KCET 2009 Mathematics Question Paper with Answer and Solution

(B) In $\triangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.
Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\circ} + B$.
Substituting these into the sine rule:
$\frac{\sin(60^{\circ} + B)}{2b} = \frac{\sin B}{b}$
$\sin(60^{\circ} + B) = 2 \sin B$
$\sin 60^{\circ} \cos B + \cos 60^{\circ} \sin B = 2 \sin B$
$\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B = 2 \sin B$
$\frac{\sqrt{3}}{2} \cos B = \frac{3}{2} \sin B$
$\tan B = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Thus,$B = 30^{\circ}$.
Then $A = 60^{\circ} + 30^{\circ} = 90^{\circ}$.
Since one angle is $90^{\circ}$,the triangle is right angled.

(C) The equation $ax + by = 1$ is a linear Diophantine equation.
According to Bezout's identity,the equation $ax + by = c$ has integer solutions for $x$ and $y$ if and only if $\gcd(a, b)$ divides $c$.
Here,$c = 1$,so $\gcd(a, b)$ must divide $1$,which implies $\gcd(a, b) = 1$. Thus,option $(d)$ is true.
Similarly,for $ax + by = 1$,we can rewrite it as $ax = 1 - by$,which implies $\gcd(a, b) = 1$ and $\gcd(a, y) = 1$ is not necessarily true,but $\gcd(a, b) = 1$ is a necessary condition.
Specifically,$\gcd(a, b) = 1$,$\gcd(a, x)$ is not necessarily $1$,$\gcd(b, x)$ is not necessarily $1$,and $\gcd(x, y)$ is not necessarily $1$.
However,looking at the options,$\gcd(b, y) = 1$ is not a required condition for the existence of solutions to $ax + by = 1$. For example,if $a=1, b=2, x=1, y=0$,then $1(1) + 2(0) = 1$. Here $\gcd(b, y) = \gcd(2, 0) = 2 \neq 1$. Thus,option $(c)$ is not necessarily true.

(B) Given equation is $x^{2}+x+1=0$.
The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^{2}$.
Let $\alpha = \omega$ and $\beta = \omega^{2}$.
We need to find $\alpha^{16} + \beta^{16} = \omega^{16} + (\omega^{2})^{16}$.
Since $\omega^{3} = 1$,we have $\omega^{16} = (\omega^{3})^{5} \cdot \omega = 1^{5} \cdot \omega = \omega$.
Similarly,$\omega^{32} = (\omega^{3})^{10} \cdot \omega^{2} = 1^{10} \cdot \omega^{2} = \omega^{2}$.
Thus,$\alpha^{16} + \beta^{16} = \omega + \omega^{2}$.
Since $1 + \omega + \omega^{2} = 0$,it follows that $\omega + \omega^{2} = -1$.

(A) To simplify the complex number,we multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\frac{1+2i}{1-i} \times \frac{1+i}{1+i} = \frac{1+i+2i+2i^2}{1^2-i^2}$
Since $i^2 = -1$,we have:
$\frac{1+3i-2}{1-(-1)} = \frac{-1+3i}{2} = -\frac{1}{2} + \frac{3}{2}i$
The real part is $-\frac{1}{2}$ (negative) and the imaginary part is $\frac{3}{2}$ (positive).
$A$ complex number with a negative real part and a positive imaginary part lies in the second quadrant.

(D) Let $z = \frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}$.
Using $\sin \theta = \cos(\frac{\pi}{2}-\theta)$ and $\cos \theta = \sin(\frac{\pi}{2}-\theta)$,let $\alpha = \frac{\pi}{2}-\frac{\pi}{8} = \frac{3\pi}{8}$.
Then $z = \frac{1+\cos \alpha + i \sin \alpha}{1+\cos \alpha - i \sin \alpha} = \frac{2 \cos^2 \frac{\alpha}{2} + 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2} - 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{\cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - i \sin \frac{\alpha}{2}} = \frac{e^{i \alpha/2}}{e^{-i \alpha/2}} = e^{i \alpha}$.
So,$z^n = e^{i n \alpha} = e^{i n (3\pi/8)} = \cos(\frac{3n\pi}{8}) + i \sin(\frac{3n\pi}{8})$.
For $z^n = 1$,we require $\sin(\frac{3n\pi}{8}) = 0$ and $\cos(\frac{3n\pi}{8}) = 1$.
This implies $\frac{3n\pi}{8} = 2k\pi$ for some integer $k$.
$3n = 16k \implies n = \frac{16k}{3}$.
For the smallest positive integer $n$,set $k=3$,which gives $n = 16$.

(A) Let $z = \sqrt{3}+i$. The modulus is $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and the argument is $\arg(z) = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.
Since $OPQ$ is an isosceles right-angled triangle with $\angle POQ = 90^{\circ}$ and $OP = OQ$,the point $Q$ can be obtained by rotating $P$ by $90^{\circ}$ in either the clockwise or counter-clockwise direction.
If we rotate by $+90^{\circ}$,the argument of $Q$ is $30^{\circ} + 90^{\circ} = 120^{\circ}$. The complex number is $2(\cos 120^{\circ} + i\sin 120^{\circ}) = 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -1 + i\sqrt{3}$.
If we rotate by $-90^{\circ}$,the argument of $Q$ is $30^{\circ} - 90^{\circ} = -60^{\circ}$. The complex number is $2(\cos(-60^{\circ}) + i\sin(-60^{\circ})) = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 1 - i\sqrt{3}$.
Thus,$Q$ represents $-1+i\sqrt{3}$ or $1-i\sqrt{3}$.

(C) The given expression is an infinite geometric series with first term $a = 1$ and common ratio $r = \sin x$.
Since the sum is $4+2 \sqrt{3}$,we use the formula $S = \frac{a}{1-r}$:
$\frac{1}{1-\sin x} = 4+2 \sqrt{3}$
$1-\sin x = \frac{1}{4+2 \sqrt{3}}$
Rationalizing the denominator:
$1-\sin x = \frac{4-2 \sqrt{3}}{(4+2 \sqrt{3})(4-2 \sqrt{3})} = \frac{4-2 \sqrt{3}}{16-12} = \frac{4-2 \sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$
Thus,$\sin x = \frac{\sqrt{3}}{2}$.
For $0 < x < \pi$,the values of $x$ satisfying $\sin x = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{3}$ and $x = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.

