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Determine: x=2 is the solution to the fractional equation <image> . ( ) | Solution: Substitute x=2 into the original equation, the left side = $$\frac{x}{x+2}$$ = $\frac{1}{2}$ = the right side,
∴ x=2 is a solution to the fractional equation $\frac{x}{x+2}$ = $\frac{1}{2}$. Correct. | Algebra |
As shown in the figure, the intersection point of the line with the y-axis is (0, -3). Therefore, when x < 0, ( ) <image> A、y < 0 B、y < -3 C、y > 0 D、y > -3 | Because the intersection point of the line y = kx + b with the y-axis is (0, -3), from the graph of the function we can see that when y > -3, x < 0.
Therefore, the answer is D. | Functions |
The graph of the function \( y = kx + b \) (where \( k \) and \( b \) are constants) is shown in the figure. Then, the solution set for the inequality \( kx + b > 0 \) is ( ) <image> A、x > 0 B、x < 0 C、x < 2 D、x > 2 | The graph of the function y = kx + b passes through the point (2, 0), and the function value y decreases as x increases.
∴ When x < 2, the function value is less than 0, i.e., the solution set of the inequality kx + b > 0 with respect to x is x < 2.
Therefore, the answer is C. | Algebra |
<image> As shown in the figure, the line \( y = kx + b \) intersects the coordinate axes at points A(-2, 0) and B(0, 3). Then the solution set of the inequality \( kx + b > 0 \) is ( ) A. x > 3 B. -2 < x < 3 C. x < -2 D. x > -2 | Since the line \( y = kx + b \) intersects the x-axis at \( A(-2, 0) \),
∴ the solution set of the inequality \( kx + b > 0 \) is \( x > -2 \),
Therefore, the answer is: D. | Plane Geometry |
The graph of the function \( y = kx + b \) (where \( k \) and \( b \) are constants) is shown in the figure. Then the solution set of the inequality \( kx + b > 0 \) is ( ) <image> A. \( x > 0 \) B. \( x < 0 \) C. \( x < 2 \) D. \( x > 2 \) | The graph of the function y = kx + b passes through the point (2, 0), and the function value y decreases as x increases,
∴ when x < 2, the function value is less than 0, i.e., the solution set of the inequality kx + b > 0 with respect to x is x < 2.
Therefore, the correct choice is C. | Algebra |
If the graph of the function \( y = kx + b \) (where \( k \) and \( b \) are constants) is as shown in the figure, then when \( y > 0 \), the range of values for \( x \) is ( ) <image> A. \( x > 1 \) B. \( x > 2 \) C. \( x < 1 \) D. \( x < 2 \) | The graph of the function y = kx + b (k, b are constants) intersects the x-axis at the point (2, 0), and y decreases as x increases.
∴ When y > 0, x < 2.
Therefore, the correct choice is D. | Functions |
If △ABC ≌ △A′B′C′, D is on BC, D′ is on B′C′, ∠BAD = ∠B′A′D′, then AD = A′D′ ( ) | Proof: As shown in the figure, since △ABC ≌ △A′B′C′,
∴ AB = A′B′, ∠B = ∠B′,
Also, since ∠BAD = ∠B′A′D′,
∴ △BAD ≌ △B′A′D′ (ASA)
∴ AD = A′D′ (corresponding sides of congruent triangles are equal).
Therefore, AD = A′D′ is definitely correct,
So the answer should be T. | Plane Geometry |
The graph of the linear function y = kx + b (k and b are constants, k ≠ 0) is shown in the figure below. Then the solution set of the inequality kx + b > 0 is ( ) <image> A. x > -2 B. x > 0 C. x < -2 D. x < 0 | From the graph, we know that the linear function y = kx + b (k, b are constants, k ≠ 0) passes through the point (-2, 0), and the function value y increases as x increases. Therefore, the solution set of the inequality kx + b > 0 is x > -2.
Hence, the answer is A. | Functions |
As shown in the figure, the line y = kx + b intersects the x-axis at the point (3, 0). The solution set for the inequality kx + b ≤ 0 is ( ) <image> A、x ≥ 3 B、x > 3 C、x ≤ 3 D、x < 3 | The line \( y = kx + b \) intersects the x-axis at the point \((3, 0)\). When \( x = 3 \), \( y = 0 \), and the function value \( y \) decreases as \( x \) increases; therefore, the solution set for the inequality \( kx + b \leq 0 \) is \( x \geq 3 \).
Hence, the answer is A. | Functions |
As shown in the figure, the system of inequalities representing the shaded region is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | Since \( x \ge 0 \) represents the part to the right of the line \( x = 0 \), \( 2x + y \le 5 \) represents the part below and to the left of the line \( y = -2x + 5 \), and \( 3x + 4y \ge 9 \) represents the part above and to the right of the line \( y = -\frac{3}{4}x + \frac{9}{4} \),
Therefore, based on the graph, the system of inequalities that satisfies the shaded region is: \(\left\{\begin{array}{l}2x+y \le 5 \\ 3x+4y \ge 9 \\ x \ge 0\end{array}\right.\).
Thus, the correct choice is D. | Algebra |
The graph of the linear function y = kx + b (k and b are constants, k ≠ 0) is shown below. Then the solution set of the inequality kx + b > 0 is ( ) <image> A、x > -2 B、x > 0 C、x < -2 D、x < 0 | From the graph, we know that the linear function \( y = kx + b \) (where \( k \) and \( b \) are constants, and \( k \neq 0 \)) passes through the point \((-2, 0)\), and the function value \( y \) increases as \( x \) increases. Therefore, the solution set of the inequality \( kx + b > 0 \) is \( x > -2 \).
Hence, the answer is A. | Functions |
Line \( l_1 \): \( y = k_1 x + b \) and Line \( l_2 \): \( y = k_2 x + c \) are shown in the same Cartesian coordinate system as shown in the figure. The solution set of the inequality \( k_1 x + b < k_2 x + c \) is ( ) <image> A、x > 1 B、x < 1 C、x > -2 D、x < -2 | From the graph, we can see that the intersection point of line \( l_{1} \) and line \( l_{2} \) in the same Cartesian coordinate system is (1, -2). When \( x < 1 \), the graph of line \( l_{1} \) is below the graph of line \( l_{2} \). Therefore, the solution set of the inequality \( k_{1}x + b < k_{2}x + c \) is: \( x < 1 \).
Hence, the correct choice is B. | Algebra |
Given the graph of the linear function y = kx + b (as shown in the figure), when x < 0, the range of y is ( ) <image> A、y > 0 B、y < 0 C、-2 < y < 0 D、y < -2 | The graph of the linear function y = kx + b passes through the point (0, -2), and the function value y increases as x increases.
∴ When x ≤ 0, the range of y is y < -2.
Therefore, the answer is D. | Plane Geometry |
<image> As shown in the figure, given the line y 1 = x + m and y 2 = kx - 1 intersect at point P(-1, 1), then the solution set of the inequality x + m > kx - 1 on the number line is ( ) A、 <image> B、 <image> C、 <image> D、 <image> | Solution: Since the line \( y_1 = x + m \) intersects with \( y_2 = kx - 1 \) at point \( P(-1, 1) \),
∴ From the graph, we can see that the solution set of the inequality \( x + m > kx - 1 \) is \( x > -1 \),
Represented on the number line as: -5 -4 -3 -2 -1 0 1 2 3 4 5,
Therefore, the answer is B. | Algebra |
As shown in the figure, the graph of the linear function \( y = kx + b \) passing through point A intersects the graph of the direct proportion function \( y = 2x \) at point B. The solution set of the inequality \( 0 < 2x < kx + b \) is ( ) <image> A. x < 1 B. x < 0 or x > 1 C. 0 < x < 1 D. x > 1 | The linear function \( y = kx + b \) passes through points A and B,
\[
\left\{ \begin{array} { l } { b = 3 } \\ { 2 = 2 k + 3 } \end{array} \right.,
\]
∴
\[
\left\{ \begin{array} { l } { k = - \frac { 1 } { 2 } } \\ { b = 3 } \end{array} \right..
\]
Solving, we get: \( k = -\frac{1}{2} \), \( b = 3 \).
