We know that eccentricity, \(e^2 = 1 - \dfrac{b^2}{a^2}\)

\(\Rightarrow 1 - \dfrac{b^2}{a^2} = \dfrac{1}{9}\) or \(\dfrac{b^2}{a^2} = \dfrac{8}{9}\)

Also equation of directrix is given by \(x = \pm \dfrac{a}{e} = \pm 3a\). Given that one of the directrices is \(x = 9\), hence \(a = 3\) and \(b^2 = 8\).

Equation of the ellipse is \(\dfrac{x^2}{9} + \dfrac{y^2}{8} = 1\)

Suppose the tower is DE of height \(h\), where D point is on the ground, then

\(AD = \dfrac{h}{\tan 30} = h\sqrt{3}\)

\(BD = \dfrac{h}{\tan 45} = h\)

\(CD = \dfrac{h}{\tan 60} = \dfrac{h}{\sqrt{3}}\)

The ratio \(\dfrac{AB}{BC} = \dfrac{h\sqrt{3} - h}{h - \dfrac{h}{\sqrt{3}}} = \dfrac{\sqrt{3}}{1}\)

We know that, \(\dfrac{\sum X}{N} = \mu\) (mean) and \(\sigma^2 = \dfrac{1}{N}\sum X^2 - \mu^2\)

The \(n^{th}\) moment about a number \(m\) is defined as \(\dfrac{1}{N}\sum_{i=1}^{N}(X - m)^k\)

Hence first moment about 2 is \(\dfrac{1}{N}\sum(X - 2) = \mu - 2\)

Given that \(\mu - 2 = 1 \Rightarrow \mu = 3\)

Second moment about 2 is \(\dfrac{1}{N}\sum(X - 2)^2 = \sigma^2 + \mu^2 - 4\mu + 4 = \sigma^2 + 1\)

Given that second moment about 2 is 16, hence \(\sigma^2 = 15\)

The integration can be written as, \(\int \dfrac{(x^2 - 1)dx}{x^5\sqrt{2 - \dfrac{2}{x^2} + \dfrac{1}{x^4}}} = \int \dfrac{(\dfrac{1}{x^3} - \dfrac{1}{x^5})dx}{\sqrt{2 - \dfrac{2}{x^2} + \dfrac{1}{x^4}}}\)

Now put \(2 - \dfrac{2}{x^2} + \dfrac{1}{x^4} = t\)

\(\Rightarrow (\dfrac{4}{x^3} - \dfrac{4}{x^5})dx = dt\)

Or \((\dfrac{1}{x^3} - \dfrac{1}{x^5})dx = \dfrac{dt}{4}\)

The integration becomes, \(\int \dfrac{1}{4}\dfrac{dt}{\sqrt{t}} = \dfrac{1}{2}\sqrt{t} + c = \dfrac{1}{2}\sqrt{2 - \dfrac{2}{x^2} + \dfrac{1}{x^4}} + C\)

Given Integration is \(\int \limits_a^b |x - a|dx + \int \limits_a^b |x - b|dx\)

\(= \int \limits_a^b (x - a)dx + \int \limits_a^b -(x - b)dx = \dfrac{x^2}{2} - ax|_a^b + bx - \dfrac{x^2}{2}|_a^b\)

\(= \dfrac{b^2}{2} - ab - (\dfrac{a^2}{2} - a^2) + b^2 - \dfrac{b^2}{2} - ab + \dfrac{a^2}{2}\)

\(= a^2 + b^2 - 2ab = (b - a)^2\)

Let us find the value of \(A_{n+1}\),

\(\alpha^{n+1} + \beta^{n+1} = (\alpha + \beta)(\alpha^n + \beta^n) - \alpha\beta(\alpha^{n-1} + \beta^{n-1})\)

\(\Rightarrow A_{n+1} = 1 \times A_n + A_{n-1}\)

Hence \(\dfrac{A_n + A_{n-1}}{2} = \dfrac{A_{n+1}}{2}\)

The temperature function, \(T(x,y) = 4x^2 - 4xy + y^2 = (2x - y)^2\)

Given that the ant traverses the circle of radius 5 and center at origin, so we have to maximize the function \(2x - y\), that satisfies the condition \(x^2 + y^2 = 25\).

