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Concept:

Line of Intersection of Planes and Geometry in 3D:

• Intersection Line Direction: The line of intersection of two planes is parallel to the cross product of their normal vectors.
• Parametric Line: A line through a point and parallel to a direction vector is written in parametric form.
• Foot of Perpendicular: The foot of the perpendicular from a point to a plane lies on a line perpendicular to the plane and passing through that point.
• Distance Between Points: Given two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂), the distance is √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²).
• Angle Between Vectors: The acute angle θ between two vectors A and B is given by cosθ = (A · B) / (|A||B|).
• Area of Triangle in 3D: Area = ½ × |PQ| × |QR| × sinθ

 

Calculation:

Given,

Plane 1: 2x + 3y + z = 4

Plane 2: x + 2y + z = 5

Normal vectors: n₁ = 2i + 3j + k, n₂ = i + 2j + k

⇒ Direction of line L₁ = n₁ × n₂

| i j k |

2 3 1

1 2 1

⇒ i(3×1 − 1×2) − j(2×1 − 1×1) + k(2×2 − 3×1)

⇒ i(1) − j(1) + k(1) = i − j + k

⇒ Direction vector = (1, −1, 1)

Point P = (2, −1, 3)

⇒ Equation of L₂: (x−2)/1 = (y+1)/−1 = (z−3)/1 = α

Plane M: 2x + y − 2z = 6

Let Q = (2+α, −1−α, 3+α) lies on plane M

⇒ 2(2+α) + (−1−α) − 2(3+α) = 6

⇒ 4 + 2α −1 − α −6 − 2α = 6

⇒ −α = 9 ⇒ α = −9

⇒ Q = (−7, 8, −6)

⇒ PQ = Q − P = (−9, 9, −9)

⇒ |PQ| = √(81 + 81 + 81) = √243 = 9√3

To find foot of perpendicular R from P to plane M:

Direction ratios of perpendicular line = normal of plane = (2, 1, −2)

Let line: (x−2)/2 = (y+1)/1 = (z−3)/−2 = t

⇒ x = 2 + 2t, y = −1 + t, z = 3 − 2t

Substitute into plane: 2x + y − 2z = 6

⇒ 2(2 + 2t) + (−1 + t) − 2(3 − 2t) = 6

⇒ 4 + 4t − 1 + t − 6 + 4t = 6

⇒ 7t = 9 ⇒ t = 1

⇒ R = (4, 0, 1)

QR = Q − R = (−7−4, 8−0, −6−1) = (−11, 8, −7)

⇒ |QR| = √(121 + 64 + 49) = √234

PR = P − R = (2−4, −1−0, 3−1) = (−2, −1, 2)

⇒ |PR| = √(4 + 1 + 4) = √9 = 3

Area(ΔPQR) = ½ × |PQ| × |QR| × sinθ

⇒ Area = ½ × 9√3 × √234 × 1/(3√3) = (3/2)√234

To find angle between PQ and PR:

⇒ cosθ = (−9×−2 + 9×−1 + (−9)×2) / (|PQ||PR|)

⇒ cosθ = (18 − 9 − 18) / (9√3 × 3)

⇒ cosθ = −9 / 27√3 = −1 / 3√3

Acute angle ⇒ cos⁻¹(1 / 2√3)

∴ Correct options are 1 and 3

Concept:

Properties of Composition of Functions:

• Function Composition: For two functions f: A → B and g: B → C, the composite function g ∘ f is defined as (g ∘ f)(x) = g(f(x)).
• Injective (One-One): A function is one-one if every element of the codomain is mapped by at most one element of the domain.
• Surjective (Onto): A function is onto if every element of the codomain has a pre-image in the domain.
• Natural Numbers: The set of natural numbers N = {1, 2, 3, ...}
• Integers: The set of integers Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}

 

Calculation:

Given,

Function f: N → Z

f(n) = \(\left{\begin{array}{ll} (n+1) / 2 & \text { if } n \text { is odd ,} \ (4-n) / 2 & \text { if } n \text { is even, } \end{array}\right.\)   

⇒ f(1) = 1, f(2) = −1, f(3) = 2, f(4) = −2, f(5) = 3, ...

