KCET 2013 Mathematics Question Paper with Answer and Solution

(D) Let $d$ be the smallest divisor of a composite number $a$ such that $1 < d < a$.
If $d > \sqrt{a}$,then the other divisor $a/d$ must also be greater than $\sqrt{a}$ because $a/d < a/\sqrt{a} = \sqrt{a}$.
This implies $a/d$ is a divisor smaller than $d$,which contradicts the assumption that $d$ is the smallest divisor greater than $1$.
Therefore,the smallest divisor $d$ must satisfy $d \leq \sqrt{a}$.

(D) Given equation is $x^{3}+a x^{2}+b x+c=0$.
Let the roots be $(\alpha, \beta, \gamma)$. Since they are in $AP$,we have $2 \beta = \alpha + \gamma$.
From the sum of roots,$\alpha + \beta + \gamma = -a$.
Substituting $\alpha + \gamma = 2 \beta$,we get $3 \beta = -a$,so $\beta = -\frac{a}{3}$.
Since $\beta$ is a root,it satisfies the equation:
$(-\frac{a}{3})^{3} + a(-\frac{a}{3})^{2} + b(-\frac{a}{3}) + c = 0$.
$-\frac{a^{3}}{27} + \frac{a^{3}}{9} - \frac{ab}{3} + c = 0$.
Multiplying by $27$,we get $-a^{3} + 3a^{3} - 9ab + 27c = 0$.
$2a^{3} - 9ab + 27c = 0$.
Therefore,$2a^{3} - 9ab = -27c$.

(C) Let $z = \frac{1+2i}{1-(1-i)^{2}}$.
First,simplify the denominator: $(1-i)^{2} = 1^{2} + i^{2} - 2i = 1 - 1 - 2i = -2i$.
Substitute this back into the expression for $z$:
$z = \frac{1+2i}{1 - (-2i)} = \frac{1+2i}{1+2i} = 1$.
Thus,$z = 1 + 0i$.
The modulus is $|z| = \sqrt{1^{2} + 0^{2}} = 1$.
The amplitude (argument) is $\theta = \tan^{-1}\left(\frac{0}{1}\right) = 0$.

(D) Given,$2x = -1 + i\sqrt{3}$.
$x = \frac{-1 + i\sqrt{3}}{2} = \omega$,where $\omega$ is a complex cube root of unity.
We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.
Substituting $x = \omega$ into the expression:
$(1 - \omega^2 + \omega)^6 - (1 - \omega + \omega^2)^6$
$= (-\omega^2 - \omega^2)^6 - (-\omega - \omega)^6$
$= (-2\omega^2)^6 - (-2\omega)^6$
$= 2^6 \cdot \omega^{12} - 2^6 \cdot \omega^6$
Since $\omega^3 = 1$,we have $\omega^6 = 1$ and $\omega^{12} = 1$.
$= 64(1) - 64(1) = 0$.

(B) The $n$-th term of the series is given by $T_n = \frac{n}{(n+1)(n+2)} \cdot 2^n$.
Using partial fractions,we can write $\frac{n}{(n+1)(n+2)} = \frac{2}{n+2} - \frac{1}{n+1}$.
Thus,$T_n = \left( \frac{2}{n+2} - \frac{1}{n+1} \right) 2^n = \frac{2^{n+1}}{n+2} - \frac{2^n}{n+1}$.
Let $f(n) = \frac{2^n}{n+1}$. Then $T_n = f(n+1) - f(n)$.
The sum $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (f(k+1) - f(k)) = f(n+1) - f(1)$.
Since $f(n+1) = \frac{2^{n+1}}{n+2}$ and $f(1) = \frac{2^1}{1+1} = 1$,we have $S_n = \frac{2^{n+1}}{n+2} - 1$.

(C) Let $n = 10^{10}$. The expression becomes $n(n+1)(n+2)$.
This is the product of $3$ consecutive integers.
The product of any $k$ consecutive integers is always divisible by $k!$.
Therefore,$n(n+1)(n+2)$ is divisible by $3! = 3 \times 2 \times 1 = 6$.
Since the expression is perfectly divisible by $6$,the remainder is $0$.

(A) We are given the sum $S = \sum_{k=0}^{n} (k+1) C_{k} = 576$.
We know that $\sum_{k=0}^{n} C_{k} x^{k} = (1+x)^{n}$.
Multiplying by $x$,we get $\sum_{k=0}^{n} C_{k} x^{k+1} = x(1+x)^{n}$.
Differentiating with respect to $x$: $\sum_{k=0}^{n} (k+1) C_{k} x^{k} = (1+x)^{n} + nx(1+x)^{n-1}$.
Setting $x=1$,we get $\sum_{k=0}^{n} (k+1) C_{k} = (1+1)^{n} + n(1)(1+1)^{n-1} = 2^{n} + n \cdot 2^{n-1}$.
This simplifies to $2^{n-1}(2+n) = 576$.
We can write $576 = 64 \times 9 = 2^{6} \times 9 = 2^{7-1}(7+2)$.
Comparing $2^{n-1}(n+2) = 2^{7-1}(7+2)$,we get $n=7$.

(D) The given expression is $P = \log(\sin 1^{\circ}) \cdot \log(\sin 2^{\circ}) \cdot \dots \cdot \log(\sin 90^{\circ}) \cdot \dots \cdot \log(\sin 179^{\circ})$.
We know that $\sin 90^{\circ} = 1$.
Therefore,the term $\log(\sin 90^{\circ}) = \log(1) = 0$.
Since the expression is a product of terms including $\log(\sin 90^{\circ})$,the entire product becomes $0$ because any number multiplied by $0$ is $0$.
Thus,the value is $0$.