Let \(2x - y = k\), then \(k\) will be the maximum when the line \(2x - y = k\) touches the circle.

\(\Rightarrow k^2 = 25(1 + 2^2) = 125\) [using \(c^2 = a^2(1 + m^2)\)]

Given mean of 25 numbers is 38, also two numbers 25, 34 are misread as 23 and 36.

Hence corrected mean = \(\dfrac{[(25 \times 38) - (23 + 36)] + (25 + 34)}{25} = 38\)

Note that sum of 25, 34 is same as 23, 36, hence there will be no change in the mean.

Eccentricity of the hyperbola is \(\sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{81/25}{144/25}} = \dfrac{15}{12} = \dfrac{5}{4}\)

Hence its focus is \((ae, 0) = (\dfrac{12}{5} \times \dfrac{5}{4}, 0) = (3, 0)\)

Given that focus of the ellipse is also same, suppose eccentricity of the ellipse is \(e'\), then

\(5e' = 3 \Rightarrow e' = \dfrac{3}{5}\)

\(\Rightarrow \sqrt{1 - \dfrac{b^2}{25}} = \dfrac{3}{5} \Rightarrow b^2 = 16\)

Since angle of elevation from the three vertices is same as \(\alpha\), the tower must be equidistant from the three vertices, and the tower is at the circumcenter of the triangle.

Distance of the circumcenter from any vertex = \(h\cot\alpha = R\) (circum-radius)

We know that \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} = \dfrac{1}{2R}\)

\(\Rightarrow h\cot\alpha = \dfrac{a}{2\sin A}\) or \(h = \dfrac{a}{2}\tan\alpha\,\text{cosec}A\)

The first two lines can be written as \(2x - y + 2 = 0, 2x - y - 2 = 0\), distance between them is

\(a = \dfrac{2 - (-2)}{\sqrt{2^2 + 1^2}} = \dfrac{4}{\sqrt{5}}\)

The next two lines can be written as \(4x - 3y - 5 = 0, 4x - 3y + \dfrac{1}{2} = 0\), distance between them is

\(b = \dfrac{|-5 - \frac{1}{2}|}{\sqrt{4^2 + 3^2}} = \dfrac{11}{10}\)

Hence \(\dfrac{a}{b} = \dfrac{4/\sqrt{5}}{11/10} = \dfrac{40}{11\sqrt{5}}\), or \(11\sqrt{5}a = 40b\)

Since \(\text{cosec}^2\theta - \cot^2\theta = 1 \Rightarrow \text{cosec}\theta - \cot\theta = \dfrac{1}{\text{cosec}\theta + \cot\theta}\)

Hence \(\text{cosec}\theta - \cot\theta = \dfrac{1}{2}\)

As we are given that \(\text{cosec}\theta - \cot\theta = 2\),

Adding the two equations, we have \(2\text{cosec}\theta = 2 + \dfrac{1}{2} \Rightarrow \text{cosec}\theta = \dfrac{5}{4}\)

Given that \(f(x) = \log(x + \sqrt{x^2 + 1})\)

\(f(-x) = \log(-x + \sqrt{(-x)^2 + 1})\)

Adding both the functions, we have

\(f(x) + f(-x) = \log[(x + \sqrt{1 + x^2})(-x + \sqrt{1 + x^2})]\)

\(= \log(-x^2 + (1 + x^2)) = \log 1 = 0\)

\(\Rightarrow f(-x) = -f(x)\) and thus it is an odd function.

