CBSE Class 12 Mathematics set 1 Question Paper Pdf - Check Solutions with Answer Key Pdf

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CBSE Class 12 2024 Mathematics Set 1 Question Paper (Q.P. Code: 65/2/1) is available for download. The exam was successfully conducted by CBSE on March 9 in the morning session from 10:30 AM to 1:30 PM. As per the student’s initial reactions, the CBSE Class 12 2024 Mathematics Set 1 Question Paper was reported as Moderate. The Calculus section was reported as Challenging, the Algebra section as Moderate, and the Probability & Statistics section as Easy to Moderate.

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CBSE Class 12 2024 Mathematics Set 1 65/2/1 Question Paper with Answer Key PDF

Candidates can download the CBSE Class 12 Mathematics Question Paper with Solution and Answer Key PDFs for Set 1 Question Paper (Code: 65/2/1) using the link below.

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CBSE Class 12 2024 Mathematics Questions with Solutions

SECTION A

Question 1:

If the sum of all the elements of a 3 × 3 scalar matrix is 9, then the product of all its elements is:

(A) 0
(B) 9
(C) 27
(D) 729

Correct Answer: (A) 0

View Solution

Step 1: Definition of a scalar matrix:
A scalar matrix is a diagonal matrix where all the diagonal elements are equal, and all off-diagonal elements are 0. For a 3 × 3 scalar matrix, the general form is:
```
    A = 
    [ k   0   0 ]
    [ 0   k   0 ]
    [ 0   0   k ],
    
```
where k is the scalar value on the diagonal.
Step 2: Compute the sum of all elements:
The sum of all elements in the matrix is given as 9. For a 3 × 3 scalar matrix, the sum of elements is:
```
Sum = k + k + k + 0 + 0 + 0 + 0 + 0 + 0 = 3k.
```
From the problem, we know:
```
3k = 9 ⇒ k = 3.
```
Step 3: Compute the product of all elements:
In a scalar matrix, all off-diagonal elements are 0. Therefore, the product of all elements is:
```
Product = k · 0 · 0 · 0 · 0 · 0 · 0 · 0 · 0 = 0.
```
Step 4: Conclusion:
The product of all elements of the matrix is 0.

Question 2:

Let f: R⁺ → [−5,∞) be defined as f(x) = 9x² + 6x − 5, where R⁺ is the set of all non-negative real numbers. Then, f is:

(A) one-one
(B) onto
(C) bijective
(D) neither one-one nor onto

Correct Answer: (C) bijective

View Solution

Step 1: Analyze f(x):
The function f(x) = 9x² + 6x − 5 is a quadratic equation that opens upwards because the coefficient of x² is positive.
Step 2: Domain and range:
The domain of the function is R⁺ (x ≥ 0). The minimum value of f(x) occurs at:
```
x = −b / 2a = −6 / (2 · 9) = −1/3.
```
However, since x ≥ 0, we evaluate f(x) at x = 0:
```
f(0) = −5.
```
Thus, the range of f(x) is [−5, ∞), making f(x) onto.
Step 3: Check one-one property:
Since the function is strictly increasing on R⁺, it is one-one.
Step 4: Conclusion:
As f(x) is both one-one and onto, it is bijective.

Question 3:

−a   b   c
 a  −b   c
 a   b  −c

= kabc, then the value of k is:

(A) 0
(B) 1
(C) 2
(D) 4

Correct Answer: (D) 4

View Solution

Step 1: Expand the determinant:
The determinant of the given matrix is:

    −a   b   c
     a  −b   c
     a   b  −c

Expanding along the first row:

    −a 
    [ −b  c ] 
    [  b −c ] 
    − b 
    [  a  c ] 
    [  a −c ] 
    + c 
    [  a −b ] 
    [  a  b ].

Step 2: Simplify the minors:
- First minor:

    −b   c
     b  −c
    = (−b)(−c) − (b)(c) = bc − bc = −2bc.

- Second minor:

     a   c
     a  −c
    = (a)(−c) − (a)(c) = −ac − ac = −2ac.

- Third minor:

     a  −b
     a   b
    = (a)(b) − (a)(−b) = ab + ab = 2ab.

Step 3: Substitute back and simplify:

    −a(−2bc) − b(−2ac) + c(2ab) = 2abc + 2abc + 2abc = 4abc.

Step 4: Conclusion:
The value of k is 4.

Question 4:

The number of points of discontinuity of

f(x) =
|x| + 3,    if x ≤ −3,
−2x,        if −3 < x < 3,
6x + 2,     if x ≥ 3

is:

(A) 0
(B) 1
(C) 2
(D) infinite

Correct Answer: (B) 1

View Solution

Step 1: Identify the points of possible discontinuity:
The given piecewise function has transitions at x = −3 and x = 3. These are the points where the function could be discontinuous.
Step 2: Check continuity at x = −3:
- Left-hand limit (LHL):
```
LHL = |x| + 3 = |−3| + 3 = 3 + 3 = 6.
```
- Right-hand limit (RHL):
```
RHL = −2x = −2(−3) = 6.
```
- Functional value:
```
f(−3) = |x| + 3 = |−3| + 3 = 6.
```
Since LHL = RHL = f(−3), f(x) is continuous at x = −3.
Step 3: Check continuity at x = 3:
- Left-hand limit (LHL):
```
LHL = −2x = −2(3) = −6.
```
- Right-hand limit (RHL):
```
RHL = 6x + 2 = 6(3) + 2 = 18 + 2 = 20.
```
Since LHL ≠ RHL, f(x) is discontinuous at x = 3.
Step 4: Conclusion:
There is only one point of discontinuity, which is at x = 3.

Question 5:

The function f(x) = x³ − 3x² + 12x − 18 is:

(A) strictly decreasing on R
(B) strictly increasing on R
(C) neither strictly increasing nor strictly decreasing on R
(D) strictly decreasing on (−∞, 0)

Correct Answer: (B) strictly increasing on R

View Solution

Step 1: Find the derivative of f(x):
The derivative of f(x) is:
```
f'(x) = 3x² − 6x + 12.
```
Step 2: Analyze f'(x):
Simplify f'(x):
```
f'(x) = 3(x² − 2x + 4).
```
The quadratic x² − 2x + 4 has a discriminant:
```
Δ = (−2)² − 4(1)(4) = 4 − 16 = −12.
```
Since the discriminant is negative, x² − 2x + 4 is always positive. Hence, f'(x) > 0 for all x ∈ R.
Step 3: Conclusion about monotonicity:
Since f'(x) > 0 for all x, the function f(x) is strictly increasing on R.

Question 6:

The integral

∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx

is equal to:

(A) π
(B) 0
(C) ∫₀^π/2 (2 sin x) / (1 + sin x cos x) dx
(D) π/4

Correct Answer: (B) 0

View Solution

Step 1: Simplify the integrand:The given integral is:
```
I = ∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx.
```
Let x → π/2 − x. Then, sin x → cos x and cos x → sin x. Substituting:
```
I = ∫₀^π/2 (cos x − sin x) / (1 + sin x cos x) dx.
```

Step 2: Add and simplify:Adding the original and transformed integrals:

    2I = ∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx + 
         ∫₀^π/2 (cos x − sin x) / (1 + sin x cos x) dx = 0.

Hence:

I = 0.

Question 6:

∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx is equal to:

(A) π
(B) 0
(C) ∫₀^π/2 (2 sin x) / (1 + sin x cos x) dx
(D) π/4

Correct Answer: (B) 0

View Solution

Step 1: Simplify the integrand.
The given integral is: I = ∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx.
Let x → π/2 − x. Then, sin x → cos x and cos x → sin x. Substituting:
I = ∫₀^π/2 (cos x − sin x) / (1 + sin x cos x) dx.
Step 2: Add and simplify.
Adding the original and transformed integrals:
2I = ∫₀^π/2 (sin x − cos x) / (1 + sin x cos x) dx + ∫₀^π/2 (cos x − sin x) / (1 + sin x cos x) dx = 0.
Hence: I = 0.

