JEE Main 2018 Mathematics Question Paper with Answer and Solution

(B) Given the equation: $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$
Simplify the left side: $\frac{x + q + x + p}{(x + p)(x + q)} = \frac{1}{r}$
Cross-multiply: $r(2x + p + q) = x^2 + (p + q)x + pq$
Rearrange into standard quadratic form: $x^2 + (p + q - 2r)x + (pq - pr - qr) = 0$
Let the roots be $\alpha$ and $\beta$. Since the roots are equal in magnitude but opposite in sign,$\alpha = -\beta$,which implies $\alpha + \beta = 0$.
From the sum of roots formula for a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$. Thus,$-(p + q - 2r) = 0$,which means $p + q = 2r$.
The sum of squares of the roots is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Since $\alpha + \beta = 0$,this simplifies to $\alpha^2 + \beta^2 = -2\alpha\beta$.
From the product of roots formula,$\alpha\beta = pq - pr - qr$.
Substituting $\alpha\beta$: $\alpha^2 + \beta^2 = -2(pq - pr - qr) = -2pq + 2pr + 2qr$.
Since $2r = p + q$,substitute $2pr + 2qr = 2r(p + q) = (p + q)(p + q) = p^2 + 2pq + q^2$.
Therefore,$\alpha^2 + \beta^2 = -2pq + (p^2 + 2pq + q^2) = p^2 + q^2$.

(B) Given the quadratic equation $9x^2 + 27x + 20 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-27 \pm \sqrt{27^2 - 4 \times 9 \times 20}}{2 \times 9} = \frac{-27 \pm \sqrt{729 - 720}}{18} = \frac{-27 \pm 3}{18}$.
Thus,the roots are $x_1 = \frac{-24}{18} = -\frac{4}{3}$ and $x_2 = \frac{-30}{18} = -\frac{5}{3}$.
Given $5 \cos A + 3 = 0$,we have $\cos A = -\frac{3}{5}$.
Then $\sec A = \frac{1}{\cos A} = -\frac{5}{3}$.
Since $\cos A$ is negative,$A$ is an obtuse angle in the second quadrant,so $\tan A$ is negative.
$\tan A = -\sqrt{\sec^2 A - 1} = -\sqrt{(-\frac{5}{3})^2 - 1} = -\sqrt{\frac{25}{9} - 1} = -\sqrt{\frac{16}{9}} = -\frac{4}{3}$.
Comparing the roots,the roots are $\sec A$ and $\tan A$.

(A) Let $f(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$.
We need the coefficient of $x^2$ in $(2 - x^2)f(x)$.
This is equal to $2 \times (\text{coefficient of } x^2 \text{ in } f(x)) - 1 \times (\text{constant term in } f(x))$.
First,find the constant term in $f(x)$:
Constant term $= (1 + 2(0) + 3(0)^2)^6 + (1 - 4(0)^2)^6 = 1^6 + 1^6 = 2$.
Next,find the coefficient of $x^2$ in $f(x)$:
For $(1 + 2x + 3x^2)^6$,using the multinomial expansion or binomial expansion $(1 + (2x + 3x^2))^6 = 1 + 6(2x + 3x^2) + \binom{6}{2}(2x)^2 + \dots = 1 + 12x + 18x^2 + 15(4x^2) + \dots = 1 + 12x + 78x^2 + \dots$
The coefficient of $x^2$ is $78$.
For $(1 - 4x^2)^6$,the expansion is $1 + 6(-4x^2) + \dots = 1 - 24x^2 + \dots$
The coefficient of $x^2$ is $-24$.
Thus,the coefficient of $x^2$ in $f(x)$ is $78 - 24 = 54$.
Finally,the coefficient of $x^2$ in $(2 - x^2)f(x)$ is $2(54) - 1(2) = 108 - 2 = 106$.

