IIT JEE 1996 Mathematics Question Paper with Answer and Solution

(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.
Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,we get:
$x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$.
For $x = 1$: $1 + 1 = A(1 - 2)(1 - 3)$ $\Rightarrow 2 = A(-1)(-2)$ $\Rightarrow 2 = 2A$ $\Rightarrow A = 1$.
For $x = 2$: $2 + 1 = B(2 - 1)(2 - 3)$ $\Rightarrow 3 = B(1)(-1)$ $\Rightarrow 3 = -B$ $\Rightarrow B = -3$.
For $x = 3$: $3 + 1 = C(3 - 1)(3 - 2)$ $\Rightarrow 4 = C(2)(1)$ $\Rightarrow 4 = 2C$ $\Rightarrow C = 2$.
Therefore,the partial fraction decomposition is $\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3}$.

(D) Given the expression: $(1 + i)^{n_1} + (1 + i^3)^{n_1} + (1 + i^5)^{n_2} + (1 + i^7)^{n_2}$.
Since $i^3 = -i$, $i^5 = i$, and $i^7 = -i$, the expression becomes:
$(1 + i)^{n_1} + (1 - i)^{n_1} + (1 + i)^{n_2} + (1 - i)^{n_2}$.
Let $z = (1 + i)^{n} + (1 - i)^{n}$.
Using the polar form $1 + i = \sqrt{2} e^{i\pi/4}$ and $1 - i = \sqrt{2} e^{-i\pi/4}$, we get:
$z = (\sqrt{2})^n (e^{in\pi/4} + e^{-in\pi/4}) = 2^{n/2} \cdot 2 \cos(n\pi/4) = 2^{n/2+1} \cos(n\pi/4)$.
Since $2^{n/2+1} \cos(n\pi/4)$ is always a real number for any positive integer $n$, the sum of two such terms is also always real.
Therefore, the expression is a real number for all positive integers $n_1$ and $n_2$.

(B) The $r^{th}$ term of the given series is $T_r = r[(r + 1) - \omega][(r + 1) - \omega^2]$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.
$T_r = r[(r + 1)^2 - (\omega + \omega^2)(r + 1) + \omega^3] = r[(r + 1)^2 + (r + 1) + 1] = r[r^2 + 2r + 1 + r + 1 + 1] = r(r^2 + 3r + 3) = r^3 + 3r^2 + 3r$.
The sum of the series is $S = \sum_{r=1}^{n-1} (r^3 + 3r^2 + 3r)$.
$S = \sum_{r=1}^{n-1} r^3 + 3 \sum_{r=1}^{n-1} r^2 + 3 \sum_{r=1}^{n-1} r$.
$S = [\frac{(n-1)n}{2}]^2 + 3 \cdot \frac{(n-1)n(2n-2+1)}{6} + 3 \cdot \frac{(n-1)n}{2}$.
$S = \frac{(n-1)^2 n^2}{4} + \frac{(n-1)n(2n-1)}{2} + \frac{3(n-1)n}{2}$.
$S = \frac{(n-1)n}{4} [(n-1)n + 2(2n-1) + 6] = \frac{(n-1)n}{4} [n^2 - n + 4n - 2 + 6] = \frac{1}{4}(n-1)n(n^2 + 3n + 4)$.

(D) Let $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.
Since $n$ is odd,the last term is ${1^3}$.
We can write $S = \sum_{k=1}^{n} k^3 - 2 \sum_{k=1}^{(n-1)/2} (2k)^3$.
Using the formula $\sum_{k=1}^{m} k^3 = \left[ \frac{m(m+1)}{2} \right]^2$,we get:
$S = \left[ \frac{n(n+1)}{2} \right]^2 - 2 \times 8 \sum_{k=1}^{(n-1)/2} k^3$
$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{\frac{n-1}{2} (\frac{n-1}{2} + 1)}{2} \right]^2$
$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{(n-1)(n+1)}{8} \right]^2$
$S = \frac{n^2(n+1)^2}{4} - 16 \frac{(n-1)^2(n+1)^2}{64}$
$S = \frac{n^2(n+1)^2}{4} - \frac{(n-1)^2(n+1)^2}{4}$
$S = \frac{(n+1)^2}{4} [n^2 - (n-1)^2]$
$S = \frac{(n+1)^2}{4} [n^2 - (n^2 - 2n + 1)]$
$S = \frac{(n+1)^2}{4} (2n - 1)$.