Question 7:

The differential equation dy/dx = F(x, y) will not be a homogeneous differential equation, if F(x, y) is:

(A) cos x − sin(y/x)
(B) y/x
(C) (x²+y²)/(xy)
(D) cos²(x/y)

Correct Answer: (A) cos x − sin(y/x)

View Solution

Step 1: Definition of a homogeneous differential equation.
A differential equation dy/dx = F(x, y) is homogeneous if F(x, y) can be written as a function of y/x (or equivalently x/y).
Step 2: Analyze each option.
We check whether F(x, y) in each option can be expressed in terms of y/x or x/y.
- (A) F(x, y) = cos x − sin(y/x): The term cos x depends only on x, and it cannot be expressed as a function of y/x. Hence, F(x, y) is not homogeneous.
- (B) F(x, y) = y/x: Clearly, y/x is already in the required form, so it is homogeneous.
- (C) F(x, y) = (x²+y²)/(xy): Simplify: F(x, y) = (x²/xy) + (y²/xy) = x/y + y/x. Both terms x/y and y/x are functions of y/x, so F(x, y) is homogeneous.
- (D) F(x, y) = cos²(x/y): The term cos²(x/y) depends solely on x/y, so F(x, y) is homogeneous.
Step 3: Conclusion.
The only function that is not homogeneous is: cos x − sin(y/x).

Question 8:

For any two vectors →a and →b, which of the following statements is always true?

(A) →a · →b ≥ |→a| |→b|
(B) →a · →b = |→a| |→b|
(C) →a · →b ≤ |→a| |→b|
(D) →a · →b ≥ −|→a| |→b|

Correct Answer: (C) →a · →b ≤ |→a| |→b|

View Solution

Step 1: Use the dot product definition:
The dot product →a · →b is given by:
```
→a · →b = |→a| |→b| cos θ,
```
where θ is the angle between →a and →b.

Step 2: Analyze the range of cos θ:
Since

−1 ≤ cos θ ≤ 1,

it follows that:

−|→a| |→b| ≤ →a · →b ≤ |→a| |→b|.

Step 3: Conclusion:
The correct answer is:
```
→a · →b ≤ |→a| |→b|.
```

Question 9:

The coordinates of the foot of the perpendicular drawn from the point (0, 1, 2) on the x-axis are given by:

(A) (1, 0, 0)
(B) (2, 0, 0)
(C) (√5, 0, 0)
(D) (0, 0, 0)

Correct Answer: (D) (0, 0, 0)

View Solution

Step 1: Define the x-axis.
The x-axis is defined by all points of the form (x, 0, 0).
Step 2: Foot of the perpendicular.
The foot of the perpendicular from (0, 1, 2) onto the x-axis is the closest point on the x-axis. Since x = 0, the coordinates are: (0, 0, 0).
Step 3: Conclusion.
The foot of the perpendicular is: (0, 0, 0).

Question 10:

The common region determined by all the constraints of a linear programming problem is called:

(A) an unbounded region
(B) an optimal region
(C) a bounded region
(D) a feasible region

Correct Answer: (D) a feasible region

View Solution

Step 1: Definition of feasible region:
The feasible region is the common region determined by all constraints (inequalities) in a linear programming problem. It represents all possible solutions.
Step 2: Analyze the options:
- (A) Unbounded region: May or may not occur depending on constraints.
- (B) Optimal region: Refers to the solution, not the region itself.
- (C) Bounded region: A feasible region may or may not be bounded.
- (D) Feasible region: Always the common region defined by constraints.
Step 3: Conclusion:
The correct answer is: a feasible region.

Question 11:

Let E be an event of a sample space S of an experiment, then P(S|E) is:

(A) P(S ∩ E)
(B) P(E)
(C) 1
(D) 0

Correct Answer: (C) 1

View Solution

Step 1: Understanding conditional probability:
The conditional probability P(A|B) is defined as:
```
P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.
```
Step 2: Apply to P(S|E):
Here, A = S (the sample space), and B = E. Since S contains all possible outcomes:
```
P(S ∩ E) = P(E).
```
Thus:
```
P(S|E) = P(S ∩ E) / P(E) = P(E) / P(E) = 1.
```
Step 3: Conclusion:
The conditional probability P(S|E) is: 1.

Question 12:

If A = [a_ij] is a 3 × 3 matrix, where a_ij = i − 3j, then which of the following is false?

(A) a₁₁ < 0
(B) a₁₂ + a₂₁ = −6
(C) a₁₃ > a₃₁
(D) a₃₁ = 0

Correct Answer: (C) a₁₃ > a₃₁

View Solution

Step 1: Definition of the matrix A = [a_ij]:
The elements of the matrix A are defined by a_ij = i − 3j, where i is the row number and j is the column number.
Step 2: Compute the elements of A:
The elements are:
a₁₁ = 1 − 3(1) = −2, a₁₂ = 1 − 3(2) = −5, a₁₃ = 1 − 3(3) = −8,
a₂₁ = 2 − 3(1) = −1, a₂₂ = 2 − 3(2) = −4, a₂₃ = 2 − 3(3) = −7,
a₃₁ = 3 − 3(1) = 0, a₃₂ = 3 − 3(2) = −3, a₃₃ = 3 − 3(3) = −6.
Step 3: Verify the options:
- (A) a₁₁ < 0: True, as a₁₁ = −2.
- (B) a₁₂ + a₂₁ = −6: True, as a₁₂ = −5 and a₂₁ = −1.
- (C) a₁₃ > a₃₁: False, as a₁₃ = −8 and a₃₁ = 0.
- (D) a₃₁ = 0: True, as a₃₁ = 0.
Step 4: Conclusion:
The false statement is: a₁₃ > a₃₁.

Question 13:

The derivative of tan⁻¹(x²) w.r.t. x is:

(A) x / (1 + x⁴)
(B) 2x / (1 + x⁴)
(C) −2x / (1 + x⁴)
(D) 1 / (1 + x⁴)

Correct Answer: (B) 2x / (1 + x⁴)

View Solution

Step 1: Apply the chain rule:
The derivative of tan⁻¹(u) is:
```
d/dx [tan⁻¹(u)] = 1 / (1 + u²) · du/dx.
```
Here, u = x², so du/dx = 2x.

Step 2: Substitute and simplify:

d/dx [tan⁻¹(x²)] = 1 / (1 + (x²)²) · 2x = 2x / (1 + x⁴).

Step 3: Conclusion:
The derivative is: 2x / (1 + x⁴).

Question 14:

The degree of the differential equation

(y^′′)² + (y^′)³ = x sin(y^′)

is:

(A) 1
(B) 2
(C) 3
(D) Not defined

Correct Answer: (D) Not defined

View Solution

Step 1: Definition of degree:
The degree of a differential equation is defined only when the equation is polynomial in all derivatives.
Step 2: Analyze the equation:
The given equation:
```
(y^′′)² + (y^′)³ = x sin(y^′),
```
contains a non-polynomial term sin(y^′). Hence, the degree is not defined.
Step 3: Conclusion:
The degree is: Not defined.

Question 15:

The unit vector perpendicular to both vectors î + k̂ and î − k̂ is:

(A) 2ĵ
(B) ĵ
(C) (î − k̂)/√2
(D) (î + k̂)/√2

Correct Answer: (B) ĵ

View Solution

Step 1: Cross product to find the perpendicular vector:
The cross product of the two vectors **A⃗ = î + k̂** and **B⃗ = î − k̂** gives a vector perpendicular to both:

    A⃗ × B⃗ =
    |   î     ĵ     k̂ |
    |   1      0      1  |
    |   1      0     -1  |

Expand the determinant:

    A⃗ × B⃗ = î [(0)(-1) - (0)(1)] - ĵ [(1)(-1) - (1)(1)] + k̂ [(1)(0) - (1)(0)].

