Solution of Global College Internal Exam Basic Mathematics Neb 11 Management
Question No. 2 (Group A)
Solution by FEEN (Focus Edge Education Network)
"If A and B are the subset of universal set U, then the symmetric difference A and B, A Δ B is the subset of..."
- A) A ∩ B
- B) A ∪ B
- C) A - B
- D) B - A
Detailed Explanation
To understand this, let's break down what Symmetric Difference (A Δ B) actually means in simple words:
- The Definition: The symmetric difference of two sets includes all elements that are in either Set A or Set B, but NOT in both at the same time.
-
The Formula: It is written as:
A Δ B = (A - B) ∪ (B - A)
Step-by-Step Logic
Step 1: Understand A ∪ B
A ∪ B (A Union B) contains every element that belongs to A, or B, or both. It is the "biggest" set formed by combining A and B.
Step 2: Compare A Δ B with A ∪ B
Since A Δ B only takes elements from A and B (excluding the middle overlapping part), every single element inside A Δ B must also exist inside the total Union (A ∪ B).
Step 3: Conclusion
In Set Theory, if every element of set X is also in set Y, then X is a subset of Y. Therefore, the Symmetric Difference is always a part (subset) of the Union.
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Question No. 3 (Group A)
Solution by FEEN (Focus Edge Education Network)
"If a and b are two real numbers, then a ≤ x ≤ b is denoted by..."
- A) x ∈ (a, b)
- B) x ∈ [a, b)
- C) x ∈ [a, b]
- D) x ∈ (a, b]
Detailed Explanation
In Mathematics, we use different types of brackets to show whether the end numbers (a and b) are included in a range. This is called Interval Notation.
- Parentheses ( , ): Used for "Open Intervals." This means the end numbers are NOT included. (Example: x > a and x < b).
- Square Brackets [ , ]: Used for "Closed Intervals." This means the end numbers ARE included. (Example: x is equal to or greater than a, and equal to or less than b).
Step-by-Step Logic
Step 1: Check the Symbols
The question uses the symbol ≤ (less than or equal to). This "equal to" part is the most important clue.
Step 2: Match Symbol to Bracket
- If the question had < or >, we would use round brackets ( ).
- Since the question has ≤ and ≤, we must use square brackets [ ].
Step 3: Final Notation
Because x can be exactly 'a' and exactly 'b', we write it as: x ∈ [a, b].
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Question No. 4 (Group A)
Solution by FEEN (Focus Edge Education Network)
"The value of i to the power 2026 (where i squared = -1) is..."
- A) 1
- B) -1
- C) i
- D) -i
Detailed Explanation
In complex numbers, the powers of 'i' follow a repeating pattern every 4 steps. To solve any high power of 'i', we just need to know this cycle:
| Power | Value |
|---|---|
| i^1 | i |
| i^2 | -1 |
| i^3 | -i |
| i^4 | 1 |
Step-by-Step Logic
Step 1: Divide the Power by 4
Take the power 2026 and divide it by 4 to see where it fits in the cycle.
2026 ÷ 4 = 506 with a remainder of 2.
Step 2: Use the Remainder
The value of i to the power 2026 is the same as i to the power of the remainder.
So, i^2026 = i^2.
Step 3: Final Calculation
We already know from the question and the cycle that i^2 = -1.
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Question No. 5 (Group A)
Solution by FEEN (Focus Edge Education Network)
"The value of √-16 × √-25 is..."
- A) -20
- B) -20i
- C) 20i
- D) 20
Detailed Explanation
A very common mistake is to multiply -16 and -25 inside the root first (which would give +400). However, in Complex Numbers, we must first convert negative square roots into 'i' (imaginary) form before multiplying.
Recall that: √-1 = i
Step-by-Step Logic
Step 1: Simplify the first term (√-16)
√-16 = √(16 × -1) = √16 × √-1 = 4i
Step 2: Simplify the second term (√-25)
√-25 = √(25 × -1) = √25 × √-1 = 5i
Step 3: Multiply the results
Now, multiply 4i and 5i:
4i × 5i = (4 × 5) × (i × i) = 20i²
Step 4: Use the property of i²
We know that i² = -1.
So, 20 × (-1) = -20.