(B) Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we can rewrite the expression as:
$\log _{n} 2 + \log _{n} 3 + \log _{n} 4 + \ldots + \log _{n} 2020$
Using the property $\log _{b} x + \log _{b} y = \log _{b} (xy)$,we get:
$\log _{n} (2 \times 3 \times 4 \times \ldots \times 2020)$
Since $n = (2020)!$,the expression becomes:
$\log _{n} (n) = 1$

(D) Let $P(n) = n^{3} + 2n$.
For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$.
For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$.
For $n = 3$,$P(3) = 3^{3} + 2(3) = 27 + 6 = 33$.
We observe that $3, 12, 33$ are all divisible by $3$.
Alternatively,$n^{3} + 2n = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$.
Since $(n-1)n(n+1)$ is the product of three consecutive integers,it is always divisible by $3! = 6$.
However,the expression $n^{3} + 2n$ is always divisible by $3$ for any positive integer $n$.

(A) Let $f(x) = (x+y)^{100} + (x-y)^{100}$.
Using the binomial expansion,$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ and $(x-y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} (-y)^k$.
Adding these,the odd terms cancel out,leaving only the even terms: $2 \sum_{k=0, 2, 4, \dots, n} \binom{n}{k} x^{n-k} y^k$.
For $n = 100$,the number of terms is $\frac{n}{2} + 1 = \frac{100}{2} + 1 = 51$.

(D) The digit in the unit place of $2009!$ is $0$ because $2009!$ contains factors $2$ and $5$,making it a multiple of $10$.
Now,consider the powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
The unit digits follow a cycle of $4$: $(3, 9, 7, 1)$.
We divide the exponent $7886$ by $4$:
$7886 = 4 \times 1971 + 2$.
Thus,the unit digit of $3^{7886}$ is the same as the unit digit of $3^2$,which is $9$.
Therefore,the unit digit of $2009! + 3^{7886}$ is $0 + 9 = 9$.

(C) Given the expression for $f(x)$ is the binomial expansion of $(1+x)^{n}$.
$f(x) = (1+x)^{n}$
Taking the first derivative with respect to $x$:
$f^{\prime}(x) = n(1+x)^{n-1}$
Taking the second derivative with respect to $x$:
$f^{\prime \prime}(x) = n(n-1)(1+x)^{n-2}$
Substituting $x = 1$:
$f^{\prime \prime}(1) = n(n-1)(1+1)^{n-2}$
$f^{\prime \prime}(1) = n(n-1) 2^{n-2}$

(C) The range of $\sin \theta$ and $\cos \theta$ is $[-1, 1]$. The range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$. The range of $\tan \theta$ is $(-\infty, \infty)$.
$(a)$ For $\sin \theta = \frac{a^{2}+b^{2}}{a^{2}-b^{2}}$,if we take $a=2, b=1$,then $\sin \theta = \frac{5}{3} > 1$,which is impossible.
$(b)$ For $\sec \theta = \frac{4}{5} = 0.8$,which is impossible since $|sec \theta| \geq 1$.
$(c)$ For $\tan \theta = 45$,this is possible as the range of $\tan \theta$ is all real numbers.
$(d)$ For $\cos \theta = \frac{7}{3} > 1$,which is impossible.
Therefore,option $(c)$ is correct.

(B) Let $f(x) = 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$.
First,simplify each term:
$3(\sin x-\cos x)^{4} = 3(1-2\sin x\cos x)^{2} = 3(1+4\sin^{2}x\cos^{2}x-4\sin x\cos x) = 3+12\sin^{2}x\cos^{2}x-12\sin x\cos x$.
$6(\sin x+\cos x)^{2} = 6(1+2\sin x\cos x) = 6+12\sin x\cos x$.
$4(\sin^{6}x+\cos^{6}x) = 4(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x) = 4(1)((\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x\cos^{2}x) = 4(1-3\sin^{2}x\cos^{2}x) = 4-12\sin^{2}x\cos^{2}x$.
Summing these up:
$f(x) = (3+12\sin^{2}x\cos^{2}x-12\sin x\cos x) + (6+12\sin x\cos x) + (4-12\sin^{2}x\cos^{2}x)$.
$f(x) = 3+6+4 + (12\sin^{2}x\cos^{2}x-12\sin^{2}x\cos^{2}x) + (-12\sin x\cos x+12\sin x\cos x) = 13$.

(B) The given equation of the pair of lines is $x^{2}+2xy-y^{2}=0$.
Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$h=1$,and $b=-1$.
We know that the lines represented by $ax^{2}+2hxy+by^{2}=0$ are perpendicular if $a+b=0$.
Here,$a+b = 1 + (-1) = 0$.
Since the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero,the lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(B) Let the vertices of the triangle be $A(6,0), B(0,6)$ and $C(6,6)$.
First,we calculate the lengths of the sides:
$AB = \sqrt{(0-6)^2 + (6-0)^2} = \sqrt{36+36} = 6\sqrt{2}$
$BC = \sqrt{(6-0)^2 + (6-6)^2} = \sqrt{36+0} = 6$
$CA = \sqrt{(6-6)^2 + (0-6)^2} = \sqrt{0+36} = 6$
Since $AB^2 = BC^2 + CA^2$ $(72 = 36 + 36)$,the triangle is a right-angled triangle at vertex $C(6,6)$.
For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $AB$.
Circumcentre $O = \left(\frac{6+0}{2}, \frac{0+6}{2}\right) = (3,3)$.
The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{6+0+6}{3}, \frac{0+6+6}{3}\right) = (4,4)$.
The distance between the circumcentre $(3,3)$ and the centroid $(4,4)$ is $\sqrt{(4-3)^2 + (4-3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.