(C) Given,$\sin x - \sin y = \frac{1}{2} \dots (i)$ and $\cos x - \cos y = 1 \dots (ii)$.
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = \frac{1}{2} \dots (iii)$
$-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = 1 \dots (iv)$
Dividing $(iv)$ by $(iii)$:
$\frac{-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)} = \frac{1}{1/2}$
$-\tan \left(\frac{x+y}{2}\right) = 2 \implies \tan \left(\frac{x+y}{2}\right) = -2$.
Using the formula $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$ where $\theta = \frac{x+y}{2}$:
$\tan(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 - \tan^2 \left(\frac{x+y}{2}\right)} = \frac{2(-2)}{1 - (-2)^2} = \frac{-4}{1 - 4} = \frac{-4}{-3} = \frac{4}{3}$.

(A) Given,$\sin x - \cos x = \sqrt{2}$
Divide both sides by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$
Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:
$\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} = 1$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\sin(x - \frac{\pi}{4}) = 1$
Since $\sin \theta = 1$ implies $\theta = 2n\pi + \frac{\pi}{2}$:
$x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}$
$x = 2n\pi + \frac{\pi}{2} + \frac{\pi}{4}$
$x = 2n\pi + \frac{3\pi}{4}$

(A) The given equation of the pair of straight lines is $(\cos^{2} \alpha - 1) x^{2} - 2 \cos^{2} \alpha \cdot xy + \sin^{2} \alpha y^{2} = 0$.
Comparing this with the general equation $ax^{2} + 2hxy + by^{2} = 0$,we get:
$a = \cos^{2} \alpha - 1 = -\sin^{2} \alpha$,
$h = -\cos^{2} \alpha$,
$b = \sin^{2} \alpha$.
Let $\theta$ be the angle between the lines. The formula for the angle is $\tan \theta = \left| \frac{2 \sqrt{h^{2} - ab}}{a + b} \right|$.
Substituting the values:
$a + b = -\sin^{2} \alpha + \sin^{2} \alpha = 0$.
Since the denominator is $0$,the value of $\tan \theta$ is undefined,which implies $\theta = 90^{\circ}$.
Thus,the lines are perpendicular.

(B) Let the coordinates of the vertices of $\triangle OAB$ be $O(0, 0)$,$A(\cos \theta, \sin \theta)$,and $B(\sin \theta, -\cos \theta)$.
Let the centroid of $\triangle OAB$ be $(h, k)$.
The formula for the centroid $(h, k)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $h = \frac{x_1 + x_2 + x_3}{3}$ and $k = \frac{y_1 + y_2 + y_3}{3}$.
Substituting the given coordinates:
$h = \frac{0 + \cos \theta + \sin \theta}{3} \Rightarrow 3h = \cos \theta + \sin \theta$ ... $(i)$
$k = \frac{0 + \sin \theta - \cos \theta}{3} \Rightarrow 3k = \sin \theta - \cos \theta$ ... (ii)
To find the locus,we eliminate $\theta$ by squaring and adding equations $(i)$ and (ii):
$(3h)^{2} + (3k)^{2} = (\cos \theta + \sin \theta)^{2} + (\sin \theta - \cos \theta)^{2}$
$9h^{2} + 9k^{2} = (\cos^{2} \theta + \sin^{2} \theta + 2\sin \theta \cos \theta) + (\sin^{2} \theta + \cos^{2} \theta - 2\sin \theta \cos \theta)$
$9h^{2} + 9k^{2} = 1 + 1 = 2$
Replacing $(h, k)$ with $(x, y)$,the locus is $9x^{2} + 9y^{2} = 2$.

(D) The given equation of the line is $3 \cos \theta \cdot x + 4 \sin \theta \cdot y = 12$.
Dividing by $12$,we get the intercept form:
$\frac{x}{4 / \cos \theta} + \frac{y}{3 / \sin \theta} = 1$.
This line intersects the coordinate axes at $A\left(\frac{4}{\cos \theta}, 0\right)$ and $B\left(0, \frac{3}{\sin \theta}\right)$.
The area of $\triangle OAB$ is given by:
$\Delta = \frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times \left| \frac{4}{\cos \theta} \right| \times \left| \frac{3}{\sin \theta} \right| = \frac{6}{|\sin \theta \cos \theta|} = \frac{12}{|\sin 2 \theta|}$.
For the area to be minimum,$|\sin 2 \theta|$ must be maximum. Since the maximum value of $|\sin 2 \theta|$ is $1$,the minimum area is:
$\Delta_{\min} = \frac{12}{1} = 12$.

(A) The given equation of the line is $6x - 7y + 8 + \lambda(3x - y + 5) = 0$.
Rearranging the terms in $x$ and $y$:
$(6 + 3\lambda)x - (7 + \lambda)y + (8 + 5\lambda) = 0$.
For a line to be parallel to the $y$-axis,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero.
Setting the coefficient of $y$ to zero:
$-(7 + \lambda) = 0$
$\lambda + 7 = 0$
$\lambda = -7$.
Checking the coefficient of $x$ for $\lambda = -7$:
$6 + 3(-7) = 6 - 21 = -15 \neq 0$.
Thus,the line is $x = \text{constant}$,which is parallel to the $y$-axis.

