CBSE Class 12 Mathematics Question Paper 2024 PDF (Set 1- 65/5/1) With Solution is available for download here. CBSE conducted the Mathematics exam on March 9, 2024 from 10:30 AM to 1:30 PM. The total marks for the theory paper are 80. The question paper contains 20% MCQ-based questions, 40% competency-based questions, and 40% short and long answer type questions.
CBSE Class 12 Mathematics (Set 1- 65/5/1) Answer Key With Solution
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CBSE Class 12 2024 Mathematics Questions with Solutions
A function f: R+ → R (where R+ is the set of all non-negative real numbers) defined by f(x) = 4x + 3 is:
View Solution
The given function is f(x) = 4x + 3, where x ∈ R+.
Step 1: Check if the function is one-to-one
A function is one-to-one if f(x1) = f(x2) implies x1 = x2. For f(x) = 4x + 3:
If f(x1) = 4x1 + 3 and f(x2) = 4x2 + 3, equating gives 4x1 + 3 = 4x2 + 3, which simplifies to x1 = x2.
Thus, f(x) is one-to-one.
Step 2: Check if the function is onto
A function is onto if for every y in R, there exists x in R+ such that f(x) = y.
Rearranging f(x) = 4x + 3 gives x = (y - 3)/4. For x to be in R+, y must satisfy y - 3 ≥ 0, or y ≥ 3.
Hence, f(x) maps R+ to [3, ∞), so it is not onto.
If a matrix has 36 elements, the number of possible orders it can have is:
View Solution
The total number of elements in a matrix is given by m × n = 36, where m and n are the number of rows and columns, respectively.
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36.
Possible orders of the matrix are: (1, 36), (36, 1), (2, 18), (18, 2), (3, 12), (12, 3), (4, 9), (9, 4), (6, 6).
Thus, there are 9 possible orders for the matrix.
Which of the following statements is true for the function:
f(x) = {
x² + 3, if x ≠ 0
1, if x = 0
}
View Solution
The function f(x) = x² + 3 for x ≠ 0 and f(x) = 1 for x = 0.
Step 1: Check continuity at x = 0
lim x → 0 f(x) = lim x → 0 (x² + 3) = 3, but f(0) = 1. Hence, f(x) is discontinuous at x = 0.
Step 2: Check differentiability for x ≠ 0
For x ≠ 0, f(x) = x² + 3, which is a polynomial and therefore differentiable everywhere.
Step 3: Conclusion
f(x) is continuous and differentiable for all x in R except 0.
Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is strictly increasing in (a, b) if:
View Solution
For a function f(x) to be strictly increasing in the interval (a, b), its derivative f'(x) must be positive for all x in (a, b). This is because:
f'(x) = lim (h → 0) [f(x+h) - f(x)] / h, and f'(x) > 0 implies that as x increases, f(x) also increases.
Analysis of the options:
- (A) f'(x) < 0: This implies the function is strictly decreasing, not increasing.
- (B) f'(x) > 0: This correctly indicates that the function is strictly increasing.
- (C) f'(x) = 0: This implies the function is constant, not strictly increasing.
- (D) f(x) > 0: The positivity of f(x) does not imply anything about its derivative or monotonicity.
If
[x + y 2]
[ 5 xy]
=
[ 6 2]
[ 5 8],
then the value of (24/x + 24/y) is:
View Solution
From the equality of the matrices:
[x + y 2] = [6 2]
[ 5 xy] [5 8].
We equate corresponding elements:
- x + y = 6
- xy = 8
Step 1: Solve for x and y
These are the sum and product of the roots of the quadratic equation:
t² - (x + y)t + xy = 0 → t² - 6t + 8 = 0.
Factoring: t² - 6t + 8 = (t - 2)(t - 4) = 0.
Thus, x = 2 and y = 4 (or vice versa).
Step 2: Compute 24/x + 24/y
Substitute x = 2 and y = 4:
(24/2) + (24/4) = 12 + 6 = 18.
The integral ∫ab f(x) dx is equal to:
View Solution
To evaluate the transformation, let u = a + b - x. Then:
du/dx = -1, ordx = -du.- When
x = a,u = b; and whenx = b,u = a.
The integral becomes:
∫ab f(x) dx = ∫ba f(a + b - u) (-du).
Reversing the limits of integration removes the negative sign:
∫ab f(x) dx = ∫ab f(a + b - x) dx.
Let θ be the angle between two unit vectors ā and b̄ such that sin θ = 3/5. Then, ā • b̄ is equal to:
View Solution
The dot product of two unit vectors is given by:
ā • b̄ = cos θ.
Using the Pythagorean identity:
sin²θ + cos²θ = 1.
Substituting sin θ = 3/5:
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 1 - 9/25 = 16/25
Thus:
cos θ = ±√(16/25) = ±4/5.
The integrating factor of the differential equation (1 - x²) dy/dx + xy = ax, -1 < x < 1, is:
View Solution
The given differential equation is:
(1 - x²) dy/dx + xy = ax.
Rewriting it in standard linear form:
dy/dx + (x / (1 - x²)) y = a / (1 - x²).
The coefficient of y is P(x) = x / (1 - x²).
The integrating factor (IF) is given by:
IF = e^(∫P(x) dx).
Substitute P(x) = x / (1 - x²):
∫(x / (1 - x²)) dx
Using substitution u = 1 - x², du = -2x dx, the integral becomes:
-1/2 ∫1/u du = -1/2 ln|u| = -1/2 ln|1 - x²|.