(C) Let the given line be $L_1: x+y=4$. The slope of $L_1$ is $m_1 = -1$.
The slope of the line $L_2$ perpendicular to $L_1$ is $m_2 = -\frac{1}{m_1} = 1$.
The equation of the line $L_2$ passing through $(2,4)$ is $y-4 = 1(x-2)$,which simplifies to $y-x=2$.
To find the foot of the perpendicular,we solve the system of equations:
$x+y=4$
$y-x=2$
Adding the two equations: $2y = 6 \Rightarrow y=3$.
Substituting $y=3$ into $x+y=4$,we get $x+3=4 \Rightarrow x=1$.
Thus,the foot of the perpendicular is $(1,3)$.

(D) For the smallest circle passing through the origin $(0, 0)$ with a diameter lying on the line $y = x + 1$,the centre of the circle must be the projection of the origin onto the line $y = x + 1$.
The line $y = x + 1$ can be written as $x - y + 1 = 0$.
The line perpendicular to $x - y + 1 = 0$ passing through the origin $(0, 0)$ has the form $x + y = k$.
Since it passes through $(0, 0)$,we have $0 + 0 = k$,so $k = 0$.
The perpendicular line is $x + y = 0$,or $y = -x$.
To find the centre,we find the intersection of $y = x + 1$ and $y = -x$.
Substituting $y = -x$ into $y = x + 1$,we get $-x = x + 1$,which implies $2x = -1$,so $x = -\frac{1}{2}$.
Then $y = -(-\frac{1}{2}) = \frac{1}{2}$.
Thus,the centre of the circle is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.

(D) The length of the arc $s$ is given by the formula: $s = 2 \pi r \times \frac{\theta}{360^{\circ}}$
Given $s = 44 \ m$ and $\theta = 72^{\circ}$.
Substituting the values: $44 = 2 \times \frac{22}{7} \times r \times \frac{72}{360^{\circ}}$
Simplifying the fraction: $\frac{72}{360} = \frac{1}{5}$
So,$44 = 2 \times \frac{22}{7} \times r \times \frac{1}{5}$
$44 = \frac{44}{35} \times r$
$r = \frac{44 \times 35}{44} = 35 \ m$
Thus,the length of the rope is $35 \ m$.

(B) Let the midpoint of the chord $AB$ be $C(x_{1}, y_{1})$. The origin is $O(0, 0)$.
Since the chord $AB$ subtends a right angle at the origin,$\angle AOB = 90^{\circ}$.
In $\Delta OAB$,$OA = OB = r = 2$ (radius of the circle).
Since $C$ is the midpoint of $AB$,$OC \perp AB$.
In $\Delta OCB$,$\angle COB = \frac{1}{2} \angle AOB = 45^{\circ}$.
Using trigonometry in $\Delta OCB$:
$\cos(45^{\circ}) = \frac{OC}{OB}$
$\frac{1}{\sqrt{2}} = \frac{\sqrt{x_{1}^{2} + y_{1}^{2}}}{2}$
Squaring both sides:
$\frac{1}{2} = \frac{x_{1}^{2} + y_{1}^{2}}{4}$
$x_{1}^{2} + y_{1}^{2} = 2$
Replacing $(x_{1}, y_{1})$ with $(x, y)$,the locus is $x^{2} + y^{2} = 2$.

(A) The points are $P = (4 \cos \theta, 4 \sin \theta)$ and $Q = (4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$.
Using the distance formula,the length of the chord $PQ$ is given by:
$PQ = \sqrt{(4 \cos (\theta+60^{\circ}) - 4 \cos \theta)^2 + (4 \sin (\theta+60^{\circ}) - 4 \sin \theta)^2}$
$PQ = 4 \sqrt{(\cos (\theta+60^{\circ}) - \cos \theta)^2 + (\sin (\theta+60^{\circ}) - \sin \theta)^2}$
$PQ = 4 \sqrt{\cos^2 (\theta+60^{\circ}) + \cos^2 \theta - 2 \cos (\theta+60^{\circ}) \cos \theta + \sin^2 (\theta+60^{\circ}) + \sin^2 \theta - 2 \sin (\theta+60^{\circ}) \sin \theta}$
Using $\sin^2 A + \cos^2 A = 1$:
$PQ = 4 \sqrt{1 + 1 - 2 (\cos (\theta+60^{\circ}) \cos \theta + \sin (\theta+60^{\circ}) \sin \theta)}$
Using $\cos (A-B) = \cos A \cos B + \sin A \sin B$:
$PQ = 4 \sqrt{2 - 2 \cos ((\theta+60^{\circ}) - \theta)}$
$PQ = 4 \sqrt{2 - 2 \cos 60^{\circ}}$
Since $\cos 60^{\circ} = \frac{1}{2}$:
$PQ = 4 \sqrt{2 - 2 \times \frac{1}{2}} = 4 \sqrt{2 - 1} = 4 \times 1 = 4$.

(C) Let the equation of the required circle be $x^{2}+y^{2}+2gx+2fy+c=0$.
Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ cut orthogonally if $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$.
For the given circles,the conditions are:
$1) -g-f = c-14 \implies -g-f-c = -14$
$2) 3g-5f = c-10 \implies 3g-5f-c = -10$
$3) -2g+3f = c-27 \implies -2g+3f-c = -27$
Subtracting $(1)$ from $(2)$: $4g-4f = 4 \implies g-f = 1 \implies g = f+1$.
Subtracting $(1)$ from $(3)$: $-g+4f = -13$.
Substituting $g = f+1$: $-(f+1)+4f = -13 \implies 3f = -12 \implies f = -4$.
Then $g = -4+1 = -3$.
From $(1)$: $-(-3)-(-4) = c-14 \implies 3+4 = c-14 \implies c = 21$.
The circle is $x^{2}+y^{2}-6x-8y+21=0$.
The radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-4)^{2}-21} = \sqrt{9+16-21} = \sqrt{4} = 2$.
The diameter is $2r = 2 \times 2 = 4$.