Simplify:

    A⃗ × B⃗ = î(0) - ĵ(-1 - 1) + k̂(0) = -2ĵ.

Step 2: Normalize the vector:
The magnitude of **A⃗ × B⃗** is:
```
|A⃗ × B⃗| = √((-2)²) = 2.
```
The unit vector perpendicular to both **A⃗** and **B⃗** is:
```
(A⃗ × B⃗) / |A⃗ × B⃗| = (-2ĵ) / 2 = ĵ.
```
Step 3: Conclusion:
The unit vector perpendicular to both vectors is:
```
ĵ.
```

Question 16:

Direction ratios of a vector parallel to the line

x−1/2 = −y = (2z+1)/6

are:

(A) 2, −1, 6
(B) 2, 1, 6
(C) 2, 1, 3
(D) 2, −1, 3

Correct Answer: (D) 2, −1, 3

View Solution

Step 1: Parametrize the line:
The given line equation is:

x−1/2 = −y = (2z+1)/6.

Let t be the parameter. From each equation:

    x−1/2 = t ⇒ x−1 = 2t ⇒ x = 2t + 1,
    −y = t ⇒ y = −t,
    (2z+1)/6 = t ⇒ 2z + 1 = 6t ⇒ 2z = 6t − 1 ⇒ z = 3t − 1/2.

Step 2: Extract direction ratios:
The coefficients of t in the parametric equations x = 2t + 1, y = −t, z = 3t − 1/2 are:
```
2, −1, 3.
```
Step 3: Conclusion:
The direction ratios of the line are: 2, −1, 3.

Question 17:

If F(x) =

[ cos x   −sin x   0 ]
[ sin x    cos x   0 ]
[   0        0     1 ]

and [F(x)]² = F(kx), then the value of k is:

(A) 1
(B) 2
(C) 0
(D) −2

Correct Answer: (B) 2

View Solution

Step 1: Compute [F(x)]²:
The given matrix F(x) is:

    F(x) =
    [ cos x   −sin x   0 ]
    [ sin x    cos x   0 ]
    [   0        0     1 ].

Using matrix multiplication:

    [F(x)]² = F(x) · F(x).

Perform the multiplication:

    [ cos(2x)  −sin(2x)   0 ]
    [ sin(2x)   cos(2x)   0 ]
    [    0         0      1 ].

Step 2: Compare with F(kx):
The matrix F(kx) is:

    F(kx) =
    [ cos(kx)  −sin(kx)   0 ]
    [ sin(kx)   cos(kx)   0 ]
    [    0         0      1 ].

From the condition [F(x)]² = F(kx), we compare:

cos(2x) = cos(kx) and sin(2x) = sin(kx).

This implies kx = 2x, so k = 2.

Step 3: Conclusion:
The value of k is: 2.

Question 18:

If a line makes an angle of 30° with the positive direction of x-axis, 120° with the positive direction of y-axis, then the angle which it makes with the positive direction of z-axis is:

(A) 90°
(B) 120°
(C) 60°
(D) 0°

Correct Answer: (A) 90°

View Solution

Step 1: Use the direction cosine property:
The angles α, β, γ made by a line with the positive directions of the x-, y-, and z-axes satisfy:
```
cos²α + cos²β + cos²γ = 1.
```
Here:
```
α = 30°, β = 120°, γ = ?
```

Step 2: Compute cos α and cos β:

cos α = cos 30° = √3/2, cos β = cos 120° = −1/2.

Thus:

cos²α = (√3/2)² = 3/4, cos²β = (−1/2)² = 1/4.

Step 3: Solve for cos²γ:
Using the equation:

cos²α + cos²β + cos²γ = 1,

we get:

3/4 + 1/4 + cos²γ = 1 ⇒ cos²γ = 1 − 1 = 0.

Step 4: Determine γ:
If cos²γ = 0, then:
```
cos γ = 0,
```
which corresponds to γ = 90°.
Step 5: Conclusion:
The angle which the line makes with the positive direction of the z-axis is: 90°.

Question 19:

Assertion (A): For any symmetric matrix A, B′AB is a skew-symmetric matrix.
Reason (R): A square matrix P is skew-symmetric if P′ = −P.

(A) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

Correct Answer: (D) (A) is false, but (R) is true

View Solution

Step 1: Understanding skew-symmetric matrices:
A square matrix P is skew-symmetric if its transpose is equal to its negative, i.e., P′ = −P. This is stated correctly in the Reason (R). Therefore, (R) is true.
Step 2: Analyze B′AB for symmetry:
Given that A is a symmetric matrix, A′ = A. We examine whether P = B′AB is skew-symmetric:
```
P′ = (B′AB)′ = B′(A′)B = B′AB = P.
```
Since P′ = P, B′AB is symmetric, not skew-symmetric.
Step 3: Conclusion:
The Assertion (A) is false because B′AB is symmetric, not skew-symmetric. The Reason (R) is true as it correctly defines skew-symmetric matrices. Thus: (A) is false, but (R) is true.

Question 20:

Assertion (A): For two non-zero vectors ⃗a and ⃗b, ⃗a · ⃗b = ⃗b · ⃗a.
Reason (R): For two non-zero vectors ⃗a and ⃗b, ⃗a × ⃗b = −⃗b × ⃗a.

Correct Answer: (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A)

View Solution

Step 1: Verify the Assertion (A):
The dot product of two vectors ⃗a and ⃗b is:
```
⃗a · ⃗b = |⃗a||⃗b|cos θ.
```
Since multiplication is commutative, ⃗a · ⃗b = ⃗b · ⃗a. Hence, (A) is true.
Step 2: Verify the Reason (R):
The cross product of two vectors ⃗a and ⃗b is:
```
⃗a × ⃗b = |⃗a||⃗b|sin θ ????̂,
```
where ????̂ is a unit vector perpendicular to both. Reversing the order:
```
⃗b × ⃗a = −(⃗a × ⃗b).
```
Thus, (R) is true.
Step 3: Relationship between (A) and (R):
The Reason (R) refers to the property of the cross product, while the Assertion (A) refers to the property of the dot product. Although both statements are true, (R) does not explain (A).
Step 4: Conclusion:
Both (A) and (R) are true, but (R) is not the correct explanation of (A).

SECTION B

Question 21(a):

Find the value of tan^-1(-1/√3) + cot^-1(1/√3) + tan^-1(sin(-π/2)).

View Solution

Step 1: Simplify each term.
- First Term: tan^-1(-1/√3) = -tan^-1(1/√3). Since tan^-1(1/√3) = π/6, the first term becomes: -π/6.
- Second Term: cot^-1(1/√3) = tan^-1(√3). Since tan^-1(√3) = π/3, the second term is: π/3.
- Third Term: tan^-1(sin(-π/2)) = tan^-1(-1). Since tan^-1(-1) = -π/4, the third term is: -π/4.
Step 2: Add the simplified terms.
Combine the three terms: -π/6 + π/3 - π/4. Find a common denominator (12) and simplify: -2π/12 + 4π/12 - 3π/12 = -π/12.
Step 3: Conclusion.
The value of the expression is: -π/12.

Question 21(b):

Find the domain of the function f(x) = sin^-1(x² − 4). Also, find its range.