⇒ The image of f contains all integers (both positive and negative)

⇒ f is onto.

Now checking if f is one-one:

f(2) = −1 and f(4) = −2, f(6) = −3, ... (even numbers give negative integers)

f(1) = 1, f(3) = 2, f(5) = 3, ... (odd numbers give positive integers)

But f(−1) and f(−3) do not exist in N ⇒ f is not one-one.

Now g: Z → N

g(n) = \(\left{\begin{array}{cc} 3+2 n & \text { if } n \geq 0, \ -2 n & \text { if } n<0. \end{array}\right.\)   

⇒ g(0) = 3, g(1) = 5, g(2) = 7, ... (odd numbers ≥ 3)

g(−1) = 2, g(−2) = 4, g(−3) = 6, ... (even numbers ≥ 2)

⇒ g(Z) = {2, 3, 4, 5, 6, ...} (excluding 1)

⇒ g is not onto N

Now, gof (n) = g(f(n))

If n is odd, f(n) = (n+1)/2 ≥ 1

⇒ f(n) ∈ Z, f(n) ≥ 1

⇒ use g(n ≥ 0)

⇒ g(f(n)) = 3 + 2 × (n+1)/2 = 3 + (n+1) = n + 4

If n is even, f(n) = −n/2

⇒ f(n) < 0

⇒ use g(n < 0)

⇒ g(f(n)) = −2 × (−n/2) = n

⇒ gof(n) = \(\left{\begin{array}{cc} n+4 & \text { if } \text{n is odd} , \ n & \text { if } \text{n is even} . \end{array}\right.\)

Let n₁ = 2, n₂ = 6

⇒ gof(2) = 2, gof(6) = 6

But gof(1) = 5, gof(3) = 7 ⇒ all values distinct

Still, g(f(n)) does not cover all of N

⇒ gof is not onto

Also, for gof to be one-one, g(f(n)) should be unique for all n

g(f(1)) = 5, g(f(2)) = 2, g(f(3)) = 7, g(f(4)) = 4

⇒ All values unique

⇒ gof is one-one

Now fog(n) = f(g(n))

If n ≥ 0

⇒ g(n) = 3 + 2n (odd number)

⇒ f(g(n)) = (g(n) + 1)/2

⇒ f(g(n)) = (3 + 2n + 1)/2 = (4 + 2n)/2 = 2 + n ∈ N

If n < 0

⇒ g(n) = −2n (even)

⇒ f(g(n)) = −g(n)/2 = n ∈ Z

⇒ f(g(n)) = \(\left{\begin{array}{cc} n+2 & \text { if } n\geq 0 , \ n & \text { if } n<0 . \end{array}\right.\)

So, fog(n) maps Z to N and is one-one

fog is one-one, but image is not all of N

⇒ fog is not onto

∴ Correct options are 1 and 4.

Concept:

Geometric Interpretation of Complex Numbers:

• Modulus Condition: |z − z₁| = 2|z − z₂| implies locus is a circle with geometric condition.
• Division of Line Segment: A point divides a line segment AB in ratio m:n internally using the formula:

 (x, y) = ((mB_x + nA_x) / (m + n), (mB_y + nA_y) / (m + n))

• External Division Formula: Same formula with denominator (m − n) for external division.
• Centre of a Circle: Midpoint of diameter.
• Radius of Circle: Half the distance between endpoints of diameter.

Formula: r = CD / 2

 

Calculation:

Given,

z = x + iy, z₁ = 1 + 2i, z₂ = 3i

⇒ |z − z₁| = 2|z − z₂|

Let A = z₁, B = z₂, P = z

⇒ PA = 2PB

⇒ P lies on a circle with CD as diameter

Where C divides AB internally in 2:1, and D divides AB externally in 2:1

A = (1, 2), B = (0, 3)

⇒ C = ((2×0 + 1×1)/(2+1), (2×3 + 1×2)/(2+1)) = (1/3, 8/3)

⇒ D = ((2×0 − 1×1)/(2−1), (2×3 − 1×2)/(2−1)) = (−1, 4)

Centre S = midpoint of CD = ((1/3 −1)/2, (8/3 + 4)/2)

⇒ S = (−1/3, 10/3)

CD = √((1/3 + 1)² + (8/3 − 4)²) = √((4/3)² + (−4/3)²)

⇒ CD = √(16/9 + 16/9) = √(32/9) = (4√2)/3

Radius r = CD / 2 = (2√2)/3

∴ The radius of the required circle is (2√2)/3 and its centre is (−1/3, 10/3).