(B) The given circles are $x^{2}+y^{2}-y=0$ and $x^{2}+y^{2}+y=0$.
For the first circle $x^{2}+y^{2}-y=0$,the center is $C_{1}(0, 1/2)$ and radius $r_{1} = \sqrt{0^{2} + (1/2)^{2} - 0} = 1/2$.
For the second circle $x^{2}+y^{2}+y=0$,the center is $C_{2}(0, -1/2)$ and radius $r_{2} = \sqrt{0^{2} + (-1/2)^{2} - 0} = 1/2$.
The distance between the centers is $C_{1}C_{2} = \sqrt{(0-0)^{2} + (1/2 - (-1/2))^{2}} = \sqrt{1^{2}} = 1$.
Since $r_{1} + r_{2} = 1/2 + 1/2 = 1$,we have $C_{1}C_{2} = r_{1} + r_{2}$.
This condition implies that the two circles touch each other externally at a single point.
When two circles touch each other externally,they have exactly $3$ common tangents (two external and one internal).

(D) Given equations are $x^{2}+y^{2}=a^{2}$ and $xy=c^{2}$.
Substituting $y = \frac{c^{2}}{x}$ into the circle equation:
$x^{2} + (\frac{c^{2}}{x})^{2} = a^{2}$
$x^{2} + \frac{c^{4}}{x^{2}} = a^{2}$
$x^{4} - a^{2}x^{2} + c^{4} = 0$
This is a biquadratic equation in $x$. The roots are $x_{1}, x_{2}, x_{3}, x_{4}$.
Comparing this with the standard form $Ax^{4} + Bx^{3} + Cx^{2} + Dx + E = 0$,we have $B=0$.
By Vieta's formulas,the sum of the roots $x_{1}+x_{2}+x_{3}+x_{4} = -\frac{B}{A} = 0$.

(D) The equation of the parabola is $y^{2} = 4x$. Comparing this with $y^{2} = 4ax$,we get $a = 1$.
For any point $P(x, y)$ on the parabola,the focal distance is given by $x + a$.
Given that the focal distance is $17$,we have $x + 1 = 17$,which implies $x = 16$.
Substituting $x = 16$ into the parabola equation: $y^{2} = 4(16) = 64$.
Thus,$y = \pm 8$.
Therefore,the required points are $(16, 8)$ or $(16, -8)$.

(B) The equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$,where $a^{2}=4$ and $b^{2}=1$.
The equation of the line is $y=mx+c$,where $m=4$.
The condition for the line $y=mx+c$ to be tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.
Substituting the values,we get $c^{2}=4(4)^{2}+1$.
$c^{2}=4(16)+1=64+1=65$.
Therefore,$c=\pm \sqrt{65}$.
Since there are two possible values for $c$,the line touches the curve for $2$ values of $c$.

(A) The equation of the parabola is $y^{2} = 12x$,which is of the form $y^{2} = 4ax$,where $a = 3$.
For any point $(x_{1}, y_{1})$,the pair of tangents is given by $SS_{1} = T^{2}$.
Here,$S = y^{2} - 12x$,$S_{1} = (2)^{2} - 12(-3) = 4 + 36 = 40$,and $T = y(2) - 6(x - 3) = 2y - 6x + 18$.
Substituting these into the formula:
$(y^{2} - 12x)(40) = (2y - 6x + 18)^{2}$
$40y^{2} - 480x = 4(y - 3x + 9)^{2}$
$10y^{2} - 120x = y^{2} + 9x^{2} + 81 - 6xy - 54x + 18y$
$9x^{2} - 9y^{2} - 6xy + 66x + 18y + 81 = 0$.
For a general second-degree equation $Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0$,the angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{H^{2} - AB}}{A + B} \right|$.
Here,$A = 9$,$B = -9$,and $H = -3$.
Since $A + B = 9 + (-9) = 0$,the lines are perpendicular.
Therefore,the angle between the tangents is $90^{\circ}$.

(C) To evaluate the limit $\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$,we divide the numerator and the denominator by $5^n$:
$\lim _{n \rightarrow \infty} \frac{3 \cdot 2 \cdot 2^{n} - 4 \cdot 5 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$
$= \lim _{n \rightarrow \infty} \frac{6 \cdot 2^{n} - 20 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$
Dividing by $5^n$:
$= \lim _{n \rightarrow \infty} \frac{6 \cdot (\frac{2}{5})^{n} - 20}{5 \cdot (\frac{2}{5})^{n} + 7}$
Since $\lim _{n \rightarrow \infty} (\frac{2}{5})^{n} = 0$,we get:
$= \frac{6 \cdot 0 - 20}{5 \cdot 0 + 7} = -\frac{20}{7}$

(D) The group $(Z_{5}, +_{5})$ is a cyclic group of prime order $p = 5$.
By Lagrange's Theorem,the order of any subgroup must divide the order of the group.
Since $5$ is a prime number,its only divisors are $1$ and $5$.
Therefore,the only subgroups are the trivial subgroup ${0}$ of order $1$ and the group itself $Z_{5}$ of order $5$.
Thus,the total number of subgroups is $2$.

(D) We need to find the negation of the statement $p \wedge (q \rightarrow \sim r)$.
Using De Morgan's Law,$\sim(A \wedge B) = \sim A \vee \sim B$:
$\sim(p \wedge (q$ $\rightarrow \sim r)) = \sim p \vee \sim(q$ $\rightarrow \sim r)$
Using the implication rule $\sim(A \rightarrow B) = A \wedge \sim B$:
$\sim p \vee (q \wedge \sim(\sim r))$
Using the double negation law $\sim(\sim r) = r$:
$\sim p \vee (q \wedge r)$
Thus,the correct option is $D$.

(C) Let $f(x) = 27^{\cos 2x} 81^{\sin 2x} = (3^3)^{\cos 2x} (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.
To find the minimum value,we analyze the exponent $g(x) = 3 \cos 2x + 4 \sin 2x$.
The expression $a \cos \theta + b \sin \theta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the range of $3 \cos 2x + 4 \sin 2x$ is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.
The minimum value of the exponent is $-5$.
Therefore,the minimum value of $f(x) = 3^{-5} = \frac{1}{3^5} = \frac{1}{243}$.