Therefore: \( y = -\frac{1}{2}x + 3 \),
Since \( 0 < 2x < -\frac{1}{2}x + 3 \),
Solving, we get: \( 0 < x < 1 \).
Thus, the answer is C. | Plane Geometry |
As shown in the figure, the graph of the linear function y = kx + b passes through points A and B. The solution set of kx + b > 0 is ( ) <image> A、x > 0 B、-3 < x < 2 C、x > 2 D、x > -3 | From the graph, the linear function $y=kx+b$ passes through points A(-3, 0) and B(0, 2),
$$\left\{ \begin{array} { l } { 0 = - 3 k + b } \\ { 2 = b } \end{array} \right.$$,
∴$\left\{ \begin{array} { l } { b = 2 } \\ { k = \frac { 2 } { 3 } } \end{array} \right.$;
Thus, b=2, k=$\frac { 2 } { 3 }$;
Therefore, the inequality is $\frac { 2 } { 3 }$x+2≥0,
Solving this, we get x≥-3.
So the answer is D. | Plane Geometry |
Student Wang Jie solved the problem "Given the coordinates of points A and B as A(3, -2) and B(6, -5), find the equation of the line A'B' which is the reflection of line AB across the x-axis" as follows: First, he established a Cartesian coordinate system (as shown in the figure), marked points A and B, and used the properties of axis symmetry to find the coordinates of A' and B' to be A'(3, 2) and B'(6, 5); then he assumed the equation of line A'B' to be y = kx + b (k ≠ 0), and substituted A'(3, 2) and B'(6, 5) into y = kx + b, resulting in the system of equations <image> , solving which gives <image> , and finally obtained the equation of line A'B' as y = x - 1. The mathematical ideas used in the problem-solving process are ( ) <image> A. Classification and transformation ideas B. Classification and equation ideas C. Combination of number and shape and holistic ideas D. Combination of number and shape and equation ideas | 【Analysis】According to the properties of axis symmetry, which belong to shape, and the coordinates of points, which belong to numbers, it can be seen that the mathematical idea of combining shape and number is used; according to solving the system of equations to find the values of unknowns, it can be seen that the idea of equations is used.
【Solution】Solution: Step 1: Establish a Cartesian coordinate system, mark points A and B, and use the properties of axis symmetry to find the coordinates of A′ and B′, which are A′(3, 2) and B′(6, 5). This is based on the properties of axis symmetry to find the coordinates (ordered real number pairs) of points, using the mathematical idea of combining shape and number;
Step 2: Assume the equation of line A′B′ is y=kx+b (k≠0), and substitute A′(3, 2) and B′(6, 5) into y=kx+b, obtaining the system of equations $$\left\{\begin{array}{l}3k+b=2 \\ 6k+b=5\end{array}\right.$$ Solving this, we get $$\left\{\begin{array}{l}k=1 \\ b=-1\end{array}\right.$$ Finally, the equation of line A′B′ is y=x-1. Here, based on the characteristics of the coordinates of points on the graph of a linear function, we set up equations to find the undetermined coefficients, using the idea of equations;
Therefore, in the problem-solving process, Wang Jie used the mathematical ideas of combining shape and number and the idea of equations.
Hence, the answer is D. | Others |
As shown in the figure, the line y = kx + b intersects the coordinate axes at points A(-3, 0) and B(0, 5). The solution set of the inequality -kx - b < 0 is ( ) <image> A、x > -3 B、x < -3 C、x > 3 D、x < 3 | Observing the graph, we can see that when \( x > -3 \), the line \( y = kx + b \) is above the x-axis,
which means the solution set of the inequality \( kx + b > 0 \) is \( x > -3 \),
∴ \(-kx - b < 0\)
∴ \(kx + b > 0\),
∴ the solution set of \(-kx - b < 0\) is \( x > -3 \).
Therefore, the answer is A. | Functions |
As shown in the figure, the graph of the linear function y = kx + b passes through points A and B. The solution set for kx + b > 0 is ( ) <image> A、x > 0 B、x > -3 C、x > 2 D、-3 < x < 2 | The graph of the linear function y = kx + b passes through point A(-3, 0), and the function value y increases as x increases; therefore, when x > -3, y = kx + b > 0; that is, the solution set for kx + b > 0 is x > -3.
Hence, the answer is B. | Plane Geometry |
As shown in the figure, the line y=kx+b passes through points A(-2, -1) and B(-3, 0). Using the graph of the function, determine the solution set of the inequality <image> < kx+b ( ) <image> A、 <image> or <image> B、 <image> C、 <image> D、 <image> or <image> | The line $y=kx+b$ passes through points A(-2, -1) and B(-3, 0),
$$\left\{ \begin{array} { l } { - 2 k + b = - 1 } \\ { - 3 k + b = 0 } \end{array} \right.$$ Solving this, we get $$\left\{ \begin{array} { l } { k = - 1 } \\ { b = - 3 } \end{array} \right.$$,
Therefore, the equation of the function is $y=-x-3$,
The x-coordinates of the intersection points of this function with $y=\frac{1}{x}$ are $\frac{-3±\sqrt{5}}{2}$,
From the graph, we can see that the solution set of the inequality $\frac{1}{x} < kx+b$ is $x < \frac{-3-\sqrt{5}}{2}$ or $\frac{-3+\sqrt{5}}{2} < x < 0$.
Therefore, the answer is D. | Algebra |
The graph of the function y = kx + b is shown in the figure. The solution set of the inequality kx + b > 0 is ( ) <image> A、x < -2 B、x > -2 C、x < -1 D、x > -1 | According to the graph: the graph of the linear function y = kx + b intersects the x-axis at the point (-2, 0),
which means when x < -2, the function value y is y > 0;
therefore, when the inequality kx + b > 0 holds, the range of x is x < -2.
Hence, the answer is A. | Functions |
As shown in the figure, the line y = kx + b passes through points A(2, 1) and B(-1, -2). The solution set for the inequality <image> x > kx + b > -2 is ( ) <image> A、x < 2 B、x > -1 C、x < 1 or x > 2 D、-1 < x < 2 | Substitute the coordinates of points A(2, 1) and B(-1, -2) into y = kx + b,
$$\left\{ \begin{array} { l l } { 2 k + b = 1 } \\ { - k + b = - 2 } \end{array} \right.$$
Solving, we get: $$\left\{ \begin{array} { l l } { k = 1 } \\ { b = - 1 . } \end{array} \right.$$
Solve the inequality system: $$\frac { 1 } { 2 } x > x - 1 > - 2$$,
We get: -1 < x < 2.
Therefore, the answer is D. | Others |
The graph of the linear function y = kx + b (k and b are constants, k ≠ 0) is shown in the figure. Then the solution set of the inequality kx + b > 0 is ( ) <image> A、x > 1 B、x < 1 C、x > 0 D、x > -1 | From the graph, we know that the linear function y = kx + b (k and b are constants, k ≠ 0) passes through the point (-1, 0), and the function value y increases as x increases. Therefore, the solution set of the inequality kx + b > 0 is x > -1.
Hence, the answer is D. | Functions |
As shown in the figure, the line y 1 = kx + b passes through point A(0, 2), and intersects with the line y 2 = mx at point P(1, m). The solution set for the inequality mx > kx + b > mx - 2 is ( ) <image> A、1 < X < 2 B、0 < X < 2 C、0 < X < 1 D、1 < X | Since the line $y_{1}=kx+b$ passes through points A(0, 2) and P(1, m),
we have: $\left\{\begin{array}{l}k+b=m \\ b=2\end{array}\right.$,
$\left\{\begin{array}{l}k=m-2 \\ b=2\end{array}\right.$.
∴ The line $y_{1}=(m-2)x+2$.