(A) Given,the centre of the circle is $C(3,4)$.
Since the circle touches the line $5x+12y-11=0$,the radius $r$ of the circle is equal to the perpendicular distance from the centre $(3,4)$ to the line.
The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$.
Substituting the values,we get:
$r = \frac{|5(3)+12(4)-11|}{\sqrt{5^2+12^2}}$
$r = \frac{|15+48-11|}{\sqrt{25+144}}$
$r = \frac{|52|}{\sqrt{169}}$
$r = \frac{52}{13} = 4$.
The area of the circle is given by $\pi r^2$.
Area $= \pi(4)^2 = 16\pi$ sq units.

(B) The equation of the circle is $2x^{2} + 2y^{2} - 3x + 4y = 0$. Dividing by $2$,we get the standard form: $x^{2} + y^{2} - \frac{3}{2}x + 2y = 0$.
Here,$AB$ represents the length of the tangent from point $B(2, 1)$ to the circle.
The length of the tangent from a point $(x_{1}, y_{1})$ to a circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ is given by $\sqrt{x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c}$.
Substituting the coordinates of $B(2, 1)$ into the circle equation:
$AB = \sqrt{(2)^{2} + (1)^{2} - \frac{3}{2}(2) + 2(1)}$
$AB = \sqrt{4 + 1 - 3 + 2}$
$AB = \sqrt{4} = 2 \text{ units}$.

(D) The given equation of the circle is $x^{2}+y^{2}+3x+2y-8=0$.
Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=\frac{3}{2}$,$f=1$,and $c=-8$.
The length of the intercept made by the $y$-axis is given by the formula $2\sqrt{f^{2}-c}$.
Substituting the values,we get:
Length $= 2\sqrt{(1)^{2}-(-8)}$
$= 2\sqrt{1+8}$
$= 2\sqrt{9}$
$= 2 \times 3 = 6$.

(A) The given equation of the circle is $x^{2}+y^{2}+2x-2y+7=0$.
Comparing this with the standard form $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=1$,$f=-1$,and $c=7$.
The radius $r$ of the circle is given by $r = \sqrt{g^{2}+f^{2}-c}$.
Substituting the values,$r = \sqrt{1^{2}+(-1)^{2}-7} = \sqrt{1+1-7} = \sqrt{-5}$.
Since the radius is $\sqrt{-5}$,which is an imaginary number,the given equation does not represent a real circle.
Therefore,there are no real circles that can cut an imaginary circle orthogonally.
Thus,the number of such real circles is $0$.

(D) Given,the equation of the parabola is $y^{2}=4x$.
Comparing with $y^{2}=4ax$,we get $a=1$.
The equation of the tangent to the parabola in slope form is $y=mx+\frac{a}{m}$.
Substituting $a=1$,we get $y=mx+\frac{1}{m} \quad (i)$.
The tangent is inclined at an angle of $\frac{\pi}{4}$ to the positive direction of the $x$-axis,so the slope $m = \tan(\frac{\pi}{4}) = 1$.
Substituting $m=1$ into equation $(i)$,we get $y = (1)x + \frac{1}{1}$.
This simplifies to $y = x + 1$,or $x - y + 1 = 0$.

(A) Given,the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
Here,$a^{2}=4$ and $b^{2}=3$.
The eccentricity of the ellipse is $e_{1}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.
Given,the equation of the hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$.
Here,$a^{2}=4$ and $b^{2}=3$.
The eccentricity of the hyperbola is $e_{2}=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{3}{4}}=\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}$.
The sum of the squares of the eccentricities is $e_{1}^{2}+e_{2}^{2} = (\frac{1}{2})^{2} + (\frac{\sqrt{7}}{2})^{2} = \frac{1}{4} + \frac{7}{4} = \frac{8}{4} = 2$.

(D) The equation of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $x^{2}+y^{2}=a^{2}$.
Area of the auxiliary circle $= \pi a^{2}$.
Area of the ellipse $= \pi ab$.
According to the problem,the area of the auxiliary circle is twice the area of the ellipse:
$\pi a^{2} = 2(\pi ab)$
$a^{2} = 2ab$
$a = 2b \Rightarrow b = \frac{a}{2}$.
The eccentricity $e$ of the ellipse is given by:
$e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 - \frac{(a/2)^{2}}{a^{2}}}$
$e = \sqrt{1 - \frac{a^{2}/4}{a^{2}}}$
$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(A) The given equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Distance between the foci $= 2ae$.
Distance between the directrices $= \frac{2a}{e}$.
Given the ratio is $3: 2$,so $\frac{2ae}{2a/e} = \frac{3}{2}$.
This simplifies to $e^{2} = \frac{3}{2}$.
We know that for a hyperbola,$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}}$.
Substituting $e^{2} = \frac{3}{2}$,we get $1 + \frac{b^{2}}{a^{2}} = \frac{3}{2}$.
$\frac{b^{2}}{a^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$.
Therefore,$\frac{b}{a} = \frac{1}{\sqrt{2}}$,which implies $\frac{a}{b} = \sqrt{2}: 1$.

