KCET 2012 Mathematics Question Paper with Answer and Solution

(D) Given: $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$
Using the property $\log_{a} m - \log_{a} n = \log_{a} (\frac{m}{n})$:
$\log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=2$
Converting to exponential form:
$\frac{9^{x-1}+7}{3^{x-1}+1}=2^{2}=4$
Let $y = 3^{x-1}$. Then $9^{x-1} = (3^{2})^{x-1} = (3^{x-1})^{2} = y^{2}$.
Substituting $y$ into the equation:
$\frac{y^{2}+7}{y+1}=4$
$y^{2}+7=4(y+1)$
$y^{2}-4y+3=0$
$(y-3)(y-1)=0$
So,$y=3$ or $y=1$.
Case $1$: $3^{x-1}=3^{1}$ $\Rightarrow x-1=1$ $\Rightarrow x=2$.
Case $2$: $3^{x-1}=3^{0}$ $\Rightarrow x-1=0$ $\Rightarrow x=1$.
Thus,the values of $x$ are $1, 2$.

(B) To find the greatest common divisor $(24, 92)$,we use the Euclidean algorithm:
$92 = 3 \times 24 + 20$
$24 = 1 \times 20 + 4$
$20 = 5 \times 4 + 0$
Thus,$(24, 92) = 4$.
Now,express $4$ as a linear combination of $24$ and $92$:
$4 = 24 - 1 \times 20$
Substitute $20 = 92 - 3 \times 24$:
$4 = 24 - 1 \times (92 - 3 \times 24)$
$4 = 24 - 92 + 3 \times 24$
$4 = 4 \times 24 - 1 \times 92$
Comparing this with $24m + 92n = 4$,we get $m = 4$ and $n = -1$.
Therefore,$(m, n) = (4, -1)$.

(D) Given the cubic equation $x^{3}+4x+2=0$.
Let the roots be $\alpha, \beta, \gamma$.
From the properties of roots of a cubic equation $ax^{3}+bx^{2}+cx+d=0$:
Sum of roots $\Sigma \alpha = -\frac{b}{a} = 0$.
Sum of roots taken two at a time $\Sigma \alpha \beta = \frac{c}{a} = 4$.
Product of roots $\alpha \beta \gamma = -\frac{d}{a} = -2$.
We know the identity $\alpha^{3}+\beta^{3}+\gamma^{3}-3 \alpha \beta \gamma = (\alpha+\beta+\gamma)(\alpha^{2}+\beta^{2}+\gamma^{2}-\alpha \beta-\beta \gamma-\gamma \alpha)$.
Since $\Sigma \alpha = 0$,the right side becomes $0$.
Therefore,$\alpha^{3}+\beta^{3}+\gamma^{3} = 3 \alpha \beta \gamma$.
Substituting the value of the product of roots:
$\alpha^{3}+\beta^{3}+\gamma^{3} = 3(-2) = -6$.

(A) Since $(x-1)$ is a factor of $x^{5}-4 x^{3}+2 x^{2}-3 x+k=0$,by the factor theorem,$x=1$ must satisfy the given equation.
Substituting $x=1$ into the equation:
$(1)^{5}-4(1)^{3}+2(1)^{2}-3(1)+k=0$
$1-4+2-3+k=0$
$-4+k=0$
$k=4$

(B) Let $z = x + iy$.
Then $z^{2} = x^{2} - y^{2} + 2ixy$.
Given $z^{2} + \overline{z} = 0$,we have $(x^{2} - y^{2} + 2ixy) + (x - iy) = 0$.
Grouping real and imaginary parts: $(x^{2} + x - y^{2}) + i(2xy - y) = 0$.
Equating real and imaginary parts to zero:
$x^{2} + x - y^{2} = 0$ $(i)$
$y(2x - 1) = 0$ (ii)
From (ii),either $y = 0$ or $x = 1/2$.
Case $1$: If $y = 0$,then $x^{2} + x = 0 \Rightarrow x(x + 1) = 0$,so $x = 0$ or $x = -1$. This gives solutions $z = 0$ and $z = -1$.
Case $2$: If $x = 1/2$,then $(1/2)^{2} + 1/2 - y^{2} = 0$ $\Rightarrow 1/4 + 1/2 = y^{2}$ $\Rightarrow y^{2} = 3/4$ $\Rightarrow y = \pm \sqrt{3}/2$. This gives solutions $z = 1/2 + i\sqrt{3}/2$ and $z = 1/2 - i\sqrt{3}/2$.
Thus,there are $4$ solutions in total.

(B) Given that the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$.
Let $z = (x + iy)(1 - 2i)$.
Then $\bar{z} = 1 + i$.
Taking the conjugate on both sides of $z = (x + iy)(1 - 2i)$,we get $\bar{z} = \overline{(x + iy)(1 - 2i)} = \overline{(x + iy)} \cdot \overline{(1 - 2i)}$.
Since $\bar{z} = 1 + i$,we have $(x - iy)(1 + 2i) = 1 + i$.
Alternatively,we can write $z = \overline{1 + i} = 1 - i$.
Thus,$(x + iy)(1 - 2i) = 1 - i$.
Dividing both sides by $(1 - 2i)$,we get $x + iy = \frac{1 - i}{1 - 2i}$.
This matches Option $(B)$.

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.
Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$.
Substituting $1 = \beta \bar{\beta}$ in the denominator,we get:
$\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| = \left|\frac{\beta-\alpha}{\beta(\bar{\beta}-\bar{\alpha})}\right|$.
Using the property of modulus $|z_1/z_2| = |z_1|/|z_2|$ and $|z_1 z_2| = |z_1||z_2|$,we have:
$= \frac{|\beta-\alpha|}{ |\beta| |\bar{\beta}-\bar{\alpha}| }$.
Since $|\beta|=1$ and $|\bar{z}| = |z|$,we know $|\bar{\beta}-\bar{\alpha}| = |\overline{\beta-\alpha}| = |\beta-\alpha|$.
Therefore,the expression becomes $\frac{|\beta-\alpha|}{1 \cdot |\beta-\alpha|} = 1$.

