JEE Mathematics Linear Equations and Graphs MCQs Set 02

Answer: (c) second quadrant

Explanation:
1. The x-coordinate is negative and the y-coordinate is positive.
2. Negative x and positive y is the second quadrant.

Teacher's Note:
1. Signs of (x, y): first (+, +), second (-, +), third (-, -), fourth (+, -).
2. Quadrants are counted anticlockwise from the positive x-axis.

Answer: (d) negative y-axis

Explanation:
1. Its x-coordinate is 0, so it is on the y-axis.
2. Its y-coordinate is negative, so it is on the negative side.

Teacher's Note:
1. Points on the axes are not in any quadrant.
2. Every point on the y-axis has x = 0.

Answer: (b) \( x=0 \)

Explanation:
1. Points such as (0, -4), (0, 2) and (0, 3) all have x = 0.
2. So the y-axis is the graph of \( x=0 \).

Teacher's Note:
1. The x-coordinate is zero everywhere on the y-axis.
2. The x-axis has equation \( y=0 \).

Answer: (d) \( y=2x \)

Explanation:
1. Substitute x = 0, y = 0 in each equation.
2. \( y=2x \) gives \( 0=0 \). The others give 0 = 1, 0 = -1 and 0 = 3, which are false.

Teacher's Note:
1. A function with no constant term passes through the origin.
2. The constant term is the y-intercept.

Answer: (a) \( \frac12 \)

Explanation:
1. Slope \( =\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{4-0}=\frac12 \).

Teacher's Note:
1. The slope is rise over run.
2. The line is \( y=\frac12x \).

Answer: (c) (0, -3)

Explanation:
1. For \( y=mx+c \), the y-intercept is c.
2. Here c = -3, so the line cuts the y-axis at (0, -3). (Put x = 0 to check.)

Teacher's Note:
1. Put x = 0 to find where a line meets the y-axis.
2. (3/2, 0) is the point where it meets the x-axis.

Answer: (a) parallel to the y-axis at a distance of 2 units from it

Explanation:
1. Every point on it has x-coordinate 2 and y can be anything.
2. So it is a vertical line, parallel to the y-axis, 2 units to the right of it.

Teacher's Note:
1. Vertical lines are x = constant.
2. They have no slope (infinite slope).

Answer: (d) parallel to the x-axis, 3 units below it

Explanation:
1. All points have y-coordinate -3.
2. So the line is horizontal, 3 units below the x-axis.

Teacher's Note:
1. Horizontal lines are y = constant.
2. A negative constant puts the line below the x-axis.

Answer: (c) Rs. 35,000

Explanation:
1. The cost equation is \( y=300x+5000 \).
2. For x = 100: \( y=30000+5000=35000 \).

Teacher's Note:
1. Do not forget the initial expense.
2. 30,000 is only the variable cost.

Answer: (c) 122

Explanation:
1. Assume \( F=h+KC \). C = 0 gives \( h=32 \); C = 100 gives \( 212=32+100K \), so \( K=\frac95 \).
2. \( F=32+\frac95C \).
3. For C = 50: \( F=32+90=122 \).

Teacher's Note:
1. The relation between the two scales is linear.
2. Rise per degree centigrade is 180/100 = 9/5 Fahrenheit degrees.

Answer: (a) 30

Explanation:
1. From \( \frac{F-32}{9}=\frac C5 \): \( \frac{86-32}{9}=6 \).
2. \( C=5\times6=30 \).

Teacher's Note:
1. Subtract 32 first, then multiply by 5/9.
2. Check: 32 + 9/5 x 30 = 86.

Answer: (d) 5

Explanation:
1. From \( 2x-y=1 \), \( y=2x-1 \). Substitute: \( 3x+4x-2=12 \), so \( x=2 \) and \( y=3 \).
2. The graphs cross at P(2, 3), and \( x+y=5 \).

Teacher's Note:
1. The solution of the pair is the common point of the two lines.
2. Check in both: 6 + 6 = 12 and 4 - 3 = 1.

Answer: (b) \( x<2 \)

Explanation:
1. \( -5x>-10 \).
2. Dividing by -5 reverses the inequality: \( x<2 \).

Teacher's Note:
1. When you multiply or divide by a negative number, the inequality sign reverses.
2. Check x = 0: 10 > 0 is true.

Answer: (a) 2.9

Explanation:
1. x must be strictly less than 3.
2. Only 2.9 satisfies this.

Teacher's Note:
1. 3 itself is not included since the sign is strict.
2. The set is all points to the left of 3 on the number line.

Answer: (c) (1, 2)

Explanation:
1. (1, 2): \( 2+2-2=2>0 \) and \( 1-4+2=-1<0 \); both hold.
2. (0, 0): -2 fails the first. (2, 0): second gives 4, fails. (-1, 0): first gives -4, fails.

Teacher's Note:
1. The common region lies above both lines.
2. Check each condition separately.

Answer: (a) \( y=3x-1 \)

Explanation:
1. Slope \( =\frac{8-2}{3-1}=3 \).
2. \( y=3x+c \): using (1, 2), \( 2=3+c \), so c = -1.
3. Check (3, 8): \( 9-1=8 \).

Teacher's Note:
1. Use one point to find the intercept and the other as a check.
2. m = 3 and c = -1.

Answer: (d) \( -\frac32 \)

Explanation:
1. Rewrite as \( y=-\frac32x+6 \).
2. The coefficient of x is the slope, \( -\frac32 \).

Teacher's Note:
1. Solve for y to read off the slope and intercept.
2. The y-intercept is 6.

Answer: (b) second and fourth quadrants

Explanation:
1. For x > 0, y is negative, so the line is in the fourth quadrant.
2. For x < 0, y is positive, so it is in the second quadrant.
3. It passes through the origin.

Teacher's Note:
1. A line through the origin with negative slope runs from the second to the fourth quadrant.
2. The line y = x lies in the first and third.

Answer: (b) 2

Explanation:
1. A straight line is fixed by two points.
2. A third point is used only as a check.

Teacher's Note:
1. Plot a third point to detect an arithmetic slip.
2. One point fits infinitely many lines.

Answer: (d) \( y=-5 \)

Explanation:
1. A line parallel to the x-axis has the form \( y=k \).
2. It passes through (4, -5), so k = -5.

Teacher's Note:
1. Horizontal means constant y.
2. \( x=4 \) would be the vertical line through the same point.

Answer: (a) (0, 6)

Explanation:
1. Put x = 0: \( y=6 \).

Teacher's Note:
1. (4, 0) is the x-intercept.
2. In the form y = mx + c, c = 6.

Answer: (c) 2

Explanation:
1. Substitute: \( 3a+2(2a)=14 \).
2. \( 7a=14 \), so a = 2.

Teacher's Note:
1. Replace x by a and y by 2a.
2. Check: (2, 4): 6 + 8 = 14.