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Solution:

$\text{Let } \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$

$a_1^2 + a_2^2 + a_3^2 = 1$

$\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}, \quad \vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}, \quad \vec{d} = \hat{k}$

$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{\vec{a} \cdot \vec{d}}{|\vec{d}|}$

Calculate magnitudes:

$|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$

$|\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$

$|\vec{d}| = \sqrt{0^2 + 0^2 + 1^2} = 1$

Set equal projections:

$\frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = \frac{a_3}{1}$

From the first equality:

$2a_1 - a_2 + 2a_3 = 3a_3 \implies 2a_1 - a_2 = a_3$

From the second equality:

$a_1 + 2a_2 - 2a_3 = 3a_3 \implies a_1 + 2a_2 = 5a_3$

Solving 

$a_1 = \frac{7}{\sqrt{155}}, \quad a_2 = \frac{9}{\sqrt{155}}, \quad a_3 = \frac{5}{\sqrt{155}}$

∴ a unit vector along  $\frac{1}{\sqrt{155}} (7\hat{i} + 9\hat{j} + 5\hat{k}) \qquad$

Hence, the correct answer is Option 3.

Solution:

R= {(f,g): f(0) = g(1) and f(1) = g(0)}

Reflexive: (f, f) € R

⇒ f(0) = f(1) and f(1) = f(0) ⇒ must hold

⇒ But this is not true for all functions

So not reflexive

Symmetric: If (f,g) € R ⇒ (g, f) € R

Now, g(0) = f(1) and g(1) = f(0) ⇒ true

∴ symmetric

Transitive: If (f,g) € R and (g,h) € R

⇒ (f,h) € R

Now (f,g) € R= f(0)= g(1) and f(1)= g(0)

⇒ (g,h) € R ⇒ g(0)= h(1) and g(1) = h(0)

⇒ For (f,h) € R we need f(0) = h(1) and f(1) = h(0)

⇒ Now f(0)= g(1) = h(0) nd f(1) = g(0) = h(1)

Hence transitive

Hence, the correct answer is Option 2

Concept:

Limits and Expansion of Elementary Functions:

• To solve limits involving indeterminate forms like 0/0, use series expansion (Maclaurin expansion) of functions such as sinx and e^x.
• Standard expansions:
• sin x = x − x³/3! + x⁵/5! − ...
• e^x = 1 + x + x²/2! + x³/3! + ...
• Compare powers of x from numerator and denominator and simplify accordingly.

 

Calculation:

Given,

lim_x→0 [ x²·sin(αx) + (γ−1)·e^x² ] / [ sin(2x) − βx ] = 3

Numerator:

x²·sin(αx) = x²·(αx − (αx)³/6 + ... ) = αx³ − α³x⁵/6 + ...

(γ − 1)·e^x² = (γ − 1)·(1 + x² + x⁴/2! + ...) = (γ − 1) + (γ − 1)x² + ...

So, numerator ≈ (γ − 1) + (γ − 1)x² + αx³ + ...

Denominator:

sin(2x) = 2x − (2x)³/6 + ... = 2x − 8x³/6 + ...

So, denominator = 2x − βx − (8x³/6) + ... = (2 − β)x − (4/3)x³ + ...

Now rewrite limit as:

lim_x→0 [ (γ−1) + (γ−1)x² + αx³ + ... ] / [ (2−β)x − (4/3)x³ + ... ] = 3

Now, use expansion up to x⁰ term after dividing:

⇒ lim_x→0 (γ−1) / (2−β)x = 0 unless (γ−1) = 0

So, γ − 1 = 0 ⇒ γ = 1

Then numerator becomes αx³ + ...

Denominator: (2 − β)x − (4/3)x³ + ...

⇒ lim = αx³ / [ (2 − β)x − (4/3)x³ ]

Now multiply numerator and denominator by x⁻³:

⇒ α / [ (2 − β)x⁻² − (4/3) ]

Now take limit x → 0, x⁻² → ∞

For limit to be finite, coefficient of x⁻² in denominator must be 0

⇒ (2 − β) = 0

⇒ β = 2

Now denominator = −(4/3),

numerator = α ⇒ α / (−4/3) = 3

⇒ α = −4

Given γ = 1, β = 2, α = −4

β + γ − α = 2 + 1 − (−4) = 7

∴ β + γ − α = 7

Calculation:

Given,

$\Delta = \begin{vmatrix} 3 & 1 & \beta \ 2 & \alpha & -1 \ 1 & 2 & 1 \end{vmatrix} = 0$

$3(\alpha + 2) - 1(2 + 1) + \beta (4 - \alpha) = 0$

$3\alpha + 6 - 3 + 4\beta - \alpha \beta = 0$

$3\alpha + 4\beta - \alpha \beta + 3 = 0$

Also,

$\Delta_3 = \begin{vmatrix} 3 & 1 & 3 \ 2 & \alpha & -3 \ 1 & 2 & 4 \end{vmatrix} = 0$