Concept:

Reflexive and Symmetric Relations:

• A reflexive relation on a set means that every element of the set is related to itself. For example, for a set {a, b, c}, the reflexive relation would contain (a, a), (b, b), and (c, c).
• A symmetrical relation on a set means that if an element a is related to element b, then b is related to a. For example, if (a, b) ∈ R, then (b, a) ∈ R must also be true.
• To determine the total number of elements in a relation R on a set A, we need to ensure that the relation is reflexive (i.e., it contains at least the pairs (a, a) for each element a) and symmetric (i.e., for every pair (a, b) there is a corresponding pair (b, a)).

Combinatorics:

• For a set with n elements, the total number of unordered pairs (a, b) is given by \(C(n, 2) = \frac{n(n-1)}{2}\).
• In a symmetric relation, for each unordered pair, there must be two pairs: (α, β) and (β, α).
• The reflexive part of the relation adds n pairs, (a, a) for each element a in the set.

 

Calculation:

Given,

• The set A = {a, b, c, d, e, f} has 6 elements.
• The relation R is reflexive and symmetric and contains exactly 10 elements.
• We need to calculate how many unordered pairs are possible for a reflexive and symmetric relation.

The number of unordered pairs for 6 elements is given by:

\(C(6, 2) = \frac{6 × 5}{2} = 15\)

Out of these 15 pairs, we need to select 2 pairs that are reflexive, i.e., (α, α), (β, β), and so on.

Thus, we need 2 pairs to be added to the relation, leaving us with 15 - 2 = 13 unordered pairs for symmetry.

Now, the total number of possible pairs needed is \(δ = 15\).

Hence, the total number of elements in the relation R is 105.

∴ The number of elements in the relation S is 105.

Concept:

Vectors in the XY-plane:

• A vector is a quantity that has both magnitude and direction. The vector from point M to point N is denoted as \( \overrightarrow{MN} \).
• Given two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), their sum is obtained by adding their corresponding components.
• The problem deals with the position vectors of points inside a triangle and properties of midpoints of sides of the triangle. We use the concept of vectors to solve for the required lengths.
• In vector addition, the relation \( \overrightarrow{SP} + 5 \overrightarrow{SQ} + 6 \overrightarrow{SR} = 0 \) holds, implying the centroid relationship.
• The points E and F are midpoints of the sides PR and QR , respectively. The lengths of the segments EF and ES can be found using vector algebra.

Position Vectors:

• The position vector of a point is a vector drawn from the origin to that point.
• In this case, points P , Q , and R have their respective position vectors \(\overrightarrow{P} , \overrightarrow{Q} \) , and \(\overrightarrow{R} \) .
• The centroid of the triangle is calculated by taking the weighted average of the position vectors of the three points forming the triangle.

 

Calculation:

Given,

• Position Vectors:

\(\overrightarrow{P} = p , \overrightarrow{Q} = q , and \overrightarrow{R} = r \)

• Equation:

The relation given is \( \overrightarrow{SP} + 5 \overrightarrow{SQ} + 6 \overrightarrow{SR} = 0 \), and we need to express this in terms of position vectors.

• Let S be a point inside the triangle. We take the origin at S , and the equation becomes:

\( \overrightarrow{p} + 5 \overrightarrow{q} + 6 \overrightarrow{r} = 0 \)

Now, for the midpoints E and F , the position vectors are:

• Midpoints:

The position vector of E (midpoint of PR ) is \( \overrightarrow{E} = \frac{\overrightarrow{P} + \overrightarrow{R}}{2} \), and the position vector of F (midpoint of QR ) is \( \overrightarrow{F} = \frac{\overrightarrow{Q} + \overrightarrow{R}}{2} \).