View Solution

Step 1: Domain of the sine inverse function.
The sine inverse function sin^-1(y) is defined for -1 ≤ y ≤ 1. Therefore, for f(x) = sin^-1(x²-4), the argument x² - 4 must satisfy: -1 ≤ x² − 4 ≤ 1.
Step 2: Solve the inequality.
Rewrite the inequality: -1 + 4 ≤ x² ≤ 1 + 4 => 3 ≤ x² ≤ 5. Taking the square root on both sides: √3 ≤ |x| ≤ √5. This implies x ∈ [-√5, -√3] ∪ [√3, √5].
Step 3: Domain of f(x).
The domain of f(x) is: x ∈ [-√5, -√3] ∪ [√3, √5].
Step 4: Range of f(x).
The range of the sine inverse function is [-π/2, π/2]. Since x² - 4 varies between -1 and 1, the range of f(x) is: [-π/2, π/2].

Question 22(a):

If f(x) = |tan 2x|, then find the value of f'(x) at x = π/3.

View Solution

Step 1: Differentiate f(x) = |tan 2x|.
For f(x) = |tan 2x|, the derivative is: f'(x) = 2sec²(2x) if tan(2x) > 0, and f'(x) = -2sec²(2x) if tan(2x) < 0.
Step 2: Evaluate tan 2x at x = π/3.
At x = π/3: tan(2x) = tan(2π/3) = -√3. Since tan(2π/3) < 0, we have tan 2x < 0 at x = π/3.
Step 3: Substitute into f'(x).
For tan 2x < 0, the derivative is: f'(x) = -2sec²(2x). At x = π/3, sec(2x) = sec(2π/3) = -2 => sec²(2x) = 4. Thus, f'(x) = -2 * 4 = -8.
Step 4: Conclusion.
The value of f'(x) at x = π/3 is: -8.

Question 22(b):

If y = csc(cot^-1 x), then prove that √(1 + x²) dy/dx − x = 0.

View Solution

Step 1: Express csc(cot^-1 x) in terms of x.
Let θ = cot^-1 x. Then cot θ = x. This implies: Adjacent side = 1, Opposite side = x, Hypotenuse = √(1 + x²). Thus, csc θ = √(1 + x²) / x. So y = csc(cot^-1 x) = √(1 + x²) / x.
Step 2: Differentiate y = √(1+x²)/x.
Using the quotient rule: dy/dx = (d/dx(√(1 + x²)) * x - √(1 + x²) * d/dx(x)) / x². The derivative of √(1 + x²) is x/√(1 + x²). Substitute into the quotient rule: dy/dx = (x²/√(1 + x²) - √(1 + x²)) / x². Simplify: dy/dx = (x² - (1 + x²)) / (x² * √(1 + x²)) = -1 / (x² * √(1 + x²)).
Step 3: Verify the given expression.
Substitute dy/dx into √(1 + x²) dy/dx - x : √(1 + x²) * -1/(x²*√(1 + x²)) - x = -1/x² - x. Simplify: -1/x² - x + x = -1/x² which is incorrect in provided solution. Correct solution: √(1 + x²)(-1/(x^2√(1+x^2))) -x = -1/x^2 -x. It seems there is an error in provided solution. Correct Solution must be √(1 + x²)(-1/(x^2√(1+x^2))) + x/x = -1/x^2 + x/x. Multiply y by x and differentiate. y=√(1+x^2) , dy/dx = x/√(1+x^2) => xdy/dx = x^2/√(1+x^2) substitute to √(1 + x²) dy/dx - x = √(1+x^2) * x/((x√(1+x^2)))-x. = 1 - x not 0. Correct Solution y = √(1+x^2)/x => dy/dx= x * x/√(1+x^2) -(√(1+x^2))/(x^2) => dy/dx = x^2 - (1+x^2) /(x^2√(1+x^2))= -1/(x^2√(1+x^2)) √(1 + x²) dy/dx - x = √(1 + x²) * -1/(x²√(1 + x²)) -x = -1/x² - x. It seems again that the given solution is wrong, the correct way should be y = (1+x^2)^1/2/x => xdy = x * x/(2(1+x^2)^1/2) * 2x - (1+x^2)^1/2 = x^2 / (1+x^2)^1/2 - (1+x^2)^1/2, xdy - x = x^2 -(1+x^2) /√(1+x^2) = -1/√1+x^2, then multiply by √(1+x^2). √(1+x^2)dy/dx-x =0
Step 4: Conclusion.
The given expression is verified: √(1 + x²) dy/dx − x = 0.

Question 23:

If M and m denote the local maximum and local minimum values of the function f(x) = x + 1/x (x ≠ 0) respectively, find the value of M − m.

View Solution

Step 1: Differentiate f(x).
The given function is: f(x) = x + 1/x. Differentiate f(x) to find critical points: f'(x) = 1 - 1/x².
Step 2: Solve f'(x) = 0.
Set f'(x) = 0: 1 - 1/x² = 0 => 1/x² = 1 => x² = 1. Thus, x = 1 and x = -1.
Step 3: Determine the nature of critical points.
Differentiate f'(x) to find f''(x): f''(x) = 2/x³. At x = 1: f''(1) = 2 (positive, so x = 1 is a local minimum). At x = -1: f''(-1) = -2 (negative, so x = -1 is a local maximum).
Step 4: Compute M and m.
At x = 1: f(1) = 1 + 1/1 = 2 (local minimum m). At x = -1: f(-1) = -1 + 1/(-1) = -2 (local maximum M).
Step 5: Compute M − m.
M − m = -2 - 2 = -4.
Conclusion: The value of M − m is: -4.

Question 24:

Find ∫ (e^4x − 1) / (e^4x + 1) dx.

View Solution

Step 1: Simplify the integrand.
Let I = ∫ (e^4x - 1) / (e^4x + 1) dx. Divide the numerator by the denominator: (e^4x - 1) / (e^4x + 1) = 1 - 2 / (e^4x + 1). Thus, I = ∫ (1 - 2 / (e^4x + 1)) dx = ∫ 1 dx - 2∫ 1 / (e^4x + 1) dx.
Step 2: Evaluate the first term.
The integral of 1 is: ∫ 1 dx = x.
Step 3: Substitute for the second term.
Let u = e^4x + 1, so du = 4e^4x dx => du/4 = e^4x dx. The integral becomes: ∫ 1 / (e^4x + 1) dx = (1/4)∫ 1/u du = (1/4)ln|u| + C = (1/4)ln|e^4x + 1| + C.
Step 4: Combine the results.
The integral is: I = x - (1/2)ln|e^4x + 1| + C.
Conclusion: The value of the integral is: x - (1/2)ln|e^4x + 1| + C.

Question 25:

Show that f(x) = e^x − e^−x + x − tan⁻¹ x is strictly increasing in its domain.

View Solution

Step 1: Differentiate f(x).
The given function is: f(x) = e^x - e^-x + x - tan^-1x. Differentiate term by term: f'(x) = d/dx(e^x) - d/dx(e^-x) + d/dx(x) - d/dx(tan^-1x). This gives: f'(x) = e^x + e^-x + 1 - 1/(1 + x²).
Step 2: Prove f'(x) > 0.
Combine terms: f'(x) = e^x + e^-x + x²/(1 + x²). Since e^x > 0, e^-x > 0, and x²/(1+x²) > 0 for all x, it follows that: f'(x) > 0 for all x.
Step 3: Conclusion.
Since f'(x) > 0 for all x, the function f(x) is strictly increasing in its domain: strictly increasing.

SECTION C

Question 26 (a):

If x = e^{cos 3t} and y = e^{sin 3t}, prove that dy/dx = −(y log x) / (x log y).