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Question No. 6 (Group A)
Solution by FEEN (Focus Edge Education Network)
"Limiting value of (x² - 5x + 6) / (x - 3) as x tends to 3 is..."
- A) 0
- B) 1
- C) 6
- D) does not exist
Detailed Explanation
When solving limits, if you directly put the value (x = 3) into the equation and get 0/0, it is called an indeterminate form. To solve this, we must first simplify the fraction by factoring the numerator.
Step-by-Step Logic
Step 1: Check for 0/0 form
If we put x = 3:
(3² - 5(3) + 6) / (3 - 3) = (9 - 15 + 6) / 0 = 0/0.
Since it is 0/0, we need to factorize.
Step 2: Factor the Numerator (x² - 5x + 6)
We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3.
x² - 5x + 6 = (x - 2)(x - 3)
Step 3: Simplify the Expression
Now, put the factors back into the limit:
Limit = [(x - 2)(x - 3)] / (x - 3)
The (x - 3) in the top and bottom cancel out!
Step 4: Apply the Limit
Now we are left with: Limit (x - 2) as x tends to 3.
Put x = 3 in the simplified version:
3 - 2 = 1
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Question No. 7 (Group A)
Solution by FEEN (Focus Edge Education Network)
"The curve in the given below is discontinuous at point x = a, because..."
- A) Limit as x approaches a from left does not exist
- B) Limit as x approaches a from right does not exist
- C) Limit as x approaches a does not exist
- D) all of them
Detailed Explanation
In Calculus, a function is continuous at a point if you can draw it without lifting your pen. If there is a break, jump, or a "hole," it is discontinuous.
Looking at the graph in the question, the curve goes towards positive infinity on the right side of 'a' and negative infinity (or stops) on the left side of 'a'. They do not meet at the same height.
Step-by-Step Logic
Step 1: Check the Left Limit
As we approach 'a' from the left side, the curve is heading toward a specific value (or the x-axis).
Step 2: Check the Right Limit
As we approach 'a' from the right side, the curve shoots straight up to infinity. This is an infinite limit.
Step 3: Compare both sides
For a limit to "exist," the left-hand limit and the right-hand limit must be equal and be a finite number. Since one side goes to infinity and the other does not, they are not equal.
Step 4: Conclusion
Because the two sides don't meet at a single point, we say the Limit as x tends to 'a' does not exist. This makes the function discontinuous.
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Question No. 8 (Group A)
Solution by FEEN (Focus Edge Education Network)
"If y = ln(sin x), then dy/dx ="
- A) cot x
- B) tan x
- C) csc x
- D) sec x
Detailed Explanation
To solve this, we use the Chain Rule. When we have a function inside another function (like sin x inside ln), we differentiate from the outside in.
Two key formulas to remember:
- The derivative of ln(u) is 1/u
- The derivative of sin x is cos x
Step-by-Step Logic
Step 1: Differentiate the Outer Function
The outer function is ln.
Derivative of ln(sin x) = 1 / (sin x)
Step 2: Differentiate the Inner Function
The inner function is sin x.
Derivative of sin x = cos x
Step 3: Multiply them together (Chain Rule)
dy/dx = [1 / sin x] * [cos x]
dy/dx = cos x / sin x
Step 4: Simplify using Trigonometry
In trigonometry, we know that:
cos x / sin x = cot x
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Question No. 9 (Group A)
Solution by FEEN (Focus Edge Education Network)
"If x² + y² = a², then dy/dx is..."
- A) x / y
- B) -x / y
- C) y / x
- D) -y / x
Detailed Explanation
This equation represents a circle. Since y is not isolated on one side, we use Implicit Differentiation. This means we differentiate both sides of the equation with respect to x, treating y as a function of x.
Geometrical representation of the slope on a circle.
Step-by-Step Logic
Step 1: Differentiate each term
- The derivative of x² is 2x.
- The derivative of y² is 2y (dy/dx) (using the chain rule).
- The derivative of a² (a constant) is 0.
Step 2: Set up the equation
After differentiating, we get:
2x + 2y (dy/dx) = 0
Step 3: Solve for dy/dx
Move 2x to the other side:
2y (dy/dx) = -2x
Step 4: Isolate dy/dx
Divide both sides by 2y:
dy/dx = -2x / 2y
Step 5: Final Simplification
The 2s cancel out, leaving us with:
dy/dx = -x / y
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Question No. 10 (Group A)
Solution by FEEN (Focus Edge Education Network)
"Anti-derivative of ∫(x² - 1/x²) dx is..."
