MHT CET 2021 Mathematics Question Paper with Answer and Solution

(D) Given curves are $\frac{x^2}{a^2} + \frac{y^2}{16} = 1$ and $y^3 = 16x$.
For the first curve,differentiating with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{16} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{16x}{a^2y} \quad (1)$
For the second curve,differentiating with respect to $x$:
$3y^2 \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{3y^2} \quad (2)$
Since the curves intersect orthogonally,the product of their slopes is $-1$:
$\left(-\frac{16x}{a^2y}\right) \left(\frac{16}{3y^2}\right) = -1$
$\frac{256x}{3a^2y^3} = 1$
Since $y^3 = 16x$,substitute this into the equation:
$\frac{256x}{3a^2(16x)} = 1$
$\frac{16}{3a^2} = 1 \Rightarrow a^2 = \frac{16}{3}$
Wait,re-evaluating the original question equation $\frac{x^2}{a^2} + \frac{y^2}{4} = 4$ which is $\frac{x^2}{4a^2} + \frac{y^2}{16} = 1$.
Using the provided solution logic for $\frac{x^2}{a^2} + \frac{y^2}{4} = 4$:
$\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{a^2y}$.
Product of slopes: $(-\frac{4x}{a^2y})(\frac{16}{3y^2}) = -1 \Rightarrow \frac{64x}{3a^2y^3} = 1$.
Substituting $y^3 = 16x$: $\frac{64x}{3a^2(16x)} = 1 \Rightarrow \frac{4}{3a^2} = 1 \Rightarrow a^2 = \frac{4}{3}$.

(D) Given curve is $y^2=2(x-3)$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.
The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{y}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -y$.
The given line is $y - 2x + 1 = 0$,which can be written as $y = 2x - 1$. The slope of this line is $2$.
Since the normal is parallel to the line,their slopes must be equal,so $-y = 2$,which gives $y = -2$.
Substituting $y = -2$ into the curve equation: $(-2)^2 = 2(x - 3) \Rightarrow 4 = 2(x - 3) \Rightarrow 2 = x - 3 \Rightarrow x = 5$.
Thus,the required point is $(5, -2)$.

(C) Let the point $P$ on the ellipse be $(5 \cos \theta, 4 \sin \theta)$ in the first quadrant.
Since the rectangle is symmetric about both axes,its vertices are $(\pm 5 \cos \theta, \pm 4 \sin \theta)$.
The length of the rectangle is $L = 2(5 \cos \theta) = 10 \cos \theta$.
The breadth of the rectangle is $B = 2(4 \sin \theta) = 8 \sin \theta$.
The area $A$ of the rectangle is $A = L \times B = (10 \cos \theta)(8 \sin \theta) = 80 \sin \theta \cos \theta = 40 \sin(2 \theta)$.
For the area to be maximum,$\sin(2 \theta)$ must be maximum,i.e.,$\sin(2 \theta) = 1$.
This implies $2 \theta = \frac{\pi}{2}$,so $\theta = \frac{\pi}{4}$.
Substituting $\theta = \frac{\pi}{4}$ into the expressions for length and breadth:
$L = 10 \cos(\frac{\pi}{4}) = 10 \times \frac{1}{\sqrt{2}} = 5 \sqrt{2}$.
$B = 8 \sin(\frac{\pi}{4}) = 8 \times \frac{1}{\sqrt{2}} = 4 \sqrt{2}$.
Thus,the dimensions of the rectangle are $5 \sqrt{2}$ and $4 \sqrt{2}$.

(D) The maximum value of ${}^6C_r$ occurs at $r = \frac{6}{2} = 3$.
The value is ${}^6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Given that the difference between the maximum value of ${}^6C_r$ and ${}^nC_3$ is $16$,we have $|20 - {}^nC_3| = 16$.
This implies ${}^nC_3 = 20 + 16 = 36$ or ${}^nC_3 = 20 - 16 = 4$.
Case $1$: ${}^nC_3 = 36$ $\Rightarrow \frac{n(n-1)(n-2)}{6} = 36$ $\Rightarrow n(n-1)(n-2) = 216$. There is no integer $n$ satisfying this.
Case $2$: ${}^nC_3 = 4$ $\Rightarrow \frac{n(n-1)(n-2)}{6} = 4$ $\Rightarrow n(n-1)(n-2) = 24$.
Testing values,for $n=4$,$4 \times 3 \times 2 = 24$.
Thus,$n = 4$.

(C) We use the Pascal's identity: ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$.
Given expression: ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C_7={ }^{14} C_{r}$.
Applying the identity to the first two terms: ${ }^{11} C_4+{ }^{11} C_5 = { }^{12} C_5$.
Now the expression becomes: ${ }^{12} C_5+{ }^{12} C_6+{ }^{13} C_7$.
Applying the identity again: ${ }^{12} C_5+{ }^{12} C_6 = { }^{13} C_6$.
Now the expression becomes: ${ }^{13} C_6+{ }^{13} C_7$.
Applying the identity one last time: ${ }^{13} C_6+{ }^{13} C_7 = { }^{14} C_7$.
Comparing this with ${ }^{14} C_{r}$,we get $r = 7$.

(D) Let the centre of the circle be $(h, k)$. Since it lies on the line $x-4y=1$,we have $h = 1+4k$. Thus,the centre is $(4k+1, k)$.
Since the circle passes through $(3,7)$ and $(5,5)$,the distances from the centre to these points are equal (equal to the radius $r$):
$(4k+1-3)^2 + (k-7)^2 = (4k+1-5)^2 + (k-5)^2$
$(4k-2)^2 + (k-7)^2 = (4k-4)^2 + (k-5)^2$
$16k^2 - 16k + 4 + k^2 - 14k + 49 = 16k^2 - 32k + 16 + k^2 - 10k + 25$
$17k^2 - 30k + 53 = 17k^2 - 42k + 41$
$12k = -12 \Rightarrow k = -1$.
Substituting $k = -1$ into $h = 1+4k$,we get $h = 1+4(-1) = -3$. So,the centre is $(-3, -1)$.
The radius $r$ is the distance between $(-3, -1)$ and $(5, 5)$:
$r^2 = (5 - (-3))^2 + (5 - (-1))^2 = 8^2 + 6^2 = 64 + 36 = 100$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$:
$(x + 3)^2 + (y + 1)^2 = 100$
$x^2 + 6x + 9 + y^2 + 2y + 1 = 100$
$x^2 + y^2 + 6x + 2y - 90 = 0$.

(A) Given the circumference of the circle is $10 \pi$ units.
We know that the circumference $C = 2 \pi r$,so $2 \pi r = 10 \pi$,which gives $r = 5$.
The center of the circle is $(h, k) = (2, -3)$.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 5^2$.
$(x - 2)^2 + (y + 3)^2 = 25$.
Expanding this,we get $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25$.
$x^2 + y^2 - 4x + 6y + 13 = 25$.
$x^2 + y^2 - 4x + 6y - 12 = 0$.

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through the origin $(0,0)$,we have $c=0$.
The circle cuts off an intercept of $-2$ on the $X$-axis,so it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.
The circle cuts off an intercept of $3$ on the $Y$-axis,so it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.
Substituting $g=1$ and $f=-\frac{3}{2}$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.
This simplifies to $x^2+y^2+2x-3y=0$.

(B) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.
We can rewrite $L_2$ as $3x - 4y - \frac{7}{2} = 0$.
Since the lines are parallel and tangent to the same circle,the distance between them is equal to the diameter $D$ of the circle.
$D = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|4 - (-7/2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|4 + 3.5|}{5} = \frac{7.5}{5} = 1.5$.
Since the diameter $D = 1.5$,the radius $r = \frac{D}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$ units.

(D) The circle $x^2+y^2=(8)^2$ has radius $r=8$ and center $(0,0)$.
The point $P$ on the circle corresponding to the parameter $\theta = \frac{2\pi}{3}$ has coordinates:
$P \equiv (8 \cos \frac{2\pi}{3}, 8 \sin \frac{2\pi}{3}) = (8 \times -\frac{1}{2}, 8 \times \frac{\sqrt{3}}{2}) = (-4, 4\sqrt{3})$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.
Substituting $(x_1, y_1) = (-4, 4\sqrt{3})$ and $r^2 = 64$:
$-4x + 4\sqrt{3}y = 64$.
Dividing by $-4$:
$x - \sqrt{3}y = -16 \Rightarrow x - \sqrt{3}y + 16 = 0$.

(B) The equation of the circle is $x^2+y^2-4x-10y+25=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C = (-g, -f) = (2, 5)$.
Let $M(1, 2)$ be the midpoint of the chord $AB$.
The chord is perpendicular to the radius $CM$ at the point $M$.
The slope of $CM$ is $m_{CM} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.
Since the chord $AB$ is perpendicular to $CM$,the slope of the chord $AB$ is $m_{AB} = -\frac{1}{m_{CM}} = -\frac{1}{3}$.
The equation of the chord passing through $M(1, 2)$ with slope $-\frac{1}{3}$ is given by:
$y - 2 = -\frac{1}{3}(x - 1)$
$3(y - 2) = -(x - 1)$
$3y - 6 = -x + 1$
$x + 3y = 7$.

(B) For the circle $x^2+y^2-4x+10y+20=0$,the center is $C_1 = (2, -5)$ and the radius is $r_1 = \sqrt{2^2 + (-5)^2 - 20} = \sqrt{4+25-20} = 3$.
For the circle $x^2+y^2+8x-6y-24=0$,the center is $C_2 = (-4, 3)$ and the radius is $r_2 = \sqrt{(-4)^2 + 3^2 - (-24)} = \sqrt{16+9+24} = 7$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2 - (-4))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = 10$.
Since $d = r_1 + r_2$ $(10 = 3 + 7)$,the circles touch each other externally at a single point.
The equation of the common tangent at the point of contact is the radical axis,given by $S_1 - S_2 = 0$.
$(x^2+y^2-4x+10y+20) - (x^2+y^2+8x-6y-24) = 0$.
$-12x + 16y + 44 = 0$.
Dividing by $-4$,we get $3x - 4y - 11 = 0$.

