CUET UG Applied Mathematics Booster Test 2 - Graphical Method, Feasible Region, and Optimal Solution -

🚀

CUET UG Applied Mathematics Booster Test 2 - Graphical Method, Feasible Region, and Optimal Solution

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Graphical Method, Feasible Region, and Optimal Solution

1 / 20

QUESTION 1 OF 20

📄 Passage

There is a factory located at each of the two places \(P\) and \(Q\). A certain commodity is distributed to three depots \(A\), \(B\), and \(C\). The weekly requirements of depots \(A\), \(B\), and \(C\) are \(5\), \(5\), and \(4\) units respectively. The production capacities of factories \(P\) and \(Q\) are \(8\) and \(6\) units respectively. Let: \(x\)= units transported from \(P\) to \(A\) \(y\)= units transported from \(P\) to \(B\) The objective is to minimize total transportation cost \(Z\).

Based on the passage, the total number of units transported from factory P to depot C, expressed in terms of x and y, is:

Answer Review

📄 Passage

Based on the passage, the total number of units transported from factory P to depot C, expressed in terms of x and y, is:

A x + y − 4

B 8 − (x + y)

C x − 7y + 190

D 5 − x

💡 Short Explanation

Factory P has a production capacity of 8 units. x units go to depot A and y units go to depot B. The remaining units are transported to depot C.

✅ Correct Answer Explanation

Factory P can produce a maximum of 8 units. Out of these: x units are sent to depot A. y units are sent to depot B. Therefore, the remaining units sent to depot C are: 8 − x − y or 8 − (x + y) This expression ensures that the total supply from factory P remains equal to its production capacity. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) x + y − 4
Incorrect because it does not represent the remaining production from factory P.
Option C) x − 7y + 190
Incorrect because it is an objective function expression, not a transportation quantity.
Option D) 5 − x
Incorrect because depot C depends on both x and y.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Subtract the quantities already transported from the total production capacity.

Final Logic:

Remaining units = 8 − (x + y); therefore, Option B is correct.

🧠 Memory Trick

"Remaining = Total − Used."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Formulation of Constraints

📄 Passage

Which inequality represents the condition that depot C receives non-negative units?

A 8 − x − y ≥ 0

B x + y ≥ 4

C x ≤ 5

D y ≤ 5

💡 Short Explanation

Depot C receives 8 − x − y units. Since depot C requires 4 units, the transportation model leads to x + y ≥ 4. This is the required constraint.

✅ Correct Answer Explanation

From the passage: Units transported from P to depot C: 8 − x − y Depot C requires a total of 4 units. The remaining requirement is supplied by factory Q. For Q's shipment to remain non-negative, 4 − (8 − x − y) ≥ 0 Simplifying, 4 − 8 + x + y ≥ 0 x + y ≥ 4 Thus, the required inequality is: x + y ≥ 4 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 8 − x − y ≥ 0
Incorrect because it only ensures that factory P does not exceed its capacity. It is not the condition derived for depot C's non-negative allocation from factory Q.
Option C) x ≤ 5
Incorrect because it represents the demand of depot A.
Option D) y ≤ 5
Incorrect because it represents the demand of depot B.

🎯 Strategy

used

Substitution

Application:

Express the remaining demand of depot C and apply the non-negativity condition.

Final Logic:

The required inequality simplifies to x + y ≥ 4; therefore, Option B is correct.

🧠 Memory Trick

"Remaining demand ≥ 0."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Transportation Constraints

3 A feasible region is bounded by:
y ≥ 0,
x − y ≥ 0,
x ≤ 4
This region is enclosed by:
y = 0
y = x
x = 4
Find the area of the region.

A 16

B 4

C 12

D 8

💡 Short Explanation

Determine the vertices of the triangular region. Calculate its base and height. Apply the triangle area formula.

✅ Correct Answer Explanation

The boundary lines are: y = 0 y = x x = 4 The vertices are: (0, 0) (4, 0) (4, 4) This forms a right-angled triangle. Base = 4 units Height = 4 units Area = (1/2) × Base × Height = (1/2) × 4 × 4 = 8 square units Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 16
Incorrect because it represents the enclosing square.
Option B) 4
Incorrect because the triangle's area is larger.
Option C) 12
Incorrect because it is not obtained using the triangle area formula.

🎯 Strategy

used

Substitution

Application:

Identify the corner points from the boundary lines and calculate the triangular area.

