KCET 2008 Mathematics Question Paper with Answer and Solution

(B) Let $x = 7^{-20}$.
Taking $\log_{10}$ on both sides:
$\log_{10} x = \log_{10} (7^{-20})$
$\log_{10} x = -20 \times \log_{10} 7$
$\log_{10} x = -20 \times 0.8451 = -16.902$.
To express this in standard form,we write:
$\log_{10} x = -16.902 = -17 + 0.098 = \overline{17}.098$.
Since the characteristic is $-17$,the first significant figure of $7^{-20}$ occurs at the $17^{th}$ decimal place.

(D) Given,$p^{2}-2q^{2}=1$ $(i)$.
Since $p$ and $q$ are prime numbers,we test small prime values.
Let $p=3$ and $q=2$.
Substituting these into the equation:
$(3)^{2}-2(2)^{2} = 9 - 2(4) = 9 - 8 = 1$.
This satisfies the given condition.
Now,we calculate the value of $p^{2}+2q^{2}$:
$p^{2}+2q^{2} = (3)^{2}+2(2)^{2} = 9 + 2(4) = 9 + 8 = 17$.

(D) Given the equation: $x^{3}-6x+9=0$
By testing integer factors of $9$,we find that for $x=-3$:
$(-3)^{3}-6(-3)+9 = -27+18+9 = 0$
Thus,$(x+3)$ is a factor of the polynomial.
Dividing $x^{3}-6x+9$ by $(x+3)$,we get:
$x^{3}+3x^{2}-3x^{2}-9x+3x+9 = x^{2}(x+3)-3x(x+3)+3(x+3) = (x+3)(x^{2}-3x+3) = 0$
For the quadratic part $x^{2}-3x+3=0$,the discriminant $D = b^{2}-4ac = (-3)^{2}-4(1)(3) = 9-12 = -3$.
Since $D < 0$,the roots of the quadratic part are imaginary.
Therefore,the only real root is $x=-3$.

(B) Let $z = x + iy$.
Given,$z = -\overline{z}$.
Substituting the values:
$x + iy = -(x - iy)$
$x + iy = -x + iy$
$2x = 0$
$x = 0$.
Since the real part $x$ is $0$,the complex number $z = iy$ is purely imaginary.

(A) The given equation is $\alpha^{2}+\alpha+1=0$.
This is the characteristic equation for the cube roots of unity,where $\alpha$ is either $\omega$ or $\omega^{2}$.
We know that $\omega^{3}=1$.
For $\alpha=\omega$,we have $\alpha^{31} = \omega^{31} = (\omega^{3})^{10} \cdot \omega = 1^{10} \cdot \omega = \omega = \alpha$.
For $\alpha=\omega^{2}$,we have $\alpha^{31} = (\omega^{2})^{31} = \omega^{62} = (\omega^{3})^{20} \cdot \omega^{2} = 1^{20} \cdot \omega^{2} = \omega^{2} = \alpha$.
In both cases,$\alpha^{31} = \alpha$.

(A) We are given the sum $S = \sum_{k=1}^{6}\left(\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7}\right)$.
Factor out $-i$ from the expression:
$S = -i \sum_{k=1}^{6}\left(\cos \frac{2 k \pi}{7} + i \sin \frac{2 k \pi}{7}\right)$.
Let $\omega = e^{i \frac{2 \pi}{7}} = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7}$. Then the sum becomes:
$S = -i \sum_{k=1}^{6} \omega^k$.
This is a geometric series with $6$ terms,where the first term is $\omega$ and the common ratio is $\omega$:
$S = -i \left( \frac{\omega(1 - \omega^6)}{1 - \omega} \right) = -i \left( \frac{\omega - \omega^7}{1 - \omega} \right)$.
Since $\omega^7 = e^{i 2 \pi} = 1$,we have:
$S = -i \left( \frac{\omega - 1}{1 - \omega} \right) = -i (-1) = i$.

(C) The $k$-th term of the series is $T_k = \frac{1}{(3k-1)(3k+2)}$.
We can write this as $T_k = \frac{1}{3} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$.
Summing up to $n$ terms:
$S_n = \sum_{k=1}^{n} T_k = \frac{1}{3} \sum_{k=1}^{n} \left( \frac{1}{3k-1} - \frac{1}{3k+2} \right)$.
This is a telescoping series:
$S_n = \frac{1}{3} \left[ \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \dots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) \right]$.
$S_n = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{3n+2} \right) = \frac{1}{3} \left( \frac{3n+2-2}{2(3n+2)} \right) = \frac{1}{3} \left( \frac{3n}{6n+4} \right) = \frac{n}{6n+4}$.