(D) Let the initial population be $P$ and the rate of growth be $r = 10 \% = 0.1$.
Assuming continuous growth,the population $P(t)$ at time $t$ is given by $P(t) = P_0 e^{rt}$.
For the population to double,$P(t) = 2P_0$.
$2P_0 = P_0 e^{0.1t} \Rightarrow 2 = e^{0.1t}$.
Taking the natural logarithm on both sides: $\ln 2 = 0.1t$.
$t = \frac{\ln 2}{0.1} = 10 \ln 2 \text{ years}$.
If the growth is compounded annually,$2P = P(1 + 0.1)^n \Rightarrow 2 = (1.1)^n$.
Taking $\log_{10}$ on both sides: $\log 2 = n \log 1.1$.
$n = \frac{\log 2}{\log 1.1} \approx \frac{0.3010}{0.0414} \approx 7.27 \text{ years}$.
Since the result $10 \ln 2$ or $7.27$ does not match the options provided,the correct choice is $D$.

(C) Given the displacement from the top of the tower is $s=48t-16t^{2}$.
The velocity is given by $v = \frac{ds}{dt} = 48-32t$.
At the greatest height,the velocity $v=0$.
$48-32t=0 \Rightarrow t = \frac{48}{32} = 1.5 \ s$.
The maximum displacement from the top of the tower is $s = 48(1.5) - 16(1.5)^{2} = 72 - 36 = 36 \ m$.
The total height above the ground is the height of the tower plus the maximum displacement: $H = 64 + 36 = 100 \ m$.

(B) For the function to be continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the standard limit $\lim_{x \to 0} \frac{\log(1+kx)}{x} = k$,we have:
$\lim_{x \to 0} \frac{\log(1+ax) - \log(1-bx)}{x} = \lim_{x \to 0} \left( \frac{\log(1+ax)}{x} - \frac{\log(1-bx)}{x} \right)$
$= \lim_{x \to 0} \frac{\log(1+ax)}{x} - \lim_{x \to 0} \frac{\log(1-bx)}{x}$
$= a - (-b) = a + b$.
Thus,the value to be assigned is $a+b$.

(C) binary operation $*$ on a set $N$ is a function $*: N \times N \to N$. This means for any $a, b \in N$,the result $a * b$ must also be in $N$.
$(A)$ $a * b = \sqrt{ab}$: If $a=1, b=2$,then $a * b = \sqrt{2} \notin N$.
$(B)$ $a * b = \frac{a-b}{a+b}$: If $a=1, b=2$,then $a * b = \frac{-1}{3} \notin N$.
$(C)$ $a * b = a + 3b$: Since $a, b \in N$,$a + 3b$ is always a natural number. Thus,this is a binary operation.
$(D)$ $a * b = 3a - 4b$: If $a=1, b=2$,then $a * b = 3(1) - 4(2) = -5 \notin N$.
Hence,option $C$ is correct.

(B) In the group $G = \{0, 1, 2, 3, 4, 5\}$ under addition modulo $6$,the identity element is $0$.
First,we find the inverse of $3$ under addition modulo $6$. Since $3 +_{6} 3 = 6 \equiv 0 \pmod{6}$,the inverse $3^{-1} = 3$.
Now,calculate the expression inside the parentheses: $2 +_{6} 3^{-1} +_{6} 4 = 2 +_{6} 3 +_{6} 4 = 9 \pmod{6} = 3$.
Finally,we find the inverse of the result $3$ under addition modulo $6$. Since $3 +_{6} 3 = 0$,the inverse of $3$ is $3$.
Thus,$(2 +_{6} 3^{-1} +_{6} 4)^{-1} = 3^{-1} = 3$.

(B) The set of fourth roots of unity is given by $S = \{1, -1, i, -i\}$.
For a set to form an additive group,it must satisfy the closure property under addition.
Consider $1 + (-1) = 0$. Since $0 \notin S$,the set is not closed under addition.
Therefore,the statement that the fourth roots of unity form an additive abelian group is false.

(A) Given,$(A+B)(A-B)=A^{2}-B^{2}$
Expanding the left side: $A^{2}-AB+BA-B^{2}=A^{2}-B^{2}$
Subtracting $A^{2}-B^{2}$ from both sides: $-AB+BA=0$
Therefore,$AB=BA$
Now,we evaluate $(ABA^{-1})^{2}$:
$(ABA^{-1})^{2} = (ABA^{-1})(ABA^{-1})$
Since $AB=BA$,we can write $AB = BA$,so $ABA^{-1} = BAA^{-1} = BI = B$
Thus,$(ABA^{-1})^{2} = B^{2}$

(D) Given,$A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 2(4 - 0) - 1(0 - 1) + 0(0 - 2) = 2(4) - 1(-1) + 0 = 8 + 1 = 9$.
We know the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the square matrix $A$.
Here,$n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A| = 9$,we get:
$|\operatorname{adj} A| = 9^2 = 81$.

(D) Given the determinant equation:
$\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$
Applying the column operation $C_{1} \rightarrow C_{1}+C_{2}$:
$\left|\begin{array}{ccc}1+\sin ^{2} \theta+\cos ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta+1+\cos ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta+\cos ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$
Since $\sin ^{2} \theta+\cos ^{2} \theta=1$,this simplifies to:
$\left|\begin{array}{ccc}2 & \cos ^{2} \theta & 4 \sin 2 \theta \\ 2 & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ 1 & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$
Applying row operations $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow 2 R_{3}-R_{1}$:
$\left|\begin{array}{ccc}2 & \cos ^{2} \theta & 4 \sin 2 \theta \\ 0 & 1 & 0 \\ 0 & \cos ^{2} \theta & 4 \sin 2 \theta-2\end{array}\right|=0$
Expanding along the first column:
$2 \times [1 \times (4 \sin 2 \theta-2) - 0] = 0$
$8 \sin 2 \theta-4=0 \Rightarrow \sin 2 \theta=\frac{1}{2}$
Now,using the identity $\cos 4 \theta=1-2 \sin ^{2} 2 \theta$:
$\cos 4 \theta=1-2\left(\frac{1}{2}\right)^{2}=1-2\left(\frac{1}{4}\right)=1-\frac{1}{2}=\frac{1}{2}$