Final Logic:

The triangle has base and height of 4 units, giving an area of 8 square units.

🧠 Memory Trick

"Triangle = ½ × Base × Height."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Feasible Region

4 Along the Y-axis from y = 0 to y = 3, the objective function is:
Z = 4y
Evaluate: \(\int_{0}^{3}\,4y dy\)

A 18

B 12

C 9

D 36

💡 Short Explanation

Integrate 4y. Apply the limits 0 and 3. The value of the integral is 18.

✅ Correct Answer Explanation

The given integral is: ∫₀³ 4y dy Integrating, ∫ 4y dy = 2y² Apply the limits: At y = 3 = 2(3²) = 18 At y = 0 = 0 Therefore, 18 − 0 = 18 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 12
Incorrect because the integration is incomplete.
Option C) 9
Incorrect because only y² is considered without multiplying by 2.
Option D) 36
Incorrect because the integral evaluates to 18.

🎯 Strategy

used

Substitution

Application:

Integrate the function and substitute the upper and lower limits.

Final Logic:

The definite integral equals 18; therefore, Option A is correct.

🧠 Memory Trick

"Integrate first, substitute later."

📚 Class XII – Mathematics – Chapter: Integrals – Topic: Definite Integrals

5 Arrange the following equations in increasing order of their Y-intercepts:
1. 5x + y = 3
2. x − 2y = −10
3. 2x + 4y = 8
4. 3x − y = −6

A 1, 2, 3, 4

B 3, 1, 4, 2

C 3, 1, 2, 4

D 1, 3, 2, 4

💡 Short Explanation

Find each y-intercept by setting x = 0. Arrange the intercepts in ascending order. Compare with the options.

✅ Correct Answer Explanation

Set x = 0 in each equation. 1. 5x + y = 3 y = 3 2. x − 2y = −10 −2y = −10 y = 5 3. 2x + 4y = 8 4y = 8 y = 2 4. 3x − y = −6 −y = −6 y = 6 The y-intercepts are: Line 3 → 2 Line 1 → 3 Line 2 → 5 Line 4 → 6 Increasing order: 3, 1, 2, 4 Therefore, the mathematically correct answer is Option C. Note: The provided answer key is incorrect.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the intercepts are not arranged in ascending order.
Option B) Incorrect because Line 4 should come after Line 2.
Option D) Incorrect because Line 1 does not have the smallest y-intercept.

🎯 Strategy

used

Substitution

Application:

Set x = 0 in each equation to obtain the y-intercepts and arrange them in increasing order.

Final Logic:

The correct order is 3, 1, 2, 4; therefore, Option C is correct, and the provided answer key is incorrect.

🧠 Memory Trick

"Y-intercept ⇒ Put x = 0."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Boundary Lines and Intercepts

6 Given:
2x + y ≤ 6
Which vector represents a point lying outside the feasible region?

A \(\hat{i}+\hat{j}\)

B \(2\hat{i}+2\hat{j}\)

C \(3\hat{i}\)

D \(2\hat{i}+3\hat{j}\)

💡 Short Explanation

Convert each vector into its corresponding point. Substitute each point into the inequality. Identify the point that does not satisfy the constraint.

✅ Correct Answer Explanation

The constraint is: 2x + y ≤ 6 Evaluate each option. Option A: î + ĵ = (1, 1) 2(1) + 1 = 3 ≤ 6 ✓ Option B: 2î + 2ĵ = (2, 2) 2(2) + 2 = 6 ≤ 6 ✓ Option C: 3î = (3, 0) 2(3) + 0 = 6 ≤ 6 ✓ Option D: 2î + 3ĵ = (2, 3) 2(2) + 3 = 7 > 6 ✗ Hence, this point lies outside the feasible region. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Correctly satisfies the inequality.
Option B) Lies exactly on the boundary line.
Option C) Also lies on the boundary line.

🎯 Strategy

used

Substitution

Application:

Replace x and y with the coordinates represented by each vector and check the inequality.

Final Logic:

Only (2, 3) violates the constraint; therefore, Option D is correct.

🧠 Memory Trick

"Substitute and check ≤."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Half-Planes and Feasible Region

7 Which statement is incorrect about feasible regions in LPP?

A The shaded region can lie in any quadrant without restriction

B A bounded feasible region forms a convex polygon

C Any point inside the region is feasible

D An unbounded region is not enclosed on all sides

💡 Short Explanation

In standard LPP, non-negativity constraints apply. Therefore, the feasible region is generally confined to the first quadrant. Hence, Option A is incorrect.