(D) Let $z = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i\sqrt{3})$:
$z = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} + (i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2}$.
We recognize this as $\omega$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is a complex cube root of unity.
Thus,$z = \omega$.
The equation becomes $\omega^n = 1$.
The smallest positive integer $n$ for which $\omega^n = 1$ is $n = 3$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$
Substituting $x = 0$,we get the indeterminate form $\frac{0}{0}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({{(27 + x)}^{\frac{1}{3}}} - 3)}}{{\frac{d}{{dx}}(9 - {{(27 + x)}^{\frac{2}{3}}})}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3}{{(27 + x)}^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}{{(27 + x)}^{ - \frac{1}{3}}}}}$
$L = \mathop {\lim }\limits_{x \to 0} -\frac{1}{2} {(27 + x)}^{ - \frac{2}{3} + \frac{1}{3}}$
$L = -\frac{1}{2} {(27)}^{ - \frac{1}{3}} = -\frac{1}{2} \times \frac{1}{3} = -\frac{1}{6}$

(D) The given circle is $x^2 + y^2 + 2x - 4y - 4 = 0.$
Its center is $O_1 = (-1, 2)$ and its radius $r_1 = \sqrt{(-1)^2 + 2^2 - (-4)} = \sqrt{1 + 4 + 4} = 3.$
Let the center of circle $C$ be $O_2 = (h, k)$ and its radius be $r_2 = 3.$
Since the circles touch externally at $P(2, 2),$ the point $P$ divides the line segment $O_1O_2$ internally in the ratio $r_1 : r_2 = 3 : 3 = 1 : 1.$
Thus,$P$ is the midpoint of $O_1O_2.$
$(2, 2) = \left( \frac{-1 + h}{2}, \frac{2 + k}{2} \right).$
Solving for $h$ and $k$:
$-1 + h = 4 \Rightarrow h = 5$
$2 + k = 4 \Rightarrow k = 2$
So,the center of circle $C$ is $(5, 2).$
The equation of circle $C$ is $(x - 5)^2 + (y - 2)^2 = 3^2,$ which simplifies to $x^2 - 10x + 25 + y^2 - 4y + 4 = 9,$ or $x^2 + y^2 - 10x - 4y + 20 = 0.$
The length of the intercept cut by this circle on the $x-$axis is given by $2\sqrt{g^2 - c},$ where $g = -5$ and $c = 20.$
Length $= 2\sqrt{(-5)^2 - 20} = 2\sqrt{25 - 20} = 2\sqrt{5}.$

(C) Let the point $P$ on the parabola $x^2 = 4y$ be $(2t, t^2)$.
The given circle is $x^2 + y^2 + 6x + 8 = 0$. Its centre $C$ is $(-3, 0)$.
For the distance $PC$ to be minimum,the line $PC$ must be the normal to the parabola at $P$.
The slope of the tangent to the parabola at $P(2t, t^2)$ is $\frac{dy}{dx} = \frac{x}{2} = t$.
Therefore,the slope of the normal at $P$ is $-\frac{1}{t}$.
The slope of the line $PC$ connecting $P(2t, t^2)$ and $C(-3, 0)$ is $\frac{t^2 - 0}{2t - (-3)} = \frac{t^2}{2t + 3}$.
Equating the slopes: $\frac{t^2}{2t + 3} = -\frac{1}{t}$.
$t^3 = -2t - 3 \Rightarrow t^3 + 2t + 3 = 0$.
By inspection,$t = -1$ is a root: $(-1)^3 + 2(-1) + 3 = -1 - 2 + 3 = 0$.
For $t = -1$,the point $P$ is $(2(-1), (-1)^2) = (-2, 1)$.
The slope of the tangent at $P$ is $t = -1$.
The equation of the tangent at $(-2, 1)$ is $y - 1 = -1(x + 2)$.
$y - 1 = -x - 2 \Rightarrow x + y + 1 = 0$.

(C) Given that the mean $\bar{x} = 9$ and standard deviation $\sigma = 0$ for $n = 5$ observations.
Since $\sigma = 0,$ all five observations must be equal to the mean.
Thus,the observations are $9, 9, 9, 9, 9.$
Let the new observation be $x_5'$ after changing one value. The sum of the other four observations is $9 \times 4 = 36.$
The new mean is given as $10,$ so $\frac{36 + x_5'}{5} = 10.$
$36 + x_5' = 50 \Rightarrow x_5' = 14.$
The new set of observations is $9, 9, 9, 9, 14.$
The new standard deviation $\sigma_{new} = \sqrt{\frac{\sum (x_i - \bar{x}_{new})^2}{n}}.$
$\sigma_{new} = \sqrt{\frac{4(9 - 10)^2 + (14 - 10)^2}{5}} = \sqrt{\frac{4(1) + 16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2.$

(D) The given series is $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \dots$
The $n$-th term $T_n$ can be written as $T_n = \frac{2^n - 1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$ for $n \ge 1$.
The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (2 - \frac{1}{2^{n-1}})$.
$S_{20} = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}}$.
$S_{20} = 2(20) - (1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{19}})$.
The second part is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 20$.
Sum $= \frac{1(1 - (1/2)^{20})}{1 - 1/2} = 2(1 - \frac{1}{2^{20}}) = 2 - \frac{1}{2^{19}}$.
Therefore,$S_{20} = 40 - (2 - \frac{1}{2^{19}}) = 38 + \frac{1}{2^{19}}$.