(C) Given,$\alpha^{2}-\alpha+1=0$.
Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3}=-1$.
Therefore,$\alpha^{6}=1$.
We need to find $\alpha^{2011}$.
Since $2011 = 6 \times 335 + 1$,we have $\alpha^{2011} = (\alpha^{6})^{335} \times \alpha^{1} = (1)^{335} \times \alpha = \alpha$.
Thus,$\alpha^{2011} = \alpha$.

(D) The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is given by the general term formula $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For $r = 12$,we have:
$T_{13} = {}^{n}C_{12} (x^{2})^{n-12} (\frac{2}{x})^{12}$
$T_{13} = {}^{n}C_{12} x^{2n-24} \cdot \frac{2^{12}}{x^{12}}$
$T_{13} = {}^{n}C_{12} \cdot 2^{12} \cdot x^{2n-24-12}$
$T_{13} = {}^{n}C_{12} \cdot 2^{12} \cdot x^{2n-36}$
Since the term is independent of $x$,the exponent of $x$ must be $0$:
$2n - 36 = 0$ $\Rightarrow 2n = 36$ $\Rightarrow n = 18$.
The divisors of $n = 18$ are $1, 2, 3, 6, 9, 18$.
The sum of the divisors is $1 + 2 + 3 + 6 + 9 + 18 = 39$.

(A) To find the last digit of $7^{886}$,we observe the pattern of the last digits of powers of $7$:
$7^{1} = 7$
$7^{2} = 49$ (last digit $9$)
$7^{3} = 343$ (last digit $3$)
$7^{4} = 2401$ (last digit $1$)
The cycle of last digits is $(7, 9, 3, 1)$ with a period of $4$.
We divide the exponent $886$ by $4$:
$886 = 4 \times 221 + 2$
Thus,$7^{886} = (7^{4})^{221} \times 7^{2}$.
The last digit of $(7^{4})^{221}$ is $1^{221} = 1$.
The last digit of $7^{2}$ is $9$.
Therefore,the last digit of $7^{886}$ is $1 \times 9 = 9$.

(C) We know that the binomial expansion is given by $(1+x)^{n} = { }^{n} C_{0} + { }^{n} C_{1} x + { }^{n} C_{2} x^{2} + \ldots + { }^{n} C_{n} x^{n}$.
Putting $x=1$ and $n=10$,we get:
$2^{10} = { }^{10} C_{0} + { }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9} + { }^{10} C_{10}$.
Since ${ }^{10} C_{0} = 1$ and ${ }^{10} C_{10} = 1$,the equation becomes:
$2^{10} = 1 + ({ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9}) + 1$.
$2^{10} = 2 + ({ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9})$.
Therefore,${ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9} = 2^{10} - 2$.

(B) We have the expression $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$.
Using the identity $\cos \theta = \sin(90^{\circ}-\theta)$ and $\sin \theta = \cos(90^{\circ}-\theta)$:
$\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}}$
Using sum-to-product formulas $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$ and $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$= \frac{2 \sin(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}{2 \cos(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}$
$= \frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}$
$= \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \tan 60^{\circ} = \sqrt{3}$.

(B) Let $z_1 = e^{i\alpha} = \cos \alpha + i \sin \alpha$,$z_2 = e^{i\beta} = \cos \beta + i \sin \beta$,and $z_3 = e^{i\gamma} = \cos \gamma + i \sin \gamma$.
Given that $z_1 + 2z_2 + 3z_3 = (\cos \alpha + 2 \cos \beta + 3 \cos \gamma) + i(\sin \alpha + 2 \sin \beta + 3 \sin \gamma) = 0 + i(0) = 0$.
Let $x = z_1$,$y = 2z_2$,and $z = 3z_3$. Then $x + y + z = 0$.
Using the identity if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Thus,$z_1^3 + (2z_2)^3 + (3z_3)^3 = 3(z_1)(2z_2)(3z_3) = 18z_1 z_2 z_3$.
Substituting the exponential forms: $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i(\alpha + \beta + \gamma)}$.
Since $\alpha + \beta + \gamma = \pi$,we have $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i\pi} = 18(\cos \pi + i \sin \pi) = -18 + 0i$.
Equating the imaginary parts,we get $\sin 3 \alpha + 8 \sin 3 \beta + 27 \sin 3 \gamma = 0$.

(D) Given,$\sin 2x = 4 \cos x$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have:
$2 \sin x \cos x = 4 \cos x$
$2 \sin x \cos x - 4 \cos x = 0$
$2 \cos x (\sin x - 2) = 0$
This implies either $\cos x = 0$ or $\sin x = 2$.
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ has no real solution.
Thus,$\cos x = 0$.
The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = n \pi + \frac{\pi}{2}$ or $x = 2n \pi \pm \frac{\pi}{2}$ for $n \in Z$.

(C) Let the slopes of the lines $ax^{2}+2hxy+by^{2}=0$ be $m_{1}$ and $m_{2}$.
According to the problem,$m_{1} = nm_{2}$.
We know that $m_{1}+m_{2} = -\frac{2h}{b}$ and $m_{1}m_{2} = \frac{a}{b}$.
Substituting $m_{1} = nm_{2}$ into the sum and product equations:
$m_{2}(n+1) = -\frac{2h}{b} \implies m_{2} = -\frac{2h}{b(n+1)}$.
$nm_{2}^{2} = \frac{a}{b}$.
Substituting the value of $m_{2}$ into the product equation:
$n \left( -\frac{2h}{b(n+1)} \right)^{2} = \frac{a}{b}$.
$n \left( \frac{4h^{2}}{b^{2}(n+1)^{2}} \right) = \frac{a}{b}$.
$4nh^{2} = ab(n+1)^{2}$.