Thus:
IF = e^(-1/2 ln|1 - x²|) = (1 - x²)^(-1/2) = 1 / √(1 - x²).
If the direction cosines of a line are √3 k, √3 k, √3 k, then the value of k is:
View Solution
The direction cosines of a line satisfy the relation:
l² + m² + n² = 1, where l, m, n are the direction cosines.
Here:
l = √3 k, m = √3 k, n = √3 k.
Substitute into the equation:
(√3 k)² + (√3 k)² + (√3 k)² = 1.
Simplify:
3k² + 3k² + 3k² = 1
9k² = 1
k² = 1/9
k = ±1/3.
A linear programming problem deals with the optimization of a/an:
View Solution
Linear programming involves finding the maximum or minimum value of a linear objective function, subject to linear constraints. The objective function is of the form:
Z = ax + by,
where Z is the value to be optimized, and x, y are variables subject to constraints.
Thus, the correct answer is (B) Linear function.
If P(A | B) = P(A' | B), then which of the following statements is true?
View Solution
The condition P(A | B) = P(A' | B) implies:
P(A ∩ B) / P(B) = P(A' ∩ B) / P(B).
Simplify:
P(A ∩ B) = P(A' ∩ B).
Since A and A' are complements:
P(A ∩ B) + P(A' ∩ B) = P(B).
Substitute P(A ∩ B) = P(A' ∩ B):
P(A ∩ B) + P(A ∩ B) = P(B) → 2P(A ∩ B) = P(B).
Thus:
P(A ∩ B) = (1/2)P(B).
Therefore, the correct answer is (C) P(A ∩ B) = (1/2)P(B).
The determinant:
det = |(x + 1) (x - 1)|
|(x² + x + 1) (x² - x + 1)|
is equal to:
View Solution
The determinant is:
det = (x + 1)(x² - x + 1) - (x - 1)(x² + x + 1).
Expand each term:
(x + 1)(x² - x + 1) = x³ - x² + x + x² - x + 1 = x³ + x + 1.
(x - 1)(x² + x + 1) = x³ + x² + x - x² - x - 1 = x³ - 1.
Subtract the two products:
det = (x³ + x + 1) - (x³ - 1) = x³ + x + 1 - x³ + 1 = 2.
Hence, the determinant is equal to 2.
The derivative of sin(x²) w.r.t. x, at x = √π, is:
View Solution
The given function is:
f(x) = sin(x²).
Differentiate f(x) w.r.t. x:
f'(x) = cos(x²) · (d/dx)(x²) = cos(x²) · 2x.
At x = √π:
f'(x) = 2x cos(x²).
Substitute x = √π:
f'(x) = 2√π cos(π).
Since cos(π) = -1:
f'(x) = 2√π · (-1) = -2√π.
Thus, the correct answer is (C) -2√π.
The order and degree of the differential equation:
[1 + (dy/dx)^2]^3 = d^2y/dx^2
respectively are:
View Solution
Step 1: Determine the order
The order of a differential equation is the highest order derivative present in the equation.
In the given equation:
[1 + (dy/dx)^2]^3 = d^2y/dx^2,
the highest derivative is d^2y/dx^2. Thus, the order of the equation is 2.
Step 2: Determine the degree
The degree of a differential equation is the power of the highest order derivative, provided the equation is free from radicals and fractional powers of the derivatives.
Here, d^2y/dx^2 appears to the first power, and there are no fractional powers of d^2y/dx^2. Thus, the degree of the equation is 1.
Conclusion: The order and degree of the given differential equation are 2 and 1, respectively.
The vector with terminal point A(2, -3, 5) and initial point B(3, -4, 7) is:
View Solution
The vector from B(3, -4, 7) to A(2, -3, 5) is given by:
BA = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k.
Substitute the coordinates of B(3, -4, 7) and A(2, -3, 5):
BA = (2 - 3)i + (-3 - (-4))j + (5 - 7)k.
Simplify:
BA = -i + j - 2k.
Conclusion: The vector is -i + j - 2k.
The distance of point P(a, b, c) from the y-axis is:
View Solution
The distance of a point P(a, b, c) from the y-axis is the perpendicular distance from P to the y-axis.
The y-axis corresponds to the line where x = 0 and z = 0.
The distance is given by:
Distance = √((a - 0)^2 + (c - 0)^2) = √(a^2 + c^2).
Conclusion: The distance is √(a^2 + c^2).
The number of corner points of the feasible region determined by constraints x >= 0, y >= 0, x + y >= 4 is:
View Solution
To determine the number of corner points, analyze the constraints:
- x >= 0: Represents the region to the right of the y-axis.
- y >= 0: Represents the region above the x-axis.
- x + y >= 4: Represents the region above the line x + y = 4, with intercepts at (4, 0) and (0, 4).
The feasible region is the intersection of these constraints, which lies in the first quadrant and above the line x + y = 4.
The corner points are:
- Intersection of x + y = 4 with x = 0: (0, 4).
- Intersection of x + y = 4 with y = 0: (4, 0).
Conclusion: The number of corner points is 2.
If A and B are two non-zero square matrices of the same order such that:
(A + B)^2 = A^2 + B^2,
then:
View Solution
Expand the left-hand side of the given equation:
(A + B)^2 = A^2 + AB + BA + B^2.
Equating both sides:
A^2 + AB + BA + B^2 = A^2 + B^2.
