IIT JEE 1988 Mathematics Question Paper with Answer and Solution

(D) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.
Given $z_1 = \sin x + i\cos 2x$ and $z_2 = \cos x - i\sin 2x$,they are conjugates if:
$1) \sin x = \cos x \implies \tan x = 1 \implies x = n\pi + \frac{\pi}{4}$
$2) \cos 2x = -(-\sin 2x) = \sin 2x \implies \tan 2x = 1 \implies 2x = m\pi + \frac{\pi}{4} \implies x = \frac{m\pi}{2} + \frac{\pi}{8}$
Comparing the two conditions:
For $n=0, x = \frac{\pi}{4} = 0.25\pi$.
For $m=0, x = \frac{\pi}{8} = 0.125\pi$.
For $m=1, x = \frac{\pi}{2} + \frac{\pi}{8} = 0.625\pi$.
There is no integer $n$ and $m$ such that $n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8}$.
Thus,there is no value of $x$ for which the complex numbers are conjugate.

(A) The cube roots of unity are $1, \omega, \omega^2$.
These roots can be represented as points on the Argand plane: $1 = (1, 0)$,$\omega = (\cos \frac{2\pi}{3}, \sin \frac{2\pi}{3})$,and $\omega^2 = (\cos \frac{4\pi}{3}, \sin \frac{4\pi}{3})$.
These points lie on the unit circle $|z| = 1$ and are separated by an angle of $\frac{2\pi}{3}$ radians $(120^\circ)$ from each other.
Since the points are equally spaced on the circle,they form the vertices of an equilateral triangle.
Alternatively,using the property that $z_1, z_2, z_3$ form an equilateral triangle if $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$,we substitute $z_1=1, z_2=\omega, z_3=\omega^2$:
$1^2 + \omega^2 + (\omega^2)^2 = 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$.
And $1(\omega) + \omega(\omega^2) + \omega^2(1) = \omega + \omega^3 + \omega^2 = \omega + 1 + \omega^2 = 0$.
Since both sides are equal to $0$,the triangle is equilateral.

(B) We use the property $|z|^2 = z \cdot \overline{z}$.
$|az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (az_1 - bz_2)(\overline{az_1 - bz_2}) + (bz_1 + az_2)(\overline{bz_1 + az_2})$
$= (az_1 - bz_2)(a\overline{z_1} - b\overline{z_2}) + (bz_1 + az_2)(b\overline{z_1} + a\overline{z_2})$
$= (a^2|z_1|^2 - ab z_1\overline{z_2} - ab \overline{z_1}z_2 + b^2|z_2|^2) + (b^2|z_1|^2 + ab z_1\overline{z_2} + ab \overline{z_1}z_2 + a^2|z_2|^2)$
$= a^2|z_1|^2 + b^2|z_2|^2 + b^2|z_1|^2 + a^2|z_2|^2$
$= (a^2 + b^2)|z_1|^2 + (a^2 + b^2)|z_2|^2$
$= (a^2 + b^2)(|z_1|^2 + |z_2|^2)$

(D) Let $\alpha$ and $\beta$ be the first and $(2n - 1)^{th}$ terms of the $A.P.$,$G.P.$,and $H.P.$ respectively.
For $A.P.$: The $n^{th}$ term is $a = \frac{\alpha + \beta}{2}$ $(i)$
For $G.P.$: The $n^{th}$ term is $b = \sqrt{\alpha \beta}$ (ii)
For $H.P.$: The $n^{th}$ term is $c = \frac{2\alpha \beta}{\alpha + \beta}$ (iii)
From $(i)$,(ii),and (iii),we observe that $a, b, c$ are the Arithmetic Mean,Geometric Mean,and Harmonic Mean of $\alpha$ and $\beta$ respectively.
We know that for any two positive numbers $\alpha$ and $\beta$,$A.M. \ge G.M. \ge H.M.$
Thus,$a \ge b \ge c$,which matches option $(a)$.
Also,$ac = \left(\frac{\alpha + \beta}{2}\right) \left(\frac{2\alpha \beta}{\alpha + \beta}\right) = \alpha \beta = b^2$.
Therefore,$ac - b^2 = 0$,which matches option $(c)$.
Hence,both $(a)$ and $(c)$ are correct.