The given function is \(f(x) = \dfrac{10^x - 10^{-x}}{10^x + 10^{-x}}\), let's assume \(f(x) = y\) and solve for \(x\),

\(y = \dfrac{10^x - 10^{-x}}{10^x + 10^{-x}}\) or \(y = \dfrac{10^{2x} - 1}{10^{2x} + 1}\)

\(\Rightarrow (10^{2x} + 1)y = 10^{2x} - 1\)

\(\Rightarrow 10^{2x} = \dfrac{1 + y}{1 - y}\)

Take log, \(2x = \log\left(\dfrac{1 + y}{1 - y}\right)\) or \(x = \dfrac{1}{2}\log\left(\dfrac{1 + y}{1 - y}\right)\)

Replace \(y\) to \(x\) and \(x\) to \(f^{-1}(x)\)

\(f^{-1}(x) = \dfrac{1}{2}\log\left(\dfrac{1 + x}{1 - x}\right)\)

Expand the triple product,

\((a \cdot c)b - (b \cdot c)a = (a \cdot c)b - (a \cdot b)c\)

\(\Rightarrow (b \cdot c)a = (a \cdot b)c\)

As \(b \cdot c\) and \(a \cdot b\) are scalar quantities, hence \(a\) and \(c\) are collinear vectors.

In the equivalent Arithmetic Series,

\(\dfrac{1}{p} = a + (q - 1)d \Rightarrow 1 = ap + (q - 1)dp\)

\(\dfrac{1}{q} = a + (p - 1)d \Rightarrow 1 = aq + (p - 1)dq\)

\(\Rightarrow ap + (q - 1)dp = aq + (p - 1)dq\)

Or \(a(p - q) = (p - q)d \Rightarrow a = d\)

Putting \(a = d\) in the first equation, we get \(d = a = \dfrac{1}{pq}\)

\(pq^{th}\) term of AP is \(a + (pq - 1)d = d + (pq - 1)d = pqd = 1\)

The given series is: \(\sqrt{5}(\dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}, \ldots)\)

Hence \(\dfrac{1}{\sqrt{5}}(3, 4, 5, \ldots)\) are in AP. Suppose \(\dfrac{13}{\sqrt{5}}\) is the \(n^{th}\) term, then

\(\dfrac{13}{\sqrt{5}} = \dfrac{1}{\sqrt{5}}[3 + (n - 1) \times 1]\)

\(\Rightarrow n = 11\)

Given equation can be written as, \(\tan^{-1}\sqrt{x^2 + x} = \dfrac{\pi}{2} - \sin^{-1}\sqrt{x^2 + x + 1}\)

\(\Rightarrow \tan^{-1}\sqrt{x^2 + x} = \cos^{-1}\sqrt{x^2 + x + 1}\)

\(\Rightarrow \cos^{-1}\dfrac{1}{\sqrt{x^2 + x + 1}} = \cos^{-1}\sqrt{x^2 + x + 1}\)

as \(\tan^{-1}x = \cos^{-1}(\dfrac{1}{\sqrt{1 + x^2}})\)

\(\Rightarrow \dfrac{1}{\sqrt{x^2 + x + 1}} = \sqrt{x^2 + x + 1}\)

Hence \(x^2 + x + 1 = 1 \Rightarrow x = 0, -1\)

Domain of \(\cos^{-1}x\) is \(-1 \le x \le 1\)

Also \([x] \ne 0\), so \(x\) cannot lie in the interval \([0, 1)\)

Domain of the function is \([-1, 0) \cup \{1\}\)

Both the circles have the same centre \((0, 1)\) and their radii are \(1, 2\). It is given that point lies in the bigger circle, then probability that it has been taken from the smaller circle is:

\(\dfrac{\text{Area of the smaller circle}}{\text{Area of the bigger circle}} = \dfrac{\pi(1)^2}{\pi(2)^2} = \dfrac{1}{4}\)

Given that, \(f(x) = \begin{cases} 0, & x \le 0 \\ 2x, & x > 0 \end{cases}\)

Now we check continuity of the given function at \(x = 0\)

\(LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0} (0) = 0\),

\(RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0} (2x) = 2 \times 0 = 0\)

And \(f(0) = 0\), hence \(LHL = RHL = f(0) = 0\)

Given function is continuous at \(x = 0\)

Hence, given function is continuous in the interval \((-\infty, \infty)\)

It is a standard result that \(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\)

The given determinant is, \(\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2+x & 1 \\ 1 & 1 & 2+y \end{vmatrix}\)

Using, \(C_1 \to C_1 - C_2, C_2 \to C_2 - C_3\)