(D) The statement $P(n)$ is $n^2 + n$ is odd.
We can write $n^2 + n = n(n + 1)$.
Since $n$ and $n + 1$ are consecutive integers,one of them must be even and the other must be odd.
The product of an even number and an odd number is always even.
Therefore,$n^2 + n$ is always even for all natural numbers $n$.
Since the statement claims $n^2 + n$ is odd,the statement $P(n)$ is false for all $n \in \mathbb{N}$.
Thus,there is no $n$ for which $P(n)$ is true.
Hence,the correct option is $(d)$.

(B) Given equation: $\tan^2 \theta + \sec 2\theta = 1$.
Using the identity $\sec 2\theta = \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta}$,the equation becomes:
$\tan^2 \theta + \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = 1$.
Let $x = \tan^2 \theta$. Then $x + \frac{1+x}{1-x} = 1$.
$x(1-x) + 1 + x = 1 - x$
$x - x^2 + 1 + x = 1 - x$
$3x - x^2 = 0 \Rightarrow x(3 - x) = 0$.
So,$\tan^2 \theta = 0$ or $\tan^2 \theta = 3$.
If $\tan^2 \theta = 0$,then $\tan \theta = 0 \Rightarrow \theta = m\pi$ for any integer $m$.
If $\tan^2 \theta = 3$,then $\tan \theta = \pm \sqrt{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{3}$ for any integer $n$.
Thus,the general solution is $\theta = m\pi, n\pi \pm \frac{\pi}{3}$.

(B) Given $a:b:c = 4:5:6$. Let $a = 4k, b = 5k, c = 6k$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{4k+5k+6k}{2} = \frac{15k}{2}$.
Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2}(\frac{15k}{2}-4k)(\frac{15k}{2}-5k)(\frac{15k}{2}-6k)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \frac{15k^2\sqrt{7}}{4}$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot \frac{15k^2\sqrt{7}}{4}} = \frac{120k^3}{15k^2\sqrt{7}} = \frac{8k}{\sqrt{7}}$.
Inradius $r = \frac{\Delta}{s} = \frac{15k^2\sqrt{7}/4}{15k/2} = \frac{k\sqrt{7}}{2}$.
Ratio $\frac{R}{r} = \frac{8k/\sqrt{7}}{k\sqrt{7}/2} = \frac{8}{\sqrt{7}} \cdot \frac{2}{\sqrt{7}} = \frac{16}{7}$.

(D) The equation of the circle is ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$.
Comparing with ${x^2} + {y^2} + 2gx + 2fy + c = 0$,we have $g = 2, f = -3, c = 9{\sin ^2}\alpha + 13{\cos ^2}\alpha$.
The centre is $C(-g, -f) = (-2, 3)$.
The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - (9{\sin ^2}\alpha + 13{\cos ^2}\alpha)} = \sqrt{13 - 9{\sin ^2}\alpha - 13{\cos ^2}\alpha} = \sqrt{13(1 - {\cos ^2}\alpha) - 9{\sin ^2}\alpha} = \sqrt{13{\sin ^2}\alpha - 9{\sin ^2}\alpha} = \sqrt{4{\sin ^2}\alpha} = 2\sin \alpha$.
Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2\alpha$,so $\angle APC = \alpha$.
In the right-angled triangle $\triangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC} = \frac{2\sin \alpha}{\sqrt{(h + 2)^2 + (k - 3)^2}}$.
Thus,$\sqrt{(h + 2)^2 + (k - 3)^2} = 2$.
Squaring both sides,$(h + 2)^2 + (k - 3)^2 = 4$.
Expanding,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$,which simplifies to $h^2 + k^2 + 4h - 6k + 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} + {y^2} + 4x - 6y + 9 = 0$.