✅ Correct Answer Explanation

In most Linear Programming Problems, x ≥ 0 y ≥ 0 These non-negativity constraints restrict the feasible region to the first quadrant. Therefore, the statement that the shaded region can lie in any quadrant without restriction is incorrect. The remaining statements are standard properties of feasible regions. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Correct because every bounded feasible region is a convex polygon.
Option C) Correct because every point inside the feasible region satisfies all constraints.
Option D) Correct because an unbounded region extends infinitely in at least one direction.

🎯 Strategy

used

Extreme Word Filter

Application:

The phrase "without restriction" is an absolute statement and contradicts the non-negativity conditions of standard LPP.

Final Logic:

Since the feasible region is generally restricted to the first quadrant, Option A is incorrect.

🧠 Memory Trick

"Standard LPP ⇒ First quadrant."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Feasible Region

8 A square region is defined by:
0 ≤ x ≤ 2,
0 ≤ y ≤ 2
A point is chosen randomly from this square. Find the probability that:
x + y ≤ 2

A 0.25

B 0.75

C 0.5

D 1.0

💡 Short Explanation

Compute the area of the square. Determine the area satisfying x + y ≤ 2. Probability equals favourable area divided by total area.

✅ Correct Answer Explanation

The square has: Side = 2 units Area of square = 2 × 2 = 4 square units The line: x + y = 2 divides the square into two congruent right triangles. Area of the favourable triangle = (1/2) × 2 × 2 = 2 square units Therefore, Probability = Favourable area / Total area = 2 / 4 = 1/2 = 0.5 Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the favourable region occupies half the square, not one-fourth.
Option B) Incorrect because the favourable region is not three-fourths of the square.
Option D) Incorrect because not every point satisfies x + y ≤ 2.

🎯 Strategy

used

Dimensional/Unit Analysis

Application:

Compare the area of the favourable region with the total area of the sample space.

Final Logic:

The favourable area is half of the square; therefore, the probability is 0.5.

🧠 Memory Trick

"Probability = Area ratio."

📚 Class XII – Mathematics – Chapter: Probability – Topic: Geometric Probability

9 Minimize:
Z = –x + 2y
Subject to:
–x + 3y ≤ 10
x + y ≤ 6
x – y ≤ 2
x, y ≥ 0
Which point gives the minimum value?

A (0, 0)

B (4, 2)

C (2, 0)

D (2, 4)

💡 Short Explanation

Evaluate Z at each given point. Check which points satisfy the constraints. Select the feasible point with the smallest value of Z.

✅ Correct Answer Explanation

The objective function is: Z = –x + 2y Evaluate each option. Option A: (0, 0) Z = 0 Feasible ✓ Option B: (4, 2) Z = –4 + 4 = 0 Feasible ✓ Option C: (2, 0) Z = –2 + 0 = –2 Feasible ✓ Option D: (2, 4) Z = –2 + 8 = 6 Not feasible because: 2 + 4 = 6 ✓ But: –2 + 12 = 10 ✓ 2 – 4 = –2 ≤ 2 ✓ Actually, this point is also feasible. Comparing objective values: (0,0) → 0 (4,2) → 0 (2,0) → –2 (2,4) → 6 The minimum value is –2 at (2,0). Therefore, Option C is the correct answer. The provided answer key is correct.

❓ Why Other Options Are Incorrect

Option A) Gives Z = 0, which is greater than –2.
Option B) Gives Z = 0, not the minimum.
Option D) Gives Z = 6, which is the largest among the feasible options.

🎯 Strategy

used

Substitution

Application:

Evaluate the objective function at each feasible point and compare the values.

Final Logic:

The smallest value is –2 at (2,0); therefore, Option C is correct.

🧠 Memory Trick

"Substitute all corner points."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Point Method

10 Assertion (A):
The optimal solution of an LPP always occurs at a corner point of the feasible region.
Reason (R):
If the feasible region is unbounded, a minimum value always exists at a corner point.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R explains A

D A is false, R is true

💡 Short Explanation

The Corner Point Theorem supports the Assertion. An unbounded region may not possess a finite minimum value. Hence, the Reason is false.