(C) We know that for any $n \ge 10$,$n!$ ends with at least two zeros,meaning $n!$ is divisible by $100$.
Specifically,$10! = 3628800$.
Calculating the sum:
$1! = 1$
$4! = 24$
$7! = 5040$
Sum $= 1 + 24 + 5040 + 10! + 12! + 13! + 15! + 16! + 17!$
Sum $= 5065 + (10! + 12! + 13! + 15! + 16! + 17!)$
Since $10!, 12!, 13!, 15!, 16!, 17!$ are all divisible by $3! = 6$,we check $5065$.
$5065$ is not divisible by $6$ (as it is odd).
However,checking divisibility by $3! = 6$:
$1! + 4! + 7! = 1 + 24 + 5040 = 5065$.
$5065 \pmod{6} \equiv 5065 \pmod{2 \times 3} \equiv 1 \pmod{2}$ and $5065 \pmod{3} \equiv 1$.
Wait,let us re-evaluate: $1! = 1, 4! = 24, 7! = 5040$.
$1 + 24 + 5040 = 5065$.
$5065$ is divisible by $5$ because it ends in $5$.
Thus,the sum is divisible by $5$.

(D) We observe the powers of $5$:
$5^{1} = 5$
$5^{2} = 25$
$5^{3} = 125$
$5^{4} = 625$
It is evident that for any positive integer $n$,the unit digit of $5^{n}$ is always $5$.
Therefore,the unit's place of $5^{834}$ is $5$.

(D) We need to find the remainder of $3^{100} \times 2^{50}$ when divided by $5$.
First,consider $3^{100} \pmod{5}$:
$3^2 = 9 \equiv 4 \equiv -1 \pmod{5}$.
So,$3^{100} = (3^2)^{50} \equiv (-1)^{50} \equiv 1 \pmod{5}$.
Next,consider $2^{50} \pmod{5}$:
$2^2 = 4 \equiv -1 \pmod{5}$.
So,$2^{50} = (2^2)^{25} \equiv (-1)^{25} \equiv -1 \pmod{5}$.
Since $-1 \equiv 4 \pmod{5}$,we have $2^{50} \equiv 4 \pmod{5}$.
Now,$3^{100} \times 2^{50} \equiv 1 \times 4 \equiv 4 \pmod{5}$.
Therefore,the remainder is $4$.

(C) Let $x = 67 \frac{1}{2}^{\circ} = 67.5^{\circ}$. We need to evaluate $\tan 67.5^{\circ} + \cot 67.5^{\circ}$.
Using the identity $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.
Here,$\theta = 67.5^{\circ}$,so $2\theta = 135^{\circ}$.
Thus,$\tan 67.5^{\circ} + \cot 67.5^{\circ} = \frac{2}{\sin 135^{\circ}}$.
Since $\sin 135^{\circ} = \sin(180^{\circ} - 45^{\circ}) = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore,the value is $\frac{2}{1/\sqrt{2}} = 2 \sqrt{2}$.

(C) Given,$\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$.
Using the identities $1+\cos A = 2\cos^{2}\frac{A}{2}$ and $1-\cos A = 2\sin^{2}\frac{A}{2}$,we get:
$\sqrt{\frac{2\cos^{2}\frac{A}{2}}{2\sin^{2}\frac{A}{2}}} = \frac{x}{y}$
$\cot\frac{A}{2} = \frac{x}{y} \implies \tan\frac{A}{2} = \frac{y}{x}$.
Now,using the formula $\tan A = \frac{2\tan\frac{A}{2}}{1-\tan^{2}\frac{A}{2}}$:
$\tan A = \frac{2(\frac{y}{x})}{1-(\frac{y}{x})^{2}} = \frac{\frac{2y}{x}}{\frac{x^{2}-y^{2}}{x^{2}}}$
$\tan A = \frac{2y}{x} \times \frac{x^{2}}{x^{2}-y^{2}} = \frac{2xy}{x^{2}-y^{2}}$.

(A) Given equation: $\sin 2x + \cos 4x = 2$
Using the identity $\cos 4x = 1 - 2\sin^2 2x$,we get:
$\sin 2x + 1 - 2\sin^2 2x = 2$
Rearranging the terms:
$2\sin^2 2x - \sin 2x + 1 = 0$
Let $t = \sin 2x$. The equation becomes $2t^2 - t + 1 = 0$.
The discriminant $D$ of this quadratic equation is $D = (-1)^2 - 4(2)(1) = 1 - 8 = -7$.
Since $D < 0$,there are no real values for $t$ that satisfy this equation.
Therefore,there are no real solutions for $x$.

(B) Given,$|\sin x| = \cos x$.
Since $|\sin x| \ge 0$,we must have $\cos x \ge 0$.
Squaring both sides,we get $\sin^2 x = \cos^2 x$.
Using the identity $\sin^2 x = 1 - \cos^2 x$,we have $1 - \cos^2 x = \cos^2 x$.
This simplifies to $2 \cos^2 x = 1$,or $\cos^2 x = \frac{1}{2}$.
Taking the square root,$\cos x = \pm \frac{1}{\sqrt{2}}$.
Since $\cos x \ge 0$,we discard the negative value,so $\cos x = \frac{1}{\sqrt{2}}$.
The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.
Here,$\cos x = \cos(\frac{\pi}{4})$,so $x = 2n\pi \pm \frac{\pi}{4}$.