(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a(1 - e) = \frac{3}{2}$.
Thus,$a - ae = \frac{3}{2}$,which implies $ae = a - \frac{3}{2}$.
The length of the latus rectum is $\frac{2b^2}{a} = 4$,so $b^2 = 2a$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $2a = a^2(1 - e^2)$,which simplifies to $1 - e^2 = \frac{2}{a}$,or $e^2 = 1 - \frac{2}{a}$.
From $ae = a - \frac{3}{2}$,squaring both sides gives $a^2e^2 = (a - \frac{3}{2})^2$.
Substituting $e^2 = 1 - \frac{2}{a}$,we get $a^2(1 - \frac{2}{a}) = a^2 - 3a + \frac{9}{4}$.
$a^2 - 2a = a^2 - 3a + \frac{9}{4}$.
$a = \frac{9}{4}$.
Now,$e^2 = 1 - \frac{2}{a} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$.
Therefore,$e = \frac{1}{3}$.

(D) Let the number of children in each family be $x$.
The total number of children in both families is $2x$.
We are distributing $3$ tickets among $2x$ children such that no child gets more than one ticket.
The total number of ways to choose $3$ children out of $2x$ is $^{2x}C_{3}$.
The number of ways to choose $3$ children from family $B$ (which has $x$ children) is $^{x}C_{3}$.
The probability that all $3$ tickets go to children of family $B$ is given by:
$P = \frac{^{x}C_{3}}{^{2x}C_{3}} = \frac{1}{12}$
Expanding the combinations:
$\frac{\frac{x(x-1)(x-2)}{3!}}{\frac{2x(2x-1)(2x-2)}{3!}} = \frac{1}{12}$
$\frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} = \frac{1}{12}$
$\frac{(x-1)(x-2)}{2(2x-1) \cdot 2(x-1)} = \frac{1}{12}$
$\frac{x-2}{4(2x-1)} = \frac{1}{12}$
$\frac{x-2}{2x-1} = \frac{4}{12} = \frac{1}{3}$
$3(x-2) = 2x-1$
$3x - 6 = 2x - 1$
$x = 5$
Thus,the number of children in each family is $5$.

(B) The implication $p \rightarrow (p \wedge \neg q)$ is false only when the antecedent $p$ is $T$ and the consequent $(p \wedge \neg q)$ is $F$.
Since $p$ is $T$,the expression $(p \wedge \neg q)$ becomes $(T \wedge \neg q)$.
For $(T \wedge \neg q)$ to be $F$,$\neg q$ must be $F$,which implies $q$ is $T$.
Therefore,the truth values are $p = T$ and $q = T$.

(D) Let the point of intersection be $(x_1, y_1)$.
For the curve $y^2 = 6x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 6$,so $\frac{dy}{dx} = \frac{3}{y_1}$. Let this be $m_1 = \frac{3}{y_1}$.
For the curve $9x^2 + by^2 = 16$,differentiating with respect to $x$ gives $18x + 2by \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{9x_1}{by_1}$. Let this be $m_2 = -\frac{9x_1}{by_1}$.
Since the curves intersect at right angles,$m_1 \times m_2 = -1$,which implies $(\frac{3}{y_1}) \times (-\frac{9x_1}{by_1}) = -1$,so $\frac{27x_1}{by_1^2} = 1$. Since $y_1^2 = 6x_1$,we have $\frac{27x_1}{b(6x_1)} = 1$,which simplifies to $\frac{27}{6b} = 1$,so $b = \frac{27}{6} = \frac{9}{2}$.