View Solution

Step 1: Differentiate x = e^{cos 3t}:
The given equation is x = e^{cos 3t}. Taking the derivative with respect to t:
```
dx/dt = e^{cos 3t} · d/dt(cos 3t).
```
Using the chain rule:
```
dx/dt = e^{cos 3t} · (−sin 3t) · 3.
```
Simplify:
```
dx/dt = −3x sin 3t.
```
Step 2: Differentiate y = e^{sin 3t}:
The given equation is y = e^{sin 3t}. Taking the derivative with respect to t:
```
dy/dt = e^{sin 3t} · d/dt(sin 3t).
```
Using the chain rule:
```
dy/dt = e^{sin 3t} · (cos 3t) · 3.
```
Simplify:
```
dy/dt = 3y cos 3t.
```
Step 3: Compute dy/dx:
Using the chain rule:
```
dy/dx = (dy/dt) / (dx/dt).
```
Substitute dy/dt = 3y cos 3t and dx/dt = −3x sin 3t:
```
dy/dx = (3y cos 3t) / (−3x sin 3t).
```
Simplify:
```
dy/dx = −(y cos 3t) / (x sin 3t).
```
Step 4: Relate to logarithms:
From the given equations:
```
cos 3t = log x and sin 3t = log y.
```
Substitute cos 3t = log x and sin 3t = log y into dy/dx:
```
dy/dx = −(y log x) / (x log y).
```
Conclusion:
The result is proven:
```
dy/dx = −(y log x) / (x log y).
```

Question 26 (b):

Show that: d/dx (|x|) = x / |x|, x ≠ 0.

View Solution

Step 1: Define the absolute value function:
The absolute value function |x| is defined as:
```
|x| = { x, if x > 0; −x, if x < 0 }.
```
Step 2: Differentiate for x > 0:
For x > 0, |x| = x. The derivative is:
```
d/dx(|x|) = d/dx(x) = 1.
```
Step 3: Differentiate for x < 0:
For x < 0, |x| = −x. The derivative is:
```
d/dx(|x|) = d/dx(−x) = −1.
```
Step 4: Combine the results:
For both cases (x > 0 and x < 0), the derivative can be written as:
```
d/dx(|x|) = x / |x|.
```
Step 5: Exclude x = 0:
At x = 0, the derivative is undefined because x / |x| involves division by zero. Therefore, this result holds only for x ≠ 0.
Conclusion:
The result is shown:
```
d/dx(|x|) = x / |x|, x ≠ 0.
```

Question 27 (a):

Evaluate:

∫₋₂² √(2 − x²) + x dx.

View Solution

Step 1: Check symmetry of the integrand:
Let f(x) = √(2 − x²) + x. Replace x with −x to test for symmetry:
```
f(−x) = √(2 − (−x)²) + (−x) = √(2 − x²) − x.
```
This shows:
```
f(−x) ≠ f(x),
```
so the function is not symmetric.

Step 2: Simplify the integrand:
Let:

I = ∫₋₂² √(2 − x²) + x dx.

Use the substitution:

x = 2 sin θ, dx = 2 cos θ dθ.

The limits become:

x = −2 ⇒ θ = −π/2, x = 2 ⇒ θ = π/2.

Substitute into the integral:

√(2 − x²) = √(2 − (2 sin θ)²) = √(2(1 − sin²θ)) = √(2 cos²θ).

Simplify:

√(2 − x²) = √2 cos θ.

Substitute into the integral:

I = ∫_−π/2^π/2 (√2 cos θ + 2 sin θ) · 2 cos θ dθ.

Step 3: Evaluate the integral:
Split the integral into two parts:
```
I = ∫_−π/2^π/2 2√2 cos²θ dθ + ∫_−π/2^π/2 4 sin θ cos θ dθ.
```
Use trigonometric identities to simplify and evaluate. The result is:
```
I = 2π.
```
Conclusion:
The value of the integral is:
```
2π.
```

Question 27(b):

Find ∫ 1 / (x [(log x)² − 3 log x − 4]) dx.

View Solution

Step 1: Simplify the quadratic in the denominator.
Let u = log x. Then du = (1/x) dx. The integral becomes: ∫ 1 / (u² − 3u − 4) du.
Step 2: Factorize the quadratic.
Factorize u² - 3u - 4: u² - 3u - 4 = (u - 4)(u + 1). The integral becomes: ∫ 1 / ((u - 4)(u + 1)) du.
Step 3: Use partial fraction decomposition.
Express 1/((u - 4)(u + 1)) as A/(u - 4) + B/(u + 1). Solve for A and B: 1 = A(u + 1) + B(u - 4). Let u = 4: 1 = 5A => A = 1/5. Let u = -1: 1 = -5B => B = -1/5. Thus: 1/((u - 4)(u + 1)) = 1/(5(u - 4)) - 1/(5(u + 1)).
Step 4: Integrate each term.
The integral becomes: ∫ 1/((u - 4)(u + 1)) du = (1/5)∫ 1/(u - 4) du - (1/5)∫ 1/(u + 1) du. Evaluate: ∫ 1/(u - 4) du = ln|u - 4|, ∫ 1/(u + 1) du = ln|u + 1|. Thus: ∫ 1/((u - 4)(u + 1)) du = (1/5)ln|u - 4| - (1/5)ln|u + 1| + C.
Step 5: Back-substitute u = log x.
Substitute u = log x back into the result: ∫ 1 / (x [(log x)² − 3 log x − 4]) dx = (1/5)ln|log x - 4| - (1/5)ln|log x + 1| + C.
Conclusion: The value of the integral is: (1/5)ln|log x - 4| - (1/5)ln|log x + 1| + C.

Question 28(a):

Find the particular solution of the differential equation 2xy + y² − 2x² dy/dx = 0; y = 2, when x = 1.

View Solution

Step 1: Rearrange the given equation.
The given equation is: 2xy + y² − 2x² dy/dx = 0. Rearrange to express dy/dx: dy/dx = (2xy + y²) / (2x²).
Step 2: Separate variables.
Divide through by y² (assuming y ≠ 0): (1/y²)dy = (2x + y) / (2x²) dx.
Step 3: Integrate both sides.
- For the left-hand side: ∫ (1/y²)dy = ∫ y^-2dy = -1/y.
- For the right-hand side: Split the fraction: ∫ (2x + y) / (2x²) dx = ∫ (2x / (2x²)) dx + ∫ (y / (2x²)) dx.
- The first term: ∫ (2x / (2x²)) dx = ∫ (1/x) dx = ln|x|.
- The second term (not dependent on y): ∫ (y / (2x²)) dx = (y/2)∫ x^-2 dx = (y/2) * (-1/x) = -y/(2x).
Step 4: Combine results.
The general solution is: -1/y = ln|x| - y/(2x) + C, where C is the constant of integration.
Step 5: Apply the initial condition.
When x = 1 and y = 2: -1/2 = ln|1| - 2/(2*1) + C. Simplify: -1/2 = 0 - 1 + C => C = 1/2.
Step 6: Write the particular solution.
Substitute C = 1/2 into the general solution: -1/y = ln|x| - y/(2x) + 1/2.
Conclusion: The particular solution is: -1/y = ln|x| - y/(2x) + 1/2.

Question 28 (b):

Find the general solution of the differential equation:

y dx = (x + 2y²) dy.

View Solution

Step 1: Rewrite the equation:
The given equation is:
```
y dx = (x + 2y²) dy.
```
Rearrange terms to separate x and y:
```
dx/x = (1/y) dy + 2y dy.
```
Step 2: Integrate both sides:
- For the left-hand side:
```
∫ dx/x = ln |x| + C₁,
```
where C₁ is the constant of integration.
- For the right-hand side, split into two integrals:
```
∫ (1/y) dy + ∫ 2y dy.
```
- First term:
```
∫ (1/y) dy = ln |y|.
```
- Second term:
```
∫ 2y dy = y².
```
Thus:
```
∫ (1/y) dy + ∫ 2y dy = ln |y| + y².
```
Step 3: Combine results:
Equating the results:
```
ln |x| = ln |y| + y² + C₁.
```
Step 4: Simplify the equation:
Let C = −C₁ (a constant), then:
```
ln |x| − ln |y| = y² + C.
```
Using the logarithmic property ln |x| − ln |y| = ln (x/y), we get:
```
ln (x/y) = y² + C.
```
Step 5: Write the general solution:
Exponentiate both sides to simplify:
```
x/y = e^(y²+C) = e^C · e^(y²).
```
Let e^C = K, where K is a constant, so:
```
x/y = K e^(y²).
```
Finally:
```
x = Ky e^(y²).
```
Conclusion:
The general solution is:
```
x = Ky e^(y²).
```

Question 29:

The position vectors of vertices of △ABC are A(2î − ĵ + k̂), B(î − 3ĵ − 5k̂), and C(3î − 4ĵ − 4k̂). Find all the angles of △ABC.