Therefore, the system of inequalities can be transformed into:
$mx > (m-2)x + 2 > mx - 2$,
$0 > -2x + 2 > -2$,
Solving this, we get: $1 < x < 2$,
Thus, the answer is A. | Plane Geometry |
As shown in the figure, given the functions y = x + b and y = ax + 3 intersect at point P, the solution set of the inequality x + b > ax + 3 is ( ) <image> A、x < 1 B、x > 1 C、x ≥ 1 D、x ≤ 1 | The graphs of the functions y=x+b and y=ax+3 intersect at point P, and the x-coordinate of point P is 1. According to the graph, when x＞1, the graph of the function y=x+b is above the graph of the function y=ax+3. Therefore, the value of the function y=x+b is greater than the value of the function y=ax+3, which means the solution set of the inequality x+b＞ax+3 is x＞1. Hence, the correct choice is B. | Algebra |
As shown in the figure, the graph of the linear function y = kx + b passes through points A and B. The solution set for kx + b > 0 is ( ) <image> A、x > 0 B、x > -3 C、x > 2 D、-3 < x < 2 | The graph of the linear function y = kx + b passes through A(-3, 0), and the function value y increases as x increases; therefore, when x > -3, y = kx + b > 0; that is, the solution set of kx + b > 0 is x > -3.
Hence, the answer is B. | Plane Geometry |
As shown in the figure, the line y = kx + b intersects the coordinate axes at points A(2, 0) and B(0, -3). The solution set for the inequality kx + b + 3 ≥ 0 is ( ) <image> A、x ≥ 0 B、x ≤ 0 C、x ≥ 2 D、x ≤ 2 | The line y = kx + b intersects the y-axis at point B(0, -3),
which means when x = 0, y = -3.
Since the function value y increases as x increases,
∴ when x ≥ 0, the function value kx + b ≥ -3,
∴ the solution set of the inequality kx + b + 3 ≥ 0 is x ≥ 0.
Therefore, the answer is A. | Plane Geometry |
As shown in the figure, the line y = kx + b (k < 0) intersects the x-axis at the point (3, 0). The solution set for the inequality kx + b > 0 is ( ) <image> A、x < 3 B、x > 3 C、x > 0 D、x < 0 | The line \( y = kx + b \) (where \( k \leq 0 \)) intersects the x-axis at the point (3, 0). When \( x = 3 \), \( y = 0 \), and the function value \( y \) decreases as \( x \) increases. According to the fact that \( y \) decreases as \( x \) increases, the solution set for the inequality \( kx + b > 0 \) is \( x < 3 \).
Therefore, the correct choice is A. | Functions |
As shown in the figure, the line y=kx+b passes through points A(1, 2) and B(-2, -1). The solution set of the inequality <image> < kx+b < 2 is ( ) <image> A、 <image> < x < 2 B、 <image> < x < 1 C、-2 < x < 1 D、 <image> < x < 1 | According to the graph, the solution set of the inequality $$\frac{1}{2}$$x < kx + b < 2 is -2 < x < 1.
Therefore, the answer is C. | Arithmetic |
As shown in the figure, the line y=kx+b (k<0) intersects the x-axis at the point (3, 0). The solution set for the inequality kx+b>0 is ( ) <image> A. x<3 B. x>3 C. x>0 D. x<0 | The line y = kx + b (k ≤ 0) intersects the x-axis at the point (3, 0). When x = 3, y = 0, and the function value y decreases as x increases; since y decreases as x increases, the solution set for the inequality kx + b > 0 with respect to x is x < 3.
Therefore, the answer is A. | Functions |
Given the graph of the linear function y = kx + b (as shown in the figure), when x < 0, the range of y is ( ) <image> A、y > 0 B、y < 0 C、-2 < y < 0 D、y < -2 | The graph of the linear function y = kx + b passes through the point (0, -2), and the function value y increases as x increases.
∴ When x ≤ 0, the range of y is y < -2.
Therefore, the answer is D. | Functions |
<image> (2007•Linyi) The line \( l_1 \): \( y = k_1 x + b \) and the line \( l_2 \): \( y = k_2 x \) are shown in the same Cartesian coordinate system as in the figure. The solution to the inequality \( k_1 x + b > k_2 x \) is ( ) A、x > -1 B、x < -1 C、x < -2 D、Cannot be determined | The solution to the inequality $k_{1}x+b＞k_{2}x$ with respect to x can be seen as the range of x values for which the line $l_{1}$ is above the line $l_{2}$: x < -1.
Therefore, the answer is B. | Algebra |
As shown in the figure, the line y=kx+b intersects the coordinate axes at two points. The solution set of the inequality kx+b<0 is ( ) <image> A、x > -2 B、x > 3 C、x < -2 D、x < 3 | From the graph, it can be seen that the values of the independent variable corresponding to the part of the function graph below the x-axis are x < -2.
∴ The solution set of the inequality kx + b < 0 is x < -2.
Therefore, the answer is C. | Plane Geometry |
Among the following four lines, the line where each point's coordinates are solutions to the linear equation x-2y=2 is ( ) A、 <image> B、 <image> C、 <image> D、 <image> | Solution: Since x-2y=2,
∴ y=$\frac{1}{2}$x-1,
∴ when x=0, y=-1, and when y=0, x=2,
∴ the linear function y=$\frac{1}{2}$x-1 intersects the y-axis at the point (0, -1) and the x-axis at the point (2, 0),
Therefore, C meets the requirements,
Hence, the answer is: C. | Functions |
As shown in the figure, the line y = kx + b intersects the coordinate axes at points A(-3, 0) and B(0, 5). The solution set of the inequality -kx - b < 0 is ( ) <image> A、x > -3 B、x < -3 C、x > 3 D、x < 3 | From the graph, we can see that when \( x > -3 \), the line \( y = kx + b \) is above the x-axis,
which means the solution set of the inequality \( kx + b > 0 \) is \( x > -3 \),
∴ \(-kx - b < 0\)
∴ \( kx + b > 0 \),
∴ the solution set of \(-kx - b < 0\) is \( x > -3 \).
Therefore, the answer is A. | Algebra |
As shown in the figure, the line y = kx + b passes through points A(-2, -1) and B(-3, 0). Using the graph of the function, determine the solution set for the inequality <image> < kx + b ( ) <image> A、 <image> or <image> B、 <image> C、 <image> D、 <image> or <image> | The line \( y = kx + b \) passes through points A(-2, -1) and B(-3, 0),
\[
\left\{ \begin{array} { l } { - 2 k + b = - 1 } \\ { - 3 k + b = 0 } \end{array} \right.
\]
Solving this, we get
\[
\left\{ \begin{array} { l } { k = - 1 } \\ { b = - 3 } \end{array} \right.
\]
Therefore, the equation of the function is \( y = -x - 3 \),
The x-coordinates of the intersection points with the function \( y = \frac{1}{x} \) are \( \frac{-3 \pm \sqrt{5}}{2} \),
From the graph, we can determine that the solution set of the inequality \( \frac{1}{x} < kx + b \) is \( x < \frac{-3 - \sqrt{5}}{2} \) or \( \frac{-3 + \sqrt{5}}{2} < x < 0 \).
Therefore, the answer is D. | Algebra |
As shown in the figure, if the line y = kx + b passes through points A(1, -2) and B(0, -4), and the line y = mx passes through point A, then the solution set of the inequality kx + b > mx is ( ) <image> A、x > 1 B、x < 1 C、0 < x < 1 D、1 < x < 2 | For the inequality kx + b > mx, for the same value of x, the graph of the function y = kx + b is above. Therefore, the solution set is: x > 1.
Hence, the answer is A. | Algebra |
The graph of the linear function y=kx+b is shown below, then the solution to the equation kx+b=0 is ( ) <image> A、x=2 B、y=2 C、x=-1 D、y=-1 | Since the graph of the linear function y = kx + b intersects the x-axis at the point (-1, 0),
∴ when kx + b = 0, x = -1.
Therefore, the answer is C. | Functions |
Given the graphs of the lines y = -x + 4 and y = x + 2 as shown in the figure, the solution to the system of equations is <image> the solution is <image> A、 <image> B、 <image> C、 <image> D、 <image> | According to the problem,
the solution to the system of linear equations $$\left\{ \begin{array} { l } { y = - x + 4 } \\ { y = x + 2 } \end{array} \right.$$ is the intersection point of the lines \( y = -x + 4 \) and \( y = x + 2 \),
and since the intersection point is (1, 3),
the solution to the original system of equations is: $$\left\{ \begin{array} { l } { x = 1 } \\ { y = 3 } \end{array} \right.$$.