(D) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Given mid-points are $F(11, 9)$,$E(2, 1)$,and $D(2, -1)$.
Since the centroid of a triangle formed by joining the mid-points of the sides of a triangle is the same as the centroid of the original triangle,we can directly calculate it.
The centroid $(G)$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Let the mid-points be $(x_a, y_a) = (11, 9)$,$(x_b, y_b) = (2, 1)$,and $(x_c, y_c) = (2, -1)$.
The centroid of the triangle formed by these mid-points is $\left(\frac{11+2+2}{3}, \frac{9+1-1}{3}\right) = \left(\frac{15}{3}, \frac{9}{3}\right) = (5, 3)$.
Thus,the centroid of the original triangle is $(5, 3)$.

(C) Let the reflection of the point $P(1, 1)$ be $Q(h, k)$.
The line of reflection is $x + y = 0$.
The mid-point of $PQ$ is $\left(\frac{h+1}{2}, \frac{k+1}{2}\right)$,which must lie on the line $x + y = 0$.
$\frac{h+1}{2} + \frac{k+1}{2} = 0 \Rightarrow h + k + 2 = 0 \quad \dots(i)$
The line $PQ$ is perpendicular to the line $x + y = 0$ (which has slope $-1$).
The slope of $PQ$ is $\frac{k-1}{h-1}$.
Since $PQ$ is perpendicular to $x + y = 0$,the product of their slopes is $-1$.
$\left(\frac{k-1}{h-1}\right) \times (-1) = -1$ $\Rightarrow \frac{k-1}{h-1} = 1$ $\Rightarrow k - 1 = h - 1$ $\Rightarrow k = h$.
Substituting $k = h$ into equation $(i)$:
$h + h + 2 = 0$ $\Rightarrow 2h = -2$ $\Rightarrow h = -1$.
Since $k = h$,we have $k = -1$.
Thus,the reflection of the point $(1, 1)$ along the line $y = -x$ is $(-1, -1)$.

(B) The equation of the line with slope $m = -1$ and $y$-intercept $c = 1$ is $y = -x + 1$,which simplifies to $x + y - 1 = 0$.
Let the circle have radius $r$. Since the circle touches both coordinate axes,its center must be $(\pm r, \pm r)$.
The distance from the center $(\pm r, \pm r)$ to the line $x + y - 1 = 0$ must be equal to the radius $r$.
Using the distance formula: $\frac{|\pm r \pm r - 1|}{\sqrt{1^2 + 1^2}} = r$.
Case $1$: Center $(r, r)$. Then $\frac{|2r - 1|}{\sqrt{2}} = r \implies |2r - 1| = r\sqrt{2}$.
$2r - 1 = r\sqrt{2} \implies r(2 - \sqrt{2}) = 1 \implies r = \frac{1}{2 - \sqrt{2}} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.
$2r - 1 = -r\sqrt{2} \implies r(2 + \sqrt{2}) = 1 \implies r = \frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$.
Both values of $r$ are positive,giving two circles in the first quadrant.
Case $2$: Center $(-r, r)$. Then $\frac{|-r + r - 1|}{\sqrt{2}} = r \implies \frac{|-1|}{\sqrt{2}} = r \implies r = \frac{1}{\sqrt{2}}$.
This gives one circle in the second quadrant.
Case $3$: Center $(r, -r)$. Then $\frac{|r - r - 1|}{\sqrt{2}} = r \implies \frac{|-1|}{\sqrt{2}} = r \implies r = \frac{1}{\sqrt{2}}$.
This gives one circle in the fourth quadrant.
Case $4$: Center $(-r, -r)$. Then $\frac{|-2r - 1|}{\sqrt{2}} = r \implies |2r + 1| = r\sqrt{2}$. Since $r > 0$,$2r + 1 > r\sqrt{2}$,so no solution for $r > 0$.
Thus,there are $2 + 1 + 1 = 4$ such circles.

(A) Given the circle equation: $x^{2} + y^{2} - 4x - 2y - 20 = 0$.
Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -2, f = -1, c = -20$.
Centre of the circle is $(-g, -f) = (2, 1)$.
Radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(-2)^{2} + (-1)^{2} - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$.
Let $P$ be the point $(10, 7)$ and $C$ be the centre $(2, 1)$.
The distance $d$ between $P$ and $C$ is $\sqrt{(10 - 2)^{2} + (7 - 1)^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$.
The least distance of the point from the circle is $d - r = 10 - 5 = 5$.
The greatest distance of the point from the circle is $d + r = 10 + 5 = 15$.
Thus,the least and greatest distances are $5$ and $15$ respectively.

(C) Given the circle equation: $x^2 + y^2 = 16x$
Rearranging the terms: $x^2 - 16x + y^2 = 0$
Completing the square: $(x - 8)^2 + y^2 = 64$
Thus,the center is $(8, 0)$ and the radius $r = 8$.
Since the line $3x + 4y - k = 0$ touches the circle,the perpendicular distance from the center $(8, 0)$ to the line must equal the radius $r = 8$.
Using the distance formula: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$
$8 = \frac{|3(8) + 4(0) - k|}{\sqrt{3^2 + 4^2}}$
$8 = \frac{|24 - k|}{\sqrt{25}}$
$8 = \frac{|24 - k|}{5}$
$|24 - k| = 40$
Case $1$: $24 - k = 40 \Rightarrow k = -16$
Case $2$: $24 - k = -40 \Rightarrow k = 64$
Therefore,the values of $k$ are $-16$ and $64$.

