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Calculation:

Given,

Number, n = 50

Prime number, p = 3

The exponent of 3 in 50! is given by,

$B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{3^2} \right\rfloor + \left\lfloor \frac{50}{3^3} \right\rfloor + \left\lfloor \frac{50}{3^4} \right\rfloor$

$B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor$

$B = 16 + 5 + 1 + 0 = 22$

∴ The maximum value of n is 22.

Concept:

Hyperbola parameters:

• Given one focus of hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is at $(\sqrt{10}, 0)$.
• The corresponding directrix is $x = \frac{9}{\sqrt{10}}$.
• Recall the relationships: $ae = c$ (distance of focus from origin), $\frac{a}{e}$ is directrix, and $c^2 = a^2 + b^2$.
• The length of the latus rectum is $l = \frac{2b^2}{a}$.

Calculation:

Given:

$ae = \sqrt{10}$ and $\frac{a}{e} = \frac{9}{\sqrt{10}}$

Then,

$a^2 = 9$ and $e = \frac{\sqrt{10}}{3}$

Now,

$(ae)^2 = a^2 + b^2 \implies 10 = 9 + b^2 \implies b^2 = 1$

$l = \frac{2b^2}{a} = \frac{2 \times 1}{3} = \frac{2}{3}$

Therefore,

$9(e^2 + l) = 9 \left( \left( \frac{\sqrt{10}}{3} \right)^2 + \frac{2}{3} \right) = 9 \left( \frac{10}{9} + \frac{2}{3} \right) = 10 + 6 = 16$

Hence, the correct answer is Option 3.

Calculation: 

11111 2 2 2 0 0 

No. of sequences $= \frac{10!}{5!3!2!} = 2520$

Note: The Sequence can start with 0.

Hence, the Correct answer is Option 3.

Calculation:

Given,

We are given the equation:

$(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$

Taking terms together:

$f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\cos x \sin y + \sin x \cos y)$

Using identities:

$f(2x+2y)\sin(x - y) = f(2x-2y)\sin(x + y)$

Rewriting,

$\frac{f(2x+2y)}{\sin(x + y)} = \frac{f(2x-2y)}{\sin(x - y)}$

Let,

$2x + 2y = m$ and $2x - 2y = n$

Then,

$\frac{f(m)}{\sin\left(\frac{m}{2}\right)} = \frac{f(n)}{\sin\left(\frac{n}{2}\right)} = K$ (constant)

$\Rightarrow f(m) = K \sin\left(\frac{m}{2}\right)$

$\therefore f(x) = K \sin\left(\frac{x}{2}\right)$

Now,

$f'(x) = \frac{K}{2} \cos\left(\frac{x}{2}\right)$

Putting $x = 0$:

$\frac{K}{2} = \frac{1}{2} \Rightarrow K = 1$

So,

$f'(x) = \frac{1}{2} \cos\left(\frac{x}{2}\right)$

$f''(x) = -\frac{1}{4} \sin\left(\frac{x}{2}\right)$

Now,

$24 f''\left(\frac{5\pi}{3}\right) = 24 \left( -\frac{1}{4} \sin\left(\frac{5\pi}{6}\right) \right)$

$= -6 \times \frac{1}{2} = -3$

∴ The correct answer is Option 2.

Solution:

$|A| = 0$

$\alpha \beta - (-1)(6) = 0$

$\alpha \beta + 6 = 0$

$\alpha \beta = -6$

$\alpha + \beta = 1$

$\implies \alpha = 3, \beta = -2$

$A = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$

$A^2 = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix}$

$\therefore A^2 = A$

$A = A^2 = A^3 = A^4 = A^5 = \cdots$

$(I + A)^8$

$= I + {}^8C_1 A + {}^8C_2 A^2 + \cdots + {}^8C_8 A^8$

$= I + A ({}^8C_1 + {}^8C_2 + \cdots + {}^8C_8)$

$= I + A (2^8 - 1)$

$= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} (256 - 1)$

$= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \ 6 & -2 \end{bmatrix} 255$

$= \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \ 1530 & -510 \end{bmatrix}$

$= \begin{bmatrix} 766 & -255 \ 1530 & -509 \end{bmatrix}$

Hence, the Correct answer is Option 4.