(C) The sum of the first $n$ terms is given by:
$S_n = \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{2^2}\right) + \left(1 - \frac{1}{2^3}\right) + \dots + \left(1 - \frac{1}{2^n}\right)$
$S_n = \sum_{k=1}^{n} \left(1 - \frac{1}{2^k}\right) = n - \sum_{k=1}^{n} \left(\frac{1}{2}\right)^k$
Using the sum formula for a geometric progression $\sum_{k=1}^{n} ar^{k-1} = a\frac{1-r^n}{1-r}$,we have:
$S_n = n - \left[ \frac{1}{2} \left( \frac{1 - (1/2)^n}{1 - 1/2} \right) \right]$
$S_n = n - \left( 1 - \frac{1}{2^n} \right) = n - 1 + 2^{-n}$
Verification for $n=1$: $S_1 = 1 - 1 + 2^{-1} = 1/2$ (Correct).
Verification for $n=2$: $S_2 = 2 - 1 + 2^{-2} = 1 + 1/4 = 5/4$ (Correct).

(B) The given series is $1^2, 2 \cdot 2^2, 3^2, 2 \cdot 4^2, 5^2, 2 \cdot 6^2, \dots$
When $n$ is even,the sum $S_n = \frac{n(n + 1)^2}{2}$.
When $n$ is odd,the $n^{th}$ term is $n^2$. The sum of the first $n$ terms is $S_n = S_{n-1} + a_n$.
Since $n$ is odd,$n-1$ is even. Using the given formula for $n-1$ terms:
$S_{n-1} = \frac{(n-1)((n-1) + 1)^2}{2} = \frac{(n-1)n^2}{2}$.
Thus,$S_n = \frac{(n-1)n^2}{2} + n^2 = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.
Verification: For $n=1$,$S_1 = 1^2 = 1$. Using formula $(b)$: $\frac{1^2(1+1)}{2} = 1$. Correct.

(B) The given equation is $|x^2 + 4x + 3| + 2x + 5 = 0$.
Case $I$: $x^2 + 4x + 3 \ge 0$,which implies $(x+1)(x+3) \ge 0$,so $x \in (-\infty, -3] \cup [-1, \infty)$.
The equation becomes $x^2 + 4x + 3 + 2x + 5 = 0$,which simplifies to $x^2 + 6x + 8 = 0$.
Factoring gives $(x+2)(x+4) = 0$,so $x = -2$ or $x = -4$.
Checking the condition: $x = -4$ satisfies $x \in (-\infty, -3]$,but $x = -2$ does not satisfy $x \in [-1, \infty)$. Thus,$x = -4$ is a solution.
Case $II$: $x^2 + 4x + 3 < 0$,which implies $x \in (-3, -1)$.
The equation becomes $-(x^2 + 4x + 3) + 2x + 5 = 0$,which simplifies to $-x^2 - 2x + 2 = 0$,or $x^2 + 2x - 2 = 0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{-2 \pm \sqrt{12}}{2} = -1 \pm \sqrt{3}$.
Checking the condition: $\sqrt{3} \approx 1.732$,so $x_1 = -1 + 1.732 = 0.732$ (not in $(-3, -1)$) and $x_2 = -1 - 1.732 = -2.732$ (in $(-3, -1)$).
Thus,$x = -1 - \sqrt{3}$ is a solution.
There are exactly $2$ real solutions.