(A) Any line passing through the point $(-5, -4)$ with slope $m$ is given by $y+4 = m(x+5)$,which simplifies to $mx - y + (5m - 4) = 0$.
For the circle $x^{2}+y^{2}+4x+6y+8=0$,the center is $(-2, -3)$ and the radius $r = \sqrt{(-2)^{2} + (-3)^{2} - 8} = \sqrt{4+9-8} = \sqrt{5}$.
Since the line is a tangent,the perpendicular distance from the center $(-2, -3)$ to the line must equal the radius:
$\frac{|m(-2) - (-3) + (5m - 4)|}{\sqrt{m^{2} + (-1)^{2}}} = \sqrt{5}$
$\frac{|3m - 1|}{\sqrt{m^{2} + 1}} = \sqrt{5}$
Squaring both sides: $(3m - 1)^{2} = 5(m^{2} + 1)$
$9m^{2} - 6m + 1 = 5m^{2} + 5$
$4m^{2} - 6m - 4 = 0$
$2m^{2} - 3m - 2 = 0$
$(2m + 1)(m - 2) = 0$
So,$m = 2$ or $m = -\frac{1}{2}$.
For $m = 2$: $2x - y + (5(2) - 4) = 0 \Rightarrow 2x - y + 6 = 0$.
For $m = -\frac{1}{2}$: $-\frac{1}{2}x - y + (5(-\frac{1}{2}) - 4) = 0$ $\Rightarrow -x - 2y - 5 - 8 = 0$ $\Rightarrow x + 2y + 13 = 0$.

(C) Given the parametric equations:
$x = t^{2} + 2$
$y = 2t$
From the second equation,we have $t = \frac{y}{2}$.
Substituting this value of $t$ into the first equation:
$x = (\frac{y}{2})^{2} + 2$
$x = \frac{y^{2}}{4} + 2$
$x - 2 = \frac{y^{2}}{4}$
$y^{2} = 4(x - 2)$
Thus,the correct option is $C$.

(D) The equation of a pair of tangents from $(\alpha, \beta)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by the condition $SS_{1}=T^{2}$.
The tangents are perpendicular if the sum of the coefficients of $x^{2}$ and $y^{2}$ in the combined equation is zero.
Expanding the condition,we get:
$\left(\frac{1}{a^{2}}\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) - \frac{\alpha^{2}}{a^{4}} + \left(\frac{1}{b^{2}}\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) - \frac{\beta^{2}}{b^{4}} = 0$
Simplifying this expression:
$\frac{\beta^{2}}{a^{2}b^{2}} - \frac{1}{a^{2}} + \frac{\alpha^{2}}{a^{2}b^{2}} - \frac{1}{b^{2}} = 0$
$\alpha^{2} + \beta^{2} = a^{2}b^{2}\left(\frac{1}{a^{2}} + \frac{1}{b^{2}}\right) = a^{2} + b^{2}$
Thus,the locus of the point $(\alpha, \beta)$ is $x^{2} + y^{2} = a^{2} + b^{2}$,which is known as the director circle.

(A) Given,$x-y=1$ $(i)$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ (ii).
Substitute $y=x-1$ from $(i)$ into (ii):
$\frac{x^{2}}{4}-\frac{(x-1)^{2}}{3}=1$
Multiply by $12$:
$3x^{2}-4(x-1)^{2}=12$
$3x^{2}-4(x^{2}-2x+1)=12$
$3x^{2}-4x^{2}+8x-4=12$
$-x^{2}+8x-16=0$
$x^{2}-8x+16=0$
$(x-4)^{2}=0$
$x=4$.
Substitute $x=4$ into $(i)$:
$4-y=1 \Rightarrow y=3$.
Therefore,the point of contact is $(4,3)$.

(A) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(2^{x}-1)}{1-\cos x}$
Since this is a $\frac{0}{0}$ form,we apply $L$'Hopital's rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(x 2^{x}-x)}{\frac{d}{dx}(1-\cos x)}$
$L = \lim _{x \rightarrow 0} \frac{2^{x} + x 2^{x} \ln 2 - 1}{\sin x}$
This is still a $\frac{0}{0}$ form. Applying $L$'Hopital's rule again:
$L = \lim _{x \rightarrow 0} \frac{2^{x} \ln 2 + 2^{x} \ln 2 + x 2^{x} (\ln 2)^{2}}{\cos x}$
Substituting $x = 0$:
$L = \frac{2^{0} \ln 2 + 2^{0} \ln 2 + 0}{\cos 0} = \frac{\ln 2 + \ln 2}{1} = 2 \ln 2$
Thus,the correct option is $A$.

(D) Given $a \equiv b \pmod{m}$,we have $m \mid (a-b)$.
$(i)$ For $(a+x) \equiv (b+x) \pmod{m}$,we check $(a+x) - (b+x) = a-b$. Since $m \mid (a-b)$,this is correct.
(ii) For $(a-x) \equiv (b-x) \pmod{m}$,we check $(a-x) - (b-x) = a-b$. Since $m \mid (a-b)$,this is correct.
(iii) For $ax \equiv bx \pmod{m}$,we check $ax - bx = x(a-b)$. Since $m \mid (a-b)$,it follows that $m \mid x(a-b)$,so this is correct.
(iv) For $(a+x) \equiv (b \div x) \pmod{m}$,this is not generally true because division is not defined in modular arithmetic in the same way as addition or multiplication,and $b \div x$ may not even be an integer. Thus,this statement is incorrect.

(B) We know that the implication $A \rightarrow B$ is logically equivalent to $\sim A \vee B$.
Applying this rule to $p \rightarrow \sim q$,we get:
$p \rightarrow \sim q \equiv \sim p \vee \sim q$.
This can also be verified using the truth table:
| $p$ | $q$ | $\sim p$ | $\sim q$ | $p \rightarrow \sim q$ | $\sim p \vee \sim q$ |
|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $F$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $T$ | $T$ |
Since the truth values of $p \rightarrow \sim q$ and $\sim p \vee \sim q$ are identical for all possible truth values of $p$ and $q$,they are logically equivalent.

(C) Let $C = A \cap B$. Since $A$ and $B$ have $n$ elements in common,the set $C$ has $n$ elements.
We know that $(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$.
Substituting $C = A \cap B$,we get $(A \times B) \cap (B \times A) = C \times C$.
The number of elements in $C \times C$ is given by $n(C) \times n(C) = n \times n = n^{2}$.
Therefore,the number of elements common to $A \times B$ and $B \times A$ is $n^{2}$.