(C) Given limit: $\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$
This is in the indeterminate form $\frac{0}{0}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\log _{e}(1+x))}{\frac{d}{dx}(3^{x}-1)}$
$= \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^{x} \cdot \log _{e} 3}$
Substituting $x = 0$:
$= \frac{\frac{1}{1+0}}{3^{0} \cdot \log _{e} 3} = \frac{1}{1 \cdot \log _{e} 3} = \frac{1}{\log _{e} 3}$
Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we get:
$= \log _{3} e$

(B) The given proposition is $(p \wedge \sim q) \rightarrow r$.
The inverse of a conditional statement $A \rightarrow B$ is defined as $\sim A \rightarrow \sim B$.
Here,$A = (p \wedge \sim q)$ and $B = r$.
Therefore,the inverse is $\sim (p \wedge \sim q) \rightarrow \sim r$.
Using De Morgan's Law,$\sim (p \wedge \sim q) \equiv \sim p \vee \sim (\sim q) \equiv \sim p \vee q$.
Thus,the inverse is $(\sim p) \vee q \rightarrow (\sim r)$.

(C) Given,$150 x \equiv 35 \pmod{31}$.
First,simplify the congruence by dividing by the common factor $5$ (since $\gcd(5, 31) = 1$):
$30 x \equiv 7 \pmod{31}$.
We can write $30$ as $-1 \pmod{31}$:
$-x \equiv 7 \pmod{31}$.
Multiplying both sides by $-1$:
$x \equiv -7 \pmod{31}$.
To find the positive residue,add $31$:
$x \equiv -7 + 31 \pmod{31} \Rightarrow x \equiv 24 \pmod{31}$.
Thus,$x = 24$ satisfies the given congruence.

(B) Given,in $\triangle ABC$,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \dots (i)$
From the sine rule,$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \dots (ii)$
Dividing Eq. $(ii)$ by Eq. $(i)$,we get $\tan A = \tan B = \tan C$.
Since $A, B, C$ are angles of a triangle,$A = B = C$.
Thus,$\triangle ABC$ is an equilateral triangle.
Since $A+B+C = 180^{\circ}$,we have $3A = 180^{\circ}$,so $A = B = C = 60^{\circ}$.
The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$.
Given $a = 2$,the area is $\frac{\sqrt{3}}{4} (2)^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3}$.

(C) The given curves are $x^{2}+y^{2}=4x$ and $x^{2}+y^{2}=8$.
For the first curve $x^{2}+y^{2}=4x$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2-x}{y}$.
At the point $(2,2)$,the slope $m_{1} = \frac{2-2}{2} = 0$.
For the second curve $x^{2}+y^{2}=8$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.
At the point $(2,2)$,the slope $m_{2} = -\frac{2}{2} = -1$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right|$.
Substituting the values,$\tan \theta = \left| \frac{0 - (-1)}{1 + (0)(-1)} \right| = \left| \frac{1}{1} \right| = 1$.
Since $\tan \theta = 1$,we have $\theta = 45^{\circ}$.
Given that $\theta = \sin^{-1} a$,we have $\sin^{-1} a = 45^{\circ}$,which implies $a = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.

(C) Let the degree of each vertex be $K$.
Since the graph is regular,every vertex has the same degree $K$.
The sum of the degrees of all vertices in a graph is given by the product of the number of vertices and the degree of each vertex.
Given that there are $15$ vertices and the sum of their degrees is $60$,we have:
$15 \times K = 60$
Dividing both sides by $15$,we get:
$K = \frac{60}{15} = 4$
Therefore,the degree of each vertex is $4$.

(A) Given that $a^{2} = e$ in the group $(G, *)$.
Multiplying both sides by $a^{-1}$ (the inverse of $a$):
$a^{-1} * (a * a) = a^{-1} * e$
$(a^{-1} * a) * a = a^{-1}$
$e * a = a^{-1}$
$a = a^{-1}$

(D) Given,in a group $(Z, *)$,the operation is defined as $a * b = a + b - n$ for all $a, b \in Z$.
First,we find the identity element $e$ such that $a * e = a$.
$a + e - n = a \implies e = n$.
Now,to find the inverse of $(-n)$,let the inverse be $x$ such that $(-n) * x = e$.
Since $e = n$,we have $(-n) * x = n$.
Using the definition of the operation:
$(-n) + x - n = n$.
$x - 2n = n$.
$x = 3n$.
Thus,the inverse of $(-n)$ is $3n$.

(C) Given that $A$ and $B$ are square matrices of order $n$ such that $A^{2}-B^{2}=(A-B)(A+B)$.
Expanding the right-hand side using the distributive property of matrix multiplication:
$(A-B)(A+B) = A(A+B) - B(A+B) = A^{2} + AB - BA - B^{2}$.
Equating this to the left-hand side:
$A^{2} - B^{2} = A^{2} + AB - BA - B^{2}$.
Subtracting $A^{2} - B^{2}$ from both sides,we get:
$0 = AB - BA$.
Therefore,$AB = BA$.
This implies that the matrices $A$ and $B$ must commute.

(D) Given,$C = \left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right] = A+B$.
We know that any square matrix $C$ can be uniquely expressed as the sum of a symmetric matrix $A$ and a skew-symmetric matrix $B$,where $A = \frac{1}{2}(C+C^T)$ and $B = \frac{1}{2}(C-C^T)$.
First,find the transpose of $C$: $C^T = \left[\begin{array}{rr}2 & 5 \\ 3 & -1\end{array}\right]$.
Now,calculate $B = \frac{1}{2}(C-C^T)$:
$B = \frac{1}{2} \left( \left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right] - \left[\begin{array}{rr}2 & 5 \\ 3 & -1\end{array}\right] \right)$
$B = \frac{1}{2} \left[\begin{array}{rr}2-2 & 3-5 \\ 5-3 & -1-(-1)\end{array}\right]$
$B = \frac{1}{2} \left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]$
$B = \left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]$.