- A) 2x + 2/x³ + C
- B) x³/3 + 1/x + C
- C) x³/3 - 1/x + C
- D) x³/3 + 2/x + C
Detailed Explanation
The "Anti-derivative" is just another name for Integration. To solve this, we use the Power Rule of Integration:
∫ xⁿ dx = [xⁿ⁺¹ / (n + 1)] + C
Step-by-Step Logic
Step 1: Rewrite the expression
We can write 1/x² as x⁻² to make it easier to use the power rule.
So the problem becomes: ∫ (x² - x⁻²) dx
Step 2: Integrate the first term (x²)
Using the power rule (n = 2):
x⁽²⁺¹⁾ / (2 + 1) = x³ / 3
Step 3: Integrate the second term (-x⁻²)
Using the power rule (n = -2):
- [x⁽⁻²⁺¹⁾ / (-2 + 1)] = - [x⁻¹ / -1]
The two minus signs cancel out to become + x⁻¹.
Step 4: Convert back to fraction
x⁻¹ is the same as 1/x.
Step 5: Final Assembly
Combine the parts and always add the constant of integration + C.
x³/3 + 1/x + C
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Question No. 11 (Group A)
Solution by FEEN (Focus Edge Education Network)
"The value of the definite integral from 0 to π/4 of 1 / (1 + cos 2x) dx is..."
- A) 1
- B) 2
- C) 1/2
- D) 1/√3
Detailed Explanation
To solve this integral easily, we first use a trigonometric identity to simplify the denominator. Integrating a single trigonometric term is much easier than integrating a fraction.
Important Identity:
We know that 1 + cos 2x = 2 cos² x
Step-by-Step Logic
Step 1: Simplify the Integral
Replace (1 + cos 2x) with (2 cos² x):
∫ [ 1 / (2 cos² x) ] dx = 1/2 ∫ (1 / cos² x) dx
Step 2: Use Reciprocal Identity
We know that 1 / cos² x = sec² x.
So the integral becomes: 1/2 ∫ sec² x dx
Step 3: Integrate
The integral of sec² x is tan x.
Result: 1/2 [ tan x ] with limits from 0 to π/4.
Step 4: Apply the Limits (Upper - Lower)
Value = 1/2 [ tan(π/4) - tan(0) ]
Step 5: Final Calculation
Since tan(π/4) = 1 and tan(0) = 0:
Value = 1/2 [ 1 - 0 ] = 1/2
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Question No. 12 (Group B)
Solution by FEEN (Focus Edge Education Network)
12. a) Logic Proof
Question: Prove that [(p ∧ q) ⇒ p] ⇒ (q ∧ ~q) is a contradiction.
Step 1: Understand the Goal
A "Contradiction" is a statement that is ALWAYS False (F) regardless of the values of p and q.
Step 2: Build the Truth Table
| p | q | p ∧ q | (p ∧ q) ⇒ p | ~q | q ∧ ~q | Final Result |
|---|---|---|---|---|---|---|
| T | T | T | T | F | F | F |
| T | F | F | T | T | F | F |
| F | T | F | T | F | F | F |
| F | F | F | T | T | F | F |
Step 3: Conclusion
Since the last column contains only F, the statement is a contradiction. Proved.
12. b) Set Theory Proof
Question: Prove that A - (B ∪ C) = (A - B) ∩ (A - C).
Step 1: Use the Property of Set Difference
Recall that X - Y = X ∩ Y' (X intersection the complement of Y).
Step 2: Expand the LHS (Left Hand Side)
LHS = A - (B ∪ C)
Using the property from Step 1:
LHS = A ∩ (B ∪ C)'
Step 3: Apply De Morgan's Law
We know that (B ∪ C)' = B' ∩ C'.
So, LHS = A ∩ (B' ∩ C')
Step 4: Use the Distributive/Associative logic
We can rewrite this as: (A ∩ B') ∩ (A ∩ C')
Step 5: Convert back to Set Difference
Since A ∩ B' = (A - B) and A ∩ C' = (A - C), we get:
(A - B) ∩ (A - C) = RHS
Conclusion: LHS = RHS. Proved.