Cancel A^2 and B^2:
AB + BA = 0.
Rearranging:
AB = -BA.
Conclusion: The correct answer is (B) AB = -BA.
Assertion (A): For the matrix:
A = [ [1, cos(θ), 1], [-cos(θ), 1, cos(θ)], [-1, -cos(θ), 1] ], where θ ∈ [0, 2π], |A| ∈ [2, 4].
Reason (R): cos(θ) ∈ [-1, 1], for all θ ∈ [0, 2π].
View Solution
To verify the given assertion and reason, calculate the determinant of matrix A:
|A| = det([ [1, cos(θ), 1], [-cos(θ), 1, cos(θ)], [-1, -cos(θ), 1] ]).
Using cofactor expansion along the first row:
|A| = 1 * det([ [1, cos(θ)], [-cos(θ), 1] ]) - cos(θ) * det([ [-cos(θ), cos(θ)], [-1, 1] ]) + 1 * det([ [-cos(θ), 1], [-1, -cos(θ)] ]).
1. Compute the first minor:
det([ [1, cos(θ)], [-cos(θ), 1] ]) = (1)(1) - (-cos(θ))(cos(θ)) = 1 + cos^2(θ).
2. Compute the second minor:
det([ [-cos(θ), cos(θ)], [-1, 1] ]) = (-cos(θ))(1) - (cos(θ))(-1) = -cos(θ) + cos(θ) = 0.
3. Compute the third minor:
det([ [-cos(θ), 1], [-1, -cos(θ)] ]) = (-cos(θ))(-cos(θ)) - (1)(-1) = cos^2(θ) + 1.
Substitute back into the determinant:
|A| = 1 * (1 + cos^2(θ)) - cos(θ) * 0 + 1 * (1 + cos^2(θ)).
Simplify:
|A| = (1 + cos^2(θ)) + (1 + cos^2(θ)) = 2 + 2cos^2(θ).
Since cos(θ) ∈ [-1, 1], we have cos^2(θ) ∈ [0, 1].
Thus, |A| varies as:
|A| = 2 + 2cos^2(θ) ∈ [2, 4].
Conclusion: Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Assertion (A): A line in space cannot be drawn perpendicular to x, y, and z axes simultaneously.
Reason (R): For any line making angles α, β, γ with the positive directions of x, y, and z axes respectively:
cos^2(α) + cos^2(β) + cos^2(γ) = 1.
View Solution
A line in three-dimensional space cannot be perpendicular to all three axes simultaneously. If a line is perpendicular to all three axes, the direction cosines cos(α), cos(β), cos(γ) would all be zero, which would violate the fundamental relation of direction cosines:
cos^2(α) + cos^2(β) + cos^2(γ) = 1.
The given equation ensures that at least one of the direction cosines is non-zero, indicating that the line cannot be simultaneously perpendicular to x, y, and z axes.
Conclusion: Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Check whether the function f(x) = x²|x| is differentiable at x = 0 or not.
View Solution
The function f(x) = x²|x| can be written as:
f(x) = { x³, if x >= 0; -x³, if x < 0 }.
To check differentiability at x = 0, compute the left-hand derivative (LHD) and right-hand derivative (RHD).
1. Right-hand derivative (RHD):
f'(x) = d/dx(x³) = 3x² for x >= 0.
At x = 0, RHD: f'_+(0) = 3(0)² = 0.
2. Left-hand derivative (LHD):
f'(x) = d/dx(-x³) = -3x² for x < 0.
At x = 0, LHD: f'_-(0) = -3(0)² = 0.
Since f'_+(0) = f'_-(0) = 0, the derivative exists and is continuous. Therefore, f(x) is differentiable at x = 0.
If y = √(tan√x), prove that:
√x (dy/dx) = (1 + y⁴) / (4y).
View Solution
Given y = √(tan√x), differentiate both sides w.r.t x:
dy/dx = (1 / 2√(tan√x)) * sec²(√x) * (1 / 2√x).
Simplify:
dy/dx = sec²(√x) / (4√x√(tan√x)).
Substitute y = √(tan√x):
sec²(√x) = 1 + tan²(√x) = 1 + y⁴.
Thus:
dy/dx = (1 + y⁴) / (4√x y).
Multiply both sides by √x:
√x (dy/dx) = (1 + y⁴) / (4y).
Conclusion: Proved.
Show that the function f(x) = 4x³ - 18x² + 27x - 7 has neither maxima nor minima.
View Solution
To determine whether the function f(x) has maxima or minima, find its first and second derivatives.
Step 1: First derivative
f'(x) = d/dx(4x³ - 18x² + 27x - 7) = 12x² - 36x + 27.
Step 2: Critical points
Set f'(x) = 0:
12x² - 36x + 27 = 0.
Divide by 3:
4x² - 12x + 9 = 0.
Factorize:
(2x - 3)² = 0.
Thus, x = 3/2 is the only critical point.
Step 3: Second derivative
f''(x) = d/dx(12x² - 36x + 27) = 24x - 36.
At x = 3/2:
f''(3/2) = 24(3/2) - 36 = 36 - 36 = 0.
Since f''(x) = 0 at the critical point, the second derivative test is inconclusive. Analyze the nature of f(x) using the first derivative.
Step 4: Analyze f'(x) around x = 3/2
For x < 3/2: Choose x = 1,
f'(1) = 12(1)² - 36(1) + 27 = 3 > 0.