(C) Let the radius of both circles be $r$. The equations of the circles are:
$S_1: (x-2)^2 + (y-3)^2 = r^2$
$S_2: (x-4)^2 + (y-5)^2 = r^2$
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x-2)^2 + (y-3)^2 - [(x-4)^2 + (y-5)^2] = r^2 - r^2$
$(x^2 - 4x + 4 + y^2 - 6y + 9) - (x^2 - 8x + 16 + y^2 - 10y + 25) = 0$
$x^2 - 4x + y^2 - 6y + 13 - x^2 + 8x - y^2 + 10y - 41 = 0$
$4x + 4y - 28 = 0$
Dividing by $4$,we get $x + y - 7 = 0$.

(A) The equation of the given circle is $x^2+y^2-10x=0$.
Substituting $y=2x$ into the circle equation:
$x^2+(2x)^2-10x=0$
$x^2+4x^2-10x=0$
$5x^2-10x=0$
$5x(x-2)=0$
So,$x=0$ or $x=2$.
If $x=0$,$y=2(0)=0$. If $x=2$,$y=2(2)=4$.
The endpoints of the chord are $(0,0)$ and $(2,4)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
Substituting the points $(0,0)$ and $(2,4)$:
$(x-0)(x-2)+(y-0)(y-4)=0$
$x(x-2)+y(y-4)=0$
$x^2-2x+y^2-4y=0$
$x^2+y^2-2x-4y=0$.

(B) Let the centre of the circle be $(h, k)$.
Since the circle passes through $(0,0), (x,0),$ and $(0,y)$,the distance from the centre to each point must be equal to the radius $R$.
$h^2 + k^2 = (h-x)^2 + k^2 = h^2 + (k-y)^2$
From $h^2 + k^2 = (h-x)^2 + k^2$,we get $h^2 = h^2 - 2hx + x^2$,which implies $2hx = x^2$,so $h = \frac{x}{2}$.
From $h^2 + k^2 = h^2 + (k-y)^2$,we get $k^2 = k^2 - 2ky + y^2$,which implies $2ky = y^2$,so $k = \frac{y}{2}$.
Thus,the coordinates of the centre are $\left(\frac{x}{2}, \frac{y}{2}\right)$.

(B) Given expression: $(1+i)^5(1-i)^7$
$= (1+i)^5(1-i)^5(1-i)^2$
$= [(1+i)(1-i)]^5(1-2i+i^2)$
$= (1-i^2)^5(1-2i-1)$
$= (1-(-1))^5(-2i)$
$= (2)^5(-2i)$
$= 32(-2i) = -64i$

(A) Given $z(2-i) = 3+i$.
$z = \frac{3+i}{2-i} = \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6+3i+2i+i^2}{4+1} = \frac{6+5i-1}{5} = \frac{5+5i}{5} = 1+i$.
Now,express $z$ in polar form: $z = 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$.
Using De Moivre's Theorem,$z^{38} = (\sqrt{2})^{38} \left( \cos \frac{38\pi}{4} + i \sin \frac{38\pi}{4} \right)$.
$z^{38} = 2^{19} \left( \cos \frac{19\pi}{2} + i \sin \frac{19\pi}{2} \right)$.
Since $\frac{19\pi}{2} = 9\pi + \frac{\pi}{2}$,we have $\cos \frac{19\pi}{2} = 0$ and $\sin \frac{19\pi}{2} = -1$.
$z^{38} = 2^{19} (0 + i(-1)) = -2^{19} i$.

(B) Given $x = 1 + 2i$,then $x - 1 = 2i$.
Squaring both sides,we get $(x - 1)^2 = (2i)^2$,which implies $x^2 - 2x + 1 = -4$,or $x^2 - 2x + 5 = 0$.
Now,we perform polynomial division of $x^3 + 7x^2 - x + 16$ by $x^2 - 2x + 5$:
$x^3 + 7x^2 - x + 16 = (x^2 - 2x + 5)(x + 9) + (12x - 29)$.
Since $x^2 - 2x + 5 = 0$,the expression simplifies to $12x - 29$.
Substituting $x = 1 + 2i$:
$12(1 + 2i) - 29 = 12 + 24i - 29 = -17 + 24i$.

(A) Let $\sqrt{-5-12i} = a+ib$,where $a, b \in \mathbb{R}$.
Squaring both sides,we get $(a+ib)^2 = -5-12i$.
$(a^2-b^2) + i(2ab) = -5-12i$.
Equating real and imaginary parts: $a^2-b^2 = -5$ and $2ab = -12$,which implies $ab = -6$.
Since $b = -6/a$,substituting into the first equation gives $a^2 - (-6/a)^2 = -5$.
$a^2 - 36/a^2 = -5 \implies a^4 + 5a^2 - 36 = 0$.
Factoring the quadratic in $a^2$: $(a^2+9)(a^2-4) = 0$.
Since $a \in \mathbb{R}$,$a^2 = 4$,so $a = \pm 2$.
If $a = 2$,$b = -6/2 = -3$. If $a = -2$,$b = -6/(-2) = 3$.
Thus,the square roots are $\pm(2-3i)$.

(C) Given: $\frac{3+2i}{1+i} = \frac{1}{2}(x+iy)$
$\therefore x+iy = \frac{2(3+2i)}{1+i} \times \frac{1-i}{1-i}$
$= \frac{2(3 - 3i + 2i - 2i^2)}{1 - i^2}$
$= \frac{2(3 - i + 2)}{1 + 1} = \frac{2(5 - i)}{2} = 5 - i$
Comparing real and imaginary parts,we get $x = 5$ and $y = -1$.
Therefore,$x - y = 5 - (-1) = 5 + 1 = 6$.

(D) Let the complex number be $z = r(\cos \theta + i \sin \theta)$.
Given that the distance from the origin is $r = 2$ and the argument is $\theta = \frac{5 \pi}{6}$.
Substituting these values,we get $z = 2 \left( \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} \right)$.
Since $\cos \frac{5 \pi}{6} = \cos(\pi - \frac{\pi}{6}) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin \frac{5 \pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2}$.
Therefore,$z = 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i$.

(C) We know that $1+\omega+\omega^2=0$,so $1+\omega=-\omega^2$.
Substituting this into the given expression:
$(1+\omega)^7 = (-\omega^2)^7$
$= (-1)^7 \times (\omega^2)^7$
$= -1 \times \omega^{14}$
Since $\omega^3=1$,we have $\omega^{14} = \omega^{12} \times \omega^2 = (\omega^3)^4 \times \omega^2 = 1^4 \times \omega^2 = \omega^2$.
Thus,$(1+\omega)^7 = -\omega^2$.
Using $1+\omega+\omega^2=0$,we have $-\omega^2 = 1+\omega$.
Comparing $1+\omega$ with $A+B\omega$,we get $A=1$ and $B=1$.

(D) We know that $1+\omega+\omega^2 = 0$,which implies $1+\omega^2 = -\omega$ and $\omega^3 = 1$.
Given expression: $E = (3+5\omega+3\omega^2)^2 + (3+3\omega+5\omega^2)^2$
Rearranging terms using $3(1+\omega+\omega^2) = 0$:
$E = (3(1+\omega+\omega^2) + 2\omega)^2 + (3(1+\omega+\omega^2) + 2\omega^2)^2$
$E = (0 + 2\omega)^2 + (0 + 2\omega^2)^2$
$E = 4\omega^2 + 4\omega^4$
Since $\omega^3 = 1$,we have $\omega^4 = \omega$.
$E = 4\omega^2 + 4\omega = 4(\omega^2 + \omega)$
Since $1+\omega+\omega^2 = 0$,we have $\omega^2 + \omega = -1$.
$E = 4(-1) = -4$.

(B) Given $z=x+iy$ and $|z+1|=1$.
Substituting $z=x+iy$ into the equation:
$|(x+1)+iy|=1$.
Squaring both sides,we get:
$(x+1)^2+y^2=1^2$.
This is the standard equation of a circle $(x-h)^2+(y-k)^2=r^2$,where the centre is $(h,k)=(-1,0)$ and the radius is $r=1$.

(B) Given the amplitude of $(z-2-3i)$ is $\frac{3\pi}{4}$ and $z=x+iy$.
Substituting $z$,we get $(x-2) + i(y-3)$.
Since $\text{arg}((x-2) + i(y-3)) = \frac{3\pi}{4}$,we have $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{3\pi}{4}$.
Taking tangent on both sides,$\frac{y-3}{x-2} = \tan\left(\frac{3\pi}{4}\right) = -1$.
This implies $y-3 = -(x-2)$,which simplifies to $y-3 = -x+2$.
Therefore,$x+y=5$.

(A) The half-life period $(T_{1/2})$ is $5$ years.
Initial amount $(N_0)$ is $64$ gms.
Total time $(t)$ is $15$ years.
The number of half-lives $(n)$ is calculated as $n = \frac{t}{T_{1/2}} = \frac{15}{5} = 3$.
The amount left $(N)$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $N = 64 \times (\frac{1}{2})^3 = 64 \times \frac{1}{8} = 8$ gms.
Therefore,the amount left after $15$ years is $8$ gms.