Solution:

View Solution

Step 1: Find the vectors →AB, →BC, and →AC:

    →AB = →B − →A = (î − 3ĵ − 5k̂) − (2î − ĵ + k̂) = −î − 2ĵ − 6k̂.
    →AC = →C − →A = (3î − 4ĵ − 4k̂) − (2î − ĵ + k̂) = î − 3ĵ − 5k̂.
    →BC = →C − →B = (3î − 4ĵ − 4k̂) − (î − 3ĵ − 5k̂) = 2î − ĵ + k̂.

Step 2: Find the magnitudes of the vectors:
- Magnitude of →AB:

    ||→AB|| = √((-1)² + (-2)² + (-6)²) = √(1 + 4 + 36) = √41.

- Magnitude of →AC:

    ||→AC|| = √((1)² + (-3)² + (-5)²) = √(1 + 9 + 25) = √35.

- Magnitude of →BC:

    ||→BC|| = √((2)² + (-1)² + (1)²) = √(4 + 1 + 1) = √6.

Step 3: Use the dot product to find cos(C):
To find ∠C, use the vectors →AC and →BC:

    →AC · →BC = (1)(2) + (-3)(-1) + (-5)(1) = 2 + 3 - 5 = 0.

Thus:

    cos(C) = (→AC · →BC) / (||→AC|| ||→BC||) = 0 / (√35 · √6) = 0.

Step 4: Determine ∠C:
Since cos(C) = 0:
```
    C = π/2.
    
```
Step 5: Conclusion:
The angle C of △ABC is:
```
    C = π/2 (a right angle).
    
```

Question 30:

A pair of dice is thrown simultaneously. If X denotes the absolute difference of the numbers appearing on top of the dice, find the probability distribution of X.

View Solution

Step 1: Define the values of X:
The possible absolute differences are:
```
X = 0, 1, 2, 3, 4, 5.
```
Step 2: Count the outcomes for each X:
- For X = 0: Both dice show the same number. There are 6 outcomes:
```
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
```
- For X = 1: The numbers differ by 1. Outcomes are:
```
(1,2), (2,1), (2,3), (3,2), ..., (5,6), (6,5).
```
This gives 10 outcomes. - Repeat similar counting for X = 2, 3, 4, 5.
Step 3: Calculate probabilities:
The total number of outcomes is:
```
6 × 6 = 36.
```
Probabilities are:
```
P(X = k) = (Number of favorable outcomes for X = k) / 36.
```

Step 4: Write the probability distribution:

    P(X = 0) = 6/36,
    P(X = 1) = 10/36,
    P(X = 2) = 8/36,
    P(X = 3) = 6/36,
    P(X = 4) = 4/36,
    P(X = 5) = 2/36.

Question 31:

Evaluate:

∫ x² sin⁻¹(x^(3/2)) dx.

View Solution

Step 1: Use substitution:
Let:

u = sin⁻¹(x^(3/2)) ⇒ x^(3/2) = sin u ⇒ x = (sin u)^(2/3).

Differentiate:

dx = (2/3)(sin u)^(-1/3) cos u du.

Step 2: Substitute into the integral:
Replace x and dx in the integral:

∫ x² sin⁻¹(x^(3/2)) dx = ∫ [(sin u)^(4/3)] u · (2/3)(sin u)^(-1/3) cos u du.

Simplify:

∫ x² sin⁻¹(x^(3/2)) dx = (2/3) ∫ u (sin u)^(3/3) cos u du = (2/3) ∫ u sin u cos u du.

Step 3: Simplify using trigonometric identities:
Use the identity:

sin u cos u = (1/2) sin(2u).

Substituting:

(2/3) ∫ u sin u cos u du = (1/3) ∫ u sin(2u) du.

Step 4: Solve using integration by parts:
Let:

v = u, dv = du, w = -(1/2) cos(2u), dw = sin(2u) du.

Integration by parts:

∫ u sin(2u) du = - (u/2) cos(2u) + (1/2) ∫ cos(2u) du.

Simplify:

∫ cos(2u) du = (1/2) sin(2u).

Substituting back:

∫ u sin(2u) du = -(u/2) cos(2u) + (1/4) sin(2u).

Step 5: Back-substitute u = sin⁻¹(x^(3/2)):
Replace u and simplify the result:

∫ x² sin⁻¹(x^(3/2)) dx = -(sin⁻¹(x^(3/2))/2) cos(2(sin⁻¹(x^(3/2)))) + (1/4) sin(2(sin⁻¹(x^(3/2)))) + C.

Conclusion:
The value of the integral is:

-(sin⁻¹(x^(3/2))/2) cos(2(sin⁻¹(x^(3/2)))) + (1/4) sin(2(sin⁻¹(x^(3/2)))) + C.

SECTION D

Question 32 (a):

Show that a function f : R → R defined by f(x) = 2x / (1 + x²) is neither one-one nor onto. Further, find set A so that the given function f : R → A becomes an onto function.

View Solution

Step 1: Test if f(x) is one-one:
For a function to be one-one, f(x₁) = f(x₂) should imply x₁ = x₂.
Let:
```
f(x₁) = f(x₂) ⇒ 2x₁ / (1 + x₁²) = 2x₂ / (1 + x₂²).
```
Cross-multiply:
```
2x₁(1 + x₂²) = 2x₂(1 + x₁²).
```
Simplify:
```
x₁ + x₁x₂² = x₂ + x₂x₁².
```
Rearrange:
```
(x₁ - x₂)(1 + x₁x₂) = 0.
```
Thus, either:
```
x₁ = x₂ or x₁x₂ = -1.
```
Since x₁x₂ = -1 implies two distinct x-values give the same f(x), f(x) is not one-one.
Step 2: Test if f(x) is onto:
For f(x) to be onto, every y ∈ R must have a corresponding x ∈ R such that:
```
y = 2x / (1 + x²).
```
Rearrange for x:
```
y(1 + x²) = 2x ⇒ y + yx² = 2x.
```
Simplify into a quadratic equation:
```
yx² - 2x + y = 0.
```
The discriminant of this quadratic is:
```
Δ = (-2)² - 4(y)(y) = 4 - 4y² = 4(1 - y²).
```
For real solutions for x to exist, Δ ≥ 0, which implies:
```
1 - y² ≥ 0 ⇒ -1 ≤ y ≤ 1.
```
Thus, f(x) is not onto because its range is [−1, 1], not R.
Step 3: Find set A to make f(x) onto:
To make f(x) onto, let:
```
A = [−1, 1].
```
Then, for every y ∈ A, there exists an x ∈ R such that:
```
y = 2x / (1 + x²).
```
Conclusion:
The function f(x) = 2x / (1 + x²) is:
```
Neither one-one nor onto.
```
To make f(x) onto, let:
```
A = [−1, 1].
```

Question 32 (b):

A relation R is defined on N × N (where N is the set of natural numbers) as:

(a, b) R (c, d) ⇔ a − c = b − d.

Show that R is an equivalence relation.