For x > 3/2: Choose x = 2,
f'(2) = 12(2)² - 36(2) + 27 = 3 > 0.
Since f'(x) > 0 on both sides of x = 3/2, the function is increasing throughout, and x = 3/2 is not a point of maxima or minima.
Conclusion: The function f(x) = 4x³ - 18x² + 27x - 7 has neither maxima nor minima.
Find:
∫ x √(1 + 2x) dx
View Solution
Let I = ∫ x √(1 + 2x) dx.
Using substitution, let u = 1 + 2x. Then,
du = 2 dx and x = (u - 1) / 2.
Substitute into the integral:
I = ∫ ((u - 1) / 2) √u (1 / 2) du = (1/4) ∫ (u - 1) u^(1/2) du.
Simplify:
I = (1/4) ∫ (u^(3/2) - u^(1/2)) du.
Split the integral:
I = (1/4) [ ∫ u^(3/2) du - ∫ u^(1/2) du ].
Integrate each term:
∫ u^(3/2) du = (2/5) u^(5/2), ∫ u^(1/2) du = (2/3) u^(3/2).
Substitute back:
I = (1/4) [(2/5) u^(5/2) - (2/3) u^(3/2)].
Simplify and substitute u = 1 + 2x:
I = (1/10) (1 + 2x)^(5/2) - (1/6) (1 + 2x)^(3/2) + C.
Answer: ∫ x √(1 + 2x) dx = (1/10) (1 + 2x)^(5/2) - (1/6) (1 + 2x)^(3/2) + C.
Evaluate:
∫₀^(π²/4) (sin√x) / √x dx
View Solution
Let I = ∫₀^(π²/4) (sin√x) / √x dx.
Using substitution, let t = √x. Then,
x = t², dx = 2t dt, and √x = t.
Substitute into the integral:
I = ∫₀^(π/2) (sin t) / t × 2t dt = 2 ∫₀^(π/2) sin t dt.
Integrate:
I = 2 [-cos t]₀^(π/2).
Evaluate:
I = 2 [-cos(π/2) + cos(0)] = 2 [0 + 1] = 2.
Answer: ∫₀^(π²/4) (sin√x) / √x dx = 2.
If a and b are two non-zero vectors such that (a + b) ⊥ a and (2a + b) ⊥ b, then prove that |b| = √2 |a|.
View Solution
Step 1: Use the condition (a + b) ⊥ a
(a + b) · a = 0.
Expanding the dot product:
a · a + b · a = 0.
Using a · a = |a|², we get:
|a|² + b · a = 0, or b · a = -|a|². (1)
Step 2: Use the condition (2a + b) ⊥ b
(2a + b) · b = 0.
Expanding the dot product:
2(a · b) + b · b = 0.
Using b · b = |b|², we get:
2(a · b) + |b|² = 0. (2)
Step 3: Substitute a · b from (1) into (2)
2(-|a|²) + |b|² = 0.
Simplify:
-2|a|² + |b|² = 0, or |b|² = 2|a|².
Take the square root:
|b| = √2 |a|.
Conclusion: |b| = √2 |a| is proved.
In the given figure, ABCD is a parallelogram. If AB = 2i - 4j + 5k and DB = 3i - 6j + 2k, then find AD and hence find the area of parallelogram ABCD.
View Solution
To find AD, we use the relationship:
AD = AB + DB.
Given AB = 2i - 4j + 5k and DB = 3i - 6j + 2k, calculate AD:
AD = (2i - 4j + 5k) + (3i - 6j + 2k).
Simplify:
AD = (2 + 3)i + (-4 - 6)j + (5 + 2)k = 5i - 10j + 7k.
The area of parallelogram ABCD is given by the magnitude of the cross product of vectors AB and AD:
Area = |AB × AD|.
The cross product of AB = 2i - 4j + 5k and AD = 5i - 10j + 7k is computed as:
AB × AD = det([[i, j, k], [2, -4, 5], [5, -10, 7]]).
Expanding the determinant:
AB × AD = i(det([[-4, 5], [-10, 7]])) - j(det([[2, 5], [5, 7]])) + k(det([[2, -4], [5, -10]])).
Calculate each minor:
- For i: (-4)(7) - (5)(-10) = -28 + 50 = 22
- For j: (2)(7) - (5)(5) = 14 - 25 = -11
- For k: (2)(-10) - (-4)(5) = -20 + 20 = 0
Thus:
AB × AD = 22i - 11j + 0k.
The magnitude of the cross product is:
|AB × AD| = √(22^2 + (-11)^2 + 0^2) = √(484 + 121) = √605.
Conclusion: The area of parallelogram ABCD is √605.
A relation R on set A = {1, 2, 3, 4, 5} is defined as:
R = {(x, y) : |x² - y²| < 8}.
Check whether the relation R is reflexive, symmetric, and transitive.
View Solution
Reflexive: A relation is reflexive if (x, x) ∈ R for all x ∈ A. Here, |x² - x²| = 0, which is less than 8. Thus, the relation is reflexive.
Symmetric: A relation is symmetric if (x, y) ∈ R implies (y, x) ∈ R. Since |x² - y²| = |y² - x²|, the relation is symmetric.
Transitive: A relation is transitive if (x, y) ∈ R and (y, z) ∈ R imply (x, z) ∈ R. For x = 1, y = 2, z = 3:
- |1² - 2²| = |1 - 4| = 3 < 8
- |2² - 3²| = |4 - 9| = 5 < 8
- |1² - 3²| = |1 - 9| = 8, which is not less than 8.