(D) The general equation of a second-degree curve is $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$.
Comparing this with the given equation $3x^{2} + xy - y^{2} - 3x + 6y + k = 0$,we get:
$a = 3, b = -1, h = \frac{1}{2}, g = -\frac{3}{2}, f = 3, c = k$.
For the equation to represent a pair of lines,the condition is $abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0$.
Substituting the values:
$3(-1)(k) + 2(3)(-\frac{3}{2})(\frac{1}{2}) - 3(3)^{2} - (-1)(-\frac{3}{2})^{2} - k(\frac{1}{2})^{2} = 0$.
$-3k - \frac{9}{2} - 27 + \frac{9}{4} - \frac{k}{4} = 0$.
Multiplying by $4$ to clear the denominators:
$-12k - 18 - 108 + 9 - k = 0$.
$-13k - 117 = 0$.
$-13k = 117$.
$k = -9$.

(A) The vertices of the triangle are $A \equiv (2,3)$,$B \equiv (8,10)$,and $C \equiv (5,5)$.
The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Substituting the given values:
$G = \left(\frac{2+8+5}{3}, \frac{3+10+5}{3}\right)$
$G = \left(\frac{15}{3}, \frac{18}{3}\right)$
$G = (5, 6)$

(D) Let the coordinates of the endpoints of the line segment intercepted between the axes be $A(a, 0)$ and $B(0, b)$.
Let the midpoint of $AB$ be $P(x, y)$.
Then,$x = \frac{a+0}{2} \Rightarrow a = 2x$ and $y = \frac{0+b}{2} \Rightarrow b = 2y$.
Given that $a+b=4$.
Substituting the values of $a$ and $b$ in the given equation,we get:
$2x + 2y = 4$
Dividing by $2$,we get:
$x + y = 2$
Thus,the locus of the midpoint is $x+y=2$.

(B) The smallest circle passing through two points $A(x_1, y_1)$ and $B(x_2, y_2)$ has the line segment $AB$ as its diameter.
Given points are $A(2,2)$ and $B(3,3)$.
The equation of the circle with diameter $AB$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the coordinates,we get $(x-2)(x-3) + (y-2)(y-3) = 0$.
Expanding this,we have $(x^2 - 5x + 6) + (y^2 - 5y + 6) = 0$.
Simplifying,we get $x^2 + y^2 - 5x - 5y + 12 = 0$.
Thus,option $B$ is correct.

(A) point $(x_1, y_1)$ lies outside the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ if $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c > 0$.
For option $(A)$,$S = x^2 + y^2 - 8x$.
Substituting $(5, -7)$:
$S_1 = 5^2 + (-7)^2 - 8(5) = 25 + 49 - 40 = 34$.
Since $34 > 0$,the point $(5, -7)$ lies outside the circle $x^2 + y^2 - 8x = 0$.

(A) Let the radius of the circle be $r$.
The perimeter of a sector is given by $P = l + 2r$,where $l = r\theta$ is the arc length and $\theta$ is the angle in radians.
The length of the arc of a semicircle is $\pi r$.
According to the problem,the perimeter of the sector equals the arc length of the semicircle:
$r\theta + 2r = \pi r$
Dividing both sides by $r$ (since $r \neq 0$):
$\theta + 2 = \pi$
$\theta = \pi - 2$
Thus,the angle at the centre of the sector is $\pi - 2$ radians.

(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1), r_{2}=\sqrt{(-\alpha)^{2}+(-1)^{2}-1} = \sqrt{\alpha^{2}} = |\alpha|$.
Since the two circles touch each other internally,the distance between their centers must be equal to the difference of their radii:
$C_{1}C_{2} = |r_{1} - r_{2}|$
$\sqrt{(-\alpha - 0)^{2} + (-1 - 0)^{2}} = |3 - |\alpha||$
$\sqrt{\alpha^{2} + 1} = |3 - |\alpha||$
Squaring both sides:
$\alpha^{2} + 1 = (3 - |\alpha|)^{2}$
$\alpha^{2} + 1 = 9 - 6|\alpha| + \alpha^{2}$
$6|\alpha| = 8$
$|\alpha| = \frac{4}{3}$
$\alpha = \pm \frac{4}{3}$

(B) The given equation of the parabola is $y^{2}=16x$.
Comparing this with the standard form $y^{2}=4ax$,we get $4a=16$,which implies $a=4$.
It is a standard property of parabolas that the tangents drawn at the extremities of any focal chord intersect at right angles on the directrix of the parabola.
The equation of the directrix for the parabola $y^{2}=4ax$ is $x=-a$.
Substituting $a=4$,the equation of the directrix is $x=-4$,which can be written as $x+4=0$.
Therefore,the tangents intersect on the line $x+4=0$.