Therefore, the answer is B. | Algebra |
As shown in the figure, the system of equations whose solution is the intersection point of two lines \( l_1 \) and \( l_2 \) is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | The line $l_{1}$ passes through (2, 3) and (0, -1), so its function expression is $y=2x-1$;
The line $l_{2}$ passes through (2, 3) and (0, 1), so its function expression is $y=x+1$;
Therefore, the system of equations whose solution is the intersection point of the two lines $l_{1}$ and $l_{2}$ is: $$\left\{ \begin{array} { l } { x - y = - 1 } \\ { 2 x - y = 1 } \end{array} \right.$$.
Hence, the answer is C. | Algebra |
When solving a system of linear equations using the graphical method, the graphs of the corresponding two linear functions are drawn in the same Cartesian coordinate system (as shown in the figure), then the system of linear equations to be solved is ( ) <image> A. <image> B. <image> C. <image> D. <image> | Based on the coordinates of the points on the given graph, (0, -1), (1, 1), (0, 2);
the equations of the two lines in the graph are y = 2x - 1, y = -x + 2,
$$\left\{ \begin{array} { l } { x + y - 2 = 0 } \\ { 2 x - y - 1 = 0 } \end{array} \right.$$.
Therefore, the system of linear equations to be solved is $$\left\{ \begin{array} { l } { x + y - 2 = 0 } \\ { 2 x - y - 1 = 0 } \end{array} \right.$$.
Hence, the correct choice is D. | Algebra |
Among the following images, the one that represents the graph of the points whose coordinates are solutions to the equation -2x + y - 2 = 0 is ( ) A、 <image> B、 <image> C、 <image> D、 <image> | In the equation -2x + y - 2 = 0,
when x = 0, y = 2;
when y = 0, x = -1.
Therefore, the answer is B. | Plane Geometry |
As shown in the figure, it is the graph of the linear function y = kx + b. Then the solution to the inequality kx + b > 0 is ( ) <image> A、x > -4 B、x < -4 C、x > 3 D、x < 3 | From the graph, we can see that when x > -4, y > 0,
which means kx + b > 0;
Therefore, the solution set for kx + b > 0 is: x > -4.
Hence, the answer is A. | Functions |
The graph of a linear function is shown in the figure. When \( kx + b < 0 \), the range of values for \( x \) is ( ) <image> A、x < 0 B、x > 0 C、x < 2 D、x > 2 | Because the intersection point of the line y = kx + b with the x-axis is (2, 0),
from the graph of the function, we can see that when x < 2, y > 0, i.e., kx + b > 0.
Therefore, the answer is: D. | Algebra |
As shown in the figure, given the linear functions y=ax+b and y=kx intersect at point P, according to the graph, the solution to the system of linear equations <image> is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | According to the problem,
the solution to the system of linear equations $$\left\{ \begin{array} { l } { y = a x + b } \\ { k x - y = 0 } \end{array} \right.$$ is the coordinates of the intersection point P of the graphs of the linear function \(y = ax + b\) and the direct proportion \(y = kx\),
From the graphs of the linear function \(y = ax + b\) and the direct proportion \(y = kx\), we get
$$\left\{ \begin{array} { l } { y = a x + b } \\ { x = - 4 } \end{array} \right.$$
The solution to the system of linear equations $$\left\{ \begin{array} { l } { k x - y = 0 } \\ { y = - 2 } \end{array} \right.$$ is $$\left\{ \begin{array} { l } { x = - 4 } \\ { y = - 2 } \end{array} \right.$$.
Therefore, the answer is A. | Others |
As shown in the figure, it is the graph of the line y = x - 3. Point P(2, m) is above this line, then the range of m is ( ) <image> A、m > -3 B、m > -1 C、m > 0 D、m < 3 | When x=2, y=2-3=-1,
Since point P(2, m) is above the line,
Therefore, m＞-1.
Hence, the answer is B. | Functions |
<image> The lines y=x and y=-x+1 divide the plane into parts Ⅰ, Ⅱ, Ⅲ, and Ⅳ (including the boundaries, as shown in the figure), then the points (x, y) that satisfy y≤x and y≥-x+1 must be in ( ) A、Part Ⅰ B、Part Ⅱ C、Part Ⅲ D、Part Ⅳ | From the graph, we can see that the points satisfying \( y \leq x \) are all below the line \( y = x \), and the points satisfying \( y \geq -x + 1 \) are all above the line \( y = -x + 1 \). Therefore, the points that satisfy both \( y \leq x \) and \( y \geq -x + 1 \) are in the overlapping region of the two. From the graph, it is known that the point \((x, y)\) must be in part Ⅱ. Hence, the answer is B. | Algebra |
The graph of the linear function y = ax + b passes through points A and B, as shown in the figure. Then the solution set of the inequality ax + b > 0 is ( ) <image> A. x < -2 B. x > -2 C. x < 1 D. x > 1 | The graph of the linear function y = ax + b passes through the point A(-2, 0), and the function value y increases as x increases,
∴ the solution set of the inequality ax + b > 0 is x > -2.
Therefore, the answer is B. | Plane Geometry |
The position relationship between the parabola and the straight line is shown in the figure. Points P(a, b) and Q(c, d) are on the parabola, and point R(e, f) is on the straight line. If -2 < a < c, e < -2, then the size relationship of b, d, f is ( ) <image> A、f > b > d B、b > d > f C、b > f > d D、d > f > b | From the graph, when -2 < a < c and e < -2, f > b, f > d.
Also, according to the properties of the parabola graph, when -2 < a < c, b > d,
∴ f > b > d,
Therefore, the answer is A. | Algebra |
Which of the following functions does not pass through the origin? ( ) A、 <image> B、y=2x 2 C、y=(x-1) 2 -1 D、 <image> | When \( x = 0 \), \( y = 0 \), which means the graph of the function must pass through the origin (0, 0). Therefore, this option is incorrect;
When \( x = 0 \), \( y = 0 \), which means the graph of the function must pass through the origin (0, 0). Therefore, this option is incorrect;
When \( x = 0 \), \( y = 0 \), which means the graph of the function must pass through the origin (0, 0). Therefore, this option is incorrect;
When \( x = 0 \), the original equation has no solution, which means the graph of the function must not pass through the origin (0, 0). Therefore, this option is correct.
Thus, the correct choice is D. | Functions |
If points A(x 1 , y 1 ) and B(x 2 , y 2 ) are two points on the line y = kx - b, and when x 1 < x 2 , y 1 < y 2 , then the graph of the function y = <image> is roughly ( ) A. <image> B. <image> C. <image> D. <image> | Since when \( x_{1} < x_{2} \), \( y_{1} < y_{2} \),
therefore \( k > 0 \),
therefore the graph of the function \( y = \frac{k}{x} \) is in the first and third quadrants, and among the four graphs, only B fits.
Hence, the answer is B. | Functions |
As shown in the figure, the graphs of the functions y = ax + b and a(x - 1) - b > 0 intersect at points (-1, 1) and (2, 2). When y 1 > y 2 , the range of x is ( ) <image> A. x < -1 B. -1 < x < 2 C. x > 2 D. x < -1 or x > 2 | Since the graphs of the functions \( y = ax + b \) and \( a(x-1) - b > 0 \) intersect at the points \((-1, 1)\) and \((2, 2)\),
we can see from the graph that when \( y_1 > y_2 \), the range of \( x \) is \( x > 2 \) or \( x < -1 \),
thus, the answer is: D. | Plane Geometry |
As shown in the figure, the system of equations whose solution is the coordinates of the intersection point of the two lines l 1 , l 2 is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | The line \( l_{1} \) passes through (2, 3) and (0, -1), so its function expression is \( y = 2x - 1 \);
The line \( l_{2} \) passes through (2, 3) and (0, 1), so its function expression is \( y = x + 1 \);
Therefore, the system of equations whose solution is the intersection point of the two lines \( l_{1} \) and \( l_{2} \) is:
\[ \left\{ \begin{array} { l } { x - y = - 1 } \\ { 2 x - y = 1 } \end{array} \right. \].