✅ Correct Answer Explanation

Assertion (A): According to the Corner Point Theorem, whenever an optimal solution exists for a Linear Programming Problem, it occurs at a corner point of the feasible region. Therefore, the Assertion is true. Reason (R): An unbounded feasible region does not always possess a finite minimum value. Whether a minimum exists depends on the direction of the objective function. Therefore, the Reason is false. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the Assertion is true.
Option C) Incorrect because the Reason is false.
Option D) Incorrect because the Assertion is true.

🎯 Strategy

used

Elimination

Application:

Evaluate the Assertion and the Reason independently using the Corner Point Theorem and properties of unbounded regions.

Final Logic:

The Assertion is true, but the Reason is false; therefore, Option B is correct.

🧠 Memory Trick

"Corner point if optimum exists."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Point Theorem

11 Evaluate:
Z = 5x + 3y
at the points:
(0, 0), (0, 3), (20/19, 45/19), (2, 0)
Find the maximum value of Z.

A \(\frac{235}{19}\)

B 10

C 9

D \(\frac{20}{19}\)

💡 Short Explanation

Evaluate Z at each point. Compare the obtained values. Select the largest value.

✅ Correct Answer Explanation

The objective function is: Z = 5x + 3y Evaluate at each point. At (0, 0): Z = 0 At (0, 3): Z = 5(0) + 3(3) = 9 At (20/19, 45/19): Z = 5(20/19) + 3(45/19) = 100/19 + 135/19 = 235/19 At (2, 0): Z = 5(2) + 3(0) = 10 Comparing the values: 0, 9, 10, 235/19 Since 235/19 ≈ 12.37, it is the maximum value. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 10
Incorrect because 235/19 is greater than 10.
Option C) 9
Incorrect because it is smaller than both 10 and 235/19.
Option D) 20/19
Incorrect because it is only the x-coordinate of one point, not the objective value.

🎯 Strategy

used

Substitution

Application:

Substitute each point into the objective function and compare the results.

Final Logic:

The largest value is 235/19; therefore, Option A is correct.

🧠 Memory Trick

"Substitute → Compare → Select maximum."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Evaluation of Objective Function

12 Match each corner point in List I with the corresponding value of the objective function
Z = x − 7y + 190

List I	List II
1. (0, 0)	a. 195
2. (5, 0)	b. 190
3. (0, 5)	c. 155
4. (5, 3)	d. 174

A 1–a, 2–b, 3–c, 4–d

B 1–d, 2–c, 3–b, 4–a

C 1–c, 2–a, 3–d, 4–b

D 1–b, 2–a, 3–c, 4–d

💡 Short Explanation

Substitute each corner point into the objective function. Compute the corresponding value. Match the values correctly.

✅ Correct Answer Explanation

The objective function is: Z = x − 7y + 190 Evaluate each point. 1. (0, 0) Z = 0 − 0 + 190 = 190 → b 2. (5, 0) Z = 5 − 0 + 190 = 195 → a 3. (0, 5) Z = 0 − 35 + 190 = 155 → c 4. (5, 3) Z = 5 − 21 + 190 = 174 → d Hence, 1–b, 2–a, 3–c, 4–d Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the first two matches are interchanged.
Option B) Incorrect because several values are mismatched.
Option C) Incorrect because the first point evaluates to 190, not 155.

🎯 Strategy

used

Substitution

Application:

Evaluate the objective function at every corner point and match the obtained values.

Final Logic:

The correct matching is 1–b, 2–a, 3–c, 4–d.

🧠 Memory Trick

"One point at a time."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Point Method

13 An operational manufacturer evaluates three parallel iso-profit lines passing through a feasible region, corresponding to profit values of ₹1000, ₹1500, and ₹2000.
If the line farthest from the origin corresponds to ₹2000, determine the arithmetic mean of these three profit values.

A 1500

B 2000

C 1000

D 4500

💡 Short Explanation

Add the three profit values. Divide the sum by 3. The result is ₹1500.

✅ Correct Answer Explanation

The three profit values are: ₹1000 ₹1500 ₹2000 Arithmetic Mean = (1000 + 1500 + 2000)/3 = 4500/3 = ₹1500 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because it is the maximum profit value.
Option C) Incorrect because it is the minimum profit value.
Option D) Incorrect because it is the total sum, not the average.

🎯 Strategy

used

Substitution

Application:

Use the arithmetic mean formula directly.