(C) Let the focus be $S(a, 0)$.
Let any point on the parabola $y^{2}=4ax$ be $P(at^{2}, 2at)$.
Let the mid-point of $SP$ be $(x, y)$.
Then,$x = \frac{a + at^{2}}{2}$ and $y = \frac{0 + 2at}{2} = at$.
From $y = at$,we get $t = \frac{y}{a}$.
Substituting $t$ into the equation for $x$:
$x = \frac{a + a(\frac{y}{a})^{2}}{2} = \frac{a + \frac{y^{2}}{a}}{2}$.
$2x = a + \frac{y^{2}}{a} \implies \frac{y^{2}}{a} = 2x - a \implies y^{2} = 2a(x - \frac{a}{2})$.
This is a parabola of the form $Y^{2} = 4AX$ where $A = \frac{a}{2}$ and $X = x - \frac{a}{2}$.
The directrix of $Y^{2} = 4AX$ is $X = -A$.
Therefore,$x - \frac{a}{2} = -\frac{a}{2}$.
$x = 0$.

(B) The given equation is $\frac{x^{2}}{2-\lambda} - \frac{y^{2}}{\lambda-5} = 1$.
For this to represent an ellipse,the equation must be of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a^{2} > 0$ and $b^{2} > 0$.
Rewriting the equation: $\frac{x^{2}}{2-\lambda} + \frac{y^{2}}{5-\lambda} = 1$.
For this to be an ellipse,both denominators must be positive:
$2 - \lambda > 0 \implies \lambda < 2$
$5 - \lambda > 0 \implies \lambda < 5$
Taking the intersection of these two conditions,we get $\lambda < 2$.

(C) By definition,an ellipse is the set of all points in a plane such that the sum of their distances from two fixed points (called foci) is a constant.
Therefore,the locus of such a point is an ellipse.

(B) The given equation of the hyperbola is $3x^{2} - 3y^{2} = 25$,which can be written as $x^{2} - y^{2} = \frac{25}{3}$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = \frac{25}{3}$ and $b^{2} = \frac{25}{3}$.
The eccentricity $e_{1}$ is given by $e_{1} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.
The equation of the conjugate hyperbola is $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$,which is $-x^{2} + y^{2} = \frac{25}{3}$.
The eccentricity $e_{2}$ of the conjugate hyperbola is given by $e_{2} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,$e_{1}^{2} + e_{2}^{2} = (\sqrt{2})^{2} + (\sqrt{2})^{2} = 2 + 2 = 4$.

(D) The equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$.
Comparing this with $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = 16$ and $b^{2} = 9$.
The point $(-4, 0)$ lies on the hyperbola since $\frac{(-4)^{2}}{16} - \frac{0^{2}}{9} = 1 - 0 = 1$.
The slope of the tangent at any point $(x_{1}, y_{1})$ is given by differentiating the hyperbola equation: $\frac{2x}{16} - \frac{2y y'}{9} = 0 \Rightarrow y' = \frac{9x}{16y}$.
At $(-4, 0)$,the tangent is vertical $(y' \to \infty)$,which means the tangent line is $x = -4$.
$A$ line perpendicular to a vertical line is a horizontal line.
Since the normal passes through $(-4, 0)$ and is perpendicular to the vertical tangent $x = -4$,the normal must be the horizontal line passing through $y = 0$.
Thus,the equation of the normal is $y = 0$.

(C) To evaluate the limit $\lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)$,let $t = \frac{1}{x}$.
As $x \rightarrow \infty$,$t \rightarrow 0$.
Substituting $x = \frac{1}{t}$ into the expression,we get:
$\lim_{t \rightarrow 0} \frac{1}{t} \sin(2t)$
$= \lim_{t \rightarrow 0} \frac{\sin(2t)}{t}$
$= \lim_{t \rightarrow 0} 2 \cdot \frac{\sin(2t)}{2t}$
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have:
$= 2 \cdot 1 = 2$

(D) The given conditional statement is $p \rightarrow \sim q$.
The contrapositive of a conditional statement $A \rightarrow B$ is $\sim B \rightarrow \sim A$.
Applying this to $p \rightarrow \sim q$,the contrapositive is $\sim(\sim q) \rightarrow \sim p$,which simplifies to $q \rightarrow \sim p$.
The converse of a conditional statement $A \rightarrow B$ is $B \rightarrow A$.
Applying this to the contrapositive $q \rightarrow \sim p$,the converse is $\sim p \rightarrow q$.