(C) To evaluate the limit $\lim _{x \rightarrow \infty} (\sqrt{x^2+5x-7}-x)$,we rationalize the expression:
$\lim _{x \rightarrow \infty} \frac{(\sqrt{x^2+5x-7}-x)(\sqrt{x^2+5x-7}+x)}{\sqrt{x^2+5x-7}+x}$
$= \lim _{x \rightarrow \infty} \frac{x^2+5x-7-x^2}{\sqrt{x^2+5x-7}+x}$
$= \lim _{x \rightarrow \infty} \frac{5x-7}{\sqrt{x^2+5x-7}+x}$
Dividing the numerator and denominator by $x$ (where $x > 0$):
$= \lim _{x \rightarrow \infty} \frac{5-\frac{7}{x}}{\sqrt{1+\frac{5}{x}-\frac{7}{x^2}}+1}$
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$:
$= \frac{5-0}{\sqrt{1+0-0}+1} = \frac{5}{1+1} = \frac{5}{2}$

(C) We evaluate the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$
First,factor the denominator: $2x^2+x-3 = (x-1)(2x+3)$
Substitute this into the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}$
Recall that $(x-1) = (\sqrt{x}-1)(\sqrt{x}+1)$,so: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)}$
Cancel the common term $(\sqrt{x}-1)$: $\lim _{x \rightarrow 1} \frac{2 x-3}{(\sqrt{x}+1)(2 x+3)}$
Now,substitute $x=1$: $\frac{2(1)-3}{(\sqrt{1}+1)(2(1)+3)} = \frac{-1}{(2)(5)} = \frac{-1}{10}$

(C) We know the standard limit formula: $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = n a^{n-1}$.
Given the expression: $\lim _{x \rightarrow 5} \frac{x^k-5^k}{x-5} = 500$.
Applying the formula,we get: $k(5)^{k-1} = 500$.
We can write $500$ as $4 \times 125 = 4 \times 5^3$.
Comparing $k(5)^{k-1}$ with $4(5)^3$,we see that $k = 4$ and $k-1 = 3$,which is consistent.
Therefore,the value of $k$ is $4$.

(C) Let $L = \lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}}$.
Since the form is $1^{\infty}$,we use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} g(x)(f(x)-1)}$.
Here,$f(x) = x-1$ and $g(x) = \frac{1}{3x-6}$.
$L = e^{\lim _{x \rightarrow 2} \frac{1}{3(x-2)} (x-1-1)}$.
$L = e^{\lim _{x \rightarrow 2} \frac{x-2}{3(x-2)}}$.
$L = e^{\lim _{x \rightarrow 2} \frac{1}{3}} = e^{\frac{1}{3}}$.

(B) Let $L = \lim _{x \rightarrow 1} \frac{a b^x - a^x b}{x^2 - 1}$.
Since the limit is of the form $\frac{0}{0}$ at $x = 1$,we apply $L$' Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(a b^x - a^x b)}{\frac{d}{dx}(x^2 - 1)}$
$L = \lim _{x \rightarrow 1} \frac{a b^x \ln b - b a^x \ln a}{2x}$
Substituting $x = 1$:
$L = \frac{a b^1 \ln b - b a^1 \ln a}{2(1)}$
$L = \frac{ab \ln b - ab \ln a}{2}$
$L = \frac{ab}{2} (\ln b - \ln a)$
$L = \frac{ab}{2} \ln \left(\frac{b}{a}\right)$

(A) Given the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$
Since the form is $\frac{0}{0}$ as $x \rightarrow 1$,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$= \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(\sqrt{x}-1)}{\frac{d}{dx}(\log x)}$
$= \lim _{x \rightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}}$
$= \lim _{x \rightarrow 1} \frac{x}{2\sqrt{x}} = \lim _{x \rightarrow 1} \frac{\sqrt{x}}{2}$
$= \frac{\sqrt{1}}{2} = \frac{1}{2}$

(C) We use the trigonometric identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.
$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} = \lim _{x \rightarrow 0} \frac{-2 \sin \frac{(m+n) x}{2} \sin \frac{(m-n) x}{2}}{x^2}$
$= -2 \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{x} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{x} \right)$
Multiplying and dividing by $\frac{m+n}{2}$ and $\frac{m-n}{2}$ respectively:
$= -2 \left( \frac{m+n}{2} \right) \left( \frac{m-n}{2} \right) \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{\frac{(m+n)x}{2}} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{\frac{(m-n)x}{2}} \right)$
$= -2 \left( \frac{m^2-n^2}{4} \right) (1)(1) = \frac{n^2-m^2}{2}$

(A) We know that $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$.
Applying this to the numerator and denominator:
$\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin^2(x^2/2)}}{2 \sin^2(x/2)} = \lim _{x \rightarrow 0} \frac{\sqrt{2} |\sin(x^2/2)|}{2 \sin^2(x/2)}$.
Since $x \rightarrow 0$,$\sin(x^2/2) > 0$,so we can remove the absolute value.
$\lim _{x \rightarrow 0} \frac{\sqrt{2} \sin(x^2/2)}{2 \sin^2(x/2)} = \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{\sin^2(x/2)}$.
Using $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$= \frac{1}{\sqrt{2}} \cdot 1 \cdot \lim _{x \rightarrow 0} \frac{x^2/2}{(\sin(x/2))^2} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot \lim _{x \rightarrow 0} \left( \frac{x/2}{\sin(x/2)} \right)^2 \cdot 4 = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot 1^2 \cdot 4 = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) First,we evaluate $a$:
$a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}$.
Next,we evaluate $b$:
$b = \lim_{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3} = \lim_{n \rightarrow \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \lim_{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2}{6} = \frac{1}{3}$.
Comparing the values,$a = \frac{1}{2}$ and $b = \frac{1}{3}$.
Therefore,$2a = 2(\frac{1}{2}) = 1$ and $3b = 3(\frac{1}{3}) = 1$.
Thus,$2a = 3b$.

(A) The system of linear inequalities is given by:
$1) x+y \geq 1$
$2) 7x+9y \leq 63$
$3) y \leq 5$
$4) x \leq 6$
$5) x \geq 0, y \geq 0$
To find the feasible region,we analyze the boundaries:
- The line $x+y=1$ passes through $(1,0)$ and $(0,1)$. The region $x+y \geq 1$ is away from the origin.
- The line $7x+9y=63$ passes through $(9,0)$ and $(0,7)$. The region $7x+9y \leq 63$ is towards the origin.
- The lines $x=6$ and $y=5$ are vertical and horizontal lines respectively,bounding the region.
- The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.
By plotting these lines and considering the inequalities,the feasible region is the bounded polygon in the first quadrant that satisfies all conditions. Comparing this with the provided options,the figure in the solution image (which matches option $A$) correctly represents the intersection of all these half-planes.

(C) To determine the correct system of linear inequalities for the shaded region,we analyze the boundary lines and the direction of the shaded area relative to the origin $(0,0)$.
$1$. The region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.
$2$. For the line $3x + 4y = 18$,the origin $(0,0)$ satisfies $3(0) + 4(0) = 0 < 18$. Since the shaded region contains the origin side of this line,the inequality is $3x + 4y \leq 18$.
$3$. For the line $x - 6y = 3$,the origin $(0,0)$ satisfies $0 - 0 = 0 < 3$. The shaded region is on the side of the origin,so the inequality is $x - 6y \leq 3$.
$4$. For the line $2x + 3y = 3$,the origin $(0,0)$ satisfies $2(0) + 3(0) = 0 < 3$. However,the shaded region is on the opposite side of the origin,so the inequality is $2x + 3y \geq 3$.
$5$. For the line $-7x + 14y = 14$,the origin $(0,0)$ satisfies $-7(0) + 14(0) = 0 < 14$. The shaded region is on the side of the origin,so the inequality is $-7x + 14y \leq 14$.
Comparing these with the given options,option $C$ matches all these conditions.

(D) To find the common region,we analyze the given inequations:
$1$. $x+y \geq 5$: This represents the region on or above the line $x+y=5$. Since $(0,0)$ does not satisfy this ($0 \geq 5$ is false),the region is on the non-origin side.
$2$. $y \leq 4$: This represents the region on or below the line $y=4$.
$3$. $x \geq 2$: This represents the region on or to the right of the line $x=2$.
$4$. $x, y \geq 0$: This restricts the region to the first quadrant.
By plotting these lines,we observe the shaded region in the figure. The region is bounded by the lines $x=2$,$y=4$,and $x+y=5$. Since the region is enclosed by these lines,it is a bounded region. Furthermore,the origin $(0,0)$ does not lie in the shaded region,so it is on the non-origin side.
Therefore,the correct option is $D$.

(B) To determine the common region,we analyze the given inequations:
$1$. $x+2y \geq 4$: The boundary line is $x+2y=4$. Testing the origin $(0,0)$,we get $0+0 \geq 4$,which is false. Thus,the region is on the non-origin side of the line.
$2$. $2x-y \leq 6$: The boundary line is $2x-y=6$. Testing the origin $(0,0)$,we get $0-0 \leq 6$,which is true. Thus,the region is on the origin side of the line.
$3$. $x, y > 0$: This restricts the region to the first quadrant.
By observing the intersection of these regions,the feasible region is not enclosed by a finite boundary,meaning it is unbounded. Since the region does not include the origin and is defined by the intersection of these half-planes in the first quadrant,it is an unbounded region on the non-origin side. Therefore,the correct option is $B$.

(B) To determine the nature of the region,we analyze the given inequalities:
$1$. $x \geq 6$: This represents the region to the right of the vertical line $x = 6$.
$2$. $y \geq 3$: This represents the region above the horizontal line $y = 3$.
$3$. $2x + y \geq 10$: This represents the region on or above the line $2x + y = 10$.
$4$. $x \geq 0, y \geq 0$: These represent the first quadrant.
By plotting these lines,we observe that the intersection of these regions starts from the point $(6, 3)$ and extends infinitely in the positive $x$ and $y$ directions.
Since the region does not have a finite area and extends infinitely,it is an unbounded region.
Therefore,the correct option is $B$.