\(\begin{vmatrix} 0 & 0 & 1 \\ -(1+x) & (1+x) & 1 \\ 0 & -(1+y) & 2+y \end{vmatrix}\)

The value of the determinant is: \((1+x)(1+y)\), hence it is divisible by both \((1+x)\) and \((1+y)\)

We know that \((1 + \tan\theta)(1 + \tan(45 - \theta)) = 2\), hence \((1 + \tan 30)(1 + \tan 15) = 2\)

Let us represent the roots by \(\alpha, \beta\), then

\((1 + \alpha)(1 + \beta) = 2 \Rightarrow 1 + \alpha + \beta + \alpha\beta = 2\)

Or \(\alpha\beta + \alpha + \beta - 1 = 0\)

\(\Rightarrow q - p - 1 = 0\) or \(p - q = -1\)

Hence \(2 + p - q = 1\)

We know that \(a^{\log_a x} = x\)

\(3^{3 - \log_3 5} = 3^3 \cdot 3^{-\log_3 5} = 3^3 \cdot 3^{\log_3(\dfrac{1}{5})}\)

\(= 27 \cdot (\dfrac{1}{5}) = \dfrac{27}{5}\)

Taking Logarithm both the sides,

\(m\log x + n\log y = (m + n)\log(x + y)\)

\(\Rightarrow \dfrac{m}{x} + \dfrac{n}{y}\dfrac{dy}{dx} = \dfrac{m+n}{x+y}(1 + \dfrac{dy}{dx})\)

\(\Rightarrow \dfrac{m}{x} - \dfrac{m+n}{x+y} = (\dfrac{m+n}{x+y} - \dfrac{n}{y})\dfrac{dy}{dx}\)

\(\Rightarrow \dfrac{my - nx}{x(x+y)} = (\dfrac{my - nx}{y(x+y)})\dfrac{dy}{dx}\)

Hence \(\dfrac{dy}{dx} = \dfrac{y}{x}\)

Power set of A is \(P[A] = 2^m\)

Power set of B is \(P[B] = 2^n\)

\(P[A] - P[B] = 112\)

\(2^m - 2^n = 112\)

\(2^n\{2^{m-n} - 1\} = 16 \times 7\)

\(2^n\{2^{m-n} - 1\} = 2^4 \times \{8 - 1\}\)

\(2^n\{2^{m-n} - 1\} = 2^4\{2^3 - 1\}\)

\(n = 4\) and \(m - n = 3\) or \(m = 7\).

Hence third choice is not correct.

\(4\cos^2 x + 6\sin^2 x = 5\)

\(4\cos^2 x + 4\sin^2 x + 2\sin^2 x = 5\)

\(4 + 2\sin^2 x = 5\) or \(2\sin^2 x = 1\)

\(\sin^2 x = \dfrac{1}{2}\) or \(\sin^2 x = (\dfrac{1}{\sqrt{2}})^2\)

\(\sin^2 x = \sin^2 \dfrac{\pi}{4}\)

\(\Rightarrow x = n\pi \pm \dfrac{\pi}{4}\)

Volume of parallelepiped is \(= [abc] = 15\)

\(\begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix} = 15\)

\(\Rightarrow 2(\alpha^2 - 4) - 3(\alpha - 2) + 4(2 - \alpha) = 15\)

Or \(2\alpha^2 - 7\alpha - 9 = 0\)

\(\Rightarrow (\alpha + 1)(2\alpha - 9) = 0\)

Hence \(\alpha = -1, \dfrac{9}{2}\)

The given parabola is: \(4y^2 - 12x - 20y + 67 = 0\)

\(\Rightarrow 4(y^2 - 5y) = -12x - 67\)

\(\Rightarrow 4[(y - \dfrac{5}{2})^2 - \dfrac{25}{4}] = -12x - 67\)

\(\Rightarrow 4(y - \dfrac{5}{2})^2 = -12x - 42\)

\(\Rightarrow (y - \dfrac{5}{2})^2 = -3(x + \dfrac{7}{2})\)

Assume \(y - \dfrac{5}{2} = Y, x + \dfrac{7}{2} = X\), so the focus is \((-\dfrac{3}{4}, 0)\)

Hence \(x + \dfrac{7}{2} = -\dfrac{3}{4} \Rightarrow x = -\dfrac{17}{4}\) and \(y = \dfrac{5}{2}\)

Given that \(P(A \cap B) = P(A)P(B)\), hence events A and B are independent.