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Question No. 13 (Group B)
Solution by FEEN (Focus Edge Education Network)
13. a) Absolute Value Verification
Question: If x = 3 and y = -4, verify:
i) |x + y| ≤ |x| + |y|
ii) |x - y| ≥ |x| - |y|
i) Verifying |x + y| ≤ |x| + |y|
Step 1 (LHS): |3 + (-4)| = |-1| = 1
Step 2 (RHS): |3| + |-4| = 3 + 4 = 7
Conclusion: Since 1 ≤ 7, the condition is Verified.
ii) Verifying |x - y| ≥ |x| - |y|
Step 1 (LHS): |3 - (-4)| = |3 + 4| = |7| = 7
Step 2 (RHS): |3| - |-4| = 3 - 4 = -1
Conclusion: Since 7 ≥ -1, the condition is Verified.
13. b) Solving Inequality
Question: Solve the inequality: 6 - 5x - x² ≤ 0
Step 1: Rearrange and Multiply by -1
To make it easier, let's write x² first. When we multiply by -1, the inequality sign flips:
x² + 5x - 6 ≥ 0
Step 2: Factorize the Quadratic
We need two numbers that multiply to -6 and add to 5. Those are +6 and -1.
(x + 6)(x - 1) ≥ 0
Step 3: Find Critical Points
Set each factor to zero: x = -6 and x = 1.
Step 4: Test Intervals
Since we want the result to be greater than or equal to zero (Positive), we look at the outer regions on the number line:
- Region 1: x ≤ -6 (Makes both factors negative, result is positive)
- Region 2: x ≥ 1 (Makes both factors positive, result is positive)
Final Solution:
x ∈ (-∞, -6] ∪ [1, ∞)
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Question No. 14 (Group B)
Solution by FEEN (Focus Edge Education Network)
14. a) Complex Number Concepts
Question: Define complex number. If z is a non-zero complex number, what conclusion can you make about z if z = z̅ (z equals its conjugate)?
Definition: A number of the form x + iy, where x and y are real numbers and i = √-1, is called a complex number. x is the real part and iy is the imaginary part.
Conclusion for z = z̅:
Let z = x + iy. Then its conjugate z̅ = x - iy.
If z = z̅, then:
x + iy = x - iy
iy = -iy
2iy = 0 → y = 0
Conclusion: Since the imaginary part (y) is zero, z is a purely real number.
14. b) Solving Equation
Question: Solve for x and y: √(7 - 24i) = x + iy. Also explain how the sign of the imaginary part affects the relationship between x and y.
Step 1: Square both sides
7 - 24i = (x + iy)²
7 - 24i = x² + 2ixy + (iy)²
7 - 24i = (x² - y²) + i(2xy)
Step 2: Compare Real and Imaginary Parts
- Real Part: x² - y² = 7 (Equation 1)
- Imaginary Part: 2xy = -24 → xy = -12 (Equation 2)
Step 3: Solve for x and y
Using the identity (x² + y²)² = (x² - y²)² + (2xy)²:
(x² + y²)² = (7)² + (-24)² = 49 + 576 = 625
x² + y² = √625 = 25 (Equation 3)
Step 4: Find x and y values
Add Eq 1 and Eq 3: 2x² = 32 → x² = 16 → x = ±4
Subtract Eq 1 from Eq 3: 2y² = 18 → y² = 9 → y = ±3
Relationship Explanation:
Since 2xy = -24 (a negative number), x and y must have opposite signs. If x is positive, y must be negative, and vice versa.
Final Answer: (x=4, y=-3) or (x=-4, y=3)
Simplifying complex concepts. FEEN.
Question No. 15 (Group B)
Solution by FEEN (Focus Edge Education Network)
15. a) Limit at Infinity
Question: Evaluate: lim (x → ∞) [ √(3x) - √(x - 1) ]
Step 1: Rationalize the Expression
When dealing with roots and infinity, we multiply and divide by the conjugate to avoid the "∞ - ∞" form.
Multiply by: [ √(3x) + √(x - 1) ] / [ √(3x) + √(x - 1) ]
Step 2: Simplify the Numerator
(a - b)(a + b) = a² - b²
(3x) - (x - 1) = 3x - x + 1 = 2x + 1
Step 3: New Expression
Limit = (2x + 1) / [ √(3x) + √(x - 1) ]
Step 4: Divide by Highest Power of x
Divide top and bottom by x. Inside the square root, we divide by x².