(D) Given,the determinant of the adjoint of a (real) matrix $A$ of order $n=3$ is $|\operatorname{adj} A| = 25$.
We know the property $|\operatorname{adj} A| = |A|^{n-1}$.
Substituting $n=3$,we get $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Thus,$|A|^2 = 25$,which implies $|A| = \pm 5$.
We need to find the determinant of the inverse of the matrix,which is $|A^{-1}|$.
Using the property $|A^{-1}| = \frac{1}{|A|}$,we get $|A^{-1}| = \frac{1}{\pm 5} = \pm 0.2$.

(C) Given,$A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = \begin{vmatrix} \alpha & 2 \\ 2 & \alpha \end{vmatrix} = \alpha^{2} - 4 \quad (i)$.
We are given that $|A^{3}| = 125$.
Using the property of determinants $|A^{n}| = |A|^{n}$,we have:
$|A|^{3} = 125$.
Taking the cube root on both sides:
$|A| = \sqrt[3]{125} = 5$.
Now,substitute $|A| = 5$ into equation $(i)$:
$5 = \alpha^{2} - 4$.
$\alpha^{2} = 5 + 4 = 9$.
Taking the square root on both sides:
$\alpha = \pm 3$.

(B) Given,$A = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$
Expanding along the first row:
$A = x(x^2 - 1) - 1(x - 1) + 1(1 - x)$
$A = x^3 - x - x + 1 + 1 - x$
$A = x^3 - 3x + 2$
Differentiating with respect to $x$:
$\frac{dA}{dx} = 3x^2 - 3$ ... $(i)$
Also,given $B = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} = x^2 - 1$
Multiplying by $3$:
$3B = 3(x^2 - 1) = 3x^2 - 3$ ... $(ii)$
From equations $(i)$ and $(ii)$,we get:
$\frac{dA}{dx} = 3B$

(B) Let the given expression be $E = \cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$.
We can rewrite the expression as $E = \cos \left(\cos ^{-1} \frac{1}{5} + \left(\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5}\right)\right)$.
Using the identity $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5} = \frac{\pi}{2}$.
Substituting this into the expression,we get $E = \cos \left(\cos ^{-1} \frac{1}{5} + \frac{\pi}{2}\right)$.
Using the trigonometric identity $\cos \left(\theta + \frac{\pi}{2}\right) = -\sin \theta$,we have $E = -\sin \left(\cos ^{-1} \frac{1}{5}\right)$.
Let $\theta = \cos ^{-1} \frac{1}{5}$,then $\cos \theta = \frac{1}{5}$.
Since $\sin \theta = \sqrt{1 - \cos^2 \theta}$,we have $\sin \theta = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2 \sqrt{6}}{5}$.
Therefore,$E = -\frac{2 \sqrt{6}}{5}$.

(A) We use the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A-B}{1+AB} \right)$.
Let $A = \frac{x}{y}$ and $B = \frac{x-y}{x+y}$.
Then,$\tan ^{-1} \left( \frac{x}{y} \right) - \tan ^{-1} \left( \frac{x-y}{x+y} \right) = \tan ^{-1} \left( \frac{\frac{x}{y} - \frac{x-y}{x+y}}{1 + \frac{x}{y} \cdot \frac{x-y}{x+y}} \right)$.
Simplifying the numerator: $\frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2 + xy - xy + y^2}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.
Simplifying the denominator: $1 + \frac{x^2 - xy}{y(x+y)} = \frac{xy + y^2 + x^2 - xy}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.
Thus,the expression becomes $\tan ^{-1} \left( \frac{\frac{x^2 + y^2}{y(x+y)}}{\frac{x^2 + y^2}{y(x+y)}} \right) = \tan ^{-1}(1)$.
Since $\tan ^{-1}(1) = \frac{\pi}{4}$,the final value is $\frac{\pi}{4}$.

(D) Given the function $f(x) = \sin [x]$ for $-\frac{\pi}{4} < x < \frac{\pi}{4}$.
We know that $\frac{\pi}{4} \approx 0.785$ and $-\frac{\pi}{4} \approx -0.785$.
The interval for $x$ is $(-0.785, 0.785)$.
For $x \in [0, 0.785)$,the greatest integer value $[x] = 0$. Thus,$f(x) = \sin(0) = 0$.
For $x \in (-0.785, 0)$,the greatest integer value $[x] = -1$. Thus,$f(x) = \sin(-1) = -\sin(1)$.
Therefore,the range of the function is $\{0, -\sin 1\}$.

(C) Given,$f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$.
For any $a \neq 0$,consider a sequence of rational numbers $r_n \to a$ and a sequence of irrational numbers $i_n \to a$.
Then $\lim_{n \to \infty} f(r_n) = \lim_{n \to \infty} 0 = 0$ and $\lim_{n \to \infty} f(i_n) = \lim_{n \to \infty} i_n = a$.
Since $0 \neq a$ for $a \neq 0$,the limit $\lim_{x \to a} f(x)$ does not exist for $a \neq 0$.
At $x=0$,$\lim_{x \to 0} f(x) = 0$ because for any sequence $x_n \to 0$,$f(x_n)$ is either $x_n$ (if $x_n$ is irrational) or $0$ (if $x_n$ is rational),both of which approach $0$.
Since $f(0) = 0$,the function is continuous only at $x=0$.