(C) Let $ED = h$ be the height of the tower. Points $A, B, C, D$ are collinear where $D$ is the foot of the tower.
In $\triangle ADE$,$\angle EAD = \alpha$,$\angle EBD = 2\alpha$,$\angle ECD = 3\alpha$.
In $\triangle ABE$,$\angle EAB = \alpha$ and $\angle EBA = 180^{\circ} - 2\alpha$.
Thus,$\angle AEB = 180^{\circ} - (\alpha + 180^{\circ} - 2\alpha) = \alpha$.
Since $\angle EAB = \angle AEB = \alpha$,$\triangle ABE$ is isosceles with $BE = AB = a$.
In $\triangle BCE$,by the sine rule:
$\frac{BE}{\sin 3\alpha} = \frac{CE}{\sin(180^{\circ} - 2\alpha)}$
$\frac{a}{\sin 3\alpha} = \frac{CE}{\sin 2\alpha} \Rightarrow CE = \frac{a \sin 2\alpha}{\sin 3\alpha}$.
In $\triangle CDE$,$\sin 3\alpha = \frac{h}{CE}$.
$h = CE \sin 3\alpha = \left(\frac{a \sin 2\alpha}{\sin 3\alpha}\right) \sin 3\alpha = a \sin 2\alpha$.

(C) Since $A, B, C$ are in $AP$,we have $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we have $A + C = 180^{\circ} - B$.
Substituting this into the $AP$ condition: $2B = 180^{\circ} - B$ $\Rightarrow 3B = 180^{\circ}$ $\Rightarrow B = 60^{\circ}$.
Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$.
Given $\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$,we have $\frac{\sin B}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$.
Substituting $B = 60^{\circ}$: $\frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$ $\Rightarrow \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$.
$\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow C = 45^{\circ}$.
Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ}$.

(D) Given the relation $R = \{(x, y) : 4x + 3y = 20\}$ on the set of natural numbers $\mathbb{N}$.
For $(x, y) \in R$,both $x$ and $y$ must be natural numbers (i.e.,$x, y \in \{1, 2, 3, \dots\}$).
Let us check the options:
$(a)$ $(-4, 12)$: $-4 \notin \mathbb{N}$.
$(b)$ $(5, 0)$: $0 \notin \mathbb{N}$.
$(c)$ $(3, 4)$: $4(3) + 3(4) = 12 + 12 = 24 \neq 20$.
$(d)$ $(2, 4)$: $4(2) + 3(4) = 8 + 12 = 20$. Since $2 \in \mathbb{N}$ and $4 \in \mathbb{N}$,$(2, 4) \in R$.
Thus,the correct option is $(d)$.

(B) Given the height function: $x = 80t - 16t^2$.
To find the time at which the stone reaches maximum height,we differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 80 - 32t$.
At maximum height,the velocity of the stone is zero,so $\frac{dx}{dt} = 0$.
Setting the derivative to zero: $80 - 32t = 0$.
$32t = 80$.
$t = \frac{80}{32} = 2.5 \text{ s}$.

(D) Given the binary operation $a * b = a + b - 5$.
First,evaluate the inner expression $(x * 3)$:
$x * 3 = x + 3 - 5 = x - 2$.
Now,substitute this into the equation $2 * (x * 3) = 5$:
$2 * (x - 2) = 5$.
Applying the definition of the operation again:
$2 + (x - 2) - 5 = 5$.
Simplify the expression:
$x - 5 = 5$.
$x = 10$.

(C) For the set of natural numbers $N$ with respect to addition,the operation is commutative $(a+b = b+a)$ and associative $((a+b)+c = a+(b+c))$.
For the set of natural numbers $N$ with respect to multiplication,the operation is associative $((a \times b) \times c = a \times (b \times c))$.
For the operation $a * b = a^{b}$,we check for commutativity: $a * b = a^{b}$ and $b * a = b^{a}$.
Since $a^{b} \neq b^{a}$ for all $a, b \in N$ (e.g.,$2 * 3 = 2^{3} = 8$ while $3 * 2 = 3^{2} = 9$),the operation $*$ is not commutative.
Therefore,option $C$ is false.

(D) Let $G = \{-1, 0, 1\}$.
For a set to be a multiplicative group,every element must have a multiplicative inverse.
The multiplicative inverse of an element $a$ is an element $b$ such that $a \times b = 1$.
For the element $0$,there is no such element $b$ in the set such that $0 \times b = 1$.
Therefore,the inverse law fails for the element $0$.

(B) We know that $A(\operatorname{adj} A) = |A| I$.
Given that $A(\operatorname{adj} A) = 5 I$,comparing both sides,we get $|A| = 5$.
For a square matrix $A$ of order $n$,the property of the adjoint matrix is $|\operatorname{adj} A| = |A|^{n-1}$.
Here,the order of the matrix is $n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\operatorname{adj} A| = (5)^2 = 25$.

(A) The characteristic roots (eigenvalues) of a square matrix $A$ are the solutions to the characteristic equation $\det(A - \lambda I) = 0$.
For a triangular matrix (either upper or lower),the determinant $\det(A - \lambda I)$ is simply the product of the diagonal elements minus $\lambda$.
Given the matrix $A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 6\end{array}\right]$,which is a lower triangular matrix,the characteristic equation is $(1 - \lambda)(3 - \lambda)(6 - \lambda) = 0$.
Setting each factor to zero,we get $\lambda = 1, 3, 6$.
Thus,the characteristic roots are $1, 3, 6$.