Top: (2 + 1/x)
Bottom: √ (3/x) + √ (1/x - 1/x²)
Step 5: Apply Limit (x → ∞)
As x becomes huge, terms like 1/x become 0.
Result = (2 + 0) / (0 + 0) = 2 / 0
Final Answer: ∞ (Infinity)
15. b) Limit with Sine
Question: Evaluate: lim (x → 1) [ (x² - 1) / sin(πx) ]
Step 1: Check Form
If x = 1, we get (1 - 1) / sin(π) = 0 / 0. We must use a substitution.
Step 2: Substitution
Let x - 1 = h. As x → 1, h → 0.
So, x = 1 + h.
Step 3: Substitute into Equation
Numerator: x² - 1 = (1 + h)² - 1 = 1 + 2h + h² - 1 = 2h + h² = h(2 + h)
Denominator: sin(π(1 + h)) = sin(π + πh) = -sin(πh)
Step 4: Rewrite Limit
lim (h → 0) [ h(2 + h) / -sin(πh) ]
Step 5: Use Standard Limit
We know lim (θ → 0) [ sin(θ) / θ ] = 1. Let's force this form:
[ h / sin(πh) ] × [ (2 + h) / -1 ]
Multiply top and bottom by π:
[ 1 / π ] × [ πh / sin(πh) ] × [ -(2 + h) ]
Step 6: Final Value
(1 / π) × (1) × -(2 + 0) = -2 / π
Navigating limits with precision. FEEN.
Question No. 16 (Group B)
Solution by FEEN (Focus Edge Education Network)
16. a) Proving Limit Does Not Exist
Question: Prove that limit of the function f(x) = |x| / x does not exist at point x = 0.
Step 1: Check Left Hand Limit (LHL)
When x approaches 0 from the left (x < 0), |x| = -x.
LHL = lim (x → 0⁻) [ -x / x ] = -1
Step 2: Check Right Hand Limit (RHL)
When x approaches 0 from the right (x > 0), |x| = x.
RHL = lim (x → 0⁺) [ x / x ] = 1
Step 3: Comparison
Since LHL (-1) is NOT equal to RHL (1), the two sides do not meet.
Conclusion: The limit does not exist at x = 0. Proved.
16. b) Continuity at x = 3
Question: Prove f(x) is discontinuous at x = 3. Can it be modified to be continuous?
Step 1: Find the Limiting Value
The function for x ≠ 3 is (x² - x - 6) / (x² - 2x - 3).
Factorize the numerator: (x - 3)(x + 2)
Factorize the denominator: (x - 3)(x + 1)
Cancel (x - 3): lim (x → 3) [ (x + 2) / (x + 1) ] = (3 + 2) / (3 + 1) = 5/4
Step 2: Check Defined Value
The question states that at x = 3, f(3) = 4/5.
Step 3: Prove Discontinuity
Since Limiting Value (5/4) ≠ Defined Value (4/5), the function has a "hole" (removable discontinuity) at x = 3.
Conclusion: It is discontinuous.
Step 4: Modification for Continuity
Yes, the function can be modified. To make it continuous, we must change the definition of f(3) so that it equals the limit.
Modification: Define f(3) = 5/4.
Bridging the gaps in your understanding. FEEN.
Question No. 17 (Group B)
Solution by FEEN (Focus Edge Education Network)
The First Principle Formula:
f'(x) = lim (h → 0) [ f(x + h) - f(x) ] / h
17. i) Derivative of f(x) = x + 1
Step 1: Set up functions
f(x) = x + 1
f(x + h) = (x + h) + 1
Step 2: Put into formula
f'(x) = lim (h → 0) [ (x + h + 1) - (x + 1) ] / h
Step 3: Simplify
f'(x) = lim (h → 0) [ x + h + 1 - x - 1 ] / h
f'(x) = lim (h → 0) [ h ] / h
Step 4: Final Value
Since h/h = 1, the limit is 1.
