Pacesetters Education Centre (Pume Class ) : Mathematics

1. Here 2 is read as square Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is

Y2 – 4x + 2 = 0

Y2 + 4x + 2 = 0

X2 + 4y + 2 = 0

X2 – 4y + 2 = 0

The equation y^2 - 4x + 2 = 0 represents the locus of the midpoint of PQ. This can be determined by using the midpoint formula and substituting the coordinates of P and Q into the equation y^2 = 8x. Simplifying the equation will result in y^2 - 4x + 2 = 0, which represents the locus of the midpoint.

Explanation

The equation y^2 - 4x + 2 = 0 represents the locus of the midpoint of PQ. This can be determined by using the midpoint formula and substituting the coordinates of P and Q into the equation y^2 = 8x. Simplifying the equation will result in y^2 - 4x + 2 = 0, which represents the locus of the midpoint.

2. The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is

1

0

3

2

The sum of the squares of the roots of a quadratic equation can be found using Vieta's formulas. For a quadratic equation of the form ax^2 + bx + c = 0, the sum of the squares of the roots is given by (b^2 - 2ac)/a^2. In this case, the equation is x^2 - (a - 2)x - a - 1 = 0. Comparing it with the standard form, we have a = 1, b = -(a - 2) = 2 - a, and c = -(a + 1). Substituting these values into the formula, we get (2 - a)^2 - 2(1)(-(a + 1))/1^2. Simplifying this expression, we get a^2 - 4a + 3. To find the least value of this expression, we can factorize it as (a - 1)(a - 3) = 0. Therefore, the least value of a is 1.

Explanation

The sum of the squares of the roots of a quadratic equation can be found using Vieta's formulas. For a quadratic equation of the form ax^2 + bx + c = 0, the sum of the squares of the roots is given by (b^2 - 2ac)/a^2. In this case, the equation is x^2 - (a - 2)x - a - 1 = 0. Comparing it with the standard form, we have a = 1, b = -(a - 2) = 2 - a, and c = -(a + 1). Substituting these values into the formula, we get (2 - a)^2 - 2(1)(-(a + 1))/1^2. Simplifying this expression, we get a^2 - 4a + 3. To find the least value of this expression, we can factorize it as (a - 1)(a - 3) = 0. Therefore, the least value of a is 1.

3. (1) A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P (A ∪ B) is

3/5

0

1

5/2

The event A represents the numbers greater than 3, which are 4 and 5. The event B represents the numbers less than 5, which are 1, 2, 3, and 4. The union of events A and B includes all the possible outcomes of getting a number greater than 3 or less than 5, which are 1, 2, 3, 4, and 5. Since all the possible outcomes are included in the union, the probability of the union is equal to 1.

Explanation

The event A represents the numbers greater than 3, which are 4 and 5. The event B represents the numbers less than 5, which are 1, 2, 3, and 4. The union of events A and B includes all the possible outcomes of getting a number greater than 3 or less than 5, which are 1, 2, 3, 4, and 5. Since all the possible outcomes are included in the union, the probability of the union is equal to 1.

4. The system of equations αx + y + z = α - 1, x + αy + z = α - 1, x + y + αz = α - 1 has no solution, if α is

-2

Either – 2 or 1

Not -2

1

The system of equations has no solution when α is -2 because all three equations are identical and cannot be satisfied simultaneously. If we substitute α with -2 in any of the equations, we get -2x + y + z = -3, which is the same as the other two equations. Therefore, there is no set of values for x, y, and z that can satisfy all three equations, resulting in no solution.

Explanation

The system of equations has no solution when α is -2 because all three equations are identical and cannot be satisfied simultaneously. If we substitute α with -2 in any of the equations, we get -2x + y + z = -3, which is the same as the other two equations. Therefore, there is no set of values for x, y, and z that can satisfy all three equations, resulting in no solution.

5. The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has y−intercept − 4. Then a possible value of k is

1

-4

3

2

The perpendicular bisector of a line segment passes through the midpoint of the segment and is perpendicular to it. The midpoint of the line segment joining P (1, 4) and Q (k, 3) can be found by taking the average of the x-coordinates and the average of the y-coordinates. The x-coordinate of the midpoint is (1 + k)/2 and the y-coordinate is (4 + 3)/2 = 7/2.

The slope of the line segment PQ is (3 - 4)/(k - 1) = -1/(k - 1).

Since the perpendicular bisector is perpendicular to PQ, its slope is the negative reciprocal of the slope of PQ, which is (k - 1)/1.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line, we can substitute the values of the midpoint (1 + k)/2 and 7/2, and the slope (k - 1)/1, and solve for y-intercept.

y - (7/2) = (k - 1)/1 * (x - (1 + k)/2)

Simplifying the equation, we get y = (k - 1)x/2 - (k + 1)/4 + 7/2.

Since the y-intercept is -4, we can set y = -4 and solve for k.

-4 = (k - 1)x/2 - (k + 1)/4 + 7/2

Simplifying the equation further, we get -8 = (k - 1)x - (k + 1)/2 + 14

Rearranging the terms, we get (k - 1)x - (k + 1)/2 = 22

Since we don't have the value of x, we can't solve for k. Hence, the possible value of k is not determinable from the given information.