(A) Given,$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
The determinant of $A$ is calculated as $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.
The adjoint of $A$,denoted as $\operatorname{adj}(A)$,is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\operatorname{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.
The inverse of a matrix is given by the formula $A^{-1} = \frac{1}{|A|} \operatorname{adj}(A)$.
Substituting the values,we get $A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.

(B) Given the determinant $A = \left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$.
We are given $B = \left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$.
First,swap the first and second rows of $B$ to get $B' = -\left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$.
Next,swap the second and third rows of $B'$ to get $B'' = -(-1) \left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right| = \left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right|$.
Since the determinant of a matrix is equal to the determinant of its transpose,$\left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right| = \left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right| = A$.
Therefore,$A = B$.

(A) Let $\theta = \sin^{-1} \sqrt{\frac{63}{65}}$. Then $\sin \theta = \sqrt{\frac{63}{65}}$.
We need to find $\sin(2\theta)$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$.
First,find $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{63}{65}} = \sqrt{\frac{2}{65}}$.
Now,substitute the values into the identity:
$\sin(2\theta) = 2 \times \sqrt{\frac{63}{65}} \times \sqrt{\frac{2}{65}}$
$\sin(2\theta) = 2 \times \frac{\sqrt{63 \times 2}}{65} = 2 \times \frac{\sqrt{126}}{65} = \frac{2 \sqrt{126}}{65}$.

(A) Given the function $f(x) = x^{3}$.
To find $f^{-1}(8)$,we need to solve the equation $f(x) = 8$ for $x$.
$x^{3} = 8$
$x^{3} - 8 = 0$
$(x - 2)(x^{2} + 2x + 4) = 0$
This gives $x = 2$ as the real solution.
Since the domain of $f$ is $R$ (the set of real numbers),we only consider the real value.
Therefore,$f^{-1}(8) = \{2\}$.

(B) Given,$f(x) = \begin{cases} 2a-x & \text{in } -a < x < a \\ 3x-2a & \text{in } a \leq x \end{cases}$
Check continuity at $x=a$:
$LHL = \lim_{x \to a^-} f(x) = \lim_{x \to a} (2a-x) = 2a-a = a$
$RHL = \lim_{x \to a^+} f(x) = \lim_{x \to a} (3x-2a) = 3a-2a = a$
$f(a) = 3(a)-2a = a$
Since $LHL = RHL = f(a)$,the function is continuous at $x=a$.
Check differentiability at $x=a$:
$LHD = \lim_{h \to 0} \frac{f(a-h)-f(a)}{-h} = \lim_{h \to 0} \frac{2a-(a-h)-a}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1$
$RHD = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{3(a+h)-2a-a}{h} = \lim_{h \to 0} \frac{3h}{h} = 3$
Since $LHD \neq RHD$,the function is not differentiable at $x=a$.
Thus,option $B$ is correct.

(A) Given,$f(x^{5}) = 5x^{3}$.
Let $x^{5} = y$,which implies $x = y^{1/5}$.
Then $x^{3} = (y^{1/5})^{3} = y^{3/5}$.
Substituting this into the function,we get $f(y) = 5y^{3/5}$.
Replacing $y$ with $x$,we have $f(x) = 5x^{3/5}$.
Now,differentiating $f(x)$ with respect to $x$:
$f'(x) = 5 \cdot \frac{3}{5} x^{(3/5 - 1)} = 3x^{-2/5}$.
This can be rewritten as $f'(x) = \frac{3}{x^{2/5}} = \frac{3}{\sqrt[5]{x^{2}}}$.

(C) Given,$f(x) = b e^{ax} + a e^{bx}$.
First,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(b e^{ax} + a e^{bx}) = b(a e^{ax}) + a(b e^{bx}) = ab e^{ax} + ab e^{bx}$.
Next,differentiate $f^{\prime}(x)$ with respect to $x$ to find the second derivative:
$f^{\prime \prime}(x) = \frac{d}{dx}(ab e^{ax} + ab e^{bx}) = ab(a e^{ax}) + ab(b e^{bx}) = a^2 b e^{ax} + ab^2 e^{bx}$.
Finally,evaluate at $x = 0$:
$f^{\prime \prime}(0) = a^2 b e^{a(0)} + ab^2 e^{b(0)} = a^2 b(1) + ab^2(1) = a^2 b + ab^2$.
Factoring out $ab$,we get:
$f^{\prime \prime}(0) = ab(a + b)$.