(C) To ensure that no two '$-$' signs come together,we first arrange the six '$+$' signs in a row: $+ + + + + +$.
This creates $7$ possible gaps (including the ends) where the '$-$' signs can be placed: $\_ + \_ + \_ + \_ + \_ + \_ \_$.
We need to choose $4$ gaps out of these $7$ available positions to place the $4$ '$-$' signs.
The number of ways to do this is given by the combination formula ${^n}C_r = \frac{n!}{r!(n-r)!}$.
Here,$n = 7$ and $r = 4$,so the number of ways is ${^7}C_4 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(A) Given $R = (5\sqrt{5} + 11)^{2n + 1}$.
Let $f' = (5\sqrt{5} - 11)^{2n + 1}$.
Since $5\sqrt{5} = \sqrt{125}$ and $11 = \sqrt{121}$,we have $0 < 5\sqrt{5} - 11 < 1$.
Thus,$0 < f' < 1$.
Consider $R + f' = (5\sqrt{5} + 11)^{2n + 1} + (5\sqrt{5} - 11)^{2n + 1}$.
By binomial expansion,all terms with odd powers of $5\sqrt{5}$ cancel out,and terms with even powers of $5\sqrt{5}$ are doubled.
Since $(5\sqrt{5})^2 = 125$,all terms are integers,so $R + f' = 2k$ for some integer $k$.
Since $R = [R] + f$,we have $[R] + f + f' = 2k$.
Since $0 < f < 1$ and $0 < f' < 1$,we have $0 < f + f' < 2$.
Because $[R] + f + f' = 2k$ is an integer,$f + f'$ must be an integer.
The only integer in the range $(0, 2)$ is $1$.
Therefore,$f + f' = 1$,which implies $f = 1 - f'$.
Then $R \cdot f = R(1 - f') = R - R \cdot f' = R - (5\sqrt{5} + 11)^{2n + 1}(5\sqrt{5} - 11)^{2n + 1}$.
$R \cdot f = R - ((5\sqrt{5})^2 - 11^2)^{2n + 1} = R - (125 - 121)^{2n + 1} = R - 4^{2n + 1}$.
Wait,the standard property is $f = 1 - f'$ if $R+f'$ is an even integer. Actually,$R \cdot f = R(1 - f') = R - R f'$.
Since $R f' = (125 - 121)^{2n + 1} = 4^{2n + 1}$,and $R = [R] + f$,we have $R f' = ([R] + f) f' = [R] f' + f f' = 4^{2n + 1}$.
Since $f = 1 - f'$,$R f = R(1 - f') = R - R f' = [R] + f - 4^{2n + 1}$.
Actually,the result is simply $4^{2n + 1}$.

(C) We use the identity $\tan \theta = \cot \theta - 2\cot 2\theta$,which implies $\cot \theta - \tan \theta = 2\cot 2\theta$.
Consider the expression $E = \tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot 8\alpha$.
Using the identity $\tan \theta = \cot \theta - 2\cot 2\theta$,we have:
$8\cot 8\alpha + 4\tan 4\alpha = 4(2\cot 8\alpha + \tan 4\alpha) = 4(\cot 4\alpha) = 4\cot 4\alpha$.
Now the expression becomes:
$E = \tan \alpha + 2\tan 2\alpha + 4\cot 4\alpha$.
Again,using $4\cot 4\alpha + 2\tan 2\alpha = 2(2\cot 4\alpha + \tan 2\alpha) = 2(\cot 2\alpha) = 2\cot 2\alpha$.
Now the expression becomes:
$E = \tan \alpha + 2\cot 2\alpha$.
Using $2\cot 2\alpha = \cot \alpha - \tan \alpha$,we get:
$E = \tan \alpha + (\cot \alpha - \tan \alpha) = \cot \alpha$.
Thus,the correct option is $C$.

(C) Given expression: $\sqrt{3} \csc 20^{\circ} - \sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} \right)}{\frac{1}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$
Using $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$:
$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{4 \sin(60^{\circ} - 20^{\circ})}{\sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$.

(D) In $\triangle ABC$,the sum of angles is $180^\circ$.
$\angle C = 180^\circ - 45^\circ - 75^\circ = 60^\circ$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin 45^\circ = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^\circ = k \frac{\sqrt{3}}{2}$.
We need to evaluate $a + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} \right) = k \frac{1 + \sqrt{3}}{\sqrt{2}}$.
Since $b = k \sin 75^\circ = k \sin(45^\circ + 30^\circ) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = k \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Therefore,$2b = 2k \frac{\sqrt{3} + 1}{2\sqrt{2}} = k \frac{\sqrt{3} + 1}{\sqrt{2}}$.
Hence,$a + c\sqrt{2} = 2b$.

(D) Let the coordinates of point $S$ be $(x, y)$.
Given the relation $SQ^2 + SR^2 = 2SP^2$.
Substituting the coordinates,we get:
$((x + 1)^2 + (y - 0)^2) + ((x - 2)^2 + (y - 0)^2) = 2((x - 1)^2 + (y - 0)^2)$
$(x^2 + 2x + 1 + y^2) + (x^2 - 4x + 4 + y^2) = 2(x^2 - 2x + 1 + y^2)$
$2x^2 - 2x + 5 + 2y^2 = 2x^2 - 4x + 2 + 2y^2$
$-2x + 5 = -4x + 2$
$2x = -3$
$x = -\frac{3}{2}$
This represents a straight line parallel to the $y$-axis.