(C) We simplify the expression using logical laws:
$(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$
Using the distributive law on the first two terms:
$\equiv ((p \vee q) \wedge (\sim q \vee q)) \vee (\sim p \wedge q)$
Since $(\sim q \vee q) \equiv T$ (Complement law):
$\equiv ((p \vee q) \wedge T) \vee (\sim p \wedge q)$
$\equiv (p \vee q) \vee (\sim p \wedge q)$
Applying the distributive law again:
$\equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$
Since $(p \vee \sim p) \equiv T$ and $(q \vee q) \equiv q$ (Idempotent law):
$\equiv (T \vee q) \wedge (p \vee q)$
Since $(T \vee q) \equiv T$:
$\equiv T \wedge (p \vee q)$
$\equiv p \vee q$
Therefore,the expression is equivalent to $p \vee q$.

(D) To simplify the logical expression $p \wedge (\sim p \vee \sim q)$,we apply the distributive law of logic:
$p \wedge (\sim p \vee \sim q) \equiv (p \wedge \sim p) \vee (p \wedge \sim q)$.
Since $(p \wedge \sim p)$ is a contradiction,it is equivalent to $F$ (False).
Therefore,the expression becomes:
$F \vee (p \wedge \sim q) \equiv p \wedge \sim q$.
Thus,the correct expression is $p \wedge \sim q$.

(B) Let $p$ be the statement '$x \in A \cap B$' and $q$ be the statement '$x \in A \text{ and } x \in B$'.
The given statement is in the form $p \rightarrow q$.
The negation of an implication $p \rightarrow q$ is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Here,$p$ is '$x \in A \cap B$' and $\sim q$ is the negation of '$x \in A \text{ and } x \in B$',which by De Morgan's Law is '$x \notin A \text{ or } x \notin B$'.
Therefore,the negation is '$x \in A \cap B \text{ and } (x \notin A \text{ or } x \notin B)$'.

(C) Given statements:
$p$: It rains today
$q$: $I$ am going to school
$r$: $I$ will meet my friend
$s$: $I$ will go to watch a movie
The statement is: "If (it does not rain today $OR$ $I$ will not go to school),then ($I$ will meet my friend $AND$ $I$ will go to watch a movie)."
Symbolically:
"It does not rain today" is $\sim p$.
"$I$ will not go to school" is $\sim q$.
"$I$ will meet my friend" is $r$.
"$I$ will go to watch a movie" is $s$.
Combining these using logical connectives:
$(\sim p \vee \sim q) \rightarrow (r \wedge s)$
Using De Morgan's Law,$\sim p \vee \sim q \equiv \sim(p \wedge q)$.
Therefore,the symbolic form is $\sim(p \wedge q) \rightarrow (r \wedge s)$.

(D) To find the negation of the implication $(p \wedge q) \rightarrow (\sim p \vee r)$,we use the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Let $A = (p \wedge q)$ and $B = (\sim p \vee r)$.
The negation is $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Substituting the values:
$(p \wedge q) \wedge \sim(\sim p \vee r)$
Using De Morgan's Law,$\sim(\sim p \vee r) \equiv \sim(\sim p) \wedge \sim r \equiv p \wedge \sim r$.
Therefore,the expression becomes:
$(p \wedge q) \wedge (p \wedge \sim r)$
By the associative and idempotent laws,this simplifies to:
$p \wedge q \wedge (\sim r)$.

(C) Given: $p = T, q = T, r = F$.
Evaluating each option:
$(A)$ $(p \vee q) \vee r \equiv (T \vee T) \vee F \equiv T \vee F \equiv T$.
$(B)$ $(p \wedge q)$ $\rightarrow r \equiv (T \wedge T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
$(C)$ $(p$ $\rightarrow r)$ $\rightarrow q \equiv (T$ $\rightarrow F)$ $\rightarrow T \equiv F$ $\rightarrow T \equiv T$.
$(D)$ $(p \leftrightarrow q)$ $\rightarrow r \equiv (T \leftrightarrow T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
Since the question asks which statement is correct,and option $(C)$ evaluates to $T$,option $(C)$ is the correct statement.

(C) The negation of a universal quantifier statement $\forall x \in S, P(x)$ is given by $\exists x \in S$ such that $\neg P(x)$.
Here,the statement is $\forall x \in \mathbb{R}, x^2+1=0$.
Applying the rule,the negation is $\exists x \in \mathbb{R}$ such that $x^2+1 \neq 0$.

(A) Given the logical statement: $(p$ $\rightarrow q) \wedge (p$ $\rightarrow \sim p)$
Using the implication law $a \rightarrow b \equiv \sim a \vee b$:
$(p \rightarrow q) \equiv \sim p \vee q$
$(p \rightarrow \sim p) \equiv \sim p \vee \sim p \equiv \sim p$
Now,the expression becomes: $(\sim p \vee q) \wedge (\sim p)$
Using the absorption law: $A \wedge (A \vee B) \equiv A$
Here,let $A = \sim p$ and $B = q$.
So,$(\sim p) \wedge (\sim p \vee q) \equiv \sim p$
Therefore,the statement is equivalent to $\sim p$.

(B) Given: $p = T, q = T, r = F, s = F$.
For the first expression $\sim(p \rightarrow q) \leftrightarrow (p \wedge s)$:
$\sim(T \rightarrow T) \leftrightarrow (T \wedge F)$
$= \sim(T) \leftrightarrow (F)$
$= F \leftrightarrow F = T$.
For the second expression $(\sim p \rightarrow q) \wedge (r \leftrightarrow s)$:
$(\sim T \rightarrow T) \wedge (F \leftrightarrow F)$
$= (F \rightarrow T) \wedge (T)$
$= T \wedge T = T$.
Thus,the truth values are $T, T$.

(C) Given $p = T, q = T, r = F, s = F$.
For $a : \sim (p \wedge \sim r) \vee (\sim q \vee s)$:
$a \equiv \sim (T \wedge \sim F) \vee (\sim T \vee F)$
$a \equiv \sim (T \wedge T) \vee (F \vee F)$
$a \equiv \sim T \vee F$
$a \equiv F \vee F$
$a \equiv F$.
For $b : (p \vee s) \leftrightarrow (q \wedge r)$:
$b \equiv (T \vee F) \leftrightarrow (T \wedge F)$
$b \equiv T \leftrightarrow F$
$b \equiv F$.
Thus,the truth values of $a$ and $b$ are $F$ and $F$ respectively.

(D) Given the curve equation is $y = 4xe^{x}$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the product rule:
$\frac{dy}{dx} = 4(e^{x} + xe^{x}) = 4e^{x}(1 + x)$.
Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:
$\left(\frac{dy}{dx}\right)_{x=-1} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.
Since the slope of the tangent is $0$,the tangent is a horizontal line parallel to the $X$-axis.
The equation of a horizontal line passing through $\left(-1, -\frac{4}{e}\right)$ is $y = y_{1}$,which is $y = -\frac{4}{e}$.

(C) Given the curve equation is $y=x^3-3x^2-9x+5$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 3x^2-6x-9$.
Since the tangent is parallel to the $X$-axis,its slope must be zero:
$\frac{dy}{dx} = 0
\Rightarrow 3x^2-6x-9 = 0$.
Dividing by $3$,we get:
$x^2-2x-3 = 0$.
Factoring the quadratic equation:
$(x-3)(x+1) = 0$.
Thus,the values of $x$ are $x=3$ and $x=-1$.

(A) Given curve: $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \sqrt{2} \cos \left(2x + \frac{\pi}{4}\right) \cdot 2 = 2\sqrt{2} \cos \left(2x + \frac{\pi}{4}\right)$.
At $x = \frac{\pi}{4}$,the slope $m$ is:
$m = \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = 2\sqrt{2} \cos \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = 2\sqrt{2} \cos \left(\frac{3\pi}{4}\right) = 2\sqrt{2} \left(-\frac{1}{\sqrt{2}}\right) = -2$.
Now,find the $y$-coordinate at $x = \frac{\pi}{4}$:
$y = \sqrt{2} \sin \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = \sqrt{2} \sin \left(\frac{3\pi}{4}\right) = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = 1$.
The point of tangency is $\left(\frac{\pi}{4}, 1\right)$.
The equation of the tangent is $(y - y_1) = m(x - x_1)$:
$(y - 1) = -2 \left(x - \frac{\pi}{4}\right)$
$y - 1 = -2x + \frac{\pi}{2}$
$2x + y - \frac{\pi}{2} - 1 = 0$.

(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point is $0$.
Substituting $P(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \quad (1)$.
The derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$. At $P(-2,0)$,$\frac{dy}{dx} = 0$: $3a(-2)^2 + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \quad (2)$.
The curve cuts the $Y$-axis at $Q(0,k)$. At $Q$,$x=0$ and the gradient is $3$: $\frac{dy}{dx}|_{x=0} = 3(a)(0)^2 + 2(b)(0) + c = 3 \Rightarrow c = 3$.
Substituting $c=3$ into $(1)$ and $(2)$:
$(1): 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \quad (3)$.
$(2): 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \quad (4)$.
Subtracting $(3)$ from $(4)$: $(12a - 8a) = -3 - (-1) \Rightarrow 4a = -2 \Rightarrow a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into $(3)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$.
Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 36 \ m^3/min$ and the radius of the base is $r = 3 \ m$.
The volume of a cylinder is given by $V = \pi r^2 h$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Substituting the given values,$36 = \pi \times (3)^2 \times \frac{dh}{dt}$.
$36 = 9 \pi \times \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{36}{9 \pi} = \frac{4}{\pi} \ m/min$.