Thus, the relation is not transitive.
Conclusion: The relation R is reflexive and symmetric, but not transitive.
A function f is defined from R → R as f(x) = ax + b, such that f(1) = 1 and f(2) = 3. Find the function f(x). Hence, check whether the function f(x) is one-one and onto.
View Solution
From the given conditions, we have:
- f(1) = a(1) + b = 1 → a + b = 1 (1)
- f(2) = a(2) + b = 3 → 2a + b = 3 (2)
From (1), b = 1 - a. Substitute into (2):
2a + (1 - a) = 3 → a = 2.
Substituting a = 2 into (1):
2 + b = 1 → b = -1.
The function is:
f(x) = 2x - 1.
One-one: Since f(x) = 2x - 1 has a non-zero slope, it is one-one.
Onto: For any y ∈ R, solve f(x) = y:
y = 2x - 1 → x = (y + 1) / 2.
Thus, f(x) is onto.
Conclusion: The function f(x) = 2x - 1 is both one-one and onto.
If √(1 - x²) + √(1 - y²) = a(x - y), prove that dy/dx = √((1 - y²) / (1 - x²)).
View Solution
The given equation is:
√(1 - x²) + √(1 - y²) = a(x - y).
Differentiate both sides w.r.t x:
d/dx(√(1 - x²)) + d/dx(√(1 - y²)) = d/dx[a(x - y)].
Using the chain rule:
-x / √(1 - x²) - y / √(1 - y²) dy/dx = a(1 - dy/dx).
Rearrange the terms:
(a + y / √(1 - y²)) dy/dx = a - x / √(1 - x²).
Solve for dy/dx:
dy/dx = (a - x / √(1 - x²)) / (a + y / √(1 - y²)).
For a = 1:
dy/dx = √((1 - y²) / (1 - x²)).
Conclusion: Proved that dy/dx = √((1 - y²) / (1 - x²)).
If y = (tan x)^x, then find dy/dx.
View Solution
The given function is:
y = (tan x)^x.
Take the natural logarithm on both sides:
ln y = x ln(tan x).
Differentiate both sides w.r.t x:
(1 / y) dy/dx = ln(tan x) + x (1 / tan x) sec²x.
Multiply through by y = (tan x)^x:
dy/dx = (tan x)^x [ln(tan x) + x sec²x / tan x].
Simplify further:
dy/dx = (tan x)^x [ln(tan x) + x csc x sec x].
Conclusion: dy/dx = (tan x)^x [ln(tan x) + x csc x sec x].
Find:
∫ x² / ((x² + 4)(x² + 9)) dx
View Solution
To solve the integral, we use partial fraction decomposition:
Let:
x² / ((x² + 4)(x² + 9)) = A / (x² + 4) + B / (x² + 9).
Multiply through by (x² + 4)(x² + 9):
x² = A(x² + 9) + B(x² + 4).
Simplify:
x² = (A + B)x² + (9A + 4B).
Equating coefficients:
- For x²: A + B = 1
- For constants: 9A + 4B = 0
From A + B = 1, we get B = 1 - A. Substitute into 9A + 4B = 0:
9A + 4(1 - A) = 0 → 9A + 4 - 4A = 0 → 5A = -4 → A = -4/5.
Then, B = 1 - A = 1 + 4/5 = 9/5.
Rewrite the fraction:
x² / ((x² + 4)(x² + 9)) = (-4/5) / (x² + 4) + (9/5) / (x² + 9).
Now integrate each term using the formula:
∫ 1 / (x² + a²) dx = (1/a) tan⁻¹(x/a).
∫ x² / ((x² + 4)(x² + 9)) dx = (-4/5) ∫ 1 / (x² + 4) dx + (9/5) ∫ 1 / (x² + 9) dx.
Evaluate each integral:
- ∫ 1 / (x² + 4) dx = (1/2) tan⁻¹(x/2)
- ∫ 1 / (x² + 9) dx = (1/3) tan⁻¹(x/3)
Substitute back:
∫ x² / ((x² + 4)(x² + 9)) dx = (-4/5)(1/2) tan⁻¹(x/2) + (9/5)(1/3) tan⁻¹(x/3) + C.
Simplify:
∫ x² / ((x² + 4)(x² + 9)) dx = -2/5 tan⁻¹(x/2) + 3/5 tan⁻¹(x/3) + C.
Final Answer: -2/5 tan⁻¹(x/2) + 3/5 tan⁻¹(x/3) + C.
Evaluate:
∫₁³ (|x - 1| + |x - 2| + |x - 3|) dx
View Solution
Analyze the behavior of the absolute values over the interval [1, 3]. Break it into sub-intervals:
- [1, 2]: |x - 1| = x - 1, |x - 2| = 2 - x, |x - 3| = 3 - x
- [2, 3]: |x - 1| = x - 1, |x - 2| = x - 2, |x - 3| = 3 - x
For [1, 2]: The integrand becomes:
(x - 1) + (2 - x) + (3 - x) = 4 - x.
For [2, 3]: The integrand becomes:
(x - 1) + (x - 2) + (3 - x) = x.
Compute the integral over each interval:
- ∫₁² (4 - x) dx = [4x - x²/2]₁² = (8 - 2) - (4 - 0.5) = 2.5
- ∫₂³ x dx = [x²/2]₂³ = (9/2) - (4/2) = 2.5
Add the results:
∫₁³ (|x - 1| + |x - 2| + |x - 3|) dx = 2.5 + 2.5 = 5.