Final Logic:

The average of the three profits is ₹1500.

🧠 Memory Trick

"Average = Sum ÷ Number."

📚 Class XII – Mathematics – Chapter: Statistics – Topic: Arithmetic Mean

14 For the iso-profit line
ax + by = k,
identify which quantities remain constant during parallel shifts of the line.
1. The slope of the line
2. The value of k
3. The coefficients a and b

A 1 and 2 only

B 1 and 3 only

C 2 and 3 only

D 1, 2, and 3

💡 Short Explanation

Parallel lines have the same slope. The coefficients remain unchanged. Only the value of k changes.

✅ Correct Answer Explanation

The equation of an iso-profit line is: ax + by = k During parallel shifting: The coefficients a and b remain fixed. Therefore, the slope remains unchanged. The value of k changes to represent different profit levels. Thus, Statement 1 is true. Statement 2 is false. Statement 3 is true. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because k changes during shifting.
Option C) Incorrect because the slope also remains constant.
Option D) Incorrect because k is not constant.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall the graphical interpretation of iso-profit lines during parallel movement.

Final Logic:

Only the slope and coefficients remain unchanged; therefore, Option B is correct.

🧠 Memory Trick

"Same slope, new profit."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-profit Method

15 Arrange the following iso-cost lines
Z = 2x + 5y
in increasing order of their perpendicular distance from the origin.
1. Z = 50
2. Z = 100
3. Z = 10
4. Z = 75

A 1, 4, 2, 3

B 2, 4, 1, 3

C 4, 1, 3, 2

D 3, 1, 4, 2

💡 Short Explanation

The perpendicular distance from the origin is proportional to the value of Z. Arrange the Z values from smallest to largest. Match the corresponding line numbers.

✅ Correct Answer Explanation

The iso-cost lines are: 2x + 5y = 10 2x + 5y = 50 2x + 5y = 75 2x + 5y = 100 The perpendicular distance of the line 2x + 5y = k from the origin is Distance = |k| / √(2² + 5²) Since the denominator remains constant, the distance increases as k increases. Therefore, 10 < 50 < 75 < 100 Corresponding order: 3, 1, 4, 2 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because it does not follow increasing values of k.
Option B) Incorrect because it starts with the largest distance.
Option C) Incorrect because the values are not in ascending order.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize that all lines are parallel, so only the constant term determines the perpendicular distance.

Final Logic:

Smaller k means a smaller distance from the origin; therefore, the correct order is 3, 1, 4, 2.

🧠 Memory Trick

"Smaller Z ⇒ Closer to origin."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-cost Method

16 Assertion (A):
In minimization problems using iso-cost lines, the optimal solution occurs at the point where the line closest to the origin still intersects the feasible region.
Reason (R):
Shifting the iso-cost line toward the origin decreases the value of the objective function Z.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R is the correct explanation of A

D A is false, R is true

💡 Short Explanation

In minimization, iso-cost lines are shifted toward the origin. Smaller values of Z correspond to lines nearer the origin. Therefore, the Reason correctly explains the Assertion.

✅ Correct Answer Explanation

Assertion (A): For a minimization problem, an iso-cost line is moved parallel to itself toward the origin. The first point where it touches the feasible region gives the minimum value of the objective function. Therefore, the Assertion is true. Reason (R): The equation of an iso-cost line is ax + by = Z. As the line is shifted toward the origin, the value of Z decreases while the slope remains unchanged. Hence, moving closer to the origin corresponds to lower cost. Therefore, the Reason is true and correctly explains the Assertion. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because both statements are true.
Option B) Incorrect because the Reason is also true.
Option D) Incorrect because the Assertion is true.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Interpret the graphical movement of iso-cost lines during minimization.

Final Logic:

The Reason correctly explains why the nearest iso-cost line gives the minimum value; therefore, Option C is correct.

🧠 Memory Trick

"Closer to origin → Lower cost."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-cost Method

17 Consider the constraints:
x + y ≤ 2,
x + y ≥ 5,
x ≥ 0,
y ≥ 0
Identify the nature of the solution.

A Unique optimal solution

B No feasible region; hence no solution

C Infinitely many optimal solutions

D Unbounded feasible region

💡 Short Explanation

The first two constraints contradict each other. No point satisfies both simultaneously. Therefore, no feasible region exists.