(A) Let $x$ and $y$ be two positive numbers.
Given that their sum is constant,$x + y = a$.
Let $z = x^3 + y^3$.
Substituting $y = a - x$,we get $z = x^3 + (a - x)^3$.
Differentiating with respect to $x$: $\frac{dz}{dx} = 3x^2 - 3(a - x)^2 = 3(x^2 - (a^2 - 2ax + x^2)) = 3(2ax - a^2) = 3a(2x - a)$.
Setting $\frac{dz}{dx} = 0$,we find $2x - a = 0$,which implies $x = \frac{a}{2}$.
Since $y = a - x$,we get $y = a - \frac{a}{2} = \frac{a}{2}$.
Checking the second derivative: $\frac{d^2z}{dx^2} = 6a$. Since $a > 0$,$\frac{d^2z}{dx^2} > 0$,confirming a minimum at $x = \frac{a}{2}$.
Thus,$x = y$,meaning the numbers are equal.

(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$
We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for $|x| < 1$.
Therefore,$y = e^{\log _{e} (1-x)^{-1}}$.
Using the property $e^{\log _{e} f(x)} = f(x)$,we get $y = (1-x)^{-1}$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x} = \frac{d}{d x} (1-x)^{-1} = -1(1-x)^{-2} \cdot \frac{d}{d x}(1-x)$
$\frac{d y}{d x} = -1(1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \frac{1}{(1-x)^{2}}$.

(D) Let $r$ be the radius and $\theta$ be the sectorial angle in radians. The perimeter $P$ is given by $P = 2r + r\theta = k$ (constant).
Thus,$r = \frac{k}{2+\theta}$.
The area $A$ of the sector is $A = \frac{1}{2}r^{2}\theta$.
Substituting $r$,we get $A = \frac{1}{2} \left( \frac{k}{2+\theta} \right)^{2} \theta = \frac{k^{2}}{2} \frac{\theta}{(2+\theta)^{2}}$.
To maximize $A$,we differentiate with respect to $\theta$:
$\frac{dA}{d\theta} = \frac{k^{2}}{2} \left[ \frac{(2+\theta)^{2}(1) - \theta(2)(2+\theta)}{(2+\theta)^{4}} \right] = \frac{k^{2}}{2} \left[ \frac{2+\theta - 2\theta}{(2+\theta)^{3}} \right] = \frac{k^{2}}{2} \frac{2-\theta}{(2+\theta)^{3}}$.
Setting $\frac{dA}{d\theta} = 0$,we get $2-\theta = 0$,so $\theta = 2^{c}$.
Since the second derivative is negative at $\theta = 2$,the area is maximum.

(C) Given the displacement equation: $s = 22t - 11t^{2}$.
To find the maximum height, we differentiate $s$ with respect to $t$ to find the velocity $v = \frac{ds}{dt}$.
$\frac{ds}{dt} = 22 - 22t$.
At the maximum height, the velocity is zero, so $22 - 22t = 0$, which gives $t = 1 \text{ s}$.
Substituting $t = 1$ into the displacement equation:
$s = 22(1) - 11(1)^{2} = 22 - 11 = 11 \text{ units}$.
Since the ball starts from $s=0$ and reaches a maximum height of $11 \text{ units}$, the total distance travelled is $11 \text{ units}$.

(A) Given curve is $xy = 25$. Let the point of tangency be $P(x_1, y_1)$. Since $P$ lies on the curve,$x_1 y_1 = 25$.
Differentiating $xy = 25$ with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
At $(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1}$.
The equation of the tangent is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$.
Multiplying by $x_1$,we get $x_1 y - x_1 y_1 = -y_1 x + x_1 y_1$,which simplifies to $x_1 y + y_1 x = 2 x_1 y_1$.
Dividing by $2 x_1 y_1$,we get $\frac{x}{2 x_1} + \frac{y}{2 y_1} = 1$.
The tangent cuts the $x$-axis at $A(2 x_1, 0)$ and the $y$-axis at $B(0, 2 y_1)$.
The area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 x_1) \times (2 y_1) = 2 x_1 y_1$.
Since $x_1 y_1 = 25$,the area is $2(25) = 50$ sq units.

(D) Given $f(x) = \left| \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & x \end{array} \right|$.
Expanding along the first row:
$f(x) = \sin x(x^3 - x) - \cos x(x^4 - 2x^2) + \tan x(x^3 - 2x^4)$.
We need to find $\lim_{x \to 0} \frac{f(x)}{x^2}$.
$\frac{f(x)}{x^2} = \frac{\sin x}{x} \cdot \frac{x^3 - x}{x} - \cos x \cdot \frac{x^4 - 2x^2}{x^2} + \frac{\tan x}{x} \cdot \frac{x^3 - 2x^4}{x}$.
$\frac{f(x)}{x^2} = \frac{\sin x}{x} (x^2 - 1) - \cos x (x^2 - 2) + \frac{\tan x}{x} (x^2 - 2x^3)$.
Taking the limit as $x \to 0$:
$= (1)(0 - 1) - (1)(0 - 2) + (1)(0 - 0)$.
$= -1 + 2 + 0 = 1$.

(A) set $G$ with a binary operation $\cdot$ is a group if it satisfies closure,associativity,identity,and inverse properties.
For $(N, \cdot)$,where $N$ is the set of natural numbers,the multiplicative inverse of an element $a \in N$ (where $a \neq 1$) is $1/a$,which is not in $N$.
Thus,$(N, \cdot)$ fails the inverse property and is not a group.
$(N, +)$ is a semi-group because it is closed and associative under addition.
$(Z, +)$ is a group as it satisfies all group axioms.
The set of even integers $2Z = \{..., -4, -2, 0, 2, 4, ...\}$ forms a group under addition because it contains the identity $0$,inverses for all elements,and is closed and associative.
Therefore,the statement $(N, \cdot)$ is a group is false.