Thus, the answer is C. | Algebra |
As shown in the figure, the graphs of the linear functions l 1 and l 2 are plotted in the same coordinate system. Let l 1 : y=k 1 x+b 1 and l 2 : y=k 2 x+b 2 . The solution to the system of equations <image> is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | From the graph, we can see that the graphs of the two linear functions pass through the points: (1, 2), (4, 1), (-1, 0), (0, -3);
Therefore, the equations of the two lines are \( y = -\frac{1}{3}x + \frac{7}{3} \) and \( y = -3x - 3 \); solving the system of equations:
\[
\left\{\begin{array}{l}
y = -\frac{1}{3}x + \frac{7}{3} \\
y = -3x - 3
\end{array}\right.
\]
we get:
\[
\left\{\begin{array}{l}
x = -2 \\
y = 3
\end{array}\right.
\]
Thus, the answer is B. | Others |
As shown in the figure, the graph of the linear function y = kx + b intersects the y-axis at the point (0, 1). Therefore, the solution set for the inequality kx + b > 1 is ( ) <image> A、x > 0 B、x < 0 C、x > 1 D、x < 1 | From the graph of the linear function, it can be seen that this function is a decreasing function,
∴ the graph of the linear function y = kx + b intersects the y-axis at the point (0,1),
∴ when x < 0, the inequality kx + b > 1 holds for x.
Therefore, the answer is B. | Algebra |
Given the graphs of the lines y = -x + 4 and y = x + 2 as shown in the figure, the solution to the system of equations <image> is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | According to the problem,
the solution to the system of linear equations $$\left\{ \begin{array} { l } { y = - x + 4 } \\ { y = x + 2 } \end{array} \right.$$ is the intersection point of the lines \( y = -x + 4 \) and \( y = x + 2 \),
Since the intersection point is (1, 3),
the solution to the original system of equations is: $$\left\{ \begin{array} { l } { x = 1 } \\ { y = 3 } \end{array} \right.$$.
Therefore, the correct choice is B. | Algebra |
When solving a system of linear equations using the graphical method, the graphs of the corresponding two linear functions are drawn in the same Cartesian coordinate system (as shown in the figure), then the system of linear equations being solved is ( ) <image> A. <image> B. <image> C. <image> D. <image> | According to the coordinates of the points on the given graph, (0, -1), (1, 1), (0, 2);
the equations of the two lines in the graph are y = 2x - 1, y = -x + 2,
$$\left\{ \begin{array} { l } { x + y - 2 = 0 } \\ { 2 x - y - 1 = 0 } \end{array} \right.$$.
Therefore, the system of linear equations to be solved is $$\left\{ \begin{array} { l } { x + y - 2 = 0 } \\ { 2 x - y - 1 = 0 } \end{array} \right.$$.
Hence, the answer is D. | Algebra |
As shown in the figure, given the intersection point of the lines \( y = 3x + b \) and \( y = ax - 2 \) has an x-coordinate of -2, there are the following 3 conclusions based on the graph: ① \( a > 0 \); ② \( b > 0 \); ③ \( x > -2 \) is the solution set of the inequality \( 3x + b > ax - 2 \). The number of correct conclusions is ( ) <image> A、0 B、1 C、2 D、3 | From the graph, we can see that a > 0, so ① is correct;
b > 0, so ② is correct;
When x > -2, the line y = 3x + b is above the line y = ax - 2, which means that when x > -2, the inequality 3x + b > ax - 2 holds, so ③ is correct.
Therefore, the answer is D. | Algebra |
<image> As shown in the figure, the line l 1 : y 1 = k 1 x + b 1 intersects with the line l 2 : y 2 = k 2 x + b 2 at the point (-1, 2). When the function value y 1 > y 2 , the range of the independent variable x is ( ) A. Equal to -1 B. Less than -1 C. Greater than -1 D. None of the above | Solution: Since the intersection point of the lines \( l_{1} \): \( y_{1}=k_{1}x+b_{1} \) and \( l_{2} \): \( y_{2}=k_{2}x+b_{2} \) is \((-1, 2)\),
∴ when \( x=-1 \), \( y_{1}=y_{2}=2 \);
and when \( y_{1} > y_{2} \), \( x > -1 \).
Therefore, the correct choice is: C. | Algebra |
As shown in the figure, the graph of a linear function passing through point Q(0, 3.5) intersects with the graph of the direct proportion function y=2x at point P. The equation that can represent this linear function is ( ) <image> A、3x-2y+3.5=0 B、3x-2y-3.5=0 C、3x-2y+7=0 D、3x+2y-7=0 | Let the equation of this linear function be y = kx + b.
$$\because$$ this line passes through points P(1, 2) and Q(0, 3.5),
$$\therefore \left\{ \begin{array} { l } { k + b = 2 } \\ { b = 3 . 5 } \end{array} \right.$$
$$\therefore \left\{ \begin{array} { l } { b = 3 . 5 } \\ { k = - 1 . 5 } \end{array} \right.$$
Solving, we get $$\left\{ \begin{array} { l } { b = 3 . 5 } \\ { k = - 1 . 5 } \end{array} \right.$$
Thus, the equation of this linear function is y = -1.5x + 3.5,
which can be written as: 3x + 2y - 7 = 0.
Therefore, the answer is D. | Plane Geometry |
As shown in the figure, the line \( l: y = x + 2 \) intersects the y-axis at point A. After rotating the line \( l \) 90° around point A, the equation of the resulting line is ( ) <image> A、y=x-2 B、y=-x+2 C、y=-x-2 D、y=-2x-1 | Since the line \( l \): \( y = x + 2 \) intersects the y-axis at point A,
∴ A(0, 2).
Let the equation of the line after rotation be: \( y = -x + b \),
then: \( 2 = 0 + b \),
solving for \( b \) gives: \( b = 2 \),
thus the equation is: \( y = -x + 2 \).
Therefore, the correct choice is B. | Others |
<image> Line \( l_1 \): \( y = k_1 x + b \) and Line \( l_2 \): \( y = k_2 x \) are shown in the same Cartesian coordinate system. The solution to the inequality \( k_1 x + b > k_2 x \) is ( ) A、x > -1 B、x < -1 C、x < -2 D、Cannot be determined | The solution to the inequality $k_{1}x+b＞k_{2}x$ with respect to x can be considered as the range of x values for which the line $l_{1}$ is above the line $l_{2}$: x < -1.
Therefore, the answer is B. | Algebra |
As shown in the figure, if the line L is translated 2 units upwards to get the line L′, then the equation of L′ is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | Let the equation of line L be y = kx + b. From the problem, we know that (-2, 0) and (0, -1) lie on line L. Therefore, -2k + b = 0 and b = -1.
Solving these, we get: k = $\frac{1}{2}$.
∴ The equation of line L is y = $\frac{1}{2}$x - 1. Shifting line L up by 2 units gives line L', so the equation of L' is y = $\frac{1}{2}$x - 1 + 2 = $\frac{1}{2}$x + 1.
Therefore, the correct answer is D. | Functions |
As shown in the figure, in the Cartesian coordinate system, the coordinates of point A are (0, 3). After triangle OAB is translated to the right along the x-axis, it becomes triangle O′A′B′. The corresponding point of point A is on the line y= <image> x. The distance between point B and its corresponding point B′ is ( ) <image> A、 <image> B、3 C、4 D、5 | As shown in the figure, connect AA′ and BB′.
Since the coordinates of point A are (0, 3), and triangle OAB is translated to the right along the x-axis to get triangle O′A′B′,
the y-coordinate of point A′ is 3.
Also, since the corresponding point of point A is on the line y=$\frac{3}{4}$x,
we have 3=$\frac{3}{4}$x, solving for x gives x=4.
Therefore, the coordinates of point A′ are (4, 3),
so AA′=4,
and according to the properties of translation, BB′=AA′=4.
Thus, the correct answer is C. | Arithmetic |
Given that the solution to the equation kx + b = 0 is x = 3, the graph of the function y = kx + b could be ( ) A、 <image> B、 <image> C、 <image> D、 <image> | Solution: Since the solution to the equation kx + b = 0 is x = 3,
∴ the line y = kx + b passes through the point (3, 0).