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 \text{ cm/sec}$.
At the instant when the radius $r = 10 \text{ cm}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$, we get:
$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 2 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi \times 10 \times 2 = 40 \pi \text{ cm}^2\text{/sec}$.
Thus, the rate of increase in the area is $40 \pi \text{ cm}^2\text{/sec}$.

(D) The position of the particle is given by $S(t) = at^2 + bt + 6$.
At $t = 0$,the initial position is $S(0) = 6 \text{ m}$.
The displacement from the starting point at time $t$ is $S(t) - S(0) = at^2 + bt$.
Given that at $t = 4 \text{ s}$,the displacement is $16 \text{ m}$,so $a(4)^2 + b(4) = 16 \Rightarrow 16a + 4b = 16 \Rightarrow 4a + b = 4$ (Equation $1$).
The velocity is $v(t) = \frac{dS}{dt} = 2at + b$.
Since the particle comes to rest at $t = 4 \text{ s}$,$v(4) = 0 \Rightarrow 2a(4) + b = 0 \Rightarrow 8a + b = 0$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(8a + b) - (4a + b) = 0 - 4 \Rightarrow 4a = -4 \Rightarrow a = -1$.
Substituting $a = -1$ into Equation $2$: $8(-1) + b = 0 \Rightarrow b = 8$.
The acceleration is $a_{acc} = \frac{dv}{dt} = 2a = 2(-1) = -2 \text{ m/s}^2$.
Re-evaluating the interpretation: If $S$ represents the total distance from the origin and the displacement is $16 \text{ m}$,then $S(4) - S(0) = 16$. If the question implies $S(4) = 16$,then $16a + 4b + 6 = 16 \Rightarrow 16a + 4b = 10 \Rightarrow 8a + 2b = 5$. Solving $8a + 2b = 5$ and $8a + b = 0$ gives $b = 5$ and $a = -5/8$. Thus,$a_{acc} = 2a = -5/4 \text{ m/s}^2$. This matches option $D$.

(D) Given that the rate of change of surface area $\frac{dA}{dt} = 2 \text{ cm}^2/\text{sec}$.
The surface area of a sphere is $A = 4\pi r^2$.
Differentiating with respect to $t$,we get $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$.
Substituting the given values: $2 = 8\pi(6) \frac{dr}{dt} \implies 2 = 48\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{24\pi} \text{ cm/sec}$.
The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{24\pi}$:
$\frac{dV}{dt} = 4\pi(6)^2 \left(\frac{1}{24\pi}\right) = 4\pi(36) \left(\frac{1}{24\pi}\right) = \frac{144\pi}{24\pi} = 6 \text{ cm}^3/\text{sec}$.

(D) The volume $V$ of a spherical snowball is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 8 \text{ cm}^3/\text{sec}$ and $r = 2 \text{ cm}$.
Substituting these values,we have $8 = 4 \pi (2)^2 \frac{dr}{dt}$.
$8 = 16 \pi \frac{dr}{dt}$.
Therefore,$\frac{dr}{dt} = \frac{8}{16\pi} = \frac{1}{2\pi} \text{ cm/sec}$.

(C) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop. We are given that the rate of evaporation is proportional to the surface area,so $\frac{dV}{dt} = -kS$,where $k > 0$ is a constant.
Since $V = \frac{4}{3}\pi r^3$,we have $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
The surface area $S = 4\pi r^2$.
Substituting these into the rate equation: $4\pi r^2 \frac{dr}{dt} = -k(4\pi r^2)$.
This simplifies to $\frac{dr}{dt} = -k$.
Integrating with respect to $t$,we get $r = -kt + c$.
At $t = 0$,$r = 3$,so $3 = -k(0) + c \Rightarrow c = 3$.
Thus,$r = -kt + 3$.
At $t = 1$,$r = 2$,so $2 = -k(1) + 3 \Rightarrow k = 1$.
Substituting $k=1$ and $c=3$ into the expression for $r$,we get $r = -t + 3$ or $r = 3 - t$.

(B) Given that the rate of change of the radius is $\frac{dr}{dt} = 0.01 \text{ cm/sec}$.
The area of a circular plate is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = \pi (2r) \frac{dr}{dt}$.
Substituting the given values $r = 12 \text{ cm}$ and $\frac{dr}{dt} = 0.01 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi (12) (0.01) = 0.24 \pi \text{ cm}^2/\text{sec}$.
Thus,the area increases at the rate of $0.24 \pi \text{ cm}^2/\text{sec}$.

(C) The distance covered by the particle is given by $s = 2 + 27t - t^3$.
To find when the particle stops,we need to find the time $t$ when its velocity $v = \frac{ds}{dt}$ is zero.
$\frac{ds}{dt} = \frac{d}{dt}(2 + 27t - t^3) = 27 - 3t^2$.
Setting the velocity to zero: $27 - 3t^2 = 0$.
$3t^2 = 27 \Rightarrow t^2 = 9$.
Since time $t > 0$,we have $t = 3 \text{ seconds}$.
Now,we calculate the distance covered at $t = 3 \text{ seconds}$:
$s(3) = 2 + 27(3) - (3)^3$.
$s(3) = 2 + 81 - 27$.
$s(3) = 56 \text{ meters}$.
Therefore,the particle will stop after covering $56 \text{ meters}$.

(A) Given function is $f(x) = \cot^{-1} x + x$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(\cot^{-1} x) + \frac{d}{dx}(x) = -\frac{1}{1+x^2} + 1$.
Simplifying the expression: $f'(x) = \frac{-1 + (1+x^2)}{1+x^2} = \frac{x^2}{1+x^2}$.
Since $x^2 \geq 0$ for all $x \in \mathbb{R}$ and $1+x^2 > 0$,it follows that $f'(x) \geq 0$ for all $x \in \mathbb{R}$.
Since $f'(x) \geq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on the entire real line $(-\infty, \infty)$.

(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.
For the function to be increasing,we must have $f'(x) \geq 0$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we calculate $f'(x)$:
$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$.
Expanding the numerator:
Numerator $= (2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)$.
Simplifying the numerator:
Numerator $= 3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x$.
Numerator $= 3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x) = 3\lambda - 12$.
Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,$f'(x) \geq 0$ implies $3\lambda - 12 \geq 0$.
Therefore,$3\lambda \geq 12$,which gives $\lambda \geq 4$.

(C) Given the function $f(x) = \log |\sin x|$ for $x \in (0, \pi)$.
Since $x \in (0, \pi)$,$\sin x$ is always positive,so we can write $f(x) = \log(\sin x)$.
To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Therefore,$\cot x > 0$.
In the interval $(0, \pi)$,$\cot x$ is positive when $x \in \left(0, \frac{\pi}{2}\right)$.
Thus,the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.

(D) Given the function $f(x) = e^{-1/x}$.
To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(e^{-1/x}) = e^{-1/x} \cdot \frac{d}{dx}(-x^{-1}) = e^{-1/x} \cdot (x^{-2}) = \frac{1}{x^2 e^{1/x}}$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Since $x^2 > 0$ for all $x \neq 0$ and $e^{1/x} > 0$ for all $x \neq 0$,the derivative $f'(x) = \frac{1}{x^2 e^{1/x}}$ is always positive for all $x$ in its domain.
The domain of the function $f(x) = e^{-1/x}$ is all real numbers except $x = 0$.
Therefore,the function is strictly increasing for all $x \in \mathbb{R} \setminus \{0\}$.

(B) Let the two parts be $x$ and $y$. Given $x + y = 20$,so $y = 20 - x$.
Let the function be $f(x) = x(20 - x)^3$.
To find the maximum,we differentiate $f(x)$ with respect to $x$:
$f'(x) = (20 - x)^3 + x \cdot 3(20 - x)^2(-1)$
$f'(x) = (20 - x)^2 [20 - x - 3x] = (20 - x)^2 (20 - 4x)$.
Setting $f'(x) = 0$,we get $x = 20$ or $x = 5$.
Since $x=20$ gives $f(x)=0$ (minimum),we check $x = 5$.
For $x = 5$,$y = 20 - 5 = 15$.
The product of the two parts is $xy = 5 \times 15 = 75$.

(B) Given the function $f(x)=2x^3-15x^2-144x-7$.
To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^3-15x^2-144x-7) = 6x^2-30x-144$.
For $f(x)$ to be strictly decreasing,$f'(x) < 0$:
$6x^2-30x-144 < 0$.
Dividing by $6$,we get $x^2-5x-24 < 0$.
Factoring the quadratic expression: $(x-8)(x+3) < 0$.
The roots are $x=8$ and $x=-3$.
The inequality holds for $x$ in the interval $(-3, 8)$.
Thus,$f(x)$ is strictly decreasing in $(-3, 8)$.

(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
Given the slant height $\ell = 3 \text{ cm}$,we have the relation $\ell^2 = r^2 + h^2$,which implies $r^2 = 9 - h^2$.
Substituting $r^2$ into the volume formula: $V = \frac{1}{3} \pi (9 - h^2) h = 3 \pi h - \frac{\pi}{3} h^3$.
To find the maximum volume,we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = 3 \pi - \pi h^2$.
Setting $\frac{dV}{dh} = 0$,we get $3 \pi = \pi h^2$,which gives $h^2 = 3$,so $h = \sqrt{3} \text{ cm}$ (since height must be positive).
Checking the second derivative: $\frac{d^2V}{dh^2} = -2 \pi h$.
At $h = \sqrt{3}$,$\frac{d^2V}{dh^2} = -2 \sqrt{3} \pi < 0$.
Since the second derivative is negative,the volume is maximum when $h = \sqrt{3} \text{ cm}$.