(A) Given,$f(x) = \begin{cases} 2a - x & \text{when } -a < x < a \\ 3x - 2a & \text{when } a \leq x \end{cases}$.
First,check for continuity at $x = a$:
Left-hand limit: $\lim_{x \to a^-} f(x) = \lim_{x \to a^-} (2a - x) = 2a - a = a$.
Right-hand limit: $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} (3x - 2a) = 3a - 2a = a$.
Value of function: $f(a) = 3(a) - 2a = a$.
Since $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$,the function is continuous at $x = a$.
Now,check for differentiability at $x = a$:
Left-hand derivative: $Lf'(a) = \frac{d}{dx}(2a - x) = -1$.
Right-hand derivative: $Rf'(a) = \frac{d}{dx}(3x - 2a) = 3$.
Since $Lf'(a) \neq Rf'(a)$,the function $f(x)$ is not differentiable at $x = a$.

(B) Given,$f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}$.
Let $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = \frac{3}{\sqrt{13}}$.
Then,$f(x)=\cos ^{-1}(\cos \alpha \cos x - \sin \alpha \sin x)$.
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$f(x) = \cos ^{-1}(\cos(x+\alpha)) = x+\alpha$.
Now,differentiating with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x+\alpha) = 1 + 0 = 1$.
Therefore,$f^{\prime}(0.5) = 1$.

(B) Given,$x+y=\tan ^{-1} y$.
Differentiating both sides with respect to $x$,we get:
$1 + \frac{dy}{dx} = \frac{1}{1+y^2} \cdot \frac{dy}{dx}$
Rearranging the terms:
$\frac{dy}{dx} \left( \frac{1}{1+y^2} - 1 \right) = 1$
$\frac{dy}{dx} \left( \frac{1 - 1 - y^2}{1+y^2} \right) = 1$
$\frac{dy}{dx} \left( \frac{-y^2}{1+y^2} \right) = 1$
$\frac{dy}{dx} = -\frac{1+y^2}{y^2} = -\left( \frac{1}{y^2} + 1 \right)$
Now,differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -\frac{d}{dx} (y^{-2} + 1)$
$\frac{d^2y}{dx^2} = -(-2y^{-3}) \cdot \frac{dy}{dx}$
$\frac{d^2y}{dx^2} = \frac{2}{y^3} \cdot \frac{dy}{dx}$
Comparing this with the given equation $\frac{d^2y}{dx^2} = f(y) \frac{dy}{dx}$,we get $f(y) = \frac{2}{y^3}$.

(A) Given,the curve is $x y^{n} = a$.
On differentiating with respect to $x$,we get:
$x \cdot n y^{n-1} \cdot \frac{dy}{dx} + y^{n} = 0$
$\Rightarrow y^{n-1} [x n \frac{dy}{dx} + y] = 0$
Since $y \neq 0$,we have $x n \frac{dy}{dx} + y = 0$,which implies $\frac{dy}{dx} = -\frac{y}{nx}$.
The length of the subtangent is given by $|\frac{y}{dy/dx}|$.
Substituting the value of $\frac{dy}{dx}$:
Length of subtangent $= |\frac{y}{-y/nx}| = |nx| = |n| |x|$.
Since the length of the subtangent is proportional to the abscissa $x$,the value $|n|$ must be a constant.
Thus,$n$ can be any non-zero real number.

(A) Given,$f(x) = \frac{x}{3} + \frac{3}{x}$.
On differentiating with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{3} - \frac{3}{x^2}$.
For the function to be decreasing,we must have $f^{\prime}(x) < 0$.
$\frac{1}{3} - \frac{3}{x^2} < 0$
$\frac{1}{3} < \frac{3}{x^2}$
$x^2 < 9$
Taking the square root on both sides,we get $|x| < 3$,which implies $x \in (-3, 3)$.
Note that at $x=0$,the function is undefined,but the interval $(-3, 3)$ is the standard solution for the inequality $x^2 < 9$.

(D) Let the length of the rectangle be $a$ and the breadth be $b$. The diagonal of the rectangle is the diameter of the circle,so $d = 2r = 2(2) = 4$.
In the right-angled triangle formed by the diagonal,$a^2 + b^2 = d^2 = 4^2 = 16$.
The area of the rectangle is $A = a \cdot b$.
Since $b = \sqrt{16 - a^2}$,we have $A = a \sqrt{16 - a^2}$.
To maximize $A$,we maximize $A^2 = f(a) = a^2(16 - a^2) = 16a^2 - a^4$.
Differentiating with respect to $a$: $f'(a) = 32a - 4a^3$.
Setting $f'(a) = 0$,we get $4a(8 - a^2) = 0$. Since $a > 0$,$a^2 = 8$,so $a = 2\sqrt{2}$.
Then $b = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.
The maximum area is $A = (2\sqrt{2})(2\sqrt{2}) = 8 \text{ sq units}$.

(C) Let $I = \int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x$.
We can rewrite the integrand as:
$I = \int \frac{\cos ^{n-1} x}{\sin ^{n-1} x \cdot \sin^2 x} d x = \int \cot^{n-1} x \cdot \csc^2 x d x$.
Let $t = \cot x$.
Then $dt = -\csc^2 x d x$,which implies $\csc^2 x d x = -dt$.
Substituting these into the integral:
$I = \int t^{n-1} (-dt) = -\int t^{n-1} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$I = -\frac{t^n}{n} + C$.
Substituting $t = \cot x$ back:
$I = -\frac{\cot^n x}{n} + C$.