Therefore, the answer is C. | Algebra |
The graph of the linear function y = ax + b passes through points A and B, as shown in the figure. Then the solution set of the inequality ax + b > 0 is ( ) <image> A、x < -2 B、x > -2 C、x < 1 D、x > 1 | The graph of the linear function y = ax + b passes through point A(-2, 0), and the function value y increases as x increases.
∴ The solution set of the inequality ax + b > 0 is x > -2.
Therefore, the answer is B. | Plane Geometry |
<image> The graph of the linear function \( y = kx + 1 - k \) is shown in the figure. The range of values for \( k \) is ( ) A、0 < k < 1 B、k < 0 C、k > 1 D、k < 1 | Solution: Since the graph passes through the second and fourth quadrants,
∴ k < 0,
Since the graph intersects the y-axis above the x-axis,
∴ 1 - k > 0, solving this gives k < 1,
∴ The range of values for k is k < 0.
Therefore, the correct choice is B. | Plane Geometry |
The following four lines, where the coordinates of each point on the line are solutions to the linear equation x-2y=2 are ( ) A. <image> B. <image> C. <image> D. <image> | $$x - 2 y = 2 ,$$
$$\therefore y = \frac { 1 } { 2 } x - 1 ,$$
$$\therefore \text { when } x = 0 , y = - 1 , \text { when } y = 0 , x = 2 ,$$
$$\therefore \text { the linear function } y = \frac { 1 } { 2 } x - 1 , \text { intersects the } y \text { axis at the point } ( 0 , - 1 ) , \text { and the } x \text { axis at the point } ( 2 , 0 ) ,$$
$$\text { thus, we can conclude that } C \text { meets the requirements, }$$
$$\text { therefore, the answer is: } C .$$ | Functions |
As shown in the figure, the graph of the linear function y=kx+b passes through points A and B. The solution set of the inequality system <image> is ( ) <image> A、x > 0 B、x > 3 C、x > 2 D、-3 < x < 0 | According to the graph, the function passes through the points (-3, 0) and (0, 2), and the function value y increases as x increases.
∴ The solution set of the system of inequalities $\left\{\begin{array}{l}kx+b＞0\\ kx+b＜2\end{array}\right.$ is: -3＜x＜0.
Therefore, the answer is D. | Plane Geometry |
As shown in the figure, the line y=kx+b intersects with the line y=mx at point A(-1, 2), and intersects the x-axis at point B(-3, 0). The solution set for the inequality 0＜kx+b＜mx is ( ) <image> A、x＞-3 B、-3＜x＜-1 C、-1＜x＜0 D、-3＜x＜0 | From the graph, we can obtain:
The solution set of \( kx + b > 0 \) is: \( x > -3 \),
The solution set of \( kx + b < mx \) is: \( x < -1 \);
∴ The solution set of the system of inequalities is: \( -3 < x < -1 \);
Therefore, the correct choice is B. | Plane Geometry |
As shown in the figure, the line y= <image> x+3 intersects the x-axis at point A. A right-angled isosceles triangle paper is placed with its right-angle vertex at the origin O, and the other two vertices M and N exactly fall on the line y= <image> x+3. If point N is in the second quadrant, then the value of tan∠AON is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | Draw OC⊥AB at C, and draw ND⊥OA at D,
Since N is on the line y=$\frac{3}{4}$x+3,
Let the coordinates of N be (x, $\frac{3}{4}$x+3),
Then DN=$\frac{3}{4}$x+3, OD=-x,
y=$\frac{3}{4}$x+3,
When x=0, y=3,
When y=0, x=-4,
Thus, A(-4,0), B(0,3),
So OA=4, OB=3,
In △AOB, by the Pythagorean theorem: AB=5,
Therefore, in △AOB, by the area formula of a triangle: AO×OB=AB×OC,
So 3×4=5OC,
OC=$\frac{12}{5}$,
Since in Rt△NOM, OM=ON, ∠MON=90°,
Thus ∠MNO=45°,
So sin45°=$\frac{OC}{ON}$=$\frac{12}{5ON}$,
Thus ON=$\frac{12\sqrt{2}}{5}$,
In Rt△NDO, by the Pythagorean theorem: ND^{2}+DO^{2}=ON^{2},
That is, ($\frac{3}{4}$x+3)^{2}+(-x)^{2}=($\frac{12\sqrt{2}}{5}$)^{2},
Solving gives: x_{1}=$\frac{24}{25}$, x_{2}=$\frac{12}{25}$,
Since N is in the second quadrant,
Thus x can only be -$\frac{24}{25}$,
$\frac{3}{4}$x+3=$\frac{12}{25}$,
So ND=$\frac{12}{25}$, OD=$\frac{84}{25}$,
tan∠AON=$\frac{ND}{OD}$=$\frac{1}{7}$.
Therefore, the answer is A. | Arithmetic |
Xiao Lan and Xiao Tan determine the position of P(x, y) by rolling two dice, A and B, respectively. They stipulate: the number Xiao Lan rolls is x, and the number Xiao Tan rolls is y. Therefore, the probability that the point determined by each roll falls on the known line y = -2x + 6 is ( ) A、 <image> B、 <image> C、 <image> D、 <image> | The list is as follows:
\begin{array}{|c|c|c|c|c|c|}
\hline (1,6) & (2,6) & (3,6) & (4,6) & (5,6) & (6,6) \\
\hline (1,5) & (2,5) & (3,5) & (4,5) & (5,5) & (6,5) \\
\hline (1,4) & (2,4) & (3,4) & (4,4) & (5,4) & (6,4) \\
\hline (1,3) & (2,3) & (3,3) & (4,3) & (5,3) & (6,3) \\
\hline (1,2) & (2,2) & (3,2) & (4,2) & (5,2) & (6,2) \\
\hline (1,1) & (2,1) & (3,1) & (4,1) & (5,1) & (6,1) \\
\hline
\end{array}
∴ There are a total of 36 cases, and the points determined by each of them landing on the known line \( y = -2x + 6 \) are (1, 4) and (2, 2).
∴ The probability that the points determined by each of them landing on the known line \( y = -2x + 6 \) is \( \frac{2}{36} = \frac{1}{18} \).
Therefore, the answer is B. | Arithmetic |
As shown in the figure, the line \( y = -2x \) is translated upwards to obtain line AB. Line AB passes through the point (m, n), and \( 2m + n = 6 \). What is the equation of line AB? ( ) <image> A. \( y = -2x - 3 \) B. \( y = -2x - 6 \) C. \( y = -2x + 3 \) D. \( y = -2x + 6 \) | The original line has a slope \( k = -2 \). After shifting upwards, a new line is obtained, so the slope of the new line is also \( k = -2 \).
∴ Line AB passes through the point \((m, n)\), and \( 2m + n = 6 \).
∴ Line AB passes through the point \((m, 6 - 2m)\).
We can assume the equation of the new line is \( y = -2x + b_1 \),
Substituting the point \((m, 6 - 2m)\) into \( y = -2x + b_1 \), we get \( b_1 = 6 \),
∴ The equation of line AB is \( y = -2x + 6 \).
Therefore, the answer is D. | Others |
The graph of the quadratic function \( y = ax^2 + bx + c \) is shown below. The graph of the linear function \( y = bx + a \) does not pass through ( ) <image> A. First Quadrant B. Second Quadrant C. Third Quadrant D. Fourth Quadrant | From the graph opening upwards, we know that a > 0,
the axis of symmetry x = $-\frac{b}{2a} < 0$, so b > 0.
Therefore, the graph of the linear function y = bx + a passes through the first, second, and third quadrants, and does not pass through the fourth quadrant.
Hence, the answer is D. | Functions |
The graph of the quadratic function y=ax 2 +bx+c is shown below, then the graph of the line y=bx+c does not pass through ( ) <image> A. First Quadrant B. Second Quadrant C. Third Quadrant D. Fourth Quadrant | From the graph, we can see that the parabola opens downwards,
∴ -a < 0,
∵ the axis of symmetry is to the right of the y-axis,
∴ the axis of symmetry x = $-\frac{b}{2a}$ > 0,
∴ b > 0;
∵ the parabola intersects the y-axis on the negative y-axis,
∴ c < 0;
∵ b > 0, c < 0
∴ the graph of the linear function y = bx + c does not pass through the second quadrant.