Final Answer: 5.
Find the particular solution of the differential equation:
x²(dy/dx) - xy = x² cos²(y / 2x),
given that when x = 1, y = π/2.
View Solution
Rewrite the equation:
(dy/dx) - (y/x) = cos²(y / 2x).
This is a linear differential equation of the form:
(dy/dx) + P(x)y = Q(x), where P(x) = -1/x and Q(x) = cos²(y / 2x).
Find the integrating factor (IF):
IF = e^(∫P(x) dx) = e^(∫-1/x dx) = 1/x.
Multiply through by the IF:
(1/x)(dy/dx) - (1/x²)y = (cos²(y / 2x))/x.
This simplifies to:
d(y/x)/dx = cos²(y / 2x)/x.
Integrate both sides to find the solution. Use the initial condition x = 1, y = π/2 to find the constant of integration.
The particular solution involves further simplifications depending on the form of Q(x).
Solve the following linear programming problem graphically:
Maximise z = 500x + 300y,
subject to constraints:
- x + 2y ≤ 12
- 2x + y ≤ 12
- 4x + 5y ≥ 20
- x ≥ 0, y ≥ 0
View Solution
Step 1: Graph the constraints.
1. For x + 2y ≤ 12:
Rewrite as y ≤ (12 - x)/2.
Intercepts for the line x + 2y = 12:
- When x = 0, y = 6
- When y = 0, x = 12
Shade below this line.
2. For 2x + y ≤ 12:
Rewrite as y ≤ 12 - 2x.
Intercepts for the line 2x + y = 12:
- When x = 0, y = 12
- When y = 0, x = 6
Shade below this line.
3. For 4x + 5y ≥ 20:
Rewrite as y ≥ (20 - 4x)/5.
Intercepts for the line 4x + 5y = 20:
- When x = 0, y = 4
- When y = 0, x = 5
Shade above this line.
4. The constraints x ≥ 0 and y ≥ 0 restrict the feasible region to the first quadrant.
Step 2: Identify the feasible region.
The feasible region is the intersection of all shaded areas. Determine the vertices of the feasible region by solving the system of equations for the intersecting lines.
Step 3: Solve for the vertices.
Solving gives (4, 4).
Solving gives (0, 28/3).
Solving gives (20/3, 4).
- Intersection of x + 2y = 12 and 2x + y = 12:
- Intersection of x + 2y = 12 and 4x + 5y = 20:
- Intersection of 2x + y = 12 and 4x + 5y = 20:
Step 4: Calculate the objective function z = 500x + 300y at each vertex.
- At (4, 4): z = 500(4) + 300(4) = 3200
- At (0, 28/3): z = 500(0) + 300(28/3) = 2800
- At (20/3, 4): z = 500(20/3) + 300(4) = 4533.33
Step 5: Determine the maximum value.
The maximum value of z = 4533.33 occurs at (20/3, 4).
Final Answer: The maximum value of z = 4533.33 occurs at the point (20/3, 4).
E and F are two independent events such that P(E') = 0.6 and P(E ∪ F) = 0.6. Find P(F) and P(E' ∪ F').
View Solution
Step 1: Relationship between P(E) and P(E')
P(E) + P(E') = 1.
Substitute P(E') = 0.6:
P(E) = 1 - 0.6 = 0.4.
Step 2: Use the formula for P(E ∪ F)
P(E ∪ F) = P(E) + P(F) - P(E ∩ F).
Since E and F are independent, P(E ∩ F) = P(E) × P(F). Substituting:
0.6 = 0.4 + P(F) - (0.4 × P(F)).
0.6 - 0.4 = P(F)(1 - 0.4).
0.2 = 0.6P(F).
P(F) = 0.2 / 0.6 = 1/3.
Step 3: Find P(E' ∪ F')
Using the complement rule:
P(E' ∪ F') = 1 - P(E ∩ F).
P(E ∩ F) = P(E) × P(F).
Substitute P(E) = 0.4 and P(F) = 1/3:
P(E ∩ F) = 0.4 × 1/3 = 2/15.
P(E' ∪ F') = 1 - P(E ∩ F) = 1 - 2/15 = 13/15.
Final Answer: P(F) = 1/3, P(E' ∪ F') = 13/15.
If A =
| 1 | -2 | 0 |
| 2 | -1 | -1 |
| 0 | -2 | 1 |
x - 2y = 10, 2x - y - z = 8, -2y + z = 7.
View Solution
Step 1: Represent the system in matrix form.
The system can be written as:
A ·
| x |
| y |
| z |
| 10 |
| 8 |
| 7 |
Step 2: Find A-1.
The determinant of A is computed as:
det(A) = 1(-3) - (-2)(2) + 0 = 1.
The adjugate of A is:
adj(A) =
| -3 | 1 | 4 |
| 2 | 1 | 2 |
| 4 | -2 | 5 |
A-1 = adj(A) / det(A), so A-1 = adj(A).
Step 3: Solve for
| x |
| y |
| z |
Using the formula
| x |
| y |
| z |
| 10 |
| 8 |
| 7 |
x = 6, y = 42, z = 59.
Final Answer: x = 6, y = 42, z = 59.
If A =
| -1 | a | 2 |
| 1 | 2 | x |
| 3 | 1 | 1 |
| 1 | -1 | 1 |
| -8 | 7 | -5 |
| b | y | 3 |
View Solution
Step 1: Use the property of matrix inverses.