(A) Given the function $f(x) = x^2 + ax + b$.
To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$.
Setting $f'(x) = 0$,we get $x = -a/2$.
Since the second derivative $f''(x) = 2 > 0$,the function has a local minimum at $x = -a/2$.
According to the problem,the minimum occurs at $x = 3$.
Therefore,$-a/2 = 3$,which implies $a = -6$.
Given that the minimum value of the function is $5$ at $x = 3$,we substitute these values into the original function:
$f(3) = (3)^2 + a(3) + b = 5$.
Substituting $a = -6$:
$9 + (-6)(3) + b = 5$.
$9 - 18 + b = 5$.
$-9 + b = 5$.
$b = 14$.
Thus,the values are $a = -6$ and $b = 14$.

(A) Let the rectangle $ABCD$ be inscribed in a circle of radius $r$ centered at the origin. Let the coordinates of vertex $B$ be $(r \cos \theta, r \sin \theta)$.
Then,the length $AB = 2r \cos \theta$ and the width $BC = 2r \sin \theta$.
The area of the rectangle $A(\theta) = AB \times BC = (2r \cos \theta)(2r \sin \theta) = 2r^2 \sin 2\theta$.
To find the maximum area,we differentiate $A(\theta)$ with respect to $\theta$:
$A'(\theta) = 4r^2 \cos 2\theta$.
Setting $A'(\theta) = 0$,we get $\cos 2\theta = 0$,which implies $2\theta = \frac{\pi}{2}$,so $\theta = \frac{\pi}{4}$.
Checking the second derivative: $A''(\theta) = -8r^2 \sin 2\theta$.
At $\theta = \frac{\pi}{4}$,$A''(\frac{\pi}{4}) = -8r^2 \sin(\frac{\pi}{2}) = -8r^2 < 0$.
Since the second derivative is negative,the area is maximum at $\theta = \frac{\pi}{4}$.
Substituting $\theta = \frac{\pi}{4}$ into the area formula:
Maximum Area $= 2r^2 \sin(2 \times \frac{\pi}{4}) = 2r^2 \sin(\frac{\pi}{2}) = 2r^2(1) = 2r^2$ sq. units.

(A) Given the function $y=x^3-\alpha x^2-\beta x+5$.
Find the derivative with respect to $x$:
$\frac{dy}{dx} = 3x^2 - 2\alpha x - \beta$.
Since $x=-2$ and $x=4$ are extreme points,the derivative must be zero at these values.
For $x=-2$:
$3(-2)^2 - 2\alpha(-2) - \beta = 0 \implies 12 + 4\alpha - \beta = 0 \implies 4\alpha - \beta = -12$ (Equation $1$).
For $x=4$:
$3(4)^2 - 2\alpha(4) - \beta = 0 \implies 48 - 8\alpha - \beta = 0 \implies 8\alpha + \beta = 48$ (Equation $2$).
Adding Equation $1$ and Equation $2$:
$(4\alpha - \beta) + (8\alpha + \beta) = -12 + 48 \implies 12\alpha = 36 \implies \alpha = 3$.
Substitute $\alpha = 3$ into Equation $1$:
$4(3) - \beta = -12 \implies 12 - \beta = -12 \implies \beta = 24$.
Thus,$\alpha=3$ and $\beta=24$.

(A) Let the two parts of $10$ be $x$ and $(10-x)$.
Define the function $f(x) = 2x + (10-x)^2$.
Expanding the function: $f(x) = 2x + 100 - 20x + x^2 = x^2 - 18x + 100$.
To find the minimum,find the first derivative: $f'(x) = 2x - 18$.
Set $f'(x) = 0$ to find the critical point: $2x - 18 = 0 \implies x = 9$.
Find the second derivative: $f''(x) = 2$.
Since $f''(9) = 2 > 0$,the function has a minimum at $x = 9$.
The first part is $x = 9$ and the second part is $10 - 9 = 1$.
Therefore,the numbers are $9$ and $1$.

(A) We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$.
To find the critical points,we calculate the derivative $f^{\prime}(x)$ using the quotient rule:
$f^{\prime}(x)=\frac{(1+x+x^2)(2x-1)-(1-x+x^2)(2x+1)}{(1+x+x^2)^2}$.
Expanding the numerator:
$f^{\prime}(x)=\frac{(2x-1+2x^2-x+2x^3-x^2)-(2x+1-2x^2-x+2x^3+x^2)}{(1+x+x^2)^2}$.
$f^{\prime}(x)=\frac{(2x^3+x^2+x-1)-(2x^3-x^2+x+1)}{(1+x+x^2)^2}$.
$f^{\prime}(x)=\frac{2x^2-2}{(1+x+x^2)^2} = \frac{2(x^2-1)}{(1+x+x^2)^2}$.
Setting $f^{\prime}(x)=0$,we get $x^2-1=0$,which implies $x=1$ or $x=-1$.
Evaluating the function at these points:
For $x=1$,$f(1)=\frac{1-1+1}{1+1+1} = \frac{1}{3}$.
For $x=-1$,$f(-1)=\frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = 3$.
Comparing the values,the minimum value of $f(x)$ is $\frac{1}{3}$.

(D) Given the function $f(x) = x \log x$.
To find the minimum value,we first find the derivative $f'(x)$.
Using the product rule: $f'(x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x = 1 + \log x$.
For critical points,set $f'(x) = 0$:
$1 + \log x = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e}$.
Now,find the second derivative $f''(x)$:
$f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$.
Evaluate $f''(x)$ at $x = \frac{1}{e}$:
$f''(\frac{1}{e}) = \frac{1}{1/e} = e$.
Since $e > 0$,the function has a local minimum at $x = \frac{1}{e}$.
The minimum value is $f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} \log(e^{-1}) = \frac{1}{e} (-1) = -\frac{1}{e}$.

(C) The required area is the sum of the absolute values of the integrals over the intervals where the function is negative and positive.
$A = \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$
Since $x < 0$ for $x \in [-1, 0]$ and $x > 0$ for $x \in [0, 2]$,we have:
$A = \int_{-1}^{0} (-x) dx + \int_{0}^{2} (x) dx$
$A = \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$A = (0 - (-\frac{(-1)^2}{2})) + (\frac{2^2}{2} - 0)$
$A = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq. units}$

(B) To find the area bounded by the curve $y=2x-x^2$ and the $x$-axis,we first find the intersection points with the $x$-axis by setting $y=0$.
$2x-x^2=0 \Rightarrow x(2-x)=0$,which gives $x=0$ and $x=2$.
The area $A$ is given by the integral:
$A = \int_0^2 (2x-x^2) dx$
$A = [x^2 - \frac{x^3}{3}]_0^2$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3} \text{ sq units}$.

(D) To find the area bounded by the parabola $y=x^2$ and the line $y=x$,we first determine the points of intersection by setting the equations equal to each other:
$x^2 = x$
$x^2 - x = 0$
$x(x - 1) = 0$
This gives $x = 0$ and $x = 1$. The points of intersection are $O(0, 0)$ and $P(1, 1)$.
In the interval $[0, 1]$,the line $y = x$ lies above the parabola $y = x^2$.
The required area $A$ is given by the integral:
$A = \int_0^1 (x - x^2) dx$
$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$
$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - (0 - 0)$
$A = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \text{ sq. units}$

(A) To find the area of the region bounded by the curve $y^2=4x$ and the line $y=x$,we first determine the points of intersection by substituting $y=x$ into the equation of the parabola:
$x^2 = 4x$
$x^2 - 4x = 0$
$x(x-4) = 0$
Thus,the points of intersection are $x=0$ and $x=4$.
For $x=0$,$y=0$,so the origin $O(0,0)$ is one point.
For $x=4$,$y=4$,so the point $P(4,4)$ is the other point.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$A = \int_0^4 (\sqrt{4x} - x) dx$
$A = 2 \int_0^4 x^{1/2} dx - \int_0^4 x dx$
$A = 2 \left[ \frac{x^{3/2}}{3/2} \right]_0^4 - \left[ \frac{x^2}{2} \right]_0^4$
$A = 2 \cdot \frac{2}{3} [x^{3/2}]_0^4 - \left[ \frac{16}{2} - 0 \right]$
$A = \frac{4}{3} (4^{3/2}) - 8$
$A = \frac{4}{3} (8) - 8$
$A = \frac{32}{3} - 8 = \frac{32-24}{3} = \frac{8}{3} \text{ sq. units.}$

(B) The region is bounded by the parabola $y^2=x$,the line $y=4$,and the $Y$-axis $(x=0)$.
To find the area,we integrate with respect to $y$ from $y=0$ to $y=4$.
The equation of the parabola is $x=y^2$.
The area $A$ is given by:
$A = \int_{0}^{4} x \, dy$
$A = \int_{0}^{4} y^2 \, dy$
$A = \left[ \frac{y^3}{3} \right]_{0}^{4}$
$A = \frac{4^3}{3} - \frac{0^3}{3}$
$A = \frac{64}{3} \text{ sq. units}$

(A) To find the area bounded between the curve $x^2=y$ and the line $y=4x$,we first find their points of intersection.
Setting $x^2 = 4x$,we get $x^2 - 4x = 0$,which implies $x(x-4) = 0$. Thus,$x=0$ and $x=4$.
For $x=0$,$y=0$,and for $x=4$,$y=16$. So the points of intersection are $(0,0)$ and $(4,16)$.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$A = \int_{0}^{4} (4x - x^2) dx$
$A = \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_{0}^{4}$
$A = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}$
$A = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - (0 - 0)$
$A = \left( 32 - \frac{64}{3} \right)$
$A = \frac{96 - 64}{3} = \frac{32}{3} \text{ sq. units}$

(A) To find the area bounded by the parabola $y^2=x$ and the line $x+y=2$ in the first quadrant,we first find the points of intersection.
Substituting $y = 2-x$ into $y^2=x$,we get $(2-x)^2 = x$,which simplifies to $x^2 - 4x + 4 = x$,or $x^2 - 5x + 4 = 0$.
Factoring gives $(x-4)(x-1) = 0$,so $x=1$ or $x=4$.
In the first quadrant,the intersection point is $(1, 1)$.
The line $x+y=2$ intersects the $x$-axis at $(2, 0)$.
The required area is the integral of the parabola from $x=0$ to $x=1$ plus the integral of the line from $x=1$ to $x=2$.
Area $= \int_0^1 \sqrt{x} \, dx + \int_1^2 (2-x) \, dx$
$= \left[ \frac{2}{3} x^{3/2} \right]_0^1 + \left[ 2x - \frac{x^2}{2} \right]_1^2$
$= \left( \frac{2}{3} - 0 \right) + \left( (4 - 2) - (2 - \frac{1}{2}) \right)$
$= \frac{2}{3} + (2 - \frac{3}{2}) = \frac{2}{3} + \frac{1}{2} = \frac{4+3}{6} = \frac{7}{6} \text{ sq. units}$.