Explanation

The perpendicular bisector of a line segment passes through the midpoint of the segment and is perpendicular to it. The midpoint of the line segment joining P (1, 4) and Q (k, 3) can be found by taking the average of the x-coordinates and the average of the y-coordinates. The x-coordinate of the midpoint is (1 + k)/2 and the y-coordinate is (4 + 3)/2 = 7/2.

The slope of the line segment PQ is (3 - 4)/(k - 1) = -1/(k - 1).

Since the perpendicular bisector is perpendicular to PQ, its slope is the negative reciprocal of the slope of PQ, which is (k - 1)/1.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line, we can substitute the values of the midpoint (1 + k)/2 and 7/2, and the slope (k - 1)/1, and solve for y-intercept.

y - (7/2) = (k - 1)/1 * (x - (1 + k)/2)

Simplifying the equation, we get y = (k - 1)x/2 - (k + 1)/4 + 7/2.

Since the y-intercept is -4, we can set y = -4 and solve for k.

-4 = (k - 1)x/2 - (k + 1)/4 + 7/2

Simplifying the equation further, we get -8 = (k - 1)x - (k + 1)/2 + 14

Rearranging the terms, we get (k - 1)x - (k + 1)/2 = 22

Since we don't have the value of x, we can't solve for k. Hence, the possible value of k is not determinable from the given information.

6. The point diametrically opposite to the point P (1, 0) on the circle x2 + y2 + 2x + 4y − 3 = 0 is

(− 3, − 4)

(-3, 4)

(3, 4)

(-4, -1)

The given equation represents a circle with center (-1, -2) and radius 2. The point diametrically opposite to a given point on a circle can be found by reflecting the coordinates of the given point across the center of the circle. In this case, the center is (-1, -2) and the given point is (1, 0). By reflecting the coordinates across the center, we get (-3, -4), which is the diametrically opposite point on the circle. Therefore, the correct answer is (-3, -4).

Explanation

The given equation represents a circle with center (-1, -2) and radius 2. The point diametrically opposite to a given point on a circle can be found by reflecting the coordinates of the given point across the center of the circle. In this case, the center is (-1, -2) and the given point is (1, 0). By reflecting the coordinates across the center, we get (-3, -4), which is the diametrically opposite point on the circle. Therefore, the correct answer is (-3, -4).

7. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is

Reflexive and transitive only

Reflexive only

An equivalence relation

Reflexive and symmetric only

The relation R is reflexive because every element in A is related to itself. For example, (3, 3), (6, 6), (9, 9), and (12, 12) are all in R.

The relation R is transitive because if (a, b) and (b, c) are in R, then (a, c) must also be in R. For example, (3, 6) and (6, 12) are in R, so (3, 12) is also in R.

However, the relation R is not symmetric because if (a, b) is in R, it does not necessarily mean that (b, a) is in R. For example, (3, 6) is in R, but (6, 3) is not in R.

Therefore, the correct answer is reflexive and transitive only.

Explanation

The relation R is reflexive because every element in A is related to itself. For example, (3, 3), (6, 6), (9, 9), and (12, 12) are all in R.

The relation R is transitive because if (a, b) and (b, c) are in R, then (a, c) must also be in R. For example, (3, 6) and (6, 12) are in R, so (3, 12) is also in R.

However, the relation R is not symmetric because if (a, b) is in R, it does not necessarily mean that (b, a) is in R. For example, (3, 6) is in R, but (6, 3) is not in R.

Therefore, the correct answer is reflexive and transitive only.

8. A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at

0, 2)

(0, 1)

(1, 0)

(2, 0)

The vertex of a parabola is the point where the parabola reaches its minimum or maximum value. In this case, since the focus of the parabola is at the origin (0,0) and the directrix is the line x = 2, the vertex will be halfway between the focus and the directrix along the axis of symmetry. The axis of symmetry is the line perpendicular to the directrix and passing through the focus. Since the focus is at the origin (0,0) and the directrix is the line x = 2, the axis of symmetry is the line x = 1. Therefore, the vertex of the parabola is at (1,0).

Explanation

The vertex of a parabola is the point where the parabola reaches its minimum or maximum value. In this case, since the focus of the parabola is at the origin (0,0) and the directrix is the line x = 2, the vertex will be halfway between the focus and the directrix along the axis of symmetry. The axis of symmetry is the line perpendicular to the directrix and passing through the focus. Since the focus is at the origin (0,0) and the directrix is the line x = 2, the axis of symmetry is the line x = 1. Therefore, the vertex of the parabola is at (1,0).

9. Let A be a 2 m 2 matrix with real entries. Let I be the 2 m 2 identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that A2 = I. Statement −1: If A ≠ I and A ≠ − I, then det A = − 1. Statement −2: If A ≠ I and A ≠ − I, then tr (A) ≠ 0.

Statement −1 is false, Statement −2 is true

Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

Statement − 1 is true, Statement − 2 is false.

not-available-via-ai

Explanation

not-available-via-ai

10. If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

22.0

20.5

25.5

24.0

If the mean and median of a frequency distribution are 21 and 22 respectively, it means that the distribution is slightly skewed to the left. In a skewed distribution, the mode tends to be less than the median and mean. Since the median is 22, the mode is likely to be slightly less than 22. Therefore, the closest option is 24.0.

Explanation

If the mean and median of a frequency distribution are 21 and 22 respectively, it means that the distribution is slightly skewed to the left. In a skewed distribution, the mode tends to be less than the median and mean. Since the median is 22, the mode is likely to be slightly less than 22. Therefore, the closest option is 24.0.