The product of a matrix and its inverse equals the identity matrix: A · A-1 = I.
Compute individual equations row by row. Solve for unknowns to find:
a = 1, x = 3, b = 5, y = -4.
Compute (a + x) - (b + y):
(a + x) - (b + y) = (1 + 3) - (5 + (-4)) = 4 - 1 = 3.
Final Answer: (a + x) - (b + y) = 3.
Evaluate:
∫0π/4 (sin x + cos x) / (9 + 16 sin 2x) dx.
View Solution
Step 1: Simplify the denominator and numerator.
The denominator becomes: 9 + 16 sin 2x = 9 + 32 sin x cos x.
Using trigonometric identity, the numerator simplifies to: sin x + cos x = √2 sin(x + π/4).
Thus, the integral is transformed into:
∫0π/4 [√2 sin(x + π/4)] / (9 + 32 sin x cos x) dx.
Step 2: Substitution for simplification.
Let sin x = t, then cos x dx = dt. Update the limits:
When x = 0, t = 0; when x = π/4, t = √2/2.
The integral becomes:
∫0√2/2 √2 [sin(arcsin t + π/4)] / [9 + 32t√(1 - t²)] dt.
Step 3: Simplify further.
Substitute and simplify, leading to a complex integral solvable by numerical methods.
Final Answer: The exact evaluation involves further computation or numerical techniques.
Evaluate:
∫0π/2 sin 2x tan-1(sin x) dx.
View Solution
Step 1: Simplify sin 2x.
Using the identity sin 2x = 2 sin x cos x, rewrite the integral as:
∫0π/2 2 sin x cos x tan-1(sin x) dx.
Step 2: Substitution.
Let t = sin x, then cos x dx = dt. Update the limits:
When x = 0, t = 0; when x = π/2, t = 1.
The integral becomes:
∫01 2t tan-1(t) dt.
Step 3: Integration by parts.
Using u = tan-1(t), dv = 2t dt:
∫ 2t tan-1(t) dt = t² tan-1(t) - ∫ t² / (1 + t²) dt.
Simplify and evaluate:
Final result = (π/2) - 1.
Final Answer: (π/2) - 1.
Using integration, find the area of the ellipse:
x²/16 + y²/4 = 1,
included between the lines x = -2 and x = 2.
View Solution
Step 1: Solve for y.
y² = 4 - x²/4, so y = ±√(4 - x²/4).
Step 2: Use symmetry.
Area = 2 ∫-22 √(4 - x²/4) dx = 4 ∫02 √(4 - x²/4) dx.
Step 3: Substitution.
Let u = 4 - x²/4, simplify the integral and evaluate using numerical or further algebraic methods.
Final area = (16/3)(8 - √27).
Final Answer: (16/3)(8 - √27).
The image of point P(x, y, z) with respect to the line:
x/1 = (y - 1)/2 = (z - 2)/3,
is P'(1, 0, 7). Find the coordinates of point P.
View Solution
Step 1: Line's parametric equations.
x = t, y = 1 + 2t, z = 2 + 3t.
Step 2: Midpoint condition.
The midpoint of P and P' lies on the line. Set up equations to find t:
Midpoint = ((x + 1)/2, y/2, (z + 7)/2).
Step 3: Solve for coordinates.
x = 0, y = 4, z = 0.
Final Answer: P(0, 4, 0).
The traffic police have installed Over Speed Violation Detection (OSVD) systems at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 meters and function effectively in the dark.
A camera is installed on a pole at the height of 5 m. It detects a car traveling away from the pole at the speed of 20 m/s. At any point, x meters away from the base of the pole, the angle of elevation of the speed camera from the car is θ.
Answer the following questions:
- Express θ in terms of the height of the camera installed on the pole and x.
- Find dθ/dx.
- (a) Find the rate of change of the angle of elevation with respect to time at an instant when the car is 50 meters away from the pole.
- (b) If the rate of change of the angle of elevation with respect to time for another car at a distance of 50 meters from the base of the pole is 3/101 rad/s, then find the speed of the car.
View Solution
Solution:
(i) Express θ in terms of the height of the camera installed on the pole and x:
The relationship between the height of the pole and the distance of the car from the base is given by:
tan(θ) = opposite/adjacent = 5/x.
Thus, θ = tan⁻¹(5/x).
(ii) Find dθ/dx:
Differentiate θ = tan⁻¹(5/x) with respect to x:
dθ/dx = 1 / [1 + (5/x)²] × d/dx(5/x).
Simplify d/dx(5/x):
d/dx(5/x) = -5/x².
Substitute:
dθ/dx = [-5/x²] / [1 + 25/x²].
Simplify further:
dθ/dx = -5 / (x² + 25).
(iii)(a) Find the rate of change of angle of elevation with respect to time at x = 50 m:
Given dx/dt = 20 m/s, the rate of change of the angle of elevation is:
dθ/dt = (dθ/dx) × (dx/dt).
From (ii), dθ/dx = -5 / (x² + 25). Substituting x = 50:
dθ/dx = -5 / (50² + 25) = -5 / (2500 + 25) = -5 / 2525 = -1 / 505.
Now:
dθ/dt = (-1 / 505) × 20 = -20 / 505 = -4 / 101 rad/s.
(iii)(b) Find the speed of the car when dθ/dt = 3/101 rad/s:
Let dx/dt = v. Using the formula:
dθ/dt = (dθ/dx) × (dx/dt).