$3(4\alpha + 6) - 1(8 + 3) + 3(4 - \alpha) = 0$

$12\alpha + 18 - 11 + 12 - 3\alpha = 0$

$9\alpha + 19 = 0 \Rightarrow \alpha = -\frac{19}{9}$

Putting in previous equation:

$3\alpha + 4\beta - \alpha \beta + 3 = 0$

$\Rightarrow \beta = \frac{6}{11}$

Now,

$22\beta - 9\alpha = 22 \left(\frac{6}{11}\right) - 9 \left(-\frac{19}{9}\right)$

$= 12 + 19 = 31$

∴ The correct answer is Option 2.

Concept:

Sequence and Series – Recurrence Relations:

• For a sequence defined by $P_n = \alpha^n + \beta^n$, the terms satisfy the linear recurrence $P_n = (\alpha+\beta),P_{n-1} - (\alpha\beta),P_{n-2}.$
• Here, $\alpha+\beta$ and $\alpha \beta$ are the sum and product of the roots of the characteristic quadratic.
• Given specific values of $P_n$, one can solve for these two parameters and hence recover the quadratic.

Calculation:

Given:

• $P_{10}=123,;P_{9}=76,;P_{8}=47,;P_{1}=1.$

Let $s = \alpha+\beta$ and $p = \alpha\beta$.

Since $P_1 = \alpha+\beta$, we have:

$s = P_1 = 1.$

Next, use the recurrence at $n=10$:

$P_{10} = s,P_9 - p,P_8$

$\Rightarrow; 123 = 1\cdot76 - 47p$

$\Rightarrow; 47p = 76 - 123 = -47 ;\Rightarrow; p = -1.$

Therefore, the characteristic equation is:

$x^2 - s,x + p = 0 \Rightarrow x^2 - x - 1 = 0.$

We want the quadratic whose roots are $\tfrac{1}{\alpha}$ and $\tfrac{1}{\beta}$.

If $\alpha,\beta$ satisfy $x^2 - x - 1=0$, then:

$\tfrac{1}{\alpha} + \tfrac{1}{\beta} = \tfrac{\alpha+\beta}{\alpha\beta} = \tfrac{s}{p} = \tfrac{1}{-1} = -1$

$\tfrac{1}{\alpha\beta} = \tfrac{1}{p} = -1$

Hence, the required equation is:

$x^2 - \left(\tfrac{1}{\alpha} + \tfrac{1}{\beta}\right)x + \tfrac{1}{\alpha\beta} = 0 \Rightarrow x^2 + x - 1 = 0.$

∴ The correct answer is Option 2.

Calculation:

 

PS + PS' = 2 × 3 √2 

⇒ b² = a² (1 - e²) ⇒ 9 = 18( 1 - e²)

⇒ e = $\frac{1}{\sqrt2}$

Diretrix x = $\frac{a}{e} = \frac{3\sqrt2}{\frac{1}{\sqrt2}{}} = 6$

⇒ PS.PS' = $PS.PS' = | \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta - 6)|$

 = $\frac{1}{2}| 18 cos^2\theta - 36|$

⇒ (PS. PS')_max = 18 and (PS .PS')_min = 9

Sum = 27

Hence, the correct answer is Option 4.

Concept:

Coordinate Geometry - Area of Triangle using Vector Approach:

• To find the area of triangle PQR, we use the coordinates of points P, Q, and R.

• Points Q and R lie on the line given by the parametric equations: $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.

• Distance QR is given as 5 units.

• Point P is fixed at (0, 2, 3).

• Area of triangle PQR is given by $\frac{1}{2} \times | \overrightarrow{PQ} \times \overrightarrow{PR}|$

• We find the point M on the line such that vector $\overrightarrow{PM}$is perpendicular to the line.

• Using the condition of perpendicularity, we solve for the parameter$\lambda$

• Calculate the length PM which is the height of the triangle from P to line QR.

• Use QR = 5 and height PM to calculate the area.