Using De Morgan's law

\(P(A \cup B)' = P(A' \cap B') = P(A')P(B')\)

This question can be solved using a set of numbers like 1, 2, 3 or 1, -2, 3.

\(n\sum a_i^2 = 3(1^2 + 2^2 + 3^2) = 42\)

\((\sum a_i)^2 = (1 + 2 + 3)^2 = 36\)

We see that \(n\sum a_i^2 \ge (\sum a_i)^2\)

To prove it algebraically, let us take three numbers a, b, c, we know that

\((a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0\)

\(\Rightarrow a^2 + b^2 + c^2 \ge ab + bc + ca\)

Or \(2(a^2 + b^2 + c^2) \ge 2(ab + bc + ca)\)

Adding \(a^2 + b^2 + c^2\) both the sides, we have

\(3(a^2 + b^2 + c^2) \ge a^2 + b^2 + c^2 + 2(ab + bc + ca)\)

\(\Rightarrow 3(a^2 + b^2 + c^2) \ge (a + b + c)^2\)

As \(\\text{cosec}^{-1}\dfrac{5}{3} = \cot^{-1}\dfrac{4}{3}\), the given sum is:

\(cot[\tan^{-1}\dfrac{3}{4} + \tan^{-1}\dfrac{2}{3}]\)

\(= cot[\tan^{-1}[\dfrac{3/4 + 2/3}{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}]]\)

\(= cot[\cot^{-1}[\dfrac{6}{17}]] = \dfrac{6}{17}\)

\(a_1, a_2, \ldots, a_n\) are in AP

Then, \(a_2 - a_1 = a_3 - a_2 = \ldots = a_n - a_{n-1} = d\)

\(\sin\,d(\dfrac{1}{\sin\,a_1 \cdot \sin\,a_2} + \dfrac{1}{\sin\,a_2 \cdot \sin\,a_3} + \ldots + \dfrac{1}{\sin\,a_{n-1} \cdot \sin\,a_n})\)

\(= (\dfrac{\sin\,d}{\sin\,a_1 \cdot \sin\,a_2} + \dfrac{\sin\,d}{\sin\,a_2 \cdot \sin\,a_3} + \ldots + \dfrac{\sin\,d}{\sin\,a_{n-1} \cdot \sin\,a_n})\)

\(= \dfrac{(\sin\,a_2\,\cos\,a_1 - \cos\,a_2\,\sin\,a_1)}{\sin\,a_1 \cdot \sin\,a_2} + \dfrac{\sin\,a_3\,\cos\,a_2 - \cos\,a_3\,\sin\,a_2}{\sin\,a_2 \cdot \sin\,a_3} + \ldots\)

\(= \cot\,a_1 - \cot\,a_2 + \cot\,a_2 - \cot\,a_3 + \ldots + \cot\,a_{n-1} - \cot\,a_n\)

\(= \cot\,a_1 - \cot\,a_n\)

It is given that

\(\tan\dfrac{A - B}{2} = \dfrac{1}{3}\tan\dfrac{A + B}{2} = \dfrac{1}{3}\cot\dfrac{C}{2}\) ....(1)

Using Napier's analogy, \(\tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2}\)

From equation (1),

\(\dfrac{a - b}{a + b}\cot\dfrac{C}{2} = \dfrac{1}{3}\cot\dfrac{C}{2} \Rightarrow \dfrac{a - b}{a + b} = \dfrac{1}{3}\)

Or \(a = 2b\) and the ratio \(a:b = 2:1\)

Let \(b = xi + yj + zk, a \cdot b = 14\)

\(\Rightarrow 2x + 2y + z = 14\)

Only first choice satisfies this condition.