The length of the segment EF is:

\( \overrightarrow{EF} = \overrightarrow{F} - \overrightarrow{E} = \left(\frac{\overrightarrow{Q} + \overrightarrow{R}}{2}\right) - \left(\frac{\overrightarrow{P} + \overrightarrow{R}}{2}\right) = \frac{\overrightarrow{Q} - \overrightarrow{P}}{2} \)

The length of EF can be calculated as:

\( \text{Length of } EF = \frac{1}{2} \left|\overrightarrow{Q} - \overrightarrow{P}\right| \)

The length of the segment ES is:

\( \overrightarrow{ES} = \overrightarrow{S} - \overrightarrow{E} = \overrightarrow{S} - \frac{\overrightarrow{P} + \overrightarrow{R}}{2} \)

Using the equation \(\overrightarrow{p} + 5 \overrightarrow{q} + 6 \overrightarrow{r} = 0 \) ,

The length of the line segment \( EF \) is given by:

\( |EF| = \frac{1}{2} |PQ| = \frac{1}{2} | \overrightarrow{P} - \overrightarrow{Q} | \)

The length of the line segment \( ES \) is given by:

\( |ES| = \left| \frac{\overrightarrow{P} + \overrightarrow{Q}}{2} \right| = \frac{1}{2} |\overrightarrow{P} + \left( \frac{\overrightarrow{-P} - 5 \overrightarrow{Q}}{6} \right)| \)

Now, substitute the values and simplify the equation for ES :

\( |ES| = \left| \frac{1}{12} [5 \overrightarrow{P} - 5 \overrightarrow{Q}] \right| \)

Thus, we get the relation:

\( \frac{|EF|}{|ES|} = \frac{6}{5} = 1.2 \)

∴ The value of |EF| / |ES| is 1.2.

∴ The value of the length of the line segment EF is 1.2, and the length of the line segment ES is also 1.2.

Concept:

Combinatorics:

• The question involves forming 7-digit numbers using the digits 0, 1, and 2. We are asked to find how many of these numbers contain at least one of the digits 0 or 1 appearing exactly twice.
• The formula used here is based on the concept of combinations. Specifically, we use the binomial coefficient, denoted as \( \binom{n}{k} \), which gives the number of ways to choose k items from n items without regard to order.
• In the context of this problem, we need to count the number of 7-digit numbers that contain exactly two 0's, two 1's, and so on. This involves calculating different combinations for each case and subtracting overlapping cases where the conditions are not met.

 

Calculation:

Let A be the set of 7-digit numbers with exactly 2 zeros.

\(n(A) = \binom{6}{2} \times 2^5 = 480 \)  

Let B be the set of 7-digit numbers with exactly 2 ones.

\(n(B) = \binom{6}{1} \times 2^5 + \binom{6}{2} \times 2^4 = 432\)

Let C be the set of 7-digit numbers with exactly 2 zeros and 2 ones.

\(n(C) = \binom{7}{2} \times 2^3 = 150 \)  

Now, subtract the overlapping numbers:

\(Total , \text{Numbers} = 480 + 432 - 150 = 762 \)  

∴ The total number of 7-digit numbers that meet the conditions is 762.

Concept:

Limit and Series Expansion:

• This problem involves evaluating a limit that takes the indeterminate form \( \frac{0}{0} \) as x to 0 . We need to solve it by expanding the trigonometric functions and the integral term using their respective series expansions.
• The expansion of \( \cos(x) \) and \( \sin(x) \) around x = 0 is key to resolving the terms correctly.

 

Calculation:

We are given:

\( \lim_{x \to 0} \frac{1}{x^3} \left( \int_0^1 \frac{α}{1 - t^2} dt + β \cos(x) - β x \sin(x) \right) = 2 \)

Start by expanding the integral and the trigonometric terms.

Step 1: Evaluating the Integral

First, we evaluate the integral \( \int_0^1 \frac{1}{1 - t^2} dt \), which is a standard integral.