(C) The implication $A \rightarrow B$ is false only when $A$ is true and $B$ is false.
Given $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p \vee q \equiv F$.
This implies $(p \wedge \sim q) \wedge (p \wedge r) \equiv T$ and $\sim p \vee q \equiv F$.
From $\sim p \vee q \equiv F$,we get $\sim p \equiv F$ and $q \equiv F$,which means $p \equiv T$ and $q \equiv F$.
Now substitute these into the first part: $(T \wedge \sim F) \wedge (T \wedge r) \equiv T$.
$(T \wedge T) \wedge (T \wedge r) \equiv T$.
$T \wedge (T \wedge r) \equiv T$.
This requires $T \wedge r \equiv T$,which implies $r \equiv T$.
Thus,the truth values are $p = T, q = F, r = T$.

(D) Given the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$.
Mean $= \frac{7+8+9+7+8+7+\lambda+8}{8} = 8$
$\Rightarrow \frac{54+\lambda}{8} = 8$
$\Rightarrow 54+\lambda = 64$
$\Rightarrow \lambda = 10$
Now,the data set is $7, 8, 9, 7, 8, 7, 10, 8$.
Variance $(\sigma^2) = \frac{1}{n} \sum (x_i - \bar{x})^2$
Variance $= \frac{(7-8)^2 + (8-8)^2 + (9-8)^2 + (7-8)^2 + (8-8)^2 + (7-8)^2 + (10-8)^2 + (8-8)^2}{8}$
Variance $= \frac{(-1)^2 + 0^2 + 1^2 + (-1)^2 + 0^2 + (-1)^2 + 2^2 + 0^2}{8}$
Variance $= \frac{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}{8} = \frac{8}{8} = 1$.

(B) The intersection point of the curves $y = x^2$ and $y = \frac{1}{x}$ is found by setting $x^2 = \frac{1}{x}$,which gives $x^3 = 1$,so $x = 1$. Thus,the intersection point is $(1, 1)$.
The area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$,the $x$-axis $(y = 0)$,and the line $x = t$ $(t > 1)$ is given by the sum of two integrals:
$\text{Area} = \int_0^1 x^2 \, dx + \int_1^t \frac{1}{x} \, dx$
Evaluating the integrals:
$\text{Area} = \left[ \frac{x^3}{3} \right]_0^1 + \left[ \ln(x) \right]_1^t$
$\text{Area} = \left( \frac{1}{3} - 0 \right) + (\ln(t) - \ln(1))$
Since $\ln(1) = 0$,we have:
$\text{Area} = \frac{1}{3} + \ln(t)$
Given that the area is $1 \, \text{sq. unit}$:
$\frac{1}{3} + \ln(t) = 1$
$\ln(t) = 1 - \frac{1}{3} = \frac{2}{3}$
Taking the exponential of both sides:
$t = e^{2/3}$

(A) If the function is continuous at $x = 0$,then $\lim_{x \to 0} f(x)$ must exist and $f(0) = \lim_{x \to 0} f(x)$.
Now,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} \right)$.
$= \lim_{x \to 0} \left( \frac{e^{2x} - 1 - (k - 1)x}{x(e^{2x} - 1)} \right)$.
Using the expansion $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$,we get:
$= \lim_{x \to 0} \frac{(1 + 2x + 2x^2 + \dots) - 1 - kx + x}{x(2x + 2x^2 + \dots)}$.
$= \lim_{x \to 0} \frac{(3 - k)x + 2x^2 + \dots}{2x^2 + 2x^3 + \dots}$.
For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $3 - k = 0$,which gives $k = 3$.
Substituting $k = 3$,the limit becomes $\lim_{x \to 0} \frac{2x^2 + \dots}{2x^2 + 2x^3 + \dots} = \frac{2}{2} = 1$.
Therefore,$f(0) = 1$,and the ordered pair is $(3, 1)$.