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Here,the point is $(0, 0)$,so $S_1 = 0^2 + 0^2 - 2r(0) - 2h(0) + h^2 = h^2$.
The equation of the tangent $T$ at $(0, 0)$ is $x(0) + y(0) - r(x + 0) - h(y + 0) + h^2 = 0$,which simplifies to $T = -rx - hy + h^2$.
Substituting these into $SS_1 = T^2$:
$h^2(x^2 + y^2 - 2rx - 2hy + h^2) = (-rx - hy + h^2)^2$.
Expanding both sides:
$h^2x^2 + h^2y^2 - 2rh^2x - 2h^3y + h^4 = r^2x^2 + h^2y^2 + h^4 + 2rhxy - 2rh^2x - 2h^3y$.
Canceling common terms $h^2y^2, h^4, -2rh^2x, -2h^3y$:
$h^2x^2 = r^2x^2 + 2rhxy$.
$(h^2 - r^2)x^2 - 2rhxy = 0$.
$x((h^2 - r^2)x - 2rhy) = 0$.
Thus,the equations are $x = 0$ and $(h^2 - r^2)x - 2rhy = 0$.

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it cuts the circle $x^2 + y^2 = p^2$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2g(0) + 2f(0) = c - p^2$,which implies $c = p^2$.
Since the circle passes through $(a, b)$,we have $a^2 + b^2 + 2ga + 2fb + p^2 = 0$.
The centre of the circle is $(-g, -f)$. Let the centre be $(x, y)$,so $g = -x$ and $f = -y$.
Substituting these into the equation,we get $a^2 + b^2 + 2(-x)a + 2(-y)b + p^2 = 0$.
This simplifies to $2ax + 2by - (a^2 + b^2 + p^2) = 0$.

(B) Given $f(x) = x\sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$.
We know that for all $x \ne 0$,$-1 \le \sin \frac{1}{x} \le 1$.
Multiplying by $|x|$,we get $-|x| \le x\sin \frac{1}{x} \le |x|$.
Applying the Squeeze Theorem,since $\lim_{x \to 0} (-|x|) = 0$ and $\lim_{x \to 0} |x| = 0$,it follows that $\lim_{x \to 0} x\sin \frac{1}{x} = 0$.
Thus,$\lim_{x \to 0} f(x) = 0$.

(B) Given the limit $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.
Since $f(9) = 9$,the expression takes the indeterminate form $\frac{0}{0}$ as $x \to 9$.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 9} \frac{\frac{d}{dx}(\sqrt{f(x)} - 3)}{\frac{d}{dx}(\sqrt{x} - 3)} = \mathop {\lim }\limits_{x \to 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x \to 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$.
Substituting the values $f(9) = 9$ and $f'(9) = 4$:
$= \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

(D) The addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Rearranging this gives $P(A \cap B) = P(A) + P(B) - P(A \cup B)$,which matches option $C$.
Since $0 \leq P(A \cup B) \leq 1$,we have $P(A \cup B) \leq 1 \implies P(A) + P(B) - P(A \cap B) \leq 1 \implies P(A \cap B) \geq P(A) + P(B) - 1$,which matches option $A$.
Also,since $P(A \cup B) \geq 0$,we have $P(A) + P(B) - P(A \cap B) \geq 0 \implies P(A \cap B) \leq P(A) + P(B)$,which matches option $B$.
Since options $A$,$B$,and $C$ are all mathematically correct,the correct choice is $D$.

(A) Let $\Delta = \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$.
Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$:
$\Delta = \left| \begin{matrix} 0 & a-b & (a^2-bc) - (b^2-ac) \\ 0 & b-c & (b^2-ac) - (c^2-ab) \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Simplify the third column elements:
$(a^2-bc) - (b^2-ac) = (a^2-b^2) + (ac-bc) = (a-b)(a+b) + c(a-b) = (a-b)(a+b+c)$
$(b^2-ac) - (c^2-ab) = (b^2-c^2) + (ab-ac) = (b-c)(b+c) + a(b-c) = (b-c)(a+b+c)$
Substituting these back:
$\Delta = \left| \begin{matrix} 0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Taking $(a-b)$ common from $R_1$ and $(b-c)$ common from $R_2$:
$\Delta = (a-b)(b-c) \left| \begin{matrix} 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Since $R_1$ and $R_2$ are identical,the value of the determinant is $0$.