(D) Given the determinant equation: $\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$
Applying the row operation $R_{1} \rightarrow R_{1} + R_{3} - 2R_{2}$:
The first row becomes: $(x+1+x+3-2(x+2), x+2+x+4-2(x+3), x+a+x+c-2(x+b))$
Simplifying the elements of the first row:
$R_{1,1} = 2x + 4 - 2x - 4 = 0$
$R_{1,2} = 2x + 6 - 2x - 6 = 0$
$R_{1,3} = 2x + a + c - 2x - 2b = a + c - 2b$
The determinant becomes: $\left|\begin{array}{ccc} 0 & 0 & a+c-2b \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right| = 0$
Expanding along the first row:
$(a+c-2b) \cdot [(x+2)(x+4) - (x+3)(x+3)] = 0$
$(a+c-2b) \cdot [x^2 + 6x + 8 - (x^2 + 6x + 9)] = 0$
$(a+c-2b) \cdot (-1) = 0$
Since $-1 \neq 0$,we must have $a+c-2b = 0$,which implies $2b = a+c$.
This is the condition for $a, b, c$ to be in Arithmetic Progression $(AP)$.

(A) Let the determinant be $\Delta = \left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$.
Using the property $\log _{a} b = \frac{\ln b}{\ln a}$,we can write the elements as:
$\Delta = \left|\begin{array}{ccc}1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\ \frac{\ln x}{\ln y} & 1 & \frac{\ln z}{\ln y} \\ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 1\end{array}\right|$.
Multiply $R_1$ by $\ln x$,$R_2$ by $\ln y$,and $R_3$ by $\ln z$:
$\Delta = \frac{1}{\ln x \ln y \ln z} \left|\begin{array}{ccc}\ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z \\ \ln x & \ln y & \ln z\end{array}\right|$.
Since all three rows are identical,the value of the determinant is $0$.

(A) Let $\sin^{-1} x = y$. Then,$x = \sin y$.
Since $-1 \leq x < 0$,we have $-\frac{\pi}{2} \leq \sin^{-1} x < 0$,which implies $-\frac{\pi}{2} \leq y < 0$.
We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.
Since $y$ is in the interval $[-\frac{\pi}{2}, 0)$,$-y$ is in the interval $(0, \frac{\pi}{2}]$.
Using the property $\cos(-y) = \cos y = \sqrt{1 - x^2}$,we get $-y = \cos^{-1}(\sqrt{1 - x^2})$.
Therefore,$y = -\cos^{-1}(\sqrt{1 - x^2})$.
Thus,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

(A) The given series is $S = \sum_{r=1}^{\infty} \cot ^{-1}(2r^2) = \sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2r^2}\right)$.
We can rewrite the argument of $\tan^{-1}$ as:
$\frac{1}{2r^2} = \frac{2}{4r^2} = \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$,we get:
$S = \sum_{r=1}^{\infty} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)]$.
Expanding the summation:
$S = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + (\tan^{-1} 7 - \tan^{-1} 5) + \dots$
This is a telescoping series where all intermediate terms cancel out.
$S = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(1)]$.
Since $\lim_{n \to \infty} \tan^{-1}(2n+1) = \frac{\pi}{2}$ and $\tan^{-1}(1) = \frac{\pi}{4}$,
$S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) Given the function $f: Z \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2}, & n \text{ is even} \\ 0, & n \text{ is odd} \end{cases}$.
$1$. Check for Injectivity (One-to-One): $A$ function is injective if $f(n_1) = f(n_2) \implies n_1 = n_2$. Consider odd integers $n_1 = 1$ and $n_2 = 3$. Then $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 \neq 3$,the function is not injective (it is many-to-one).
$2$. Check for Surjectivity (Onto): $A$ function is surjective if for every $y \in Z$,there exists an $n \in Z$ such that $f(n) = y$. If we choose $y = 5$,there is no integer $n$ such that $f(n) = 5$ because even numbers map to $n/2$ (which would require $n=10$) and odd numbers map to $0$. However,the range of $f$ is the set of all integers $Z$ because for any $y \in Z$,we can choose $n = 2y$ (which is even),then $f(2y) = \frac{2y}{2} = y$. Thus,for every $y \in Z$,there exists $n = 2y$ such that $f(n) = y$. Therefore,the function is surjective.
Conclusion: The function is surjective but not injective.

(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.
Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e}(\log _{e} x)}{\log _{e} x}$.
Applying the quotient rule,$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot \frac{d}{dx}(\log _{e}(\log _{e} x)) - \log _{e}(\log _{e} x) \cdot \frac{d}{dx}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot (\frac{1}{\log _{e} x} \cdot \frac{1}{x}) - \log _{e}(\log _{e} x) \cdot \frac{1}{x}}{(\log _{e} x)^{2}} \right]$.
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{1}{x} \log _{e}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.
At $x = e$,$\log _{e} e = 1$ and $\log _{e}(\log _{e} e) = \log _{e} 1 = 0$.
Substituting these values,$f^{\prime}(e) = \frac{1}{2} \left[ \frac{\frac{1}{e} - \frac{1}{e} \cdot 0}{(1)^{2}} \right] = \frac{1}{2e}$.

(D) Given,$y = \sin^{n} x \cos nx$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = \sin^{n} x \frac{d}{dx}(\cos nx) + \cos nx \frac{d}{dx}(\sin^{n} x)$
$\frac{dy}{dx} = \sin^{n} x (-n \sin nx) + \cos nx (n \sin^{n-1} x \cos x)$
$\frac{dy}{dx} = n \sin^{n-1} x (\cos x \cos nx - \sin x \sin nx)$
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\frac{dy}{dx} = n \sin^{n-1} x \cos (nx + x)$
$\frac{dy}{dx} = n \sin^{n-1} x \cos (n+1) x$

(D) Given the function: $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$.
Simplifying the expression,we get: $f(x) = \frac{g(x) + g(-x)}{2} + 2[h(x) + h(-x)]$.
Differentiating both sides with respect to $x$ using the chain rule:
$f^{\prime}(x) = \frac{g^{\prime}(x) - g^{\prime}(-x)}{2} + 2[h^{\prime}(x) - h^{\prime}(-x)]$.
Now,substitute $x = 0$ into the derivative:
$f^{\prime}(0) = \frac{g^{\prime}(0) - g^{\prime}(0)}{2} + 2[h^{\prime}(0) - h^{\prime}(0)]$.
$f^{\prime}(0) = 0 + 2[0] = 0$.