(D) subset $H$ of a group $G$ is a subgroup if it is itself a group under the same operation.
For $H = \{4^{n} \mid n \in \mathbb{Z}\}$,we note that $4^{n} = (2^{2})^{n} = 2^{2n}$. Since $2n \in \mathbb{Z}$ for all $n \in \mathbb{Z}$,$H \subset G$.
$1.$ Closure: $4^{n} \cdot 4^{m} = 4^{n+m} \in H$.
$2.$ Identity: $4^{0} = 1 = 2^{0} \in H$.
$3.$ Inverse: For $4^{n} \in H$,the inverse is $4^{-n} \in H$.
Thus,$\{4^{n} \mid n \in \mathbb{Z}\}$ is a subgroup of $G$.

(A) We are given the equation $4 \otimes_{7} x = 5$ in the group $G = \{1, 2, 3, 4, 5, 6\}$ under multiplication modulo $7$.
To find $x$,we test the elements of the group:
$4 \otimes_{7} 1 = 4$
$4 \otimes_{7} 2 = 8 \equiv 1 \pmod{7}$
$4 \otimes_{7} 3 = 12 \equiv 5 \pmod{7}$
$4 \otimes_{7} 4 = 16 \equiv 2 \pmod{7}$
$4 \otimes_{7} 5 = 20 \equiv 6 \pmod{7}$
$4 \otimes_{7} 6 = 24 \equiv 3 \pmod{7}$
Comparing this with the given equation $4 \otimes_{7} x = 5$,we find that $x = 3$.

(D) The characteristic equation of a $3 \times 3$ matrix $A$ is given by $|A - \lambda I| = 0$,which results in $-\lambda^3 + \text{tr}(A)\lambda^2 - (\dots)\lambda + |A| = 0$.
Given the characteristic equation $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$,we compare this with the standard form $\lambda^3 - \text{tr}(A)\lambda^2 + (\dots)\lambda - |A| = 0$.
Comparing the constant terms,we get $-|A| = 2$,which implies $|A| = -2$.
We know that for a square matrix $A$ of order $n$,$|\text{adj}(A)| = |A|^{n-1}$.
Here,$n = 3$,so $|\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\text{adj}(A)| = (-2)^2 = 4$.

(D) Let $A$ be a diagonal matrix of order $n \times n$ defined as $A = \text{diag}(a_1, a_2, \dots, a_n)$,where $a_i \neq 0$ for all $i$ because the matrix is non-singular.
The inverse of a diagonal matrix $A$ is given by $A^{-1} = \text{diag}(1/a_1, 1/a_2, \dots, 1/a_n)$.
Since all off-diagonal elements of $A^{-1}$ remain $0$,the resulting matrix $A^{-1}$ is also a diagonal matrix.
Thus,the inverse of a diagonal non-singular matrix is a diagonal matrix.

(B) Given the equation: $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|$.
To find the value of $e$,we set $x = 0$ on both sides of the equation.
Substituting $x = 0$ into the determinant,we get:
$e = \left|\begin{array}{ccc}0+3(0) & 0-1 & 0+3 \\ 0+1 & -2(0) & 0-4 \\ 0-3 & 0+4 & 3(0)\end{array}\right| = \left|\begin{array}{ccc}0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0\end{array}\right|$.
Expanding the determinant along the first row:
$e = 0(0 - (-16)) - (-1)(0 - 12) + 3(4 - 0)$
$e = 0(16) + 1(-12) + 3(4)$
$e = 0 - 12 + 12 = 0$.
Thus,$e = 0$.

(C) Let $\cos ^{-1} x = y$,then $x = \cos y$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $0 \leq y \leq \pi$.
The given equation is $2y = \sin ^{-1} (2 \cos y \sin y) = \sin ^{-1} (\sin 2y)$.
For $\sin ^{-1} (\sin 2y) = 2y$,we must have $-\frac{\pi}{2} \leq 2y \leq \frac{\pi}{2}$,which implies $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$.
Combining $0 \leq y \leq \pi$ and $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$,we get $0 \leq y \leq \frac{\pi}{4}$.
Since $y = \cos ^{-1} x$,we have $0 \leq \cos ^{-1} x \leq \frac{\pi}{4}$.
Taking cosine on all sides (noting that cosine is a decreasing function in $[0, \pi]$),we get $\cos(0) \geq x \geq \cos(\frac{\pi}{4})$.
Thus,$1 \geq x \geq \frac{1}{\sqrt{2}}$,or $\frac{1}{\sqrt{2}} \leq x \leq 1$.

(C) Given equation is $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$.
For the domain,we must have $x(x+1) \ge 0$ and $0 \le x^2+x+1 \le 1$.
The condition $x^2+x+1 \le 1$ implies $x^2+x \le 0$,which means $x(x+1) \le 0$.
Combining $x(x+1) \ge 0$ and $x(x+1) \le 0$,we get $x(x+1) = 0$,so $x=0$ or $x=-1$.
If $x=0$,the equation becomes $\tan ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
If $x=-1$,the equation becomes $\tan ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is also a solution.
Thus,there are $2$ real solutions.

(D) The general term of the series is $T_r = \tan^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}\right)$ for $r=1, 2, \dots, n$.
We can rewrite the argument as:
$\frac{1}{1 + (x+r)(x+r-1)} = \frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}$.
Thus,$T_r = \tan^{-1}(x+r) - \tan^{-1}(x+r-1)$.
Summing these terms from $r=1$ to $n$:
$y = \sum_{r=1}^{n} [\tan^{-1}(x+r) - \tan^{-1}(x+r-1)]$.
This is a telescoping series:
$y = (\tan^{-1}(x+1) - \tan^{-1}(x)) + (\tan^{-1}(x+2) - \tan^{-1}(x+1)) + \dots + (\tan^{-1}(x+n) - \tan^{-1}(x+n-1))$.
$y = \tan^{-1}(x+n) - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1+(x+n)^2} - \frac{1}{1+x^2}$.
Evaluating at $x=0$:
$y'(0) = \frac{1}{1+n^2} - \frac{1}{1+0^2} = \frac{1}{1+n^2} - 1 = \frac{1 - 1 - n^2}{1+n^2} = -\frac{n^2}{1+n^2}$.