✅ Correct Answer Explanation

The given constraints are: x + y ≤ 2 x + y ≥ 5 A point cannot simultaneously satisfy: x + y ≤ 2 and x + y ≥ 5 Hence, there is no common region satisfying all constraints. Without a feasible region, the Linear Programming Problem has no feasible solution. Therefore, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because a unique optimum requires a feasible region.
Option C) Incorrect because no feasible point exists.
Option D) Incorrect because there is no feasible region to be unbounded.

🎯 Strategy

used

Elimination

Application:

Check whether all constraints can be satisfied simultaneously.

Final Logic:

Since the constraints are contradictory, Option B is correct.

🧠 Memory Trick

"Contradictory constraints = No feasible region."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Infeasible Region

18 Evaluate the improper integral: \(\int_{2}^{\infty }\,\frac{1}{x^{2}} dx\)

A 2

B 1

C ∞

D 1/2

💡 Short Explanation

Rewrite the integrand as x⁻². Evaluate the improper integral using limits. The value obtained is 1/2.

✅ Correct Answer Explanation

The given integral is: ∫₂^∞ (1/x²) dx Rewrite: = ∫₂^∞ x⁻² dx Integrating, = [−1/x]₂^∞ Now apply the limits. As x → ∞, −1/x → 0 Therefore, 0 − (−1/2) = 1/2 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the definite integral evaluates to 1/2.
Option B) Incorrect because the area under the curve is smaller than 1.
Option C) Incorrect because the improper integral converges.

🎯 Strategy

used

Substitution

Application:

Evaluate the antiderivative and apply the infinite limit.

Final Logic:

The improper integral converges to 1/2; therefore, Option D is correct.

🧠 Memory Trick

"1/x² from a to ∞ = 1/a."

📚 Class XII – Mathematics – Chapter: Integrals – Topic: Improper Integrals

19 Given the point
C(20/19, 45/19),
let the position vector be r⃗.
Compute: \(\vec{r}⋅\hat{i}\)

A \(\frac{20}{19}\)

B \(\frac{45}{19}\)

C \(\frac{65}{19}\)

D 0

💡 Short Explanation

The dot product with î gives the x-component. The x-coordinate of C is 20/19. Therefore, the answer is 20/19.

✅ Correct Answer Explanation

The position vector of C(20/19, 45/19) is r⃗ = (20/19)î + (45/19)ĵ Taking the dot product with î, r⃗ · î = (20/19)(1) + (45/19)(0) = 20/19 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because it is the y-component.
Option C) Incorrect because the vector components are not added in a dot product with î.
Option D) Incorrect because the x-component is non-zero.

🎯 Strategy

used

Substitution

Application:

Write the vector in component form and evaluate its dot product with the unit vector î.

Final Logic:

The dot product with î equals the x-coordinate; therefore, Option A is correct.

🧠 Memory Trick

"Dot with î = x-coordinate."

📚 Class XI – Mathematics – Chapter: Vector Algebra – Topic: Scalar (Dot) Product

20 Case Study:
Maximize the objective function:
Z = 15x + 10y
Subject to the constraints:
4x + 6y ≤ 360,
3x ≤ 180,
5y ≤ 200,
x, y ≥ 0
The optimal solution occurs at:
(x, y) = (60, 20)
Determine the maximum value of Z.

A 900

B 1000

C 1100

D 1200

💡 Short Explanation

Substitute the optimal point into the objective function. Evaluate Z. The maximum value obtained is 1100.

✅ Correct Answer Explanation

The objective function is: Z = 15x + 10y Substitute the optimal point: x = 60 y = 20 Therefore, Z = 15(60) + 10(20) = 900 + 200 = 1100 Since the point (60, 20) is the optimal solution, the corresponding objective value is the maximum value. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 900
Incorrect because it considers only the contribution from x.
Option B) 1000
Incorrect because it is not obtained by substituting the given coordinates.
Option D) 1200
Incorrect because the correct calculation gives 1100.

🎯 Strategy

used

Substitution

Application:

Replace x and y with the coordinates of the optimal point in the objective function.

Final Logic:

Substituting (60, 20) gives Z = 1100; therefore, Option C is correct.

🧠 Memory Trick

"Optimal point → Substitute into Z."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Evaluation of Objective Function at the Optimal Point

CUET UG Applied Mathematics Booster Test 2 - Graphical Method, Feasible Region, and Optimal Solution

Test Complete!

Answer Review