(B) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx}$.
If the tangent is perpendicular to the $x$-axis,it is a vertical line.
For a vertical line,the slope is undefined,which means $\frac{dy}{dx} \to \infty$.
This is equivalent to saying that the slope of the tangent with respect to $y$,which is $\frac{dx}{dy}$,must be $0$.

(A) Given,$x=a(t+\sin t)$ and $y=a(1-\cos t)$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = a(1+\cos t) = 2a \cos^2 \frac{t}{2}$
$\frac{dy}{dt} = a \sin t = 2a \sin \frac{t}{2} \cos \frac{t}{2}$
Now,the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \sin \frac{t}{2} \cos \frac{t}{2}}{2a \cos^2 \frac{t}{2}} = \tan \frac{t}{2}$.
The length of the subtangent is defined as $\left| \frac{y}{dy/dx} \right|$.
Substituting the values,we get:
Length of subtangent $= \frac{a(1-\cos t)}{\tan \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{\tan \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{\sin \frac{t}{2} / \cos \frac{t}{2}} = 2a \sin \frac{t}{2} \cos \frac{t}{2} = a \sin t$.

(A) Let $I = \int e^{\tan ^{-1} x} \left(1 + \frac{x}{1+x^2}\right) dx$.
We can split the integral as $I = \int e^{\tan ^{-1} x} dx + \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx$.
Now,apply integration by parts to the first integral $\int e^{\tan ^{-1} x} dx$ by taking $u = e^{\tan ^{-1} x}$ and $dv = dx$.
Then $du = e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$.
Using the formula $\int u dv = uv - \int v du$,we get:
$\int e^{\tan ^{-1} x} dx = x e^{\tan ^{-1} x} - \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx$.
Substituting this back into the expression for $I$:
$I = \left(x e^{\tan ^{-1} x} - \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx\right) + \int \frac{x e^{\tan ^{-1} x}}{1+x^2} dx + c$.
The integral terms cancel out,leaving $I = x e^{\tan ^{-1} x} + c$.

(C) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx$.
Multiply and divide by $\sin a$:
$I = \int \frac{\sin a}{\sin a \sin(x-a) \sin x} \, dx$.
Since $\sin a = \sin(x - (x-a))$,we have:
$I = \frac{1}{\sin a} \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} \, dx$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} \, dx$.
$I = \frac{1}{\sin a} \int \left( \frac{\cos(x-a)}{\sin(x-a)} - \frac{\cos x}{\sin x} \right) \, dx$.
$I = \frac{1}{\sin a} \int (\cot(x-a) - \cot x) \, dx$.
Integrating,we get:
$I = \frac{1}{\sin a} [\log |\sin(x-a)| - \log |\sin x|] + c$.
$I = \frac{1}{\sin a} \log \left| \frac{\sin(x-a)}{\sin x} \right| + c$.
This can be written as $\frac{1}{\sin a} \log |\sin(x-a) \operatorname{cosec} x| + c$.

(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 1+3 = 4$.
Substituting $x$ with $(4-x)$:
$I = \int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} dx$
$I = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} dx \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{1}^{3} \left( \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} + \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} \right) dx$
$2I = \int_{1}^{3} \frac{\sqrt{4-x}+\sqrt{x}}{\sqrt{x}+\sqrt{4-x}} dx$
$2I = \int_{1}^{3} 1 dx$
$2I = [x]_{1}^{3} = 3 - 1 = 2$
$I = \frac{2}{2} = 1$

(D) Given,$\int_{0}^{1} f(x) dx = 5$.
Let $I = 100 \int_{0}^{1} x^{9} f(x^{10}) dx$.
Substitute $x^{10} = t$,then $10x^{9} dx = dt$,which implies $x^{9} dx = \frac{dt}{10}$.
When $x = 0, t = 0$ and when $x = 1, t = 1$.
Therefore,$I = 100 \int_{0}^{1} f(t) \frac{dt}{10} = 10 \int_{0}^{1} f(t) dt$.
Since $\int_{0}^{1} f(t) dt = \int_{0}^{1} f(x) dx = 5$,we have $I = 10 \times 5 = 50$.
Thus,the required value is $\int_{0}^{1} f(x) dx + I = 5 + 50 = 55$.

(B) Given,$f(x) = \int_{-1}^{x} |t| dt$.
Since $x \geq 0$,we can split the integral at $t = 0$:
$f(x) = \int_{-1}^{0} |t| dt + \int_{0}^{x} |t| dt$.
For $t \in [-1, 0]$,$|t| = -t$,and for $t \in [0, x]$,$|t| = t$.
So,$f(x) = \int_{-1}^{0} (-t) dt + \int_{0}^{x} t dt$.
Evaluating the integrals:
$f(x) = -\left[ \frac{t^{2}}{2} \right]_{-1}^{0} + \left[ \frac{t^{2}}{2} \right]_{0}^{x}$.
$f(x) = -\left( 0 - \frac{(-1)^{2}}{2} \right) + \left( \frac{x^{2}}{2} - 0 \right)$.
$f(x) = -\left( -\frac{1}{2} \right) + \frac{x^{2}}{2} = \frac{1}{2} + \frac{x^{2}}{2} = \frac{1}{2}(1 + x^{2})$.