We can solve the question in detail, using \(a \times b = 3i - j - 8k\)

\(a \times b = \begin{vmatrix} i & j & k \\ 2 & 2 & 1 \\ x & y & z \end{vmatrix} = (2z - y)i - (2z - x)j + (2y - 2x)k\)

Comparing this with the given product \(a \times b\), we have

\(2z - y = 3, 2z - x = -1, 2y - 2x = -8\)

Using equation (1), \(x = 5, y = 1, z = 2\)

We know that velocity, \(v = \dfrac{dx}{dt}\) and acceleration \(a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = \dfrac{dv}{dx} \cdot v\)

Given that \(a = v \cdot \dfrac{dv}{dx} = x - x^2\)

\(\Rightarrow \int v\,dv = \int (x - x^2)dx\)

\(\Rightarrow \dfrac{v^2}{2} = \dfrac{x^2}{2} - \dfrac{x^3}{3} + c\)

Since particle is at rest at origin, it means at \(x = 0, v = 0\), therefore \(c = 0\).

Again the particle comes to rest when \(\dfrac{x^2}{2} - \dfrac{x^3}{3} = 0 \Rightarrow x = \dfrac{3}{2}\)

If a parallelogram is made by the set of 2 parallel lines \(\ell_1, \ell_2\) and \(\ell_3, \ell_4\), then area of the parallelogram is given by the formula \(\dfrac{d_1 d_2}{\sin\theta}\), where \(d_1\) is the distance between the lines \(\ell_1, \ell_2\) and \(d_2\) is the distance between the lines \(\ell_3, \ell_4\) and \(\theta\) is the angle between the lines.

Here \(d_1 = \dfrac{1}{\sqrt{4^2 + 1^2}} = \dfrac{1}{\sqrt{17}}\), \(d_2 = \dfrac{1}{\sqrt{1^2 + 1^2}} = \dfrac{1}{\sqrt{2}}\)

\(\tan\theta = |\dfrac{4 - (-1)}{1 - 4}| = \dfrac{5}{3}\) or \(\sin\theta = \dfrac{5}{\sqrt{34}}\)

Area = \(\dfrac{\dfrac{1}{\sqrt{17}} \times \dfrac{1}{\sqrt{2}}}{\dfrac{5}{\sqrt{34}}} = \dfrac{1}{5}\)

Alternately, we can solve all the four intersection points and calculate the area of parallelogram.

Given that the vectors are coplanar, hence

\(\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0\)

Using \(C_1 = C_1 - C_2\)

\(\begin{vmatrix} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{vmatrix} = 0 \Rightarrow -1(ab - c^2) = 0\)

Hence c is the geometric mean of a and b.

Given that \(\cos^{-1}\dfrac{x}{2} + \cos^{-1}\dfrac{y}{3} = \phi\)

\(\Rightarrow \cos^{-1}(\dfrac{xy}{6} - \sqrt{1 - \dfrac{x^2}{4}}\sqrt{1 - \dfrac{y^2}{9}}) = \phi\)

\(\Rightarrow \dfrac{xy}{6} - \dfrac{\sqrt{(4 - x^2)(9 - y^2)}}{6} = \cos\phi\)

\(\Rightarrow (4 - x^2)(9 - y^2) = (6\cos\phi - xy)^2\)

\(\Rightarrow 9x^2 - 12xy\cos\phi + 4y^2 = 36\sin^2\phi\)

Plot the graph \(|x| + |y| = 2\), we get a square having vertices (2, 0), (-2, 0), (0, 2) and (0, -2).

Area of the square = 4 × Area of one triangle = 4 × 2 = 8

The function can be written as \(\lim_{n \to \infty} \dfrac{n^a \sum_{r=1}^{n}(\dfrac{r}{n})^a}{(n+1)^{a-1} n \sum_{r=1}^{n}(a + \dfrac{r}{n})} = \dfrac{1}{60}\)

\(\Rightarrow \lim_{n \to \infty} (\dfrac{n}{n+1})^{a-1} \dfrac{\sum_{r=1}^{n}(\dfrac{r}{n})^a}{\sum_{r=1}^{n}(a + \dfrac{r}{n})} = \lim_{n \to \infty} \dfrac{\sum_{r=1}^{n}(\dfrac{r}{n})^a}{\sum_{r=1}^{n}(a + \dfrac{r}{n})} = \dfrac{\int_0^1 x^a dx}{\int_0^1 (a + x)dx} = \dfrac{2}{(2a + 1)(a + 1)}\)

Given that \(\dfrac{2}{(2a + 1)(a + 1)} = \dfrac{1}{60}\), solving this we get \(a = 7, -\dfrac{17}{2}\)

Using the result \(\lim_{x \to 0} (1 + kx)^{1/x} = e^k\).