(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.
The coefficient matrix is:
$A = \begin{bmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{bmatrix}$
Setting the determinant to zero:
$\begin{vmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$1(-k - 3) - k(3 - (-1)) - 1(-9 - (-k)) = 0$
$1(-k - 3) - k(4) - 1(-9 + k) = 0$
$-k - 3 - 4k + 9 - k = 0$
$-6k + 6 = 0$
$6k = 6$
$k = 1$

(B) Given that $A$ and $B$ are square matrices of order $n = 3$.
We know the property of determinants that $|AB| = |A| \times |B|$.
So,$|AB| = (-1) \times (3) = -3$.
We also know the property $|kA| = k^n |A|$,where $n$ is the order of the matrix.
Here,$|3AB| = 3^3 |AB|$.
Substituting the value of $|AB|$,we get $|3AB| = 27 \times (-3) = -81$.

(A) Let the given determinant be $\Delta = 0$.
Applying row operations $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$:
$\Delta = \left| \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$
Expanding along $R_1$:
$1(1 + 4 \sin 4 \theta + \cos^2 \theta) - 0 + (-1)(0 - \sin^2 \theta) = 0$
$1 + 4 \sin 4 \theta + \cos^2 \theta + \sin^2 \theta = 0$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$1 + 4 \sin 4 \theta + 1 = 0$
$2 + 4 \sin 4 \theta = 0$
$4 \sin 4 \theta = -2$
$\sin 4 \theta = -\frac{1}{2}$
Given $0 < \theta < \frac{\pi}{2}$,so $0 < 4 \theta < 2 \pi$.
The values of $4 \theta$ for which $\sin 4 \theta = -\frac{1}{2}$ in the interval $(0, 2 \pi)$ are $4 \theta = \frac{7 \pi}{6}$ and $4 \theta = \frac{11 \pi}{6}$.
Therefore,$\theta = \frac{7 \pi}{24}$ and $\theta = \frac{11 \pi}{24}$.

(C) Let $\vec{a} = l\hat{i} + m\hat{j} + n\hat{k}$,where $l^2 + m^2 + n^2 = 1$.
Since $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis,$n = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Thus,$l^2 + m^2 + (\frac{1}{\sqrt{2}})^2 = 1 \implies l^2 + m^2 = \frac{1}{2} \dots (i)$.
Given that $\vec{a} + \hat{i} + \hat{j}$ is a unit vector,we have $|(l+1)\hat{i} + (m+1)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}| = 1$.
Squaring both sides,$(l+1)^2 + (m+1)^2 + (\frac{1}{\sqrt{2}})^2 = 1^2$.
$l^2 + 2l + 1 + m^2 + 2m + 1 + \frac{1}{2} = 1$.
$(l^2 + m^2) + 2(l+m) + 2.5 = 1$.
Substituting $(i)$,$\frac{1}{2} + 2(l+m) + 2.5 = 1 \implies 2(l+m) = -2 \implies l+m = -1$.
Since $l^2 + m^2 = \frac{1}{2}$ and $l+m = -1$,we have $(l+m)^2 = l^2 + m^2 + 2lm = 1 \implies \frac{1}{2} + 2lm = 1 \implies 2lm = \frac{1}{2} \implies lm = \frac{1}{4}$.
Solving $l+m = -1$ and $lm = \frac{1}{4}$ gives $l = m = -\frac{1}{2}$.
Therefore,$\vec{a} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.

(C) Given expression: $3\overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC}$
We can rewrite the expression by splitting $3\overrightarrow{OD}$ into three parts:
$= \overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{OD} + \overrightarrow{DB} + \overrightarrow{OD} + \overrightarrow{DC}$
Using the triangle law of vector addition,$\overrightarrow{OX} + \overrightarrow{XY} = \overrightarrow{OY}$:
$\overrightarrow{OD} + \overrightarrow{DA} = \overrightarrow{OA}$
$\overrightarrow{OD} + \overrightarrow{DB} = \overrightarrow{OB}$
$\overrightarrow{OD} + \overrightarrow{DC} = \overrightarrow{OC}$
Substituting these into the expression:
$= \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$
Thus,the correct option is $C$.

