CALCU-(1) 曲线 y=x (e+1x-1) 的渐近线方程为( )(A) y=x+e(B)

(1) 曲线

y=x \ln \left(e+\frac{1}{x-1}\right)

的渐近线方程为( ) (A)

y=x+e

(B)

y=x+\frac{1}{e}

(C)

y=x

(D)

y=x-\frac{1}{e}

Generated Graph

Solution by Steps

step 2

For vertical asymptotes, we set the denominator of the fraction inside the logarithm to zero:

x - 1 = 0

, which gives us

x = 1

. So there is a vertical asymptote at

x = 1

step 3

For horizontal asymptotes, we examine the behavior of

y

as

x

approaches infinity. The term

\frac{1}{x - 1}

approaches 0, so

\ln(e + \frac{1}{x - 1})

approaches

\ln(e)

, which is 1

step 4

Therefore, the function

y = x \ln\left(e + \frac{1}{x - 1}\right)

is asymptotic to

y = x \cdot 1 = x

. Additionally, since

\ln(e) = 1

, the constant term

\frac{1}{e}

is the horizontal shift of the asymptote, resulting in the asymptote

y = x + \frac{1}{e}

B

Key Concept

Asymptotes of a function

Explanation

To find the horizontal asymptote of the function

y = x \ln\left(e + \frac{1}{x - 1}\right)

, we evaluate the limit of the logarithmic expression as

x

approaches infinity. The term

\frac{1}{x - 1}

becomes negligible, and the logarithm of

e

is 1, leading to the asymptote

y = x + \frac{1}{e}

.

(2) 若微分方程

y^{n \prime}+a y^{\prime}+b y=0

的解在

(-\infty,+\infty)

上有界, 则( ) (A) a<0, b>0 (B) a>0, b>0 (C) a=0, b>0 (D) a=0, b<0

Solution by Steps

step 2

For the solution to be bounded, the characteristic equation associated with the differential equation must have roots with non-positive real parts. The characteristic equation is

r^n + a r + b = 0

step 3

If a < 0 and b > 0 , the roots of the characteristic equation will have non-positive real parts, ensuring that the solution

y

is bounded

step 4

Therefore, the correct answer is that

a

must be less than 0 and

b

must be greater than 0 for the solution to be bounded

A

Key Concept

Characteristic Equation and Bounded Solutions

Explanation

For a linear homogeneous differential equation, the solution is bounded if the roots of the characteristic equation have non-positive real parts. This occurs when a < 0 and b > 0 in the given equation.

（3）设函数

y=f(x)

由

\left\{\begin{array}{l}x=2 t+|t| \\ y=|t| \sin t\end{array}\right.

确定, 则() (A)

f(x)

连续,

f^{\prime}(0)

不存在 (B)

f^{\prime}(0)

存在,

f^{\prime}(x)

在

x=0

处不连续 (C)

f^{\prime}(x)

连续,

f^{\prime \prime}(0)

不存在 (D)

f^{\prime \prime}(0)

存在,

f^{\prime \prime}(x)

在

x=0

处不连续

Solution by Steps

step 2

For

t \geq 0

,

x = 3t

and

y = t \sin(t)

; for t < 0 ,

x = t

and

y = -t \sin(t)

..

step 3

To find

f'(0)

, we need to find the derivatives of

y

with respect to

t

and

x

with respect to

t

, and then use the chain rule

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

..

step 4

The derivative

\frac{dx}{dt}

is

3

for

t \geq 0

and

1

for t < 0 ; the derivative

\frac{dy}{dt}

is

\sin(t) + t\cos(t)

for

t \geq 0

and

-\sin(t) - t\cos(t)

for t < 0 ..

step 5

At

t = 0

,

\frac{dy}{dt} = 0

and

\frac{dx}{dt} = 3

for

t \geq 0

, and

\frac{dy}{dt} = 0

and

\frac{dx}{dt} = 1

for t < 0 , so

f'(0)

exists and is equal to

0

..

step 6

To determine the continuity of

f'(x)

at

x = 0

, we need to check if the left-hand and right-hand derivatives at

x = 0

are equal..

step 7

Since

f'(0)

exists and is the same from both sides,

f'(x)

is continuous at

x = 0

..

step 8

To find

f''(0)

, we would need to differentiate

f'(x)

with respect to

x

again, but since

f'(x)

involves absolute value and sine functions, it may not be differentiable at

t = 0

..

step 9

Therefore,

f(x)

is continuous,

f'(0)

exists, and

f'(x)

is continuous at

x = 0

, but

f''(0)

may not exist..

C

Key Concept

Continuity and Differentiability of Parametric Functions

Explanation

To determine the continuity and differentiability of a function defined parametrically, we can use the derivatives with respect to the parameter and apply the chain rule. The existence of the first derivative at a point implies continuity at that point, but the existence of the second derivative requires further investigation.

(4) 已知 a_{n}<b_{n}(n=1,2, \cdots), 若级数

\sum_{n=1}^{\infty} a_{n}

与

\sum_{n=1}^{\infty} b_{n}

均收玫, 则 “

\sum_{n=1}^{\infty} a_{n}

绝对收玫”是 “

\sum_{n=1}^{\infty} b_{n}

绝对收敛的” ( ) (A) 充分必要条件 (B) 充分不必要条件 (C) 必要不充分条件 (D) 既不充分也不必要条件

Solution by Steps

step 2

Absolute convergence of a series means that the series formed by taking the absolute values of its terms also converges

step 3

If

\sum_{n=1}^{\infty} b_n

converges absolutely, it implies that

\sum_{n=1}^{\infty} |b_n|

converges

step 4

Since a_n < b_n , it follows that

|a_n| \leq |b_n|

because absolute values are non-negative

step 5

By the Comparison Test, if

\sum_{n=1}^{\infty} |b_n|

converges, then

\sum_{n=1}^{\infty} |a_n|

must also converge, because the terms of

|a_n|

are less than or equal to the corresponding terms of

|b_n|

step 6

Therefore, the absolute convergence of

\sum_{n=1}^{\infty} b_n

implies the absolute convergence of

\sum_{n=1}^{\infty} a_n

. However, the converse is not necessarily true;

\sum_{n=1}^{\infty} a_n

could converge absolutely without

\sum_{n=1}^{\infty} b_n

converging absolutely

C

Key Concept

Comparison Test for Absolute Convergence

Explanation

The absolute convergence of the larger series

\sum_{n=1}^{\infty} b_n

is a sufficient condition for the absolute convergence of the smaller series

\sum_{n=1}^{\infty} a_n

, but it is not necessary. Therefore, the statement is a necessary but not sufficient condition.

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