The value of the function \(f(x) = (1 + 2x)^{1/x}\) when \(x \to 0\), is: \(e^2\). Also given that \(f(0) = e^2\)

To check differentiability, let \(y = (1 + 2x)^{1/x} \Rightarrow \log y = \dfrac{1}{x}\log(1 + 2x) = \dfrac{1}{x}(2x - \dfrac{4x^2}{2} + ...) = 2 - 2x + ...\)

Hence \(y = e^{2 - 2x + ...} = e^2(1 - 2x + ...)\)

Right hand derivative = \(\dfrac{e^2(1 - 2h + ...) - e^2}{h} = -2e^2\). Similarly LHD = \(-2e^2\). Therefore differentiable at \(x = 0\)

The fourth choice is not correct as the limit \(\displaystyle \lim_{x \to 0^+} \dfrac{\cos x}{1 + 2x} = 1\)

Given that \([\hat{a}\hat{b}\hat{c}] = 7\)

\(\Rightarrow \begin{vmatrix} \lambda & 1 & -2 \\ 1 & \lambda & -2 \\ 1 & 1 & 1 \end{vmatrix} = 7\)

\(\Rightarrow \lambda(\lambda + 2) - 1(1 + 2) - 2(1 - \lambda) = 7\) or \(\lambda^2 + 4\lambda - 12 = 0\)

\(\Rightarrow (\lambda - 2)(\lambda + 6) = 0 \Rightarrow \lambda = 2, -6\)

We arrange the data in ascending order: 23, 29, 32, 49, 118 (n = 5 observations)

\(P_{50} = 50\dfrac{(n + 1)}{100} = \dfrac{50}{100} \times 6 = 3^{rd}\) observation, so \(P_{50} = 32\)

\(P_{10} = \dfrac{10}{100} \times 6 = 0.6 = 1^{st}\) observation, so \(P_{10} = 23\)

Population who like tea \(n(T) = \dfrac{60}{100} \times 200 = 120\), Population who like coffee \(n(C) = \dfrac{72}{100} \times 200 = 144\)

\(n(T \cup C) = n(T) + n(C) - n(T \cap C) = 120 + 144 - n(T \cap C)\), so \(n(T \cap C) = 264 - n(T \cup C)\)

As \(144 \le n(T \cup C) \le 200\), maximum and minimum values of \(n(T \cap C) = 120, 64\) and difference = 56.

Differentiating first equation: \(\dfrac{dy}{dx} = \dfrac{-b^2}{a^2}\dfrac{x}{y}\). Differentiating second: \(\dfrac{dy}{dx} = \dfrac{x}{y}\)

Since curves cut at right angles: \(-(\dfrac{b^2}{a^2}\dfrac{x}{y})(\dfrac{x}{y}) = -1 \Rightarrow \dfrac{x}{y} = \dfrac{a}{b}\)

Put \(x = ak, y = bk\) in both equations and eliminate k, we get \(a^2 - b^2 = 2c^2\)

Total no. of ways = \(^5C_4 \times 4! = 120\)

For divisibility by 3, sum of digits must be multiple of 3. Only case (1, 2, 4, 5) works.

Favourable cases = 4! = 24. Probability = \(\dfrac{24}{120} = \dfrac{1}{5}\)

Since mid-point A is (4, 5), line intercepts at x axis at (8, 0) and y axis at (0, 10).

Hence equation: \(\dfrac{x}{8} + \dfrac{y}{10} = 1 \Rightarrow 5x + 4y = 40\)