 

Calculation:

Given,

$M(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$

Directions of PM:

$\implies 5\lambda - 3, 2\lambda + 1 - 2, 3\lambda - 4 - (-3) \implies 5\lambda - 3, 2\lambda - 1, 3\lambda - 1$

Directions of line L:

$5, 2, 3$

Since $\overrightarrow{PM}$ is perpendicular  $\overrightarrow{L}$,

$(5\lambda - 3) \times 5 + (2\lambda - 1) \times 2 + (3\lambda - 1) \times 3 = 0$

Simplifying,

$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 3 = 0$

$38\lambda - 20 = 0$

$\lambda = \frac{20}{38} = \frac{10}{19}$

Coordinates of M:

$M\left(5 \times \frac{10}{19} - 3, 2 \times \frac{10}{19} + 1, 3 \times \frac{10}{19} - 4 \right) = M\left(\frac{50 - 57}{19}, \frac{20 + 19}{19}, \frac{30 - 76}{19}\right) = M\left(-\frac{7}{19}, \frac{39}{19}, -\frac{46}{19}\right)$

Calculate length PM:

$PM = \sqrt{\left(-\frac{7}{19} - (-3)\right)^2 + \left(\frac{39}{19} - 2\right)^2 + \left(-\frac{46}{19} - (-3)\right)^2}$

$PM = \sqrt{\left(\frac{50}{19}\right)^2 + \left(\frac{1}{19}\right)^2 + \left(\frac{11}{19}\right)^2} = \sqrt{\frac{2500 + 1 + 121}{361}} = \frac{\sqrt{2622}}{19}$

Calculate area of triangle:

$\text{Area} = \frac{1}{2} \times QR \times PM = \frac{1}{2} \times 5 \times \frac{\sqrt{2622}}{19} = \frac{5 \sqrt{2622}}{38} = \frac{m}{n}$

Given relation involving m) and n:

$2m - 5 \sqrt{21} n = 0$

Hence, the correct answer is Option 2.

Concept:

Area of Tetrahedron from Areas of Perpendicular Triangles:

• In a tetrahedron where edges AB, AC, and AD are mutually perpendicular, the area of the face opposite to vertex A (i.e., triangle BCD) is related to the areas of the triangles ABC, ACD, and ADB.
• These three triangles form right-angled faces and their areas can be used to find the area of the face BCD using Pythagorean relation in three dimensions.

 

The relation for the area of triangle BCD is,

$\text{Ar}(\triangle BCD) = \sqrt{ \text{Ar}(\triangle ABC)^2 + \text{Ar}(\triangle ACD)^2 + \text{Ar}(\triangle ADB)^2 }$

 

Calculation:

Given,

$\text{Ar}(\triangle ABC) = 5, \quad \text{Ar}(\triangle ACD) = 6, \quad \text{Ar}(\triangle ADB) = 7$

⇒ $\text{Ar}(\triangle BCD) = \sqrt{5^2 + 6^2 + 7^2} = \sqrt{25 + 36 + 49} = \sqrt{110}$

∴ The area of triangle BCD is $\sqrt{110}$, hence the correct answer is Option 3.

A +  I = $\begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix}$

⇒A = $\begin{bmatrix} 0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1 \end{bmatrix}$

Given det(A)= −4.

det(A)= 0⋅(0. 1 − 0. 1)−a. (2. 1−0. a)+1. (2. 1−0. a)

= - a. (2)+1. (2) =−2a + 2

Set equal to −4−4-4 -4

⇒ 2a + 2 =  -4 ⟹ −2a = −6 ⟹a = 3:

Also

det((a+1) adj ((a−1)A)) =2^m3ⁿ

• For a 3 × 3 Matrix, det(adj)(B) = (det(B))²
• For a scalar k det(kB) = k³det(B)

Let’s plug in values:

⇒ det((a−1) A) = (a−1)³.det(A) = (3−1)³.(−4)=2³.(−4) = 8.(−4) = −32

⇒ det(adj(a - 1)A)) = (-32)² = 1024

⇒ (a +1)³ = (3 + 1)³ = 64

⇒ det((a+1)adj((a−1)A)) = 64⋅1024 = 65536

Thus, 65536 = 2¹⁶3⁰

⇒ Then m = 16,n = 0

⇒ m + n  = 16 + 0 = 16

Concept:

Focal Chord of Parabola and Circle Properties:

• The focal chord of the parabola $y^2 = 4ax$passes through points on the parabola making a certain angle with the x-axis.

• The circle with one diameter as the focal chord touches the y-axis at a certain point.

• The tangent of the given angle and the coordinates of the point P on the parabola can be used to find the parameter t.

• The equation of the circle with diameter PS and its intersection with the y-axis is used to find $\alpha$

 

Calculation:

Given,

The parabola: $y^2 = 4x$

Focal chord PQ makes an angle of $60^\circ$ with positive x-axis.

Using slope$m = \tan 60^\circ = \sqrt{3}$

$\tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies t = \sqrt{3}$

Coordinates of P: $(3, 2\sqrt{3})$

Circle with diameter PS and focus S:

$(x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0$

At x = 0,

$(-1)(-3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0 \implies (y - \sqrt{3})^2 = 0 \implies y = \sqrt{3} = \alpha$

Calculate $5\alpha^2:$

$5 \alpha^2 = 5 (\sqrt{3})^2 = 5 \times 3 = 15$

∴ The correct answer is Option 1.