17. ii) Derivative of f(x) = tan 2x
Step 1: Set up functions
f(x) = tan 2x
f(x + h) = tan 2(x + h) = tan(2x + 2h)
Step 2: Put into formula and convert to Sin/Cos
f'(x) = lim (h → 0) [ tan(2x + 2h) - tan 2x ] / h
Using tan θ = sin θ / cos θ:
f'(x) = lim (h → 0) [ sin(2x + 2h)/cos(2x + 2h) - sin 2x/cos 2x ] / h
Step 3: Find Common Denominator
f'(x) = lim (h → 0) [ sin(2x + 2h)cos 2x - cos(2x + 2h)sin 2x ] / [ h · cos(2x + 2h)cos 2x ]
Step 4: Use Identity sin(A - B)
The numerator is in the form sinA cosB - cosA sinB, which is sin(A - B).
sin(2x + 2h - 2x) = sin 2h
Step 5: Use Standard Limit
f'(x) = lim (h → 0) [ (sin 2h / h) ] × [ 1 / (cos(2x + 2h)cos 2x) ]
To make sin(θ)/θ, multiply and divide by 2:
f'(x) = lim (h → 0) [ 2 · (sin 2h / 2h) ] × [ 1 / (cos(2x + 2h)cos 2x) ]
Step 6: Apply Limit
f'(x) = 2 · (1) · [ 1 / (cos 2x · cos 2x) ]
f'(x) = 2 / cos² 2x = 2 sec² 2x
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Question No. 18 (Group B)
Solution by FEEN (Focus Edge Education Network)
18. i) Parametric Differentiation
Question: Find dy/dx if x = 2a/t and y = a/t².
Step 1: Differentiate x with respect to t
x = 2at⁻¹
dx/dt = 2a(-1)t⁻² = -2a/t²
Step 2: Differentiate y with respect to t
y = at⁻²
dy/dt = a(-2)t⁻³ = -2a/t³
Step 3: Apply Parametric Formula
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (-2a/t³) / (-2a/t²)
Step 4: Simplify
The -2a cancels out. We are left with (1/t³) / (1/t²).
dy/dx = t² / t³ = 1/t
18. ii) Inverse Trigonometric Derivative
Question: Find dy/dx if y = cos⁻¹ [ (1 - x²) / (1 + x²) ]
Step 1: Use Substitution
Let x = tan θ. Then θ = tan⁻¹ x.
Step 2: Substitute into the function
y = cos⁻¹ [ (1 - tan² θ) / (1 + tan² θ) ]
We know from trigonometry that (1 - tan² θ) / (1 + tan² θ) = cos 2θ.
Step 3: Simplify
y = cos⁻¹ (cos 2θ) = 2θ
Step 4: Convert back to x
Since θ = tan⁻¹ x, our function is now: y = 2 tan⁻¹ x
Step 5: Differentiate
We know the derivative of tan⁻¹ x is 1 / (1 + x²).
dy/dx = 2 / (1 + x²)
Simplifying advanced calculus for you. FEEN.
Question No. 19 (Group B)
Solution by FEEN (Focus Edge Education Network)
19. a) Derivative vs. Anti-derivative
Question: If d/dx [g(x)] = f(x), then write the relationship between f(x) and g(x) from an anti-derivative point of view.
Explanation: Differentiation and Integration are inverse processes of each other. If differentiating one function gives you another, then integrating the result brings you back to the original function (plus a constant).
The Relationship:
If d/dx [g(x)] = f(x), then the anti-derivative (integral) of f(x) with respect to x is g(x).
Mathematically: ∫ f(x) dx = g(x) + C
Where C is the constant of integration.