Therefore, the answer is B. | Others |
Given that \(a\), \(b\), and \(c\) are positive real numbers, and they satisfy <image> = <image> = <image> = \(k\), the graph of the linear function \(y = kx + (1 - k)\) must pass through ( ) A、First, Second, Third Quadrants B、First, Second, Fourth Quadrants C、First, Third, Fourth Quadrants D、Second, Third, Fourth Quadrants | Let's consider two cases:
When $a+b+c \neq 0$, according to the property of proportional equality, we get: $k=\frac{2(a+b+c)}{a+b+c}=2$. In this case, the line is $y=2x-1$, which passes through the first, third, and fourth quadrants.
When $a+b+c=0$, i.e., $a+b=-c$,
Since $a$, $b$, and $c$ are positive real numbers,
This situation does not exist.
Therefore, the correct choice is C. | Arithmetic |
The line has the equation y = -2x + b and the hyperbola y = <image> The positions in the Cartesian coordinate system are shown in the figure below. The following conclusions: ① k > 0; ② b > 0; ③ k < 0; ④ b < 0. Which of the following is correct? ( ) <image> A、①② B、②③ C、③④ D、①④； | From the graph of the line y = -2x + b, we can see that b < 0. From the graph of the hyperbola y = $\frac{k}{x}$ in the second and fourth quadrants, we can see that k < 0.
Therefore, the answer is C. | Functions |
Given <image> = <image> = <image> = p, then the graph of the line y = px + p must pass through ( ) A. 1st, 2nd, and 3rd quadrants B. 2nd and 3rd quadrants C. 2nd, 3rd, and 4th quadrants D. 2nd and 4th quadrants | When \( a + b + c = 0 \), \( a + b = -c \)
∴ \( p = -1 \), if \( p = -1 \), it passes through the second, third, and fourth quadrants. When \( a + b + c \neq 0 \), according to the properties of proportions, we can get:
\[
\frac{(a+b)+(b+c)+(a+c)}{a+b+c}=p
\]
∴ \( p = 2 \)
When \( p = 2 \), the equation of the line is: \( y = 2x + 2 \), passing through the first, second, and third quadrants.
Therefore, the line must pass through the second and third quadrants.
Hence, the answer is B. | Arithmetic |
It is known that the graph of the linear function y = kx + b is as shown in the figure. Which of the following conclusions is correct? ( ) <image> A、k > 0, b > 0 B、k > 0, b < 0 C、k < 0, b > 0 D、k < 0, b < 0 | As shown in the figure, the graph of the linear function y = kx + b, y increases as x increases, so k > 0. The line intersects the negative y-axis, so b < 0. Therefore, the answer is B. | Functions |
<image> (2010•Benxi) Given the graph of the linear function y=(a-1)x+b as shown, the range of values for a is ( ) A、a > 1 B、a < 1 C、a > 0 D、a < 0 | From the graph, it can be seen that: y increases as x increases,
∴ -a-1 > 0,
∴ -a > 1.
Therefore, the answer is A. | Plane Geometry |
Given the approximate graph of the quadratic function \( y = ax^2 + bx + 1 \) as shown in the figure, the graph of the function \( y = ax + b \) does not pass through ( ) <image> A. First quadrant B. Second quadrant C. Third quadrant D. Fourth quadrant | From the graph of the quadratic function opening downwards, we can get $a < 0$,
and from the axis of symmetry $x = \frac{b}{-2a} < 0$, we can get $b < 0$,
so the graph of the function $y = ax + b$ passes through the second, third, and fourth quadrants,
therefore, the graph does not pass through the first quadrant.
Hence, the answer is A. | Functions |
In the Cartesian coordinate system, the line \( l \) passes through the second, third, and fourth quadrants and goes through the point (-3, -2). The points (-2, a), (0, b), (c, 1), and (d, -1) are all on the line \( l \). Which of the following statements is correct? ( ) <image> A. a = -3 B. b > -2 C. c < -3 D. d = -2 | Let the expression of a linear function be y = kx + b (k ≠ 0),
Since line l passes through the points (-3, -2), (-2, a), (0, b), (c, 1), (d, -1),
The slope k = $\frac{a+2}{-2+3}$ = $\frac{b+2}{0+3}$ = $\frac{1+2}{c+3}$ = $\frac{-1+2}{d+3}$, i.e., k = a + 2 = $\frac{b+2}{3}$ = $\frac{3}{c+3}$ = $\frac{1}{d+3}$,
Since l passes through the second, third, and fourth quadrants,
k < 0,
Therefore, a < -2, b < -2, c < -3, d < -3.
Thus, the answer is C. | Plane Geometry |
As shown in the figure, the line L is translated 2 units to the right along the positive x-axis to get the line L′. The equation of the line L′ is ( ) <image> A、y=2x+1 B、y=-2x+2 C、y=2x-4 D、y=-2x-2 | We can find two points on line L: (0, 0) and (1, 2). Shifting these points 2 units to the right results in the points (2, 0) and (3, 2). Then, shifting line L 2 units to the right along the positive x-axis results in line L' with the equation y = kx + b, where 2k + b = 0 and k + b = 2.
Solving these equations, we get: k = 2, b = -4.
∴ The function equation is: y = 2x - 4.
Therefore, the answer is C. | Algebra |
As shown in the figure, the equation of the line \( l \) is <image> , and it intersects the x-axis and y-axis at points A and B, respectively. A circle with a radius of 1.5, centered at C, starts at the point (0, 1.5) and moves downward along the y-axis at a speed of 0.5 units per second. When the circle is tangent to the line \( l \), the time the circle has been moving is ( ) <image> A、6 seconds or 10 seconds B、6 seconds or 16 seconds C、3 seconds or 16 seconds D、3 seconds or 6 seconds | Let x=0, then y=-4;
Let y=0, solving gives x=3;
∴A(3,0), B(0,-4),
∴AB=5,
∵DE⊥l, GF⊥l,
∴△BDE∽△BOA, △BFG∽△BAO,
$$\frac{DE}{OA}$$=$$\frac{BE}{AB}$$, $$\frac{GF}{OA}$$=$$\frac{BF}{AB}$$,
which means $$\frac{1.5}{3}$$=$$\frac{BE}{5}$$, $$\frac{1.5}{3}$$=$$\frac{BF}{5}$$,
solving gives BE=2.5, BF=2.5,
∴The distance the circle moves is 3 or 8,
∵The center C of the circle starts from the point (0,1.5) and moves downward along the y-axis at a speed of 0.5 units per second,
∴The time of movement is 6s or 16s.
Therefore, the answer is B. | Others |
If (2, k) is a point on the hyperbola <image> , then the graph of the function y=(k-1)x+k does not pass through ( ) A. First quadrant B. Second quadrant C. Third quadrant D. Fourth quadrant | Since (2, k) is a point on the hyperbola \( y = \frac{1}{x} \),
∴ 2k = 1, solving for k gives k = \(\frac{1}{2}\),
∴ The equation of the linear function is: \( y = -\frac{1}{2}x + \frac{1}{2} \),
∴ The graph of this function passes through the first, second, and fourth quadrants, but does not pass through the third quadrant.
Therefore, the answer is C. | Functions |
As shown in the figure, the line l passes through the first, second, and fourth quadrants. The equation of line l is y = (m - 3)x + m + 2. The range of values for m on the number line is ( ) <image> A、 <image> B、 <image> C、 <image> D、 <image> | 【Analysis】First, determine the range of values for \( m \) based on the position of the function's graph, then represent it on the number line to determine the correct option.
【Solution】Since line \( l \) passes through the first, second, and fourth quadrants,
\[
\left\{ \begin{array} { l } { m - 3 < 0 } \\ { m + 2 > 0 } \end{array} \right.
\],
Solving this, we get: -2 < m < 3,
Therefore, the correct choice is C. | Algebra |
The graph of the quadratic function y=ax 2 +bx+c is shown in the figure. Then the graph of the linear function y=bx+a does not pass through ( ) <image> A. First Quadrant B. Second Quadrant C. Third Quadrant D. Fourth Quadrant | From the graph opening upwards, we know that a > 0,
the axis of symmetry x = $-\frac{b}{2a} < 0$, which implies b > 0.