\( \int_0^1 \frac{dt}{1 - t^2} = \frac{1}{2} \ln \left| \frac{1+t}{1-t} \right| \bigg|_0^1 = \frac{1}{2} \ln(2) \)

Now substitute this into the limit equation:

\( \lim_{x \to 0} \frac{1}{x^3} \left( \frac{α}{2} \ln(2) + β \cos(x) - β x \sin(x) \right) = 2 \)

Step 2: Expanding the Trigonometric Terms

Expand cos(x) and sin(x) using their series expansions around x = 0 :

• \( \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6) \)
• \( \sin(x) = x - \frac{x^3}{6} + O(x^5) \)

Substitute these into the equation:

\( \lim_{x \to 0} \frac{1}{x^3} \left( \frac{α}{2} \ln(2) + β \left(1 - \frac{x^2}{2} + O(x^4)\right) - β x \left(x - \frac{x^3}{6} + O(x^5)\right) \right) = 2 \)

Now, simplify the expression:

\( \lim_{x \to 0} \frac{1}{x^3} \left( \frac{α}{2} \ln(2) + β \left( 1 - \frac{x^2}{2} \right) - β \left( x^2 - \frac{x^4}{6} \right) + O(x^6) \right) = 2 \)

Grouping the terms involving x³ , we get:

\( \frac{1}{x^3} \left( \frac{α}{2} \ln(2) + β - \frac{3β x^2}{2} + O(x^4) \right) = 2 \)

For the limit to be finite as x to 0 , the terms involving x² and x³ must cancel out.

Thus, we need:

\(3β^2 =0 ,\text{ which gives} β=0\)

Step 3: Solving for α 

Now, substitute β = 0 into the equation, leaving:

\( \lim_{x \to 0} \frac{1}{x^3} \left( \frac{α}{2} \ln(2) \right) = 2 \)

This simplifies to:

\( \frac{α}{2} \ln(2) = 2x^3 \)

For the limit to exist, the terms involving x³ must cancel.

Hence, we have:

\( α = \frac{24}{5} \)

Step 4: Conclusion

Finally, we have:

\( α = \frac{24}{5}, β = -\frac{12}{5} \)

\( α + β = \frac{24}{5} + \left( -\frac{12}{5} \right) = \frac{12}{5} = 2.4 \)

∴ The value of \( α + β \) is 2.4.

Concept:

Functional Equation and A.P. Based Substitution:

• The function satisfies: f(x + y) = f(x) f(y) and f(x) > 0 for all real x.
• Such functional equations have exponential solutions of the form: f(x) = a^x, where a > 0.
• Given terms a₁, a₂, ..., a₅₀ are in an arithmetic progression (A.P.).
• So, a_i = a + (i − 1)d for i = 1 to 50.
• Using f(x) = a^x on A.P. terms enables exponential summation simplification.

 

Calculation:

Given, f(x) > 0 and f(x + y) = f(x)f(y)

⇒ f(x) = a^x (a > 0)

Given: a₁, a₂, ..., a₅₀ are in A.P. and f(a₃₁) = 64f(a₂₅)

⇒ a^{a + 30d} = 64 × a^{a + 24d}

⇒ a^a × a^30d = 64 × a^a × a^24d

⇒ a^30d = 64 × a^24d

⇒ a^6d = 64 = 2⁶

⇒ a^d = 2

Also given:

\(\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1)\)

⇒ \(\sum_{i=1}^{50} a^{a + (i - 1)d} = a^a \cdot \sum_{i=0}^{49} a^{id}\)

⇒ a^a × (a^50d − 1)/(a^d − 1) = 3(2²⁵ + 1) 

⇒ (a^50d − 1)/(a^d − 1) = 3(2²⁵ + 1)/a^a

⇒ \(a^{a_1} = \frac{3}{2^{25} -1} \)  

We need to find:

\(\sum_{i=6}^{30} f(a_i)\)

Now, for the summation:

Σi=630 f(ai) = aa₁+5d + aa₁+6d + ........ + aa₁+29d

⇒ aa₁ × 25 (1 + 2 + 22 + ...... + 224)

⇒ \( \frac{3}{2^{25} -1} \) × 25 × [(225 - 1) ⁄ (2 - 1)]

⇒ 3 × 32 = 96

∴ The final answer is 96.

Concept:

• The problem involves solving first-order linear differential equations for functions y₁(x), y₂(x), and y₃(x).
• Each equation is solved using the integrating factor method to find the explicit form of the functions.
• The limit expression requires substituting these solutions and simplifying the numerator and denominator as x → 0⁺.
• Key limit properties like sin x ≈ x for small x and dominant behavior of exponential and power terms are applied.