(A) Given the relation $2\overrightarrow{AC} = 3\overrightarrow{CB}$.
We express the vectors in terms of the origin $O$ as $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$ and $\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC}$.
Substituting these into the given equation:
$2(\overrightarrow{OC} - \overrightarrow{OA}) = 3(\overrightarrow{OB} - \overrightarrow{OC})$
$2\overrightarrow{OC} - 2\overrightarrow{OA} = 3\overrightarrow{OB} - 3\overrightarrow{OC}$
Rearranging the terms to isolate $2\overrightarrow{OA} + 3\overrightarrow{OB}$:
$2\overrightarrow{OA} + 3\overrightarrow{OB} = 2\overrightarrow{OC} + 3\overrightarrow{OC}$
$2\overrightarrow{OA} + 3\overrightarrow{OB} = 5\overrightarrow{OC}$

(B) The component of vector $a$ along vector $b$ is the projection of $a$ on $b$,which is given by $a \cos \theta = \frac{a \cdot b}{|b|}$.
The component of vector $a$ perpendicular to vector $b$ is given by $a \sin \theta$. Since $|a \times b| = |a||b| \sin \theta$,we have $a \sin \theta = \frac{|a \times b|}{|b|}$.
Thus,the components are $\frac{a \cdot b}{|b|}$ and $\frac{|a \times b|}{|b|}$.

(D) Given $p = \frac{b \times c}{[a, b, c]}$,$q = \frac{c \times a}{[a, b, c]}$,and $r = \frac{a \times b}{[a, b, c]}$.
We calculate each term separately:
$(a+b) \cdot p = a \cdot p + b \cdot p = a \cdot \frac{b \times c}{[a, b, c]} + b \cdot \frac{b \times c}{[a, b, c]} = \frac{[a, b, c]}{[a, b, c]} + 0 = 1$.
$(b+c) \cdot q = b \cdot q + c \cdot q = b \cdot \frac{c \times a}{[a, b, c]} + c \cdot \frac{c \times a}{[a, b, c]} = \frac{[b, c, a]}{[a, b, c]} + 0 = 1$.
$(c+a) \cdot r = c \cdot r + a \cdot r = c \cdot \frac{a \times b}{[a, b, c]} + a \cdot \frac{a \times b}{[a, b, c]} = \frac{[c, a, b]}{[a, b, c]} + 0 = 1$.
Summing these results: $1 + 1 + 1 = 3$.

(D) First,we rewrite the function $f(x)$ by analyzing the absolute value $|x - 3|$:
$|x - 3| = x - 3$ if $x \ge 3$ and $-(x - 3) = 3 - x$ if $x < 3$.
Thus,the function is defined as:
$f(x) = \begin{cases} \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x < 1 \\ 3 - x, & 1 \le x < 3 \\ x - 3, & x \ge 3 \end{cases}$
Check continuity at $x = 1$:
$f(1) = 3 - 1 = 2$.
$LHL = \lim_{x \to 1^-} (\frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}) = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = \frac{14}{4} - \frac{6}{4} = 2$.
$RHL = \lim_{x \to 1^+} (3 - x) = 2$.
Since $LHL = RHL = f(1)$,the function is continuous at $x = 1$.
Check continuity at $x = 3$:
$f(3) = 3 - 3 = 0$.
$LHL = \lim_{x \to 3^-} (3 - x) = 0$.
$RHL = \lim_{x \to 3^+} (x - 3) = 0$.
Since $LHL = RHL = f(3)$,the function is continuous at $x = 3$.
Check differentiability at $x = 1$:
$LHD = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{\frac{1}{4}(1+h)^2 - \frac{3}{2}(1+h) + \frac{13}{4} - 2}{h} = \lim_{h \to 0^-} \frac{\frac{1}{4}(1+2h+h^2) - \frac{3}{2} - \frac{3}{2}h + \frac{13}{4} - 2}{h} = \lim_{h \to 0^-} \frac{\frac{1}{4}h^2 + \frac{1}{2}h - \frac{3}{2}h}{h} = \lim_{h \to 0^-} (\frac{1}{4}h - 1) = -1$.
$RHD = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{3 - (1+h) - 2}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1$.
Since $LHD = RHD$,the function is differentiable at $x = 1$.
Therefore,all statements are correct.