(C) Given relations on $N$ are:
$R_1 = \{(x,y) \in N \times N : 2x + y = 10\}$
$R_2 = \{(x,y) \in N \times N : x + 2y = 10\}$
For $R_1$,we solve $y = 10 - 2x$ for $x, y \in N$:
If $x=1, y=8$; if $x=2, y=6$; if $x=3, y=4$; if $x=4, y=2$.
So,$R_1 = \{(1,8), (2,6), (3,4), (4,2)\}$.
The range of $R_1$ is $\{2, 4, 6, 8\}$. Thus,option $D$ is incorrect.
$R_1$ is not symmetric because $(1,8) \in R_1$ but $(8,1) \notin R_1$.
$R_1$ is not transitive because $(3,4) \in R_1$ and $(4,2) \in R_1$,but $(3,2) \notin R_1$.
For $R_2$,we solve $x = 10 - 2y$ for $x, y \in N$:
If $y=1, x=8$; if $y=2, x=6$; if $y=3, x=4$; if $y=4, x=2$.
So,$R_2 = \{(8,1), (6,2), (4,3), (2,4)\}$.
The range of $R_2$ is $\{1, 2, 3, 4\}$.
Thus,option $C$ is correct.
$R_2$ is not symmetric because $(8,1) \in R_2$ but $(1,8) \notin R_2$.
$R_2$ is not transitive because $(4,3) \in R_2$ and $(3,y) \notin R_2$ (as no $y$ exists for $x=3$ in $R_2$).

(A) Let $I = \int \frac{\tan x}{1 + \tan x + \tan^2 x} dx$.
We can rewrite the numerator as $\frac{1}{2} [ (1 + \tan x + \tan^2 x) - (1 - \tan x + \tan^2 x) ]$ is not ideal,so let's use the identity: $\tan x = \frac{1}{2} (1 + \tan x + \tan^2 x) - \frac{1}{2} (1 - \tan x + \tan^2 x)$.
Alternatively,note that $\int \frac{\tan x}{1 + \tan x + \tan^2 x} dx = \int \frac{\tan x + 1 + \tan^2 x - (1 + \tan^2 x)}{1 + \tan x + \tan^2 x} dx = \int 1 dx - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$.
Let $I = x - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$.
Substitute $\tan x = t$,so $\sec^2 x dx = dt$.
$I = x - \int \frac{dt}{t^2 + t + 1} = x - \int \frac{dt}{(t + 1/2)^2 + 3/4}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = x - \frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{t + 1/2}{\sqrt{3}/2} \right) + C = x - \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2t + 1}{\sqrt{3}} \right) + C$.
Substituting $t = \tan x$,we have $I = x - \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2 \tan x + 1}{\sqrt{3}} \right) + C$.
Comparing this with the given form $x - \frac{K}{\sqrt{A}} \tan^{-1} \left( \frac{K \tan x + 1}{\sqrt{A}} \right) + C$,we find $K = 2$ and $A = 3$.
Thus,the ordered pair $(K, A)$ is $(2, 3)$.

(A) The number of one-one functions $\alpha$ from a set $X$ $(|X|=5)$ to a set $Y$ $(|Y|=7)$ is given by $P(7, 5) = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
Alternatively,$\alpha = {}^{7}C_{5} \times 5! = 21 \times 120 = 2520$.
For onto functions $\beta$ from $Y$ $(|Y|=7)$ to $X$ $(|X|=5)$,we use the formula for the number of onto functions: $m! \times S(n, m)$,where $S(n, m)$ is the Stirling number of the second kind.
$\beta = 5! \times S(7, 5) = 120 \times \frac{1}{2!} \sum_{k=0}^{5} (-1)^{5-k} {}^{5}C_{k} k^{7} = 120 \times 140 = 16800$.
Now,calculate $\frac{1}{5!}(\beta - \alpha) = \frac{16800 - 2520}{120} = \frac{14280}{120} = 119$.

(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.
Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = \frac{4x}{\sin x}$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
Substituting the values,$y \sin x = \int \frac{4x}{\sin x} \cdot \sin x dx + C = \int 4x dx + C = 2x^2 + C$.
Thus,$y = \frac{2x^2 + C}{\sin x}$.
Given $y(\frac{\pi}{2}) = 0$,we have $0 = \frac{2(\frac{\pi}{2})^2 + C}{\sin(\frac{\pi}{2})} = \frac{\pi^2}{2} + C$,so $C = -\frac{\pi^2}{2}$.
The particular solution is $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$.
Evaluating at $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}}{\sin(\frac{\pi}{6})} = \frac{\frac{2\pi^2}{36} - \frac{\pi^2}{2}}{1/2} = 2 \left( \frac{\pi^2}{18} - \frac{\pi^2}{2} \right) = 2 \left( \frac{\pi^2 - 9\pi^2}{18} \right) = 2 \left( \frac{-8\pi^2}{18} \right) = -\frac{8}{9} \pi^2$.