(B) Let $I = \int \frac{x-1}{(x+1)^{3}} e^{x} d x$.
We can rewrite the numerator as $(x+1)-2$:
$I = \int \frac{(x+1)-2}{(x+1)^{3}} e^{x} d x$
$I = \int \left( \frac{x+1}{(x+1)^{3}} - \frac{2}{(x+1)^{3}} \right) e^{x} d x$
$I = \int \frac{e^{x}}{(x+1)^{2}} d x - \int \frac{2 e^{x}}{(x+1)^{3}} d x$.
Now,apply integration by parts to the first integral $\int \frac{e^{x}}{(x+1)^{2}} d x$,taking $u = \frac{1}{(x+1)^{2}}$ and $dv = e^{x} dx$:
$du = -2(x+1)^{-3} dx$ and $v = e^{x}$.
$\int \frac{e^{x}}{(x+1)^{2}} d x = \frac{e^{x}}{(x+1)^{2}} - \int e^{x} \left( -\frac{2}{(x+1)^{3}} \right) d x$
$= \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x$.
Substituting this back into the expression for $I$:
$I = \left( \frac{e^{x}}{(x+1)^{2}} + \int \frac{2 e^{x}}{(x+1)^{3}} d x \right) - \int \frac{2 e^{x}}{(x+1)^{3}} d x$
$I = \frac{e^{x}}{(x+1)^{2}} + C$.

(D) Given,$\frac{(x+1)^{2}}{x^{3}+x} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1} \dots (i)$
Multiplying both sides by $x(x^{2}+1)$,we get:
$(x+1)^{2} = A(x^{2}+1) + (Bx+C)x$
$x^{2} + 2x + 1 = Ax^{2} + A + Bx^{2} + Cx$
$x^{2} + 2x + 1 = (A+B)x^{2} + Cx + A$
Comparing the coefficients of $x^{2}$,$x$,and the constant term on both sides:
$A+B = 1$
$C = 2$
$A = 1$
Substituting $A=1$ into $A+B=1$,we get $1+B=1$,so $B=0$.
Now,we need to calculate $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$:
$\sin^{-1}(1) + \tan^{-1}(0) + \sec^{-1}(2)$
$= \frac{\pi}{2} + 0 + \frac{\pi}{3}$
$= \frac{3\pi + 2\pi}{6} = \frac{5\pi}{6}$

(D) We are given $I_{1} = \int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2} = \int_{0}^{\pi / 2} x \cos x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
For $I_{1}$,let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$.
$I_{1} = [x(-\cos x)]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} (-\cos x) \, dx = [0 - 0] + [\sin x]_{0}^{\pi / 2} = 1 - 0 = 1$.
For $I_{2}$,let $u = x$ and $dv = \cos x \, dx$. Then $du = dx$ and $v = \sin x$.
$I_{2} = [x \sin x]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \, dx = [\frac{\pi}{2} \sin(\frac{\pi}{2}) - 0] - [-\cos x]_{0}^{\pi / 2} = \frac{\pi}{2} - [-(0 - 1)] = \frac{\pi}{2} - 1$.
Adding the two results,$I_{1} + I_{2} = 1 + (\frac{\pi}{2} - 1) = \frac{\pi}{2}$.
Thus,the correct option is $D$.

(B) Let $I = \int_{-1}^{2} \frac{|x|}{x} d x$.
We know that $|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$.
Splitting the integral at $x = 0$,we get:
$I = \int_{-1}^{0} \frac{-x}{x} d x + \int_{0}^{2} \frac{x}{x} d x$
$I = \int_{-1}^{0} (-1) d x + \int_{0}^{2} (1) d x$
$I = -[x]_{-1}^{0} + [x]_{0}^{2}$
$I = -(0 - (-1)) + (2 - 0)$
$I = -(1) + 2$
$I = 1$.

(B) Let $I = \int_{0}^{\pi} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$.
Since $f(\pi - x) = \frac{\cos^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{(-\cos x)^4}{(\sin x)^4 + (-\cos x)^4} = \frac{\cos^4 x}{\sin^4 x + \cos^4 x} = f(x)$,we use the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x \dots (i)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 2 \int_{0}^{\pi / 2} \frac{\cos ^{4}(\frac{\pi}{2}-x)}{\sin ^{4}(\frac{\pi}{2}-x)+\cos ^{4}(\frac{\pi}{2}-x)} d x = 2 \int_{0}^{\pi / 2} \frac{\sin ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x \dots (ii)$.
Adding $(i)$ and $(ii)$:
$2I = 2 \int_{0}^{\pi / 2} \frac{\cos^4 x + \sin^4 x}{\sin^4 x + \cos^4 x} dx = 2 \int_{0}^{\pi / 2} 1 dx = 2[x]_{0}^{\pi / 2} = 2(\frac{\pi}{2}) = \pi$.
Therefore,$I = \frac{\pi}{2}$.

(C) The required area is given by the integral $\int_{0}^{3\pi} y \, dx = \int_{0}^{3\pi} \sin \left(\frac{x}{3}\right) \, dx$.
Let $t = \frac{x}{3}$,then $dx = 3 \, dt$.
When $x = 0$,$t = 0$.
When $x = 3\pi$,$t = \pi$.
Substituting these into the integral,we get:
Area $= \int_{0}^{\pi} \sin(t) \cdot 3 \, dt = 3 \int_{0}^{\pi} \sin(t) \, dt$.
$= 3 [-\cos(t)]_{0}^{\pi}$.
$= -3 [\cos(\pi) - \cos(0)]$.
$= -3 [-1 - 1] = -3(-2) = 6$.
Thus,the area is $6$ square units.