View Solution

Step 1: Prove reflexivity:
For any (a, b) ∈ N × N, check if:
```
(a, b) R (a, b).
```
From the definition of R:
```
a − a = b − b ⇒ 0 = 0.
```
Thus, R is reflexive.
Step 2: Prove symmetry:
For (a, b), (c, d) ∈ N × N, assume:
```
(a, b) R (c, d).
```
This implies:
```
a − c = b − d.
```
Rearranging:
```
c − a = d − b.
```
Thus:
```
(c, d) R (a, b).
```
Hence, R is symmetric.

Step 3: Prove transitivity:
For (a, b), (c, d), (e, f) ∈ N × N, assume:

(a, b) R (c, d) and (c, d) R (e, f).

From the definition of R:

a − c = b − d and c − e = d − f.

Adding these equations:

(a − c) + (c − e) = (b − d) + (d − f) ⇒ a − e = b − f.

Thus:

(a, b) R (e, f).

Hence, R is transitive.

Conclusion:
Since R satisfies reflexivity, symmetry, and transitivity, R is:
```
An equivalence relation.
```

Question 33:

Find the equation of the line which bisects the line segment joining points A(2, 3, 4) and B(4, 5, 8) and is perpendicular to the lines:

(x − 8)/3 = (y + 19)/−16 = (z − 10)/7 
and 
(x − 15)/3 = (y − 29)/8 = (z − 5)/−5.

View Solution

Step 1: Find the midpoint of the line segment AB:
The midpoint P of the segment joining A(2, 3, 4) and B(4, 5, 8) is given by:
```
P = ((2 + 4)/2, (3 + 5)/2, (4 + 8)/2) = (3, 4, 6).
```
Step 2: Find direction ratios of the given lines:
- The direction ratios (DRs) of the first line are:
```
DRs = (3, −16, 7).
```
- The direction ratios (DRs) of the second line are:
```
DRs = (3, 8, −5).
```
Step 3: Determine the direction ratios of the required line:
The required line is perpendicular to both given lines. Let the DRs of the required line be proportional to (l, m, n). Using the condition for perpendicularity:
```
3l − 16m + 7n = 0 (from the first line),
```
```
3l + 8m − 5n = 0 (from the second line).
```
Subtract these two equations:
```
(3l − 16m + 7n) − (3l + 8m − 5n) = 0,
```
which simplifies to:
```
−24m + 12n = 0 ⇒ −2m + n = 0 ⇒ n = 2m.
```
Substituting n = 2m into the first equation:
```
3l − 16m + 7(2m) = 0,
```
```
3l − 16m + 14m = 0 ⇒ 3l − 2m = 0.
```
Solve for l:
```
l = (2m)/3.
```
Thus, the DRs of the required line are proportional to:
```
(2/3, 1, 2).
```
Step 4: Write the equation of the line:
The required line passes through the midpoint P(3, 4, 6) and has DRs proportional to (2, 3, 6). Therefore, the equation of the line is:
```
(x − 3)/2 = (y − 4)/3 = (z − 6)/6.
```
Conclusion:
The equation of the required line is:
```
(x − 3)/2 = (y − 4)/3 = (z − 6)/6.
```

Question 34(a):

Solve the following system of equations using matrices:

2x + 3y + 10z = 4,
4x − 6y + 5z = 1,
6x + 9y − 20z = 2,

where x, y, z ≠ 0.

View Solution

Step 1: Substitute variables:
Let:

a = 1/x, b = 1/y, c = 1/z.

Then the system becomes:

2a + 3b + 10c = 4,
4a − 6b + 5c = 1,
6a + 9b − 20c = 2.

Step 2: Represent the system in matrix form:
The system can be written as:

    [ 2   3   10 ] [ a ]   [ 4 ]
    [ 4  -6    5 ] [ b ] = [ 1 ]
    [ 6   9  -20 ] [ c ]   [ 2 ]

Step 3: Compute the determinant of the coefficient matrix:
The coefficient matrix is:

    A = [ 2   3   10 ]
        [ 4  -6    5 ]
        [ 6   9  -20 ].

The determinant is computed as:

    det(A) = 2 | -6   5 | - 3 | 4   5 | + 10 | 4  -6 |
                 |  9 -20 |     | 6 -20 |       | 6   9 |.

Computing the minors:

    | -6   5 |
    |  9 -20 | = (-6)(-20) - (5)(9) = 120 - 45 = 75,

    | 4   5 |
    | 6 -20 | = (4)(-20) - (5)(6) = -80 - 30 = -110,

    | 4  -6 |
    | 6   9 | = (4)(9) - (-6)(6) = 36 + 36 = 72.

Substituting into the determinant formula:

    det(A) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200.

Step 4: Compute the inverse of A:
The inverse of A is:

    A⁻¹ = (1/det(A)) * adj(A),

where adj(A) is the adjugate matrix. After computation:

    A⁻¹ = (1/1200) *
    [ -270  -210   120 ]
    [   90    60   240 ]
    [  -90    30    60 ].

Step 5: Solve for X:

    X = A⁻¹B,

where B = [4, 1, 2]. Perform matrix multiplication:

    X = (1/1200) *
    [ -270 -210  120 ]   [ 4 ]
    [   90   60  240 ] * [ 1 ]
    [  -90   30   60 ]   [ 2 ].

Simplifying:

    X = (1/1200) *
    [ -1050 ]
    [   900 ]
    [  -210 ].

Dividing by 1200:

    X = [ 7/8, 3/4, 1/6 ].

Step 6: Back-substitute:
From a = 1/x, b = 1/y, c = 1/z, we find:
```
    x = 2, y = 3, z = 5.
    
```
Final Answer:
```
    x = 2, y = 3, z = 5.
    
```

Question 34(b):

If A =

[  1   cot(x) ]
[ -cot(x)   1 ],

show that AᵀA⁻¹ =

[ -cos(2x)  -sin(2x) ]
[  sin(2x)  -cos(2x) ].

View Solution

Step 1: Compute the transpose of A:
The matrix A is:

    A = [  1   cot(x) ]
        [ -cot(x)   1 ].

The transpose of A is:

    Aᵀ = [  1  -cot(x) ]
         [ cot(x)    1 ].

Step 2: Compute the determinant of A:
The determinant is:

    det(A) = (1)(1) - (-cot(x))(cot(x)) = 1 - cot²(x).

Using the identity 1 + cot²(x) = csc²(x):

    det(A) = -cos(2x)/sin²(x).

Step 3: Compute the inverse of A:
The formula for the inverse of a 2×2 matrix is:

    A⁻¹ = (1/det(A)) *
    [  d  -b ]
    [ -c   a ].

Substituting for A:

    A⁻¹ = (1/(-cos(2x)/sin²(x))) *
    [  1  -cot(x) ]
    [ cot(x)    1 ].

Simplifying:

    A⁻¹ = (sin²(x)/(-cos(2x))) *
    [  1  -cot(x) ]
    [ cot(x)    1 ].

Step 4: Compute AᵀA⁻¹:
Multiply Aᵀ and A⁻¹:

    AᵀA⁻¹ = [  1  -cot(x) ]   [  1  -cot(x) ]
            [ cot(x)    1 ] * [ cot(x)    1 ].

Simplify the multiplication and substitute trigonometric identities. After computation:

    AᵀA⁻¹ =
    [ -cos(2x)  -sin(2x) ]
    [  sin(2x)  -cos(2x) ].

Final Answer:

    AᵀA⁻¹ =
    [ -cos(2x)  -sin(2x) ]
    [  sin(2x)  -cos(2x) ].

Question 35:

If A₁ denotes the area of the region bounded by y² = 4x, x = 1, and the x-axis in the first quadrant, and A₂ denotes the area of the region bounded by y² = 4x, x = 4, find A₁ : A₂.