(A) By definition,a function $f(x)$ is even if $f(-x) = f(x)$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$
$f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$
$-f^{\prime}(-x) = f^{\prime}(x)$
$f^{\prime}(-x) = -f^{\prime}(x)$
Since $f^{\prime}(-x) = -f^{\prime}(x)$,the derivative $f^{\prime}(x)$ is an odd function.

(A) Let $x_{1}, x_{2} \in R$.
For injectivity,let $f(x_{1}) = f(x_{2})$.
$2x_{1} + 3 = 2x_{2} + 3$
$2x_{1} = 2x_{2}$
$x_{1} = x_{2}$.
Hence,$f$ is injective.
For surjectivity,let $y \in \text{codomain } R$.
Let $y = f(x) = 2x + 3$.
$y - 3 = 2x$
$x = \frac{y-3}{2}$.
Since for every $y \in R$,there exists an $x = \frac{y-3}{2} \in R$,$f$ is surjective.
Since $f$ is both injective and surjective,$f^{-1}$ exists.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x-3}{2}$.

(B) For a function $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x \to 2$ must equal $f(2)$.
Given $f(2) = 2$.
We evaluate the limit: $\lim_{x \to 2} \frac{x^2 - (a+2)x + a}{x-2}$.
Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ at $x = 2$ for the limit to be finite.
Substituting $x = 2$ in the numerator: $2^2 - (a+2)(2) + a = 0$.
$4 - 2a - 4 + a = 0$.
$-a = 0 \implies a = 0$.
Alternatively,using $L$'Hospital's rule: $\lim_{x \to 2} \frac{2x - (a+2)}{1} = 2(2) - a - 2 = 2 - a$.
Equating to $f(2)$: $2 - a = 2 \implies a = 0$.

(B) We know that $\pi^{2} \approx 9.86$.
Since $[\cdot]$ is the greatest integer function:
$[\pi^{2}] = [9.86] = 9$.
$[-\pi^{2}] = [-9.86] = -10$.
Substituting these values into the function $f(x)$:
$f(x) = \sin(9x) + \cos(-10x)$.
Since $\cos(-\theta) = \cos(\theta)$,we have:
$f(x) = \sin(9x) + \cos(10x)$.
Now,differentiating $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin(9x)) + \frac{d}{dx}(\cos(10x))$.
Using the chain rule:
$f'(x) = 9 \cos(9x) - 10 \sin(10x)$.

(B) Given the curve $x^{n} y^{m}=a^{m+n}$,where $m, n > 0$. Taking the natural logarithm on both sides:
$n \ln x + m \ln y = (m+n) \ln a$.
Differentiating with respect to $x$:
$\frac{n}{x} + \frac{m}{y} \frac{dy}{dx} = 0$.
Solving for the derivative:
$\frac{dy}{dx} = -\frac{n}{m} \cdot \frac{y}{x}$.
The length of the subtangent is defined as $|\frac{y}{dy/dx}|$.
Substituting the derivative:
$|\frac{y}{-(n/m)(y/x)}| = |-\frac{m}{n} x| = \frac{m}{n} |x|$.
Thus,at the point $(x_{1}, y_{1})$,the length of the subtangent is $\frac{m}{n} |x_{1}|$.

(C) Let the curve be $y = f(x)$. The length of the subtangent is given by $|y \frac{dx}{dy}|$ and the length of the subnormal is given by $|y \frac{dy}{dx}|$.
Let the ordinate be $y$.
Then,the product of the subtangent and the subnormal is:
$\text{Subtangent} \times \text{Subnormal} = |y \frac{dx}{dy}| \times |y \frac{dy}{dx}| = |y^2| = y^2$.
Since the product of the first and third terms equals the square of the second term (the ordinate),the three quantities form a Geometric Progression $(GP)$.

(B) Let $y = x e^{-x}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.
For maximum or minimum values,we set $\frac{dy}{dx} = 0$:
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any real $x$,we have $1 - x = 0$,which implies $x = 1$.
Now,we find the second derivative to check for maxima:
$\frac{d^2y}{dx^2} = e^{-x}(-1) + (1 - x)(-e^{-x}) = -e^{-x} - e^{-x} + x e^{-x} = e^{-x}(x - 2)$.
At $x = 1$:
$\frac{d^2y}{dx^2} = e^{-1}(1 - 2) = -\frac{1}{e} < 0$.
Since the second derivative is negative at $x = 1$,the function has a local maximum at $x = 1$.
The maximum value is $y(1) = 1 \cdot e^{-1} = \frac{1}{e}$.

(C) Given,$\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$
Multiplying both sides by $(x-1)(x+3)$,we get:
$3x+1 = A(x+3) + B(x-1)$
Expanding the right side:
$3x+1 = (A+B)x + (3A-B)$
Comparing the coefficients of $x$ and the constant terms:
$A+B = 3$ $(i)$
$3A-B = 1$ (ii)
Adding equations $(i)$ and (ii):
$(A+B) + (3A-B) = 3+1$
$4A = 4 \Rightarrow A = 1$
Substituting $A=1$ into equation $(i)$:
$1+B = 3 \Rightarrow B = 2$
Now,calculate $\sin^{-1} \frac{A}{B}$:
$\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$

(B) Given $I_{n}=\int(\log x)^{n} dx$.
Using integration by parts,let $u = (\log x)^{n}$ and $dv = dx$.
Then $du = n(\log x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$.
Applying the formula $\int u dv = uv - \int v du$:
$I_{n} = x(\log x)^{n} - \int x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} dx$
$I_{n} = x(\log x)^{n} - n \int (\log x)^{n-1} dx$
$I_{n} = x(\log x)^{n} - n I_{n-1}$
Rearranging the terms,we get:
$I_{n} + n I_{n-1} = x(\log x)^{n} + C$.