(C) Let $u = \sin(x^{3})$ and $v = \cos(x^{3})$.
By differentiating both with respect to $x$,we get:
$\frac{du}{dx} = \cos(x^{3}) \cdot 3x^{2}$
$\frac{dv}{dx} = -\sin(x^{3}) \cdot 3x^{2}$
Now,the derivative of $u$ with respect to $v$ is given by the chain rule:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{3x^{2} \cos(x^{3})}{-3x^{2} \sin(x^{3})}$
$\frac{du}{dv} = -\frac{\cos(x^{3})}{\sin(x^{3})} = -\cot(x^{3})$

(B) Given that $f(x)$ is an even function,we have $f(-x) = f(x)$.
Differentiating both sides with respect to $x$,we get $f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$.
This implies $f^{\prime}(-x) = -f^{\prime}(x)$,which means $f^{\prime}(x)$ is an odd function.
Substituting $x = e$,we get $f^{\prime}(-e) = -f^{\prime}(e)$.
Therefore,$f^{\prime}(e) + f^{\prime}(-e) = f^{\prime}(e) - f^{\prime}(e) = 0$.

(C) Let the rectangle have sides $x$ and $y$ inscribed in a circle of radius $r = 2 \text{ unit}$.
The diagonal of the rectangle is equal to the diameter of the circle,so $d = 2r = 2 \times 2 = 4 \text{ unit}$.
By the Pythagorean theorem,$x^2 + y^2 = d^2 = 4^2 = 16$.
The area of the rectangle is $A = xy$.
To maximize $A$,we maximize $A^2 = x^2 y^2$.
Since $y^2 = 16 - x^2$,we have $A^2 = x^2(16 - x^2) = 16x^2 - x^4$.
Let $f(x) = 16x^2 - x^4$. To find the maximum,set $f'(x) = 32x - 4x^3 = 0$.
$4x(8 - x^2) = 0$,which gives $x^2 = 8$ (since $x > 0$).
Then $y^2 = 16 - 8 = 8$,so $x = y = \sqrt{8} = 2\sqrt{2}$.
The rectangle is a square with side length $2\sqrt{2}$.
The maximum area is $A = x \times y = \sqrt{8} \times \sqrt{8} = 8 \text{ sq unit}$.

(C) Let $y = \frac{\log x}{x}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For maxima,set $\frac{dy}{dx} = 0$:
$\frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e$.
Since $e \approx 2.718$,$x = e$ lies in the interval $(2, \infty)$.
Now,check the second derivative:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2x \log x - 3x}{x^4}$.
At $x = e$,$\frac{d^2y}{dx^2} = \frac{2e(1) - 3e}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} < 0$.
Since the second derivative is negative,the function has a maximum at $x = e$.
The maximum value is $y = \frac{\log e}{e} = \frac{1}{e}$.

(B) Given that $\int f(x) dx = g(x)$.
This implies that $f(x) = g'(x)$.
We need to evaluate the integral $\int f(x) g(x) dx$.
Substituting $f(x) = g'(x)$,we get $\int g'(x) g(x) dx$.
Let $g(x) = u$,then $g'(x) dx = du$.
The integral becomes $\int u du = \frac{u^2}{2} + C$.
Substituting back $u = g(x)$,we get $\frac{1}{2} g^2(x) + C$.

(A) Let $I = \int \frac{\sec x}{\sec x+\tan x} dx$.
Multiplying the numerator and denominator by $(\sec x - \tan x)$,we get:
$I = \int \frac{\sec x(\sec x - \tan x)}{\sec^2 x - \tan^2 x} dx$.
Since $\sec^2 x - \tan^2 x = 1$,the integral simplifies to:
$I = \int (\sec^2 x - \sec x \tan x) dx$.
Integrating term by term,we obtain:
$I = \tan x - \sec x + C$.

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$.
Substitute $\sin^{2} x = t$.
Differentiating both sides with respect to $x$,we get $2 \sin x \cos x dx = dt$,which implies $\sin x \cos x dx = \frac{1}{2} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{2 \sqrt{1-t^{2}}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1} t + C$,we obtain $I = \frac{1}{2} \sin^{-1} t + C$.
Finally,substituting $t = \sin^{2} x$ back,we get $I = \frac{1}{2} \sin^{-1}(\sin^{2} x) + C$.

(C) We use the formula for integration by parts: $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.
Let $I = \int e^{x} \cdot x^{5} \, dx$.
Using the generalized integration by parts formula $\int e^{x} f(x) \, dx = e^{x} [f(x) - f'(x) + f''(x) - f'''(x) + f^{(4)}(x) - f^{(5)}(x)] + C$.
Here,$f(x) = x^{5}$.
Then,$f'(x) = 5x^{4}$,$f''(x) = 20x^{3}$,$f'''(x) = 60x^{2}$,$f^{(4)}(x) = 120x$,and $f^{(5)}(x) = 120$.
Substituting these into the formula:
$I = e^{x} [x^{5} - 5x^{4} + 20x^{3} - 60x^{2} + 120x - 120] + C$.