(C) Given ${y^2} = p(x)$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = p'(x)$.
Thus,$\frac{dy}{dx} = \frac{p'(x)}{2y}$.
Differentiating again with respect to $x$,we get $2 \left( \frac{dy}{dx} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.
Substituting $\frac{dy}{dx} = \frac{p'(x)}{2y}$,we have $2 \left( \frac{p'(x)}{2y} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.
$2y \frac{d^2y}{dx^2} = p''(x) - \frac{(p'(x))^2}{2y^2} = p''(x) - \frac{(p'(x))^2}{2p(x)}$.
Multiplying by $y^2$,we get $2y^3 \frac{d^2y}{dx^2} = p(x) p''(x) - \frac{1}{2} (p'(x))^2$.
Now,differentiating $2y^3 \frac{d^2y}{dx^2}$ with respect to $x$:
$\frac{d}{dx} \left( 2y^3 \frac{d^2y}{dx^2} \right) = \frac{d}{dx} \left( p(x) p''(x) - \frac{1}{2} (p'(x))^2 \right)$.
$= p'(x) p''(x) + p(x) p'''(x) - \frac{1}{2} \cdot 2 p'(x) p''(x)$.
$= p'(x) p''(x) + p(x) p'''(x) - p'(x) p''(x) = p(x) p'''(x)$.

(B) To evaluate the integral $I = \int_0^{1.5} [x^2] \, dx$,we break the interval $[0, 1.5]$ based on the values where $x^2$ is an integer.
$x^2 = 1 \implies x = 1$
$x^2 = 2 \implies x = \sqrt{2} \approx 1.414$
Thus,we split the integral as follows:
$I = \int_0^1 [x^2] \, dx + \int_1^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{1.5} [x^2] \, dx$
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$1 \le x^2 < 2$,so $[x^2] = 1$.
For $\sqrt{2} \le x < 1.5$,$2 \le x^2 < 2.25$,so $[x^2] = 2$.
Substituting these values:
$I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx$
$I = 0 + [x]_1^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{1.5}$
$I = (\sqrt{2} - 1) + 2(1.5 - \sqrt{2})$
$I = \sqrt{2} - 1 + 3 - 2\sqrt{2}$
$I = 2 - \sqrt{2}$

(A) Let $R_1$ be the event that a red ball is drawn from $A$ and placed in $B$,and $B_1$ be the event that a black ball is drawn from $A$ and placed in $B$.
$P(R_1) = \frac{6}{10} = \frac{3}{5}$,$P(B_1) = \frac{4}{10} = \frac{2}{5}$.
After transferring a ball,urn $B$ has $11$ balls.
If $R_1$ occurred,$B$ has $5$ red and $6$ black balls. $P(R_2|R_1) = \frac{5}{11}$,$P(B_2|R_1) = \frac{6}{11}$.
If $B_1$ occurred,$B$ has $4$ red and $7$ black balls. $P(R_2|B_1) = \frac{4}{11}$,$P(B_2|B_1) = \frac{7}{11}$.
After transferring back to $A$,urn $A$ has $10$ balls.
If $R_1$ and $R_2$ occurred,$A$ has $6$ red balls. $P(R|R_1, R_2) = \frac{6}{10}$.
If $R_1$ and $B_2$ occurred,$A$ has $5$ red balls. $P(R|R_1, B_2) = \frac{5}{10}$.
If $B_1$ and $R_2$ occurred,$A$ has $7$ red balls. $P(R|B_1, R_2) = \frac{7}{10}$.
If $B_1$ and $B_2$ occurred,$A$ has $6$ red balls. $P(R|B_1, B_2) = \frac{6}{10}$.
The total probability is $P(R) = P(R_1)P(R_2|R_1)P(R|R_1, R_2) + P(R_1)P(B_2|R_1)P(R|R_1, B_2) + P(B_1)P(R_2|B_1)P(R|B_1, R_2) + P(B_1)P(B_2|B_1)P(R|B_1, B_2)$.
$P(R) = (\frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}) + (\frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}) + (\frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}) + (\frac{4}{10} \times \frac{7}{11} \times \frac{6}{10})$.
$P(R) = \frac{180 + 180 + 112 + 168}{1100} = \frac{640}{1100} = \frac{64}{110} = \frac{32}{55}$.

(D) Let $X$ be the number of heads,which follows a binomial distribution $B(n, p)$ with $n = 100$.
The probability of getting $k$ heads is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.
We are given that $P(X=50) = P(X=51)$.
Substituting the values,we get:
${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$.
Dividing both sides by $p^{50} (1-p)^{49}$,we get:
${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$.
Rearranging to solve for $p$:
$\frac{1-p}{p} = \frac{{}^{100}C_{51}}{{}^{100}C_{50}} = \frac{100!}{51! 49!} \times \frac{50! 50!}{100!} = \frac{50}{51}$.
Thus,$51(1-p) = 50p$.
$51 - 51p = 50p$.
$101p = 51$.
$p = \frac{51}{101}$.