(D) The given differential equation is $\left(y^{\prime \prime}\right)^{5}+4 \cdot \frac{\left(y^{\prime \prime}\right)^{3}}{y^{\prime \prime \prime}}+y^{\prime \prime \prime}=\sin x$.
To find the order and degree,we must eliminate the fraction by multiplying the entire equation by $y^{\prime \prime \prime}$.
Multiplying by $y^{\prime \prime \prime}$,we get: $\left(y^{\prime \prime}\right)^{5} \cdot y^{\prime \prime \prime} + 4 \cdot \left(y^{\prime \prime}\right)^{3} + \left(y^{\prime \prime \prime}\right)^{2} = \sin x \cdot y^{\prime \prime \prime}$.
The highest order derivative present is $y^{\prime \prime \prime}$,so the order $m = 3$.
The degree $n$ is the power of the highest order derivative after the equation is expressed as a polynomial in derivatives.
The term with the highest order derivative is $\left(y^{\prime \prime \prime}\right)^{2}$,so the degree $n = 2$.

(D) Given,$f^{\prime \prime}(x)+f(x)=0$ ... $(i)$
And $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$.
Differentiating $g(x)$ with respect to $x$:
$g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)f^{\prime \prime}(x)$
Substitute $f^{\prime \prime}(x) = -f(x)$ from equation $(i)$:
$g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)(-f(x))$
$g^{\prime}(x) = 2f(x)f^{\prime}(x) - 2f(x)f^{\prime}(x) = 0$.
Since the derivative of $g(x)$ is $0$,$g(x)$ is a constant function.
Given $g(3) = 8$,therefore $g(x) = 8$ for all $x$.
Thus,$g(8) = 8$.

(C) Given differential equation is $\sqrt{1-x^{2} y^{2}} \cdot dx = y \cdot dx + x \cdot dy$.
We know that $d(xy) = y \cdot dx + x \cdot dy$.
Substituting this into the equation,we get $\sqrt{1-(xy)^{2}} \cdot dx = d(xy)$.
Rearranging the terms,we have $dx = \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.
Integrating both sides,we get $\int dx = \int \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.
This results in $x = \sin^{-1}(xy) + C$.
Taking the sine of both sides,we get $\sin(x - C) = xy$,which is equivalent to $\sin(x + C) = xy$ (where $C$ is an arbitrary constant).

(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$
Assume $f(x) = e^{kx}$. Then $f'(x) = ke^{kx}$,$f''(x) = k^2e^{kx}$,$f'''(x) = k^3e^{kx}$,and so on.
Substituting these into the given equation:
$e^{kx} = ke^{kx} + k^2e^{kx} + k^3e^{kx} + \ldots$
Dividing by $e^{kx}$ (since $e^{kx} \neq 0$):
$1 = k + k^2 + k^3 + \ldots$
This is an infinite geometric series with first term $a = k$ and common ratio $r = k$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$ for $|r| < 1$.
So,$1 = \frac{k}{1-k}$.
$1 - k = k \implies 2k = 1 \implies k = \frac{1}{2}$.
Thus,$f(x) = Ce^{x/2}$.
Given $f(0) = 1$,we have $Ce^0 = 1$,which implies $C = 1$.
Therefore,$f(x) = e^{x/2}$.

(C) Given that $a \perp b$,therefore the dot product $a \cdot b = 0$.
Since $(a+b) \perp (a+mb)$,their dot product must be zero:
$(a+b) \cdot (a+mb) = 0$
Expanding the dot product:
$a \cdot a + m(a \cdot b) + (b \cdot a) + m(b \cdot b) = 0$
Substituting $a \cdot b = 0$ and $b \cdot a = 0$:
$|a|^{2} + m(0) + 0 + m|b|^{2} = 0$
$|a|^{2} + m|b|^{2} = 0$
Solving for $m$:
$m|b|^{2} = -|a|^{2}$
$m = -\frac{|a|^{2}}{|b|^{2}}$

(B) Given that $a, b$ and $c$ are unit vectors.
Therefore,$|a| = |b| = |c| = 1 \dots (i) $
We are given the equation $a+b+c = 0$.
Squaring both sides,we get:
$|a+b+c|^2 = |0|^2$
$|a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
Substituting the values from $(i)$:
$1 + 1 + 1 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$3 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$2(a \cdot b + b \cdot c + c \cdot a) = -3$
$a \cdot b + b \cdot c + c \cdot a = -\frac{3}{2}$

(B) Given,$a$ is perpendicular to both $b$ and $c$.
Then,$a \cdot b = 0$ and $a \cdot c = 0$ ... $(i)$
Now,using the vector triple product formula:
$a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$
Substituting the values from Eq. $(i)$:
$a \times (b \times c) = (0) b - (0) c$
$a \times (b \times c) = 0 - 0 = 0$
Therefore,the correct option is $B$.

(B) Given vectors are $a = (1, 2, 3)$,$b = (2, -1, 1)$,and $c = (3, 2, 1)$.
Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
First,calculate the dot products:
$a \cdot c = (1)(3) + (2)(2) + (3)(1) = 3 + 4 + 3 = 10$.
$a \cdot b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3$.
Substituting these into the formula:
$a \times (b \times c) = 10b - 3c$.
We are given $a \times (b \times c) = \alpha a + \beta b + \gamma c$.
Therefore,$0a + 10b - 3c = \alpha a + \beta b + \gamma c$.
Comparing the coefficients of $a, b,$ and $c$,we get:
$\alpha = 0, \beta = 10, \gamma = -3$.