Therefore, the graph of the linear function y = bx + a passes through the first, second, and third quadrants, and does not pass through the fourth quadrant.
Hence, the answer is D. | Functions |
Given the quadratic function \( y = ax^2 + bx + 1 \) with the approximate graph shown below, the graph of the function \( y = ax - b \) does not pass through ( ) <image> A. First Quadrant B. Second Quadrant C. Third Quadrant D. Fourth Quadrant | The parabola opens downwards,
$$\therefore a<0$$
$$\because$$ the axis of symmetry is to the left of the y-axis,
$$\therefore$$ a and b have the same sign, i.e., b < 0,
$$\therefore$$ the line y = ax - b does not pass through the third quadrant,
Therefore, the answer is C. | Functions |
In the coordinate plane shown in the figure, there is a line L passing through the point (-3, -2). If the four points (-2, a), (0, b), (c, 0), and (d, -1) lie on L, which of the following statements is correct ( ) <image> A、a=3 B、b＞-2 C、c＜-3 D、d=2 | From the given information: this function is a decreasing function,
A、-2 > -3, so a < -2, hence this option is incorrect;
B、-3 < 0, so -2 > b, hence this option is incorrect;
C、0 > -2, so c < -3, hence this option is correct;
D、-1 > -2, so d < -3, hence this option is incorrect.
Therefore, the correct choice is C. | Plane Geometry |
Given that the graph of \( y = ax^2 + bx \) is as shown, then the graph of \( y = ax - b \) must pass through ( ) <image> A. First, Second, Third Quadrant B. First, Second, Fourth Quadrant C. Second, Third, Fourth Quadrant D. First, Third, Fourth Quadrant | ∵ the parabola opens downwards
∴ a＜0
∵ the axis of symmetry of the parabola x=$\frac{b}{-2a}＞0$,
∴ b＞0
∴ in y=ax-b, a＜0, b＜0
∴ the graph passes through the second, third, and fourth quadrants.
Therefore, the answer is C. | Functions |
Given real numbers a, b, c satisfy <image> = <image> = <image> = k and abc ≠ 0, then the graph of the linear function y = kx + k must pass through ( ) A、First and Second Quadrants B、Second and Third Quadrants C、Third and Fourth Quadrants D、First and Fourth Quadrants | Given $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$, we have:
a + b - c = ck, ①
a - b + c = bk, ②
-a + b + c = ak, ③
By adding ①, ②, and ③, we get:
a + b + c = k(a + b + c),
(1) When a + b + c ≠ 0, k = 1;
Therefore, the equation of the linear function y = kx + k is: y = x + 1,
Thus, the function passes through the first, second, and third quadrants;
(2) When a + b + c = 0, b + c = -a, ④
Substituting ④ into ③, we get:
-2a = ak;
Since abc ≠ 0,
a ≠ 0,
Therefore, k = -2,
Thus, the equation of the linear function y = kx + k is: y = -2x - 2;
Therefore, the function passes through the second, third, and fourth quadrants;
In summary, the quadrants that the linear function must pass through are the second and third quadrants;
Therefore, the correct choice is B. | Algebra |
Given the graph of \( y = ax^2 + bx + c \) as shown, the graph of \( y = ax + b \) must pass through ( ) <image> A. First, Second, Third Quadrants B. First, Second, Fourth Quadrants C. Second, Third, Fourth Quadrants D. First, Third, Fourth Quadrants | The parabola opens upwards,
∴ a > 0,
∵ the axis of symmetry is x = $-\frac{b}{2a} > 0$,
and since a > 0,
∴ b < 0,
The intersection point of the linear function with the y-axis is (0, b), and the intersection point with the x-axis is (-$\frac{b}{a}$, 0),
Therefore, the graph must pass through the first, third, and fourth quadrants.
Hence, the answer is D. | Functions |
If the graphs of the linear functions y=k 1 x+b and y=k 2 x in the same Cartesian coordinate system are as shown in the figure, then the solution to the inequality y=k 1 x+b > k 2 x is ( ) <image> A、x > -2 B、x < -2 C、x > -1 D、x < -1 | The coordinates of the intersection point of the two lines are (-1, -2), and when x < -1, the line y = k_{2}x is below the line y = k_{1}x + b. Therefore, the solution set of the inequality k_{1}x + b > k_{2}x is x < -1.
Hence, the answer is: D. | Algebra |
The graphs of the functions y=ax+b and y=ax 2 +bx+c are shown in the figure below. Which of the following options is correct? ( ) <image> A. ab > 0, c > 0 B. ab < 0, c > 0 C. ab > 0, c < 0 D. ab < 0, c < 0 | ∵The graph of the linear function passes through the first, third, and fourth quadrants, ∴a＞0, b＜0,
∴The graph of the quadratic function intersects the y-axis below the x-axis, ∴c＜0,
then ab＜0, c＜0.
Therefore, the answer is D. | Algebra |
Given that the graph of the linear function y = kx + b passes through the first, second, and fourth quadrants, the graph of the inverse proportion function y = <image> is approximately ( ) A. <image> B. <image> C. <image> D. <image> | Since the graph of the linear function y = kx + b passes through the first, second, and fourth quadrants,
∴ k < 0, b > 0,
∴ kb < 0,
∴ the graph of the inverse proportional function y = $\frac{kb}{x}$ is in the second and fourth quadrants.
Therefore, the answer is A. | Functions |
As shown in the figure, the line L is translated 2 units to the right along the positive x-axis to get the line L′. The equation of the line L′ is ( ) <image> A、y=2x+1 B、y=-2x+2 C、y=2x-4 D、y=-2x-2 | We can find two points on line L: (0, 0) and (1, 2). Shifting these points 2 units to the right, we get the points (2, 0) and (3, 2). Then, translating line L 2 units to the right along the positive x-axis, we obtain the equation of line L' as y = kx + b, where 2k + b = 0 and k + b = 2. Solving these equations, we get k = 2 and b = -4. Therefore, the equation of the function is y = 2x - 4. Hence, the correct answer is C. | Plane Geometry |
Given that the graph of the linear function y = kx - 1 does not pass through the second quadrant, the graph of the inverse proportion function y = <image> could be ( ) A. <image> B. <image> C. <image> D. <image> | ∵The graph of the linear function y = kx - 1 does not pass through the second quadrant,
∴it passes through the first, third, and fourth quadrants,
∴k > 0,
∴-k < 0,
∴the graph of the inverse proportion function y = $\frac{-k}{x}$ is in the second and fourth quadrants,
Therefore, the answer is: A. | Functions |
If the graph of the quadratic function y = ax 2 + bx + c (a ≠ 0) is as shown, then the line y = bx - c does not pass through ( ) <image> A. First Quadrant B. Second Quadrant C. Third Quadrant D. Fourth Quadrant | According to the direction of the opening of the quadratic function graph being downwards, we know that $a > 0$.
Since the axis of symmetry $x = -\frac{b}{2a} > 0$,
we have $b < 0$.
Also, since the parabola intersects the y-axis at the positive half-axis,
we have $c > 0$, thus $-c < 0$,
so the line $y = bx - c$ passes through the first, third, and fourth quadrants, i.e., it does not pass through the second quadrant.
Therefore, the answer is B. | Functions |
The graph shown is of the linear function y = kx + b. The signs of k and b are ( ) <image> A. k > 0, b < 0 B. k < 0, b > 0 C. k < 0, b < 0 D. k > 0, b > 0 | ∵The graph of the linear function y = kx + b passes through the first, second, and third quadrants,
∴k > 0,
∴The intersection point of the graph with the y-axis is on the positive y-axis,
∴b > 0.
Therefore, the answer is D. | Functions |
It is known that the graph of the linear function y = kx + b is as shown in the figure. Which of the following conclusions is correct? ( ) <image> A. k > 0, b > 0 B. k > 0, b < 0 C. k < 0, b > 0 D. k < 0, b < 0 | As shown in the figure, the graph of the linear function y = kx + b, y increases as x increases, so k > 0. The line intersects the negative y-axis, so b < 0. Therefore, the answer is B. | Functions |
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