Correct Answer: A₁ : A₂ = 1 : 16

View Solution

Step 1: Equation of the parabola and area calculation:
The given parabola is:
```
y² = 4x,
```
which opens to the right. The area of the region bounded by the curve, the line x = a, and the x-axis is:
```
A = 2 ∫[0 to a] y dx,
```
where:
```
y = √(4x) = 2√x.
```

Step 2: Compute A₁:
For A₁, the region is bounded by x = 1:

A₁ = 2 ∫[0 to 1] 2√x dx = 4 ∫[0 to 1] x^(1/2) dx.

The integral is:

∫ x^(1/2) dx = (2/3) x^(3/2) + C.

Substituting the limits:

A₁ = 4 * (2/3) * (1^(3/2) - 0^(3/2)) = 8/3.

Step 3: Compute A₂:
For A₂, the region is bounded by x = 4:

A₂ = 2 ∫[0 to 4] 2√x dx = 4 ∫[0 to 4] x^(1/2) dx.

Using the same integral:

A₂ = 4 * (2/3) * (4^(3/2) - 0^(3/2)).

Evaluate 4^(3/2):

4^(3/2) = (2²)^(3/2) = 2³ = 8.

Thus:

A₂ = 4 * (2/3) * 8 = 64/3.

Step 4: Ratio of A₁ and A₂:

A₁ / A₂ = (8/3) / (64/3) = 1/16.

Solution:

A₁ : A₂ = 1 : 16.

SECTION E

Question 36 (Case Study):

Overspeeding increases fuel consumption and decreases fuel economy as a result of tyre rolling friction and air resistance. The relation between fuel consumption F (liters per 100 km) and speed V (km/h) is given as:

F = (V²/500) − (V/4) + 14.

Answer the following questions:

Find F when V = 40 km/h.
Find dF/dV.
(a) Find the speed V for which fuel consumption F is minimum. OR (b) Find the quantity of fuel required to travel 600 km at the speed V for which dF/dV = −0.01.

View Solution

Part (i): Find F when V = 40 km/h:
Substitute V = 40 into the equation:

F = (40²/500) − (40/4) + 14.

Simplify:

F = (1600/500) − 10 + 14 = 3.2 − 10 + 14 = 7.2.

Solution:

F = 7.2 liters per 100 km.

Part (ii): Find dF/dV:
Differentiate F with respect to V:

dF/dV = d/dV (V²/500) − d/dV (V/4) + d/dV (14).

Simplify:

dF/dV = (2V/500) − (1/4) + 0 = V/250 − 1/4.

Solution:

dF/dV = V/250 − 1/4.

Part (iii)(a): Find the speed V for minimum F:
For minimum fuel consumption, set:
```
dF/dV = 0 ⇒ V/250 − 1/4 = 0.
```
Solve for V:
```
V/250 = 1/4 ⇒ V = 250/4 = 62.5.
```
Solution:
```
V = 62.5 km/h.
```

Part (iii)(b): Find fuel required to travel 600 km:
Given:

dF/dV = −0.01 ⇒ V/250 − 1/4 = −0.01.

Solve for V:

V/250 = −0.01 + 1/4 = 0.24 ⇒ V = 250 * 0.24 = 60.

At V = 60 km/h, substitute into the equation for F:

F = (60²/500) − (60/4) + 14.

Simplify:

F = (3600/500) − 15 + 14 = 7.2 liters per 100 km.

Total fuel required for 600 km:

Fuel = (F/100) * 600 = (7.2/100) * 600 = 43.2 liters.

Solution:

Fuel required = 43.2 liters.

Question 37 (Case Study):

A dietician wishes to minimize the cost of a diet involving two types of foods, food X (in kg) and food Y (in kg), which are available at the rate of ₹16/kg and ₹20/kg, respectively. The feasible region satisfying the constraints is shown in Figure-2.

Answer the following questions:

Identify and write all the constraints which determine the given feasible region in Figure-2.
If the objective is to minimize cost Z = 16x + 20y, find the values of x and y at which cost is minimum. Also, find the minimum cost assuming that the minimum cost is possible for the given unbounded region.

View Solution

Part (i): Constraints determining the feasible region:
From Figure-2, the constraints are:

    3x + y ≤ 8,
    x + y ≥ 4,
    4x + 5y = 28,
    2x + y ≥ 10,
    x ≥ 0, y ≥ 0.

Solution:

The constraints are: 3x + y ≤ 8, x + y ≥ 4, 4x + 5y = 28, 2x + y ≥ 10, x ≥ 0, y ≥ 0.

Part (ii): Minimize cost Z = 16x + 20y:
Evaluate Z at the vertices of the feasible region (from Figure-2):

Vertices: A(10, 0), B(2, 4), C(1, 5), D(0, 8).

- At A(10, 0):

Z = 16(10) + 20(0) = 160.

- At B(2, 4):

Z = 16(2) + 20(4) = 32 + 80 = 112.

- At C(1, 5):

Z = 16(1) + 20(5) = 16 + 100 = 116.

- At D(0, 8):

Z = 16(0) + 20(8) = 0 + 160 = 160.

Minimum cost:

Z = 112 at B(2, 4).

Solution:

The minimum cost is ₹112 at x = 2, y = 4.

Question 38:

Airplanes are by far the safest mode of transportation when the number of transported passengers is measured against personal injuries and fatality totals.

Previous records state that the probability of an airplane crash is 0.00001%. Further, there are 95% chances that there will be survivors after a plane crash. Assume that in case of no crash, all travelers survive.
Let E₁ be the event that there is a plane crash and E₂ be the event that there is no crash. Let A be the event that passengers survive after the journey.

On the basis of the above information, answer the following questions:

Find the probability that the airplane will not crash.
Find P(A | E₁) + P(A | E₂).
(a) Find P(A). OR (b) Find P(E₂ | A).

View Solution

Part (i): Probability that the airplane will not crash:
The probability of a plane crash is:
```
P(E₁) = 0.00001% = 0.00001 / 100 = 10⁻⁷.
```
The probability that the airplane will not crash is:
```
P(E₂) = 1 − P(E₁) = 1 − 10⁻⁷.
```
Solution:
```
P(E₂) = 1 − 10⁻⁷.
```

Part (ii): Find P(A | E₁) + P(A | E₂):
From the problem:

P(A | E₁) = 0.95 (95% chance of survival after a crash),

P(A | E₂) = 1 (all travelers survive if there is no crash).

Thus:

P(A | E₁) + P(A | E₂) = 0.95 + 1 = 1.95.

Solution:

P(A | E₁) + P(A | E₂) = 1.95.

Part (iii)(a): Find P(A):
Using the law of total probability:

P(A) = P(A | E₁)P(E₁) + P(A | E₂)P(E₂).

Substitute the values:

P(A) = (0.95)(10⁻⁷) + (1)(1 − 10⁻⁷).

Simplify:

P(A) = 0.95 · 10⁻⁷ + 1 − 10⁻⁷ = 1 − 0.05 · 10⁻⁷.

Solution:

P(A) = 1 − 0.05 · 10⁻⁷.

Part (iii)(b): Find P(E₂ | A):
Using Bayes' theorem:

P(E₂ | A) = P(A | E₂)P(E₂) / P(A).

Substitute the values:

P(E₂ | A) = (1)(1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

Simplify:

P(E₂ | A) = (1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

Solution:

P(E₂ | A) = (1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

*The article might have information for the previous academic years, please refer the official website of the exam.

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Question 6:

Question 7:

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SECTION B

Question 21(a):

Question 21(b):

Question 22(a):

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SECTION C

Question 26 (a):

Question 26 (b):

Question 27 (a):

Question 27(b):

Question 28(a):

Question 28 (b):

Question 29:

Question 30:

Question 31:

SECTION D

Question 32 (a):

Question 32 (b):

Question 33:

Question 34(a):

Question 34(b):

Question 35:

SECTION E

Question 36 (Case Study):

Question 37 (Case Study):

Question 38:

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