(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \frac{y+4}{2}$ into the parabola equation:
$y^{2} = 4\left(\frac{y+4}{2}\right)$
$y^{2} = 2(y+4)$
$y^{2} - 2y - 8 = 0$
$(y-4)(y+2) = 0$
Thus,the intersection points occur at $y = 4$ and $y = -2$.
The required area is given by the integral of the difference between the line and the parabola with respect to $y$:
$\text{Area} = \int_{-2}^{4} \left( \frac{y+4}{2} - \frac{y^{2}}{4} \right) dy$
$= \left[ \frac{y^{2}}{4} + 2y - \frac{y^{3}}{12} \right]_{-2}^{4}$
$= \left( \frac{16}{4} + 8 - \frac{64}{12} \right) - \left( \frac{4}{4} - 4 - \frac{-8}{12} \right)$
$= \left( 4 + 8 - \frac{16}{3} \right) - \left( 1 - 4 + \frac{2}{3} \right)$
$= \left( 12 - \frac{16}{3} \right) - \left( -3 + \frac{2}{3} \right)$
$= \frac{20}{3} - \left( -\frac{7}{3} \right) = \frac{27}{3} = 9 \text{ sq unit}$.

(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a parameter.
From this equation,we have $2h = \frac{x^{2} + y^{2}}{x}$.
Differentiating the equation $x^{2} + y^{2} - 2hx = 0$ with respect to $x$,we get:
$2x + 2y \frac{dy}{dx} - 2h = 0$.
Substituting the value of $2h$ from the first equation:
$2x + 2y \frac{dy}{dx} - \left( \frac{x^{2} + y^{2}}{x} \right) = 0$.
Multiplying the entire equation by $x$:
$2x^{2} + 2xy \frac{dy}{dx} - x^{2} - y^{2} = 0$.
Simplifying the expression:
$x^{2} - y^{2} + 2xy \frac{dy}{dx} = 0$,which can be rewritten as $y^{2} = x^{2} + 2xy \frac{dy}{dx}$.

(B) We know that the dot product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos \theta$,where $\theta$ is the angle between them.
Given that $\overrightarrow{a} \cdot \overrightarrow{b} = -|\overrightarrow{a}||\overrightarrow{b}|$.
Substituting the formula,we get $|\overrightarrow{a}||\overrightarrow{b}| \cos \theta = -|\overrightarrow{a}||\overrightarrow{b}|$.
Assuming $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-zero vectors,we can divide both sides by $|\overrightarrow{a}||\overrightarrow{b}|$.
This gives $\cos \theta = -1$.
The value of $\theta$ for which $\cos \theta = -1$ is $\theta = 180^{\circ}$.

(D) Given,$\overrightarrow{a}+2 \overrightarrow{b}+3 \overrightarrow{c}=\overrightarrow{0} \quad \dots(i)$
Taking the cross product of equation $(i)$ with $\overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} + 2(\overrightarrow{b} \times \overrightarrow{b}) + 3(\overrightarrow{c} \times \overrightarrow{b}) = \overrightarrow{0} \times \overrightarrow{b}$
Since $\overrightarrow{b} \times \overrightarrow{b} = \overrightarrow{0}$ and $\overrightarrow{c} \times \overrightarrow{b} = -(\overrightarrow{b} \times \overrightarrow{c})$:
$\overrightarrow{a} \times \overrightarrow{b} - 3(\overrightarrow{b} \times \overrightarrow{c}) = \overrightarrow{0} \implies \overrightarrow{a} \times \overrightarrow{b} = 3(\overrightarrow{b} \times \overrightarrow{c})$
Taking the cross product of equation $(i)$ with $\overrightarrow{c}$:
$\overrightarrow{a} \times \overrightarrow{c} + 2(\overrightarrow{b} \times \overrightarrow{c}) + 3(\overrightarrow{c} \times \overrightarrow{c}) = \overrightarrow{0} \times \overrightarrow{c}$
Since $\overrightarrow{c} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{a} \times \overrightarrow{c} = -(\overrightarrow{c} \times \overrightarrow{a})$:
$-(\overrightarrow{c} \times \overrightarrow{a}) + 2(\overrightarrow{b} \times \overrightarrow{c}) = \overrightarrow{0} \implies \overrightarrow{c} \times \overrightarrow{a} = 2(\overrightarrow{b} \times \overrightarrow{c})$
Now,substituting these into the expression $\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a}$:
$= 3(\overrightarrow{b} \times \overrightarrow{c}) + (\overrightarrow{b} \times \overrightarrow{c}) + 2(\overrightarrow{b} \times \overrightarrow{c})$
$= 6(\overrightarrow{b} \times \overrightarrow{c})$

(A) Given,the volume of the parallelepiped with coterminous edges $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the scalar triple product $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 40 \text{ cubic units}$.
The volume of the parallelepiped with coterminous edges $\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}$ is given by the scalar triple product $[\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}]$.
Using the property of the scalar triple product:
$[\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{c}+\overrightarrow{a}, \overrightarrow{a}+\overrightarrow{b}] = (\overrightarrow{b}+\overrightarrow{c}) \cdot ((\overrightarrow{c}+\overrightarrow{a}) \times (\overrightarrow{a}+\overrightarrow{b}))$
$= (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{c} \times \overrightarrow{a} + \overrightarrow{c} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,this simplifies to:
$= (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{c} \times \overrightarrow{a} + \overrightarrow{c} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b})$
$= \overrightarrow{b} \cdot (\overrightarrow{c} \times \overrightarrow{a}) + \overrightarrow{b} \cdot (\overrightarrow{c} \times \overrightarrow{b}) + \overrightarrow{b} \cdot (\overrightarrow{a} \times \overrightarrow{b}) + \overrightarrow{c} \cdot (\overrightarrow{c} \times \overrightarrow{a}) + \overrightarrow{c} \cdot (\overrightarrow{c} \times \overrightarrow{b}) + \overrightarrow{c} \cdot (\overrightarrow{a} \times \overrightarrow{b})$
$= [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}] + 0 + 0 + 0 + 0 + [\overrightarrow{c} \overrightarrow{a} \overrightarrow{b}]$
$= [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] + [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 2[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$
$= 2 \times 40 = 80 \text{ cubic units}$.