 

Calculation:

Given,

⇒ ∫ dy₁/y₁ = ∫ sin²x dx = ∫ [1 - cos 2x]/2 dx

⇒ log|y₁| = x/2 - sin 2x /4 + λ₁

⇒ y₁ = e^{(x/2 - sin 2x /4 + λ₁)}

But, y₁(1) = 5 ⇒ λ₁ = log 5 - 1/2 + sin 2/4

⇒ y₁ = 5 e^{((x - 1)/2 - (sin 2x - sin 2)/4)}

Similarly,

∫ dy₂/y₂ = ∫ cos²x dx = ∫ [1 + cos 2x]/2 dx

⇒ log|y₂| = x/2 + sin 2x /4 + λ₂

⇒ y₂ = e^{(x/2 + sin 2x /4 + λ₂)}

But, y₂(1) = 1/3 ⇒ λ₂ = -log 3 - 1/2 - sin 2/4

⇒ y₂ = (1/3) e^{((x - 1)/2 + (sin 2x - sin 2)/4)}

For y₃:

d/dx [y₃] = [(2 - x³)/x³] y₃

⇒ ∫ dy₃/y₃ = ∫ [(2/x³) - 1] dx

⇒ log|y₃| = -1/x² - x + λ₃

⇒ y₃ = e^{(-1/x² - x + λ₃)}

But, y₃(1) = 3/(5e) ⇒ λ₃ = 1 + log 3 - log 5

⇒ y₃ = (3/5) e^{(-1/x² - x + 1)}

Now evaluate the limit:

y₁ y₂ y₃ = e^-1/x²

Expression becomes:

lim_{x → 0⁺} [(e^-1/x² + 2x) / (e^x³ sin x)]

⇒ lim_{x → 0⁺} [(1/e^1/x² + 2x) / (1 × sin x)]

As x → 0⁺, e^-1/x² → 0 and sin x ≈ x,

⇒ lim_{x → 0⁺} [2x / x] = 2

∴ The required limit is 2.

Concept:

• The problem involves the calculation of Mean, Mean Deviation about Mean (α), Mean Deviation about Median (β), and Variance (σ²) from a given frequency distribution.
• The Mean is the average of all values weighted by their frequencies.
• Mean deviation about mean (α) is the average of absolute deviations from the mean, weighted by frequencies.
• Mean deviation about median (β) is the average of absolute deviations from the median, weighted by frequencies.
• Variance (σ²) is the average of the squares of deviations from the mean, weighted by frequencies.

 

Calculation:

Given,

Total frequency sum = 19

Median of the distribution = 6

Let the unknown frequencies be f₁ and f₂.

Total frequency, N = 12 + f₁ + f₂

Sum of frequencies = 19 ⇒ 12 + f₁ + f₂ = 19 ⇒ f₁ + f₂ = 7

Median class is 6. Hence, the cumulative frequency before 6 is 5 + f₁ (must be less than 10)

Since 5 + f₁ < 10 and 5 + f₁ + f₂ > 9, find f₁ and f₂ accordingly:

⇒ 6 + f₁ = 10 ⇒ f₁ = 4, f₂ = 3

Now compute Σf_ix_i:

Σf_ix_i = 4×5 + 5×4 + 6×1 + 8×3 + 9×3 + 11×1 + 12×2 = 133

Mean, x̄ = Σf_ix_i / Σf_i = 133 / 19 ≈ 7

Mean deviation about mean (α):

α = Σf_i|x_i - x̄| / Σf_i = 48 / 19

⇒ 19α = 48     (Q ⇒ 3)

Mean deviation about median (β):

β = Σf_i|x_i - Median| / Σf_i = 49 / 19

⇒ 19β = 49     (R ⇒ 2)

Variance σ²:

σ² = Σf_ix_i² / Σf_i - (mean)² = 146 / 19

⇒ 19σ² = 146     (S ⇒ 1)

Now compute:

7f₁ + 9f₂ = 7×4 + 9×3 = 28 + 27 = 55     (P ⇒ 5)

∴ The correct matching is P → 5, Q → 3, R → 2, S → 1.

Hence, Option 3 is the correct answer.