Calculation:

$\left(\frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x-1}{x-x^{1/2}}\right)^{10}, ; x>1$

Simplify the terms inside the bracket:

$\frac{x+1}{x^{2/3}+1-x^{1/3}} = x^{1/3}+1$

$\frac{x-1}{x-x^{1/2}} = 1 + x^{-1/2}$

Subtracting:

$(x^{1/3}+1)-(1+x^{-1/2}) = x^{1/3}-x^{-1/2}$

So the expression reduces to

$(x^{1/3}-x^{-1/2})^{10}$

General term:

$T_{r+1} = \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^{r}$

$= \binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac{r}{2}}$

For the term independent of x:

$\frac{10-r}{3}-\frac{r}{2}=0$

$2(10-r)-3r=0 \Rightarrow 20-5r=0 \Rightarrow r=4$

Required term:

$T_5=\binom{10}{4}(-1)^4=\binom{10}{4}=210$

The correct answer is 210.

Solution:

$2\sqrt{2} \cos^2 \theta + 2 \cos \theta - \sqrt{6} \cos \theta - \sqrt{3} = 0$

$\ (2 \cos \theta - \sqrt{3})(\sqrt{2} \cos \theta + 1) = 0$

$\ \cos \theta = \frac{\sqrt{3}}{2}, -\frac{1}{\sqrt{2}}$

Number of solutions = 8.

Hence, the correct answer is Option 3.

Concept:

Arithmetic Progression (A.P.):

• An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant, called the common difference (d).
• The sum of first n terms of an A.P. is given by $S_n = \frac{n}{2} [2a + (n-1)d]$, where a is the first term.

 

Calculation:

Sum of first 12 odd terms in A.P. is

 

$S_{12} = \frac{12}{2} [2a + (12-1) \times 2d] = 6[2a + 22d]$

Equate to the given sum,

$6[2a + 22d] = -\frac{72}{5} a \Rightarrow 12a + 132d = -\frac{72}{5} a$

Multiply both sides by 5,

$60a + 660d = -72a \Rightarrow 132a + 660d = 0 \Rightarrow 132a + 132 \times 5 d = 0 \Rightarrow a = -5d$

Now, sum of first n terms $S_n$ = 0

$\frac{n}{2} (2a + (n-1)d) = 0 \Rightarrow \frac{n}{2} (2(-5d) + (n-1)d) = 0$

$\Rightarrow \frac{n}{2} (-10d + nd - d) = 0 \Rightarrow \frac{n}{2} (nd - 11d) = 0 \Rightarrow \frac{nd(n-11)}{2} = 0$

Since d ≠ 0 and n ≠ 0  ,

$n - 11 = 0 \Rightarrow n = 11$

Hence, the correct answer is Option 1.

Concept:

Finding local maxima and minima of a cubic function:

• To find local maxima and minima of a cubic function, first differentiate the function and find the critical points by setting the derivative equal to zero.
• The critical points correspond to values of xx

where the function attains local maxima or minima.

• Use given conditions on the critical points to find the parameters and evaluate the function at required points.

 

Calculation:

Given,

$f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1$

First derivative,

$f'(x) = 6x^2 - 18ax + 12a^2$

Set $f'(x) = 0$ to find critical points:

$6x^2 - 18ax + 12a^2 = 0$

Dividing by 6:

$x^2 - 3ax + 2a^2 = 0$

Let the roots be $p$ and $q$. Then, by Vieta's formulas,

• $p + q = 3a$
• $pq = 2a^2$

Given condition,

$p^2 = q$

Substitute $q = p^2$:

$p + p^2 = 3a$

$p \cdot p^2 = p^3 = 2a^2$

From sum,

$a = \frac{p + p^2}{3}$

From product,

$a^2 = \frac{p^3}{2}$

Squaring the sum and equating to product:

$\left( \frac{p + p^2}{3} \right)^2 = \frac{p^3}{2}$

Solving for $p$ gives $p = 2$.

Then,

$a = \frac{2 + 4}{3} = 2$

Finally, evaluate $f(3)$:

$f(3) = 2(3)^3 - 9(2)(3)^2 + 12(2)^2 (3) + 1$

$= 54 - 162 + 144 + 1 = 37$

∴ The correct answer is Option 4.

Calculation:

$\frac{2 + k^2 z}{k + \bar{z}} = kz$

$|z|^2 = 1$

$k = 2$

Point P = (2, 4), center of circle C = (1, 2) 

Equation of a circle: $(x - 1)^2 + (y - 2)^2 = 1$

Maximum distance from point Pto the circle is given by:

$OP + r = \sqrt{(2 - 1)^2 + (4 - 2)^2} + 1 = \sqrt{1 + 4} + 1 = \sqrt{5} + 1$

Hence, the correct answer is Option 1.