(C) Let $I = \int_{-2}^{2}(a x^{3} + b x + c) d x$.
We can split the integral as:
$I = \int_{-2}^{2} a x^{3} d x + \int_{-2}^{2} b x d x + \int_{-2}^{2} c d x$.
We know that for an odd function $f(x)$,$\int_{-k}^{k} f(x) d x = 0$.
Since $a x^{3}$ and $b x$ are odd functions,$\int_{-2}^{2} a x^{3} d x = 0$ and $\int_{-2}^{2} b x d x = 0$.
Thus,$I = 0 + 0 + \int_{-2}^{2} c d x = \int_{-2}^{2} c d x = [c x]_{-2}^{2} = c(2 - (-2)) = 4c$.
Therefore,the value of the integral depends only on the value of $c$.

(B) The given curve is $y=2x-x^{2}$.
To find the points where the curve intersects the $x$-axis,we set $y=0$:
$2x-x^{2}=0 \implies x(2-x)=0$,which gives $x=0$ and $x=2$.
Thus,the curve intersects the $x$-axis at $(0,0)$ and $(2,0)$.
The required area is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=2$:
$\text{Area} = \int_{0}^{2} (2x-x^{2}) dx$
$= \left[ x^{2} - \frac{x^{3}}{3} \right]_{0}^{2}$
$= \left( 2^{2} - \frac{2^{3}}{3} \right) - (0 - 0)$
$= 4 - \frac{8}{3}$
$= \frac{12-8}{3} = \frac{4}{3} \text{ sq unit}$.

(B) Given differential equation is $y \frac{dy}{dx} + x = c$.
Separating the variables,we get $y \, dy = (c - x) \, dx$.
Integrating both sides,we have $\int y \, dy = \int (c - x) \, dx$.
This gives $\frac{y^2}{2} = cx - \frac{x^2}{2} + k$,where $k$ is the constant of integration.
Multiplying by $2$,we get $y^2 = 2cx - x^2 + 2k$,which simplifies to $x^2 - 2cx + y^2 = 2k$.
Completing the square for $x$,we get $(x^2 - 2cx + c^2) + y^2 = 2k + c^2$.
Thus,$(x - c)^2 + y^2 = R^2$,where $R^2 = 2k + c^2$.
This is the equation of a family of circles with centres at $(c, 0)$,which lie on the $x$-axis.

(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$.
We are given $|\overrightarrow{a} + \overrightarrow{b}| = 1$.
Squaring both sides,we get $|\overrightarrow{a} + \overrightarrow{b}|^2 = 1^2$.
Using the property $|\overrightarrow{x} + \overrightarrow{y}|^2 = |\overrightarrow{x}|^2 + |\overrightarrow{y}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b})$,we have:
$|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1$.
Substituting the values,$1^2 + 1^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1$.
$2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) = 1 \Rightarrow 2(\overrightarrow{a} \cdot \overrightarrow{b}) = -1$.
Now,we need to find $|\overrightarrow{a} - \overrightarrow{b}|$.
$|\overrightarrow{a} - \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2(\overrightarrow{a} \cdot \overrightarrow{b})$.
Substituting the values,$|\overrightarrow{a} - \overrightarrow{b}|^2 = 1^2 + 1^2 - (-1) = 1 + 1 + 1 = 3$.
Therefore,$|\overrightarrow{a} - \overrightarrow{b}| = \sqrt{3}$.

(D) Let $\overrightarrow{a} = \hat{i} + \hat{j}$ and $\overrightarrow{b} = \hat{j} + \hat{k}$.
The vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by the cross product $\overrightarrow{a} \times \overrightarrow{b}$.
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$.
The magnitude of this vector is $|\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The required unit vector is $\pm \frac{\overrightarrow{a} \times \overrightarrow{b}}{|\overrightarrow{a} \times \overrightarrow{b}|} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.
Comparing this with the given options,the correct option is $\frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.

(C) Let $\overrightarrow{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$.
Given that $\overrightarrow{a} \cdot \hat{i} = 1$,we have $a_{1} = 1$.
Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j}) = 1$,we have $a_{1} + a_{2} = 1$. Substituting $a_{1} = 1$,we get $1 + a_{2} = 1$,which implies $a_{2} = 0$.
Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$,we have $a_{1} + a_{2} + a_{3} = 1$. Substituting $a_{1} = 1$ and $a_{2} = 0$,we get $1 + 0 + a_{3} = 1$,which implies $a_{3} = 0$.
Therefore,$\overrightarrow{a} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i}$.

(C) Given vectors are $\overrightarrow{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ and $\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$.
The formula for the projection of vector $\overrightarrow{a}$ on vector $\overrightarrow{b}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}$.
First,calculate the dot product $\overrightarrow{a} \cdot \overrightarrow{b}$:
$\overrightarrow{a} \cdot \overrightarrow{b} = (3)(2) + (-1)(3) + (5)(1) = 6 - 3 + 5 = 8$.
Next,calculate the magnitude of vector $\overrightarrow{b}$:
$|\overrightarrow{b}| = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
Therefore,the projection is $\frac{8}{\sqrt{14}}$.