19. b) Solving Integrals
i) Integrate: ∫ [ 1 / (√(x+2) - √x) ] dx
Step 1: Rationalize the denominator
Multiply top and bottom by (√(x+2) + √x):
∫ [ (√(x+2) + √x) / ((x+2) - x) ] dx
= ∫ [ (√(x+2) + √x) / 2 ] dx
Step 2: Split and Integrate
1/2 [ ∫ (x+2)^(1/2) dx + ∫ x^(1/2) dx ]
= 1/2 [ (x+2)^(3/2) / (3/2) + x^(3/2) / (3/2) ] + C
Step 3: Simplify
= 1/2 * 2/3 [ (x+2)^(3/2) + x^(3/2) ] + C
= 1/3 [ (x+2)^(3/2) + x^(3/2) ] + C
ii) Integrate: ∫ [ 1 / (sin²x · cos²x) ] dx
Step 1: Use the identity 1 = sin²x + cos²x
Replace '1' in the numerator:
∫ [ (sin²x + cos²x) / (sin²x · cos²x) ] dx
Step 2: Split the fraction
∫ [ sin²x / (sin²x · cos²x) + cos²x / (sin²x · cos²x) ] dx
= ∫ [ 1/cos²x + 1/sin²x ] dx
Step 3: Convert to Secant and Cosecant
= ∫ [ sec²x + csc²x ] dx
Step 4: Final Integration
Integral of sec²x is tan x
Integral of csc²x is -cot x
= tan x - cot x + C
Mastering the inverse laws of calculus. FEEN.
Question No. 20 (Group C)
Solution by FEEN (Focus Edge Education Network)
20. a) Commutative Property
Question: Does commutative property of real number hold under addition operation? Justify with an example.
Answer: Yes, the commutative property holds under addition for real numbers.
Justification: The property states that for any two real numbers a and b, a + b = b + a. The order of numbers does not change the sum.
Example: Let a = 5 and b = 10.
LHS: 5 + 10 = 15
RHS: 10 + 5 = 15
Since LHS = RHS, it is justified.
20. b) Set Operations on Intervals
Question: If U = (-∞, ∞), A = (-1, 3] and B = [0, 5), find A ∩ B and A - B.
and [0, 5)]i) Find A ∩ B (Intersection):
Intersection means the common area. A starts at -1 and ends at 3. B starts at 0 and ends at 5.
The common range starts at 0 (included in B) and ends at 3 (included in A).
A ∩ B = [0, 3]
ii) Find A - B (Difference):
A - B means elements in A that are NOT in B.
A goes from -1 to 3. B starts at 0. So we remove everything from 0 onwards.
What remains is the part from -1 to 0. Since 0 is in B, it is removed from A.
A - B = (-1, 0)
20. c) Complex Value Calculations
i) Find Conjugate and Absolute Value of [(2+3i)(3+2i)] / (5-12i)
Step 1: Simplify Numerator
(2+3i)(3+2i) = 6 + 4i + 9i + 6i² = 6 + 13i - 6 = 13i
Step 2: Simplify Fraction
13i / (5-12i) × (5+12i)/(5+12i) = (65i + 156i²) / (25 + 144) = (-156 + 65i) / 169
= -156/169 + (65/169)i (Simplify by 13: -12/13 + (5/13)i)
Step 3: Conjugate and Modulus
Conjugate (z̅): -12/13 - (5/13)i
Absolute Value (|z|): √[(-12/13)² + (5/13)²] = √[144/169 + 25/169] = √[169/169] = 1
ii) Verify z̅w̅ = (zw)̅ and (z+w)̅ = z̅+w̅
Given z = 1+2i, w = 2-i. Conjugates are z̅ = 1-2i, w̅ = 2+i.
Addition Verification:
z+w = (1+2) + (2-1)i = 3+i. Its conjugate is 3-i.
z̅+w̅ = (1-2i) + (2+i) = 3-i. (Verified!)
High-scoring solutions with FEEN.
