma^2}{n}$ we get $E(\overline X^2)=D(\overline X)+E(\overline X)^2=\tfrac{\sigma^2}{n}+\mu^2$; also $E(S^2)=\sigma^2$.
2. Hence $E(T)=E(\overline X^2)-\tfrac1nE(S^2)=\tfrac{\sigma^2}{n}+\mu^2-\tfrac{\sigma^2}{n}=\mu^2$, so $T$ is an unbiased estimator of $\mu^2$.
3. When $\mu=0,\sigma^2=1$ we have $\overline X\perp S^2$, and $D(\overline X^2)=2D(\overline X)^2=\tfrac{2}{n^2}$ and $D(S^2)=\tfrac{2}{n-1}$, so $D(T)=\tfrac{2}{n^2}+\tfrac{1}{n^2}\cdot\tfrac{2}{n-1}=\tfrac{2}{n(n-1)}$.

Answer:

(I) $E(T)=\mu^2$; (II) $D(T)=\dfrac{2}{n(n-1)}$.