(B) The given parabolas are $y^2=8x$ and $x^2=8y$.
To find the points of intersection,substitute $y = \frac{x^2}{8}$ into $y^2=8x$:
$\left(\frac{x^2}{8}\right)^2 = 8x \Rightarrow \frac{x^4}{64} = 8x \Rightarrow x^4 = 512x \Rightarrow x(x^3 - 512) = 0$.
Thus,$x=0$ or $x=8$. The intersection points are $O(0,0)$ and $P(8,8)$.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=8$:
$A = \int_0^8 (\sqrt{8x} - \frac{x^2}{8}) dx = \int_0^8 (2\sqrt{2}\sqrt{x} - \frac{x^2}{8}) dx$.
$A = [2\sqrt{2} \cdot \frac{x^{3/2}}{3/2}]_0^8 - [\frac{x^3}{24}]_0^8$.
$A = [\frac{4\sqrt{2}}{3} \cdot x^{3/2}]_0^8 - \frac{512}{24}$.
$A = \frac{4\sqrt{2}}{3} \cdot (8\sqrt{8}) - \frac{64}{3} = \frac{4\sqrt{2}}{3} \cdot (16\sqrt{2}) - \frac{64}{3} = \frac{128}{3} - \frac{64}{3} = \frac{64}{3} \text{ sq. units}$.

(A) The parabola is $y^2=4ax$. The latus-rectum is the line $x=a$.
To find the area,we integrate with respect to $x$ from $0$ to $a$.
Since the parabola is symmetric about the $x$-axis,the total area $A$ is twice the area in the first quadrant.
$A = 2 \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx$
$A = 2 \int_{0}^{a} 2\sqrt{a} \sqrt{x} \, dx = 4\sqrt{a} \int_{0}^{a} x^{1/2} \, dx$
$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a} = 4\sqrt{a} \cdot \frac{2}{3} \cdot a^{3/2}$
$A = \frac{8}{3} \sqrt{a} \cdot a \sqrt{a} = \frac{8}{3} a^2 \text{ sq. units}$

(B) Since $f(x)$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.
Let $x - \pi = h$. As $x \rightarrow \pi$,$h \rightarrow 0$.
Then $f(\pi) = \lim_{h \rightarrow 0} \frac{4^h + 4^{-h} - 2}{h^2}$.
We can rewrite the numerator as $(4^h - 1) - (1 - 4^{-h}) = (4^h - 1) - (1 - \frac{1}{4^h}) = (4^h - 1) - \frac{4^h - 1}{4^h} = (4^h - 1)(1 - 4^{-h})$.
Alternatively,using the expansion $a^h = e^{h \ln a} = 1 + h \ln a + \frac{(h \ln a)^2}{2!} + \dots$,we have:
$4^h = 1 + h \ln 4 + \frac{(h \ln 4)^2}{2} + O(h^3)$
$4^{-h} = 1 - h \ln 4 + \frac{(h \ln 4)^2}{2} + O(h^3)$
Adding these: $4^h + 4^{-h} = 2 + (h \ln 4)^2 + O(h^4)$.
Thus,$f(\pi) = \lim_{h \rightarrow 0} \frac{2 + (h \ln 4)^2 - 2}{h^2} = \lim_{h \rightarrow 0} \frac{h^2 (\ln 4)^2}{h^2} = (\ln 4)^2$.
Since $\ln 4 = \ln(2^2) = 2 \ln 2$,we have $f(\pi) = (2 \ln 2)^2 = 4(\ln 2)^2$.
Wait,checking the sign: $4^h + 4^{-h} - 2 = (2^{h/2} - 2^{-h/2})^2$.
As $h \rightarrow 0$,$\frac{(2^{h/2} - 2^{-h/2})^2}{h^2} = \left( \frac{2^{h/2} - 2^{-h/2}}{h} \right)^2 = \left( \frac{1}{2} \cdot \frac{2^{h/2} - 1}{h/2} - \frac{1}{2} \cdot \frac{2^{-h/2} - 1}{h/2} \right)^2 = \left( \frac{1}{2} \ln 2 - \frac{1}{2} (-\ln 2) \right)^2 = (\ln 2)^2$.

(D) For $f(x)$ to be continuous at $x = 1$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow 1^{-}} (1 + \sin \frac{\pi x}{2}) = 1 + \sin \frac{\pi}{2} = 1 + 1 = 2$.
$\lim_{x \rightarrow 1^{+}} (ax + b) = a(1) + b = a + b$.
Thus,$a + b = 2$ --- $(1)$
For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow 3^{-}} (ax + b) = 3a + b$.
$\lim_{x \rightarrow 3^{+}} (6 \tan \frac{x \pi}{12}) = 6 \tan \frac{3 \pi}{12} = 6 \tan \frac{\pi}{4} = 6(1) = 6$.
Thus,$3a + b = 6$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(3a + b) - (a + b) = 6 - 2$
$2a = 4 \implies a = 2$.
Substituting $a = 2$ into equation $(1)$:
$2 + b = 2 \implies b = 0$.
Therefore,the values are $a = 2$ and $b = 0$.

(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit,i.e.,$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x)$.
First,calculate the left-hand limit:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}$
Multiply the numerator and denominator by the conjugate $\sqrt{1+px} + \sqrt{1-px}$:
$= \lim_{x \rightarrow 0^-} \frac{(\sqrt{1+px}-\sqrt{1-px})(\sqrt{1+px}+\sqrt{1-px})}{x(\sqrt{1+px}+\sqrt{1-px})}$
$= \lim_{x \rightarrow 0^-} \frac{(1+px)-(1-px)}{x(\sqrt{1+px}+\sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{2px}{x(\sqrt{1+px}+\sqrt{1-px})}$
$= \lim_{x \rightarrow 0^-} \frac{2p}{\sqrt{1+px}+\sqrt{1-px}} = \frac{2p}{1+1} = p$.
Next,calculate the right-hand limit:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = \frac{-1}{2}$.
Equating the two limits,we get $p = \frac{-1}{2}$.

(B) For $x \leq -3$,$f(x) = |x| + 3 = -x + 3$.
At $x = -3$:
Left-hand limit: $\lim_{x \to -3^-} f(x) = -(-3) + 3 = 6$.
Right-hand limit: $\lim_{x \to -3^+} f(x) = -2(-3) = 6$.
Value of function: $f(-3) = -(-3) + 3 = 6$.
Since $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x) = f(-3)$,$f(x)$ is continuous at $x = -3$.
At $x = 3$:
Left-hand limit: $\lim_{x \to 3^-} f(x) = -2(3) = -6$.
Right-hand limit: $\lim_{x \to 3^+} f(x) = 6(3) + 2 = 20$.
Since $\lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x)$,$f(x)$ is discontinuous at $x = 3$.

(A) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \frac{\pi}{4} + a \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} + a$
$\lim_{x \rightarrow \frac{\pi}{4}^+} (2x \cot x + b) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b$
Equating them: $\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad \dots(1)$
For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow \frac{\pi}{2}^-} (2x \cot x + b) = 2\left(\frac{\pi}{2}\right) \lim_{x \rightarrow \frac{\pi}{2}^-} \cot x + b = \pi(0) + b = b$
$\lim_{x \rightarrow \frac{\pi}{2}^+} (a \cos 2x - b \sin x) = a \cos(\pi) - b \sin\left(\frac{\pi}{2}\right) = -a - b$
Equating them: $b = -a - b \implies a + 2b = 0 \implies a = -2b \quad \dots(2)$
Substitute $(2)$ into $(1)$: $-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.

(D) For a function to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3ax + b) = 3a + b$
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (5ax - 2b) = 5a - 2b$
Given $f(1) = 11$,we have:
$3a + b = 11$ (Equation $1$)
$5a - 2b = 11$ (Equation $2$)
From Equation $1$,$b = 11 - 3a$. Substituting this into Equation $2$:
$5a - 2(11 - 3a) = 11$
$5a - 22 + 6a = 11$
$11a = 33 \implies a = 3$
Substituting $a = 3$ into $b = 11 - 3a$:
$b = 11 - 3(3) = 11 - 9 = 2$
Thus,$a = 3$ and $b = 2$.