(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)}{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x) + \cos ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)} d x$
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{2} - x)}{\sin ^{3}(\frac{\pi}{2} - x) + \cos ^{3}(\frac{\pi}{2} - x)} d x$
Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x + \sin ^{3} x} d x$ (ii)
Adding $(i)$ and (ii):
$2I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x + \cos ^{3} x}{\sin ^{3} x + \cos ^{3} x} d x$
$2I = \int_{\pi / 6}^{\pi / 3} 1 d x$
$2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(D) The integral $\int_{0}^{11} [x] dx$ can be split into intervals of unit length where $[x]$ is constant:
$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} \int_{k}^{k+1} [x] dx$
Since for $x \in [k, k+1)$,$[x] = k$,we have:
$\int_{k}^{k+1} [x] dx = \int_{k}^{k+1} k dx = k(k+1 - k) = k$
Therefore,the sum becomes:
$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} k = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$
Using the sum formula for the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$\sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55$
Thus,the value of the integral is $55$.

(A) Given,$m \sin ^{-1} x = \log _{e} y$
$\Rightarrow y = e^{m \sin ^{-1} x}$
Differentiating with respect to $x$,we get:
$y' = e^{m \sin ^{-1} x} \times \frac{m}{\sqrt{1 - x^{2}}}$
$\Rightarrow \sqrt{1 - x^{2}} \cdot y' = m y$
Squaring both sides,we get:
$(1 - x^{2}) (y')^{2} = m^{2} y^{2}$
Differentiating again with respect to $x$:
$(1 - x^{2}) \cdot 2 y' \cdot y'' + (y')^{2} (-2 x) = m^{2} \cdot 2 y y' $
Dividing both sides by $2 y'$,we get:
$(1 - x^{2}) y'' - x y' = m^{2} y$

(D) Given the differential equation $e^{dy/dx} = x$.
Taking the natural logarithm on both sides,we get:
$\frac{dy}{dx} = \log x$
Integrating both sides with respect to $x$:
$\int dy = \int \log x \, dx$
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log x$ and $dv = dx$:
$y = x \log x - \int x \cdot \frac{1}{x} \, dx$
$y = x \log x - \int 1 \, dx$
$y = x \log x - x + C$
$y = x(\log x - 1) + C$ (Equation $i$)
Given the initial conditions $x = 1$ and $y = 0$,substitute these into Equation $i$:
$0 = 1(\log 1 - 1) + C$
$0 = 1(0 - 1) + C$
$0 = -1 + C$
$C = 1$
Substituting $C = 1$ back into Equation $i$,we get:
$y = x(\log x - 1) + 1$

(D) Let the adjacent sides of the parallelogram be $\vec{a} = \hat{i}+\hat{j}-\hat{k}$ and $\vec{b} = 2\hat{i}-3\hat{j}+\hat{k}$.
The diagonals of the parallelogram are given by $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$.
First diagonal: $\vec{d_1} = (\hat{i}+\hat{j}-\hat{k}) + (2\hat{i}-3\hat{j}+\hat{k}) = 3\hat{i}-2\hat{j}$.
Length of the first diagonal: $|\vec{d_1}| = \sqrt{3^2 + (-2)^2 + 0^2} = \sqrt{9+4} = \sqrt{13}$.
Second diagonal: $\vec{d_2} = (\hat{i}+\hat{j}-\hat{k}) - (2\hat{i}-3\hat{j}+\hat{k}) = -\hat{i}+4\hat{j}-2\hat{k}$.
Length of the second diagonal: $|\vec{d_2}| = \sqrt{(-1)^2 + 4^2 + (-2)^2} = \sqrt{1+16+4} = \sqrt{21}$.
Thus,the lengths of the diagonals are $\sqrt{13}$ and $\sqrt{21}$.

(C) We know that for any two vectors $a$ and $b$,$(a \times b)^{2} + (a \cdot b)^{2} = |a|^{2} |b|^{2} \sin^{2} \theta + |a|^{2} |b|^{2} \cos^{2} \theta$.
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,this simplifies to $|a|^{2} |b|^{2}$.
Given the equation $(a \times b)^{2} + (a \cdot b)^{2} = 144$,we have $|a|^{2} |b|^{2} = 144$.
Given $|a| = 4$,we substitute $|a|^{2} = 16$ into the equation:
$16 |b|^{2} = 144$.
Dividing both sides by $16$,we get $|b|^{2} = 9$.
Taking the square root,we find $|b| = 3$.

(B) The volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 4$.
We need to find the value of $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]$.
Using the property of the scalar triple product of cross products: $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \vec{b} \vec{c}]^2$.
Substituting the given value: $[\vec{a} \vec{b} \vec{c}]^2 = (4)^2 = 16$.
Thus,the value is $16$.

(D) Given $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$.
We need to evaluate the expression $E = (a+b) \cdot p + (b+c) \cdot q + (c+a) \cdot r$.
Substituting the values of $p, q, r$:
$E = (a+b) \cdot \frac{b \times c}{[a b c]} + (b+c) \cdot \frac{c \times a}{[a b c]} + (c+a) \cdot \frac{a \times b}{[a b c]}$
$E = \frac{1}{[a b c]} [a \cdot (b \times c) + b \cdot (b \times c) + b \cdot (c \times a) + c \cdot (c \times a) + c \cdot (a \times b) + a \cdot (a \times b)]$
Using the property of scalar triple product $[x y z] = x \cdot (y \times z)$:
$E = \frac{1}{[a b c]} [[a b c] + 0 + [b c a] + 0 + [c a b] + 0]$
Since $[a b c] = [b c a] = [c a b]$:
$E = \frac{[a b c] + [a b c] + [a b c]}{[a b c]} = \frac{3[a b c]}{[a b c]} = 3$.