Question No. 21 (Group C)
Solution by FEEN (Focus Edge Education Network)
21. a) Evaluating Trigonometric Limit
Question: Evaluate: lim (h → 0) [ sin²(x+h) - sin²x ] / h
Step 1: Use Algebraic Identity
Use a² - b² = (a - b)(a + b):
Numerator = [sin(x+h) - sin x] · [sin(x+h) + sin x]
Step 2: Use Sine Difference Formula
Recall: sin C - sin D = 2 cos((C+D)/2) sin((C-D)/2)
[sin(x+h) - sin x] = 2 cos(x + h/2) sin(h/2)
Step 3: Combine and Apply Limit
Limit = lim (h → 0) [ 2 cos(x + h/2) sin(h/2) / h ] · [sin(x+h) + sin x]
= lim (h → 0) [ cos(x + h/2) · (sin(h/2) / (h/2)) ] · [sin(x+h) + sin x]
Step 4: Final Value
As h → 0: cos(x) · (1) · (sin x + sin x)
= cos x · 2 sin x
= sin 2x
21. b) Finding dy/dx
Question: Find dy/dx if y = (x/2)√(x²-9) - (9/2) ln[x + √(x²-9)]
Step 1: Differentiate first term (Product Rule)
d/dx [(x/2)√(x²-9)] = (1/2)√(x²-9) + (x/2) · [2x / 2√(x²-9)]
= (1/2)√(x²-9) + x² / [2√(x²-9)]
Step 2: Differentiate second term (Log Rule)
d/dx [(9/2) ln(u)] = (9/2) · [1 / (x + √(x²-9))] · [1 + x/√(x²-9)]
After simplifying the bracket: (9/2) · [1 / √(x²-9)]
Step 3: Subtract the results
dy/dx = [(x² - 9 + x²) / 2√(x²-9)] - [9 / 2√(x²-9)]
dy/dx = (2x² - 18) / 2√(x²-9) = 2(x² - 9) / 2√(x²-9)
= √(x² - 9)
21. c) Integration by Substitution
Question: Integrate: ∫ (2x + 3)(5x + 6)⁷ dx
Step 1: Substitution
Let u = 5x + 6. Then x = (u - 6)/5 and dx = du/5.
Step 2: Substitute x in (2x + 3)
2[(u - 6)/5] + 3 = (2u - 12 + 15) / 5 = (2u + 3) / 5
Step 3: Rewrite Integral
∫ [ (2u + 3)/5 ] · u⁷ · (du/5) = (1/25) ∫ (2u⁸ + 3u⁷) du
Step 4: Integrate
(1/25) [ (2u⁹/9) + (3u⁸/8) ] + C
Step 5: Final Answer (Substitute u back)
= (1/25) [ 2(5x+6)⁹/9 + 3(5x+6)⁸/8 ] + C
Step-by-step clarity with FEEN.
Question No. 22 (Group C)
Solution by FEEN (Focus Edge Education Network)
22. a) Analyzing f(x) = 4x³ - 15x² + 12x + 7
i) Find f'(x) and f''(x)
f'(x) = d/dx (4x³ - 15x² + 12x + 7) = 12x² - 30x + 12
f''(x) = d/dx (12x² - 30x + 12) = 24x - 30
ii) Show Increasing/Decreasing behavior
A function increases if f'(x) > 0 and decreases if f'(x) < 0.
- At x = 0: f'(0) = 12(0)² - 30(0) + 12 = 12. Since 12 > 0, it is increasing.
- At x = 1: f'(1) = 12(1)² - 30(1) + 12 = 12 - 30 + 12 = -6. Since -6 < 0, it is decreasing.
iii) Find Absolute Maximum and Minimum on [-1, 3]
First, find critical points by setting f'(x) = 0:
12x² - 30x + 12 = 0 → 2x² - 5x + 2 = 0 → (2x - 1)(x - 2) = 0.
Critical points: x = 0.5 and x = 2.
Now test the endpoints and critical points in original f(x):
- f(-1) = 4(-1) - 15(1) - 12 + 7 = -24
- f(0.5) = 4(0.125) - 15(0.25) + 12(0.5) + 7 = 0.5 - 3.75 + 6 + 7 = 9.75
- f(2) = 4(8) - 15(4) + 12(2) + 7 = 32 - 60 + 24 + 7 = 3
- f(3) = 4(27) - 15(9) + 12(3) + 7 = 108 - 135 + 36 + 7 = 16
Absolute Max: 16 (at x=3). Absolute Min: -24 (at x=-1).
22. b) Maximum Area of Rectangular Plot
Question: Find the maximum area of a rectangular plot enclosed by a rope of length 60 metres.
Step 1: Perimeter Formula
Perimeter = 2(Length + Width) = 60
L + W = 30 → W = 30 - L
Step 2: Area Function
Area (A) = L × W = L(30 - L) = 30L - L²
Step 3: Maximize using Derivatives
dA/dL = 30 - 2L
Set dA/dL = 0 → 30 - 2L = 0 → L = 15
If L = 15, then W = 30 - 15 = 15.
Step 4: Final Calculation
Maximum Area = 15m × 15m = 225 square metres.
(Note: For a fixed perimeter, a square always provides the maximum area.)
*** The End of Solution Set ***
Brought to you by FEEN (Focus Edge Education Network)
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