(B) Given $f(x) = \frac{(1 + \cos x) - \sin x}{(1 + \cos x) + \sin x}$.
Since $f(x)$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.
Using trigonometric identities $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$\lim_{x \rightarrow \pi} f(x) = \lim_{x \rightarrow \pi} \frac{2 \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}$
$= \lim_{x \rightarrow \pi} \frac{2 \cos \frac{x}{2} (\cos \frac{x}{2} - \sin \frac{x}{2})}{2 \cos \frac{x}{2} (\cos \frac{x}{2} + \sin \frac{x}{2})}$
$= \lim_{x \rightarrow \pi} \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$= \lim_{x \rightarrow \pi} \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \lim_{x \rightarrow \pi} \tan \left( \frac{\pi}{4} - \frac{x}{2} \right)$
$= \tan \left( \frac{\pi}{4} - \frac{\pi}{2} \right) = \tan \left( -\frac{\pi}{4} \right) = -1$.

(C) For $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at $x = -\pi/2$ and $x = \pi/2$.
At $x = -\pi/2$:
$\lim_{x \to -\pi/2^-} f(x) = -2 \sin(-\pi/2) = -2(-1) = 2$.
$\lim_{x \to -\pi/2^+} f(x) = a \sin(-\pi/2) + b = -a + b$.
Since it is continuous,$-a + b = 2$ (Equation $1$).
At $x = \pi/2$:
$\lim_{x \to \pi/2^-} f(x) = a \sin(\pi/2) + b = a + b$.
$\lim_{x \to \pi/2^+} f(x) = \cos(\pi/2) = 0$.
Since it is continuous,$a + b = 0$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(-a + b) + (a + b) = 2 + 0 \implies 2b = 2 \implies b = 1$.
Substituting $b = 1$ in Equation $2$: $a + 1 = 0 \implies a = -1$.
Now,calculate $(3a + 2b)^3$:
$(3(-1) + 2(1))^3 = (-3 + 2)^3 = (-1)^3 = -1$.

(C) Given $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$.
For continuity at $x = 0$:
Left Hand Limit $(LHL)$ = $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$.
Right Hand Limit $(RHL)$ = $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$.
Value of function $f(0) = 0$.
Since $LHL$ = $RHL$ = $f(0)$,the function is continuous at $x = 0$.
For differentiability at $x = 0$:
Left Hand Derivative $(LHD)$ = $\lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{(0-h) - 0}{-h} = \lim_{h \to 0^+} \frac{-h}{-h} = 1$.
Right Hand Derivative $(RHD)$ = $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0$.
Since $LHD$ $\neq$ $RHD$,the function is not differentiable at $x = 0$.

(B) Let $I=\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}$.
Substitute $\tan \frac{x}{2}=t$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dt$,so $dx = \frac{2 dt}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$.
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$.
$I = \int_0^1 \frac{1}{5+4(\frac{2t}{1+t^2})} \cdot \frac{2 dt}{1+t^2} = 2 \int_0^1 \frac{dt}{5+5t^2+8t} = \frac{2}{5} \int_0^1 \frac{dt}{t^2+\frac{8}{5}t+1}$.
Completing the square: $t^2+\frac{8}{5}t+1 = (t+\frac{4}{5})^2 + (1-\frac{16}{25}) = (t+\frac{4}{5})^2 + (\frac{3}{5})^2$.
$I = \frac{2}{5} \int_0^1 \frac{dt}{(t+\frac{4}{5})^2 + (\frac{3}{5})^2} = \frac{2}{5} \cdot \frac{1}{3/5} [\tan^{-1}(\frac{t+4/5}{3/5})]_0^1 = \frac{2}{3} [\tan^{-1}(\frac{5t+4}{3})]_0^1$.
$I = \frac{2}{3} [\tan^{-1}(3) - \tan^{-1}(\frac{4}{3})] = \frac{2}{3} \tan^{-1}(\frac{3-4/3}{1+3(4/3)}) = \frac{2}{3} \tan^{-1}(\frac{5/3}{5}) = \frac{2}{3} \tan^{-1}(\frac{1}{3})$.
Comparing with $A \tan^{-1} B$,we get $A=\frac{2}{3}$ and $B=\frac{1}{3}$.
Therefore,$A+B = \frac{2}{3} + \frac{1}{3} = 1$.

(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$.
Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$.
When $x = 2$,$t = \log 2$. When $x = e$,$t = 1$.
Substituting these into the integral:
$I = \int_{\log 2}^{1} \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt$.
We know that $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$.
Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$.
Therefore,$I = \left[ e^t \cdot \frac{1}{t} \right]_{\log 2}^{1}$.
$I = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log 2} \cdot \frac{1}{\log 2} \right)$.
$I = e - \frac{2}{\log 2}$.
Comparing this with $a + \frac{b}{\log 2}$,we get $a = e$ and $b = -2$.

(A) Let $I = \int_0^a \sqrt{\frac{a-x}{x}} dx$.
Substitute $x = a \sin^2 \theta$,then $dx = 2a \sin \theta \cos \theta d\theta$.
When $x = 0$,$\theta = 0$. When $x = a$,$\theta = \frac{\pi}{2}$.
$I = \int_0^{\frac{\pi}{2}} \sqrt{\frac{a - a \sin^2 \theta}{a \sin^2 \theta}} (2a \sin \theta \cos \theta) d\theta$
$I = \int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} (2a \sin \theta \cos \theta) d\theta$
$I = 2a \int_0^{\frac{\pi}{2}} \cos^2 \theta d\theta$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$I = 2a \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} d\theta = a \int_0^{\frac{\pi}{2}} (1 + \cos 2\theta) d\theta$
$I = a \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\frac{\pi}{2}} = a \left( \frac{\pi}{2} + 0 - 0 - 0 \right) = \frac{a\pi}{2}$.
Given $\int_0^a \sqrt{\frac{a-x}{x}} dx = \frac{k}{2}$,we have $\frac{a\pi}{2} = \frac{k}{2}$.
Therefore,$k = \pi a$.

(D) coin is tossed $4$ times. The total number of outcomes is $n(S) = 2^4 = 16$.
Let $X$ be the number of heads. The possible values for $X$ are $0, 1, 2, 3, 4$.
We need to find $P[X < 3] = P(X=0) + P(X=1) + P(X=2)$.
The number of ways to get $r$ heads in $n$ tosses is given by $\binom{n}{r}$.
For $X=0$: $\binom{4}{0} = 1$ way.
For $X=1$: $\binom{4}{1} = 4$ ways.
For $X=2$: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ ways.
Total favorable outcomes = $1 + 4 + 6 = 11$.
Therefore,$P[X < 3] = \frac{11}{16}$.

(B) Let $I = \int_0^\pi \frac{1}{4+3 \cos x} dx$.
Using the substitution $t = \tan(\frac{x}{2})$,we have $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2}{1+t^2} dt$.
When $x=0$,$t=0$,and when $x=\pi$,$t \to \infty$.
Substituting these into the integral:
$I = \int_0^{\infty} \frac{1}{4+3(\frac{1-t^2}{1+t^2})} \cdot \frac{2}{1+t^2} dt$
$I = \int_0^{\infty} \frac{2}{4(1+t^2) + 3(1-t^2)} dt$
$I = \int_0^{\infty} \frac{2}{4+4t^2+3-3t^2} dt = \int_0^{\infty} \frac{2}{7+t^2} dt$
$I = 2 \int_0^{\infty} \frac{1}{(\sqrt{7})^2 + t^2} dt$
Using the formula $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = 2 \left[ \frac{1}{\sqrt{7}} \tan^{-1}(\frac{t}{\sqrt{7}}) \right]_0^{\infty}$
$I = \frac{2}{\sqrt{7}} [\tan^{-1}(\infty) - \tan^{-1}(0)] = \frac{2}{\sqrt{7}} [\frac{\pi}{2} - 0] = \frac{\pi}{\sqrt{7}}$.

(C) Let $I = \int_0^{\pi / 4} \log (1+\tan x) d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$.
Since $\tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x}$,the integral becomes:
$I = \int_0^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x = \int_0^{\pi / 4} \log \left(\frac{1+\tan x + 1 - \tan x}{1+\tan x}\right) d x$.
$I = \int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$.
Using the property $\log(\frac{a}{b}) = \log a - \log b$,we get:
$I = \int_0^{\pi / 4} \log 2 d x - \int_0^{\pi / 4} \log (1+\tan x) d x$.
$I = \int_0^{\pi / 4} \log 2 d x - I$.
$2I = \log 2 \int_0^{\pi / 4} d x = \log 2 [x]_0^{\pi / 4} = \frac{\pi}{4} \log 2$.
Therefore,$I = \frac{\pi}{8} \log 2$.

(D) We need to evaluate the integral $I = \int_1^4 (|x-1|+|x-2|+|x-3|) dx$.
Since the absolute value functions change their behavior at $x=1, 2, 3$,we split the integral into three intervals:
$I = \int_1^2 ((x-1) + (2-x) + (3-x)) dx + \int_2^3 ((x-1) + (x-2) + (3-x)) dx + \int_3^4 ((x-1) + (x-2) + (x-3)) dx$
Simplifying the integrands:
$I = \int_1^2 (4-x) dx + \int_2^3 x dx + \int_3^4 (3x-6) dx$
Now,integrate each part:
$\int_1^2 (4-x) dx = [4x - \frac{x^2}{2}]_1^2 = (8-2) - (4-0.5) = 6 - 3.5 = 2.5$
$\int_2^3 x dx = [\frac{x^2}{2}]_2^3 = \frac{9}{2} - \frac{4}{2} = 4.5 - 2 = 2.5$
$\int_3^4 (3x-6) dx = [\frac{3x^2}{2} - 6x]_3^4 = (24-24) - (13.5-18) = 0 - (-4.5) = 4.5$
Adding these values: $I = 2.5 + 2.5 + 4.5 = 9.5 = \frac{19}{2}$.