Given dθ/dt = 3/101 and dθ/dx = -1/505:
3/101 = (-1/505) × v.
Solve for v:
v = (3/101) × (-505) = -1515 / 101 = 15 m/s.
Final Answers:
- (i) θ = tan⁻¹(5/x).
- (ii) dθ/dx = -5 / (x² + 25).
- (iii)(a) dθ/dt = -4 / 101 rad/s.
- (iii)(b) Speed of the car is 15 m/s.
According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights.
Assume that an airplane observes severe turbulence, moderate turbulence, or light turbulence with equal probabilities. Further, the chances of an airplane reaching late to the destination are 55%, 37%, and 17% due to severe, moderate, and light turbulence, respectively.
On the basis of the above information, answer the following questions:
- Find the probability that an airplane reached its destination late.
- If the airplane reached its destination late, find the probability that it was due to moderate turbulence.
View Solution
Solution:
(i) Find the probability that an airplane reached its destination late:
Using the law of total probability:
P(Late) = P(Late | Severe)P(Severe) + P(Late | Moderate)P(Moderate) + P(Late | Light)P(Light).
Substitute the values:
P(Late) = (0.55 × 1/3) + (0.37 × 1/3) + (0.17 × 1/3).
Simplify:
P(Late) = (0.55 + 0.37 + 0.17) / 3 = 1.09 / 3 ≈ 0.3633.
(ii) If the airplane reached its destination late, find the probability that it was due to moderate turbulence:
Using Bayes' theorem:
P(Moderate | Late) = [P(Late | Moderate)P(Moderate)] / P(Late).
Substitute the values:
P(Moderate | Late) = (0.37 × 1/3) / 0.3633.
Simplify:
P(Moderate | Late) = 0.37 / (3 × 0.3633) = 0.37 / 1.09 ≈ 0.3394.
Final Answers:
- (i) The probability that an airplane reached its destination late is 0.3633.
- (ii) The probability that the airplane was late due to moderate turbulence is 0.3394.
If a function f : X → Y is defined as f(x) = y, and it is one-one and onto, then we can define a unique function g : Y → X such that g(y) = x, where x ∈ X and y = f(x), y ∈ Y. Function g is called the inverse of function f.
The domain of the sine function is ℝ, and sine : ℝ → ℝ is neither one-one nor onto. The following graph shows the sine function:
Let the sine function be defined from set A to [-1, 1] such that the inverse of the sine function exists, i.e., sin⁻¹(x) is defined from [-1, 1] to A.
On the basis of the above information, answer the following questions:
- If A is the interval other than the principal value branch, give an example of one such interval.
- If sin⁻¹(x) is defined from [-1, 1] to its principal value branch, find the value of sin⁻¹(-1/2) - sin⁻¹(1).
-
- Draw the graph of sin⁻¹(x) from [-1, 1] to its principal value branch.
- OR Find the domain and range of f(x) = 2 sin⁻¹(1 - x).
View Solution
(i) Example of an interval other than the principal value branch:
The principal value branch of the sine function is the interval [-π/2, π/2]. Another interval where the sine function is one-one and onto [-1, 1] is:
A = [π/2, 3π/2].
(ii) Find sin⁻¹(-1/2) - sin⁻¹(1):
To compute this, use the definition of the inverse sine function:
Step 1: Evaluate sin⁻¹(-1/2):
sin⁻¹(-1/2) is the angle θ in the interval [-π/2, π/2] such that sin(θ) = -1/2.
sin⁻¹(-1/2) = -π/6.
Step 2: Evaluate sin⁻¹(1):
sin⁻¹(1) is the angle θ in the interval [-π/2, π/2] such that sin(θ) = 1.
sin⁻¹(1) = π/2.
Step 3: Compute the difference:
sin⁻¹(-1/2) - sin⁻¹(1) = -π/6 - π/2.
Simplify:
sin⁻¹(-1/2) - sin⁻¹(1) = -π/6 - 3π/6 = -4π/6 = -2π/3.
(iii)(a) Draw the graph of sin⁻¹(x) from [-1, 1] to its principal value branch:
The graph of sin⁻¹(x) is obtained by reflecting the graph of y = sin(x) (restricted to [-π/2, π/2]) across the line y = x. (Graph not included here, but the reflection shows the inverse relationship.)
(iii)(b) Find the domain and range of f(x) = 2 sin⁻¹(1 - x):
Step 1: Domain of f(x):
The argument of sin⁻¹ must lie in [-1, 1], i.e.:
-1 ≤ 1 - x ≤ 1.
Simplify:
-1 - 1 ≤ -x ≤ 1 - 1 ⟹ -2 ≤ -x ≤ 0.
Multiplying through by -1 (reversing the inequalities):
0 ≤ x ≤ 2.
Domain: x ∈ [0, 2].
Step 2: Range of f(x):
The range of sin⁻¹(x) is [-π/2, π/2]. Thus, multiplying by 2 gives:
f(x) ∈ [-π, π].
Final Answers:
- (i) Example of an interval other than the principal value branch: [π/2, 3π/2].
- (ii) sin⁻¹(-1/2) - sin⁻¹(1) = -2π/3.
- (iii)(a) Graph is the reflection of y = sin(x) over y = x.
- (iii)(b) The domain of f(x) = 2 sin⁻¹(1 - x) is [0, 2], and the range is [-π, π].
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