(A) The equation of any circle passing through the intersection of the circle $x^2 + y^2 - 2x = 0$ and the line $y - x = 0$ is given by the family of circles equation:
$x^2 + y^2 - 2x + \lambda(y - x) = 0$
$x^2 + y^2 - (2 + \lambda)x + \lambda y = 0$
The center of this circle is $\left( \frac{2 + \lambda}{2}, -\frac{\lambda}{2} \right)$.
Since $AB$ is the diameter,the center of the circle must lie on the line $y = x$.
Substituting the center coordinates into $y = x$:
$-\frac{\lambda}{2} = \frac{2 + \lambda}{2}
-\lambda = 2 + \lambda
2\lambda = -2
\lambda = -1$
Substituting $\lambda = -1$ back into the family equation:
$x^2 + y^2 - (2 - 1)x - 1y = 0$
$x^2 + y^2 - x - y = 0$.

(A) The common tangents to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$ are $x = 1$ and $x = -1$.
Out of these, $x = 1$ is nearer to the point $P\left( \frac{1}{2}, 1 \right)$.
Thus, the directrix of the required ellipse is $x = 1$.
Let $Q(x, y)$ be any point on the ellipse.
By the definition of an ellipse, the distance from the focus $P$ is $QP = e \times (\text{distance from directrix})$.
$QP = \sqrt{(x - 1/2)^2 + (y - 1)^2}$ and the distance from the directrix $x = 1$ is $|x - 1|$.
Given $e = 1/2$, we have $\sqrt{(x - 1/2)^2 + (y - 1)^2} = \frac{1}{2} |x - 1|$.
Squaring both sides: $(x - 1/2)^2 + (y - 1)^2 = \frac{1}{4}(x - 1)^2$.
$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}(x^2 - 2x + 1)$.
$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}x^2 - \frac{1}{2}x + 1/4$.
$\frac{3}{4}x^2 - \frac{1}{2}x + (y - 1)^2 = 0$.
Multiply by $4/3$: $x^2 - \frac{2}{3}x + \frac{4}{3}(y - 1)^2 = 0$.
$(x^2 - \frac{2}{3}x + 1/9) + \frac{4}{3}(y - 1)^2 = 1/9$.
$(x - 1/3)^2 + \frac{4}{3}(y - 1)^2 = 1/9$.
Dividing by $1/9$: $\frac{(x - 1/3)^2}{1/9} + \frac{(y - 1)^2}{1/12} = 1$.

(A) We use the standard limit formula: $\mathop {\lim }\limits_{x \to 0} (1 + f(x))^{1/g(x)} = e^{\mathop {\lim }\limits_{x \to 0} \frac{f(x)}{g(x)}}$.
Given the expression $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}}$,we can rewrite the base as $1 + \left( \frac{1 + 5x^2}{1 + 3x^2} - 1 \right) = 1 + \frac{2x^2}{1 + 3x^2}$.
Thus,the limit becomes $e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \cdot \frac{2x^2}{1 + 3x^2}}$.
Simplifying the exponent: $\mathop {\lim }\limits_{x \to 0} \frac{2}{1 + 3x^2} = \frac{2}{1 + 0} = 2$.
Therefore,the result is $e^2$.

(A) The number of solutions to $x_1 + x_2 + ... + x_k = n$ with $x_i \ge i$ is the coefficient of $t^n$ in the product $(t^1 + t^2 + ...) (t^2 + t^3 + ...) ... (t^k + t^{k+1} + ...)$.
This is the coefficient of $t^n$ in $t^{1+2+...+k} (1 + t + t^2 + ...)^k$.
Let $r = 1 + 2 + ... + k = \frac{k(k+1)}{2}$.
The expression becomes the coefficient of $t^{n-r}$ in $(1-t)^{-k}$.
Using the binomial expansion $(1-t)^{-k} = \sum_{j=0}^{\infty} \binom{k+j-1}{k-1} t^j$,the coefficient of $t^{n-r}$ is $\binom{k+(n-r)-1}{k-1}$.
Let $m = k + n - r - 1 = k + n - \frac{k(k+1)}{2} - 1 = \frac{2k + 2n - k^2 - k - 2}{2} = \frac{2n - k^2 + k - 2}{2}$.
Thus,the number of solutions is $\binom{m}{k-1}$.

(C) Let $\frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}$.
Multiplying both sides by $(x - 1)(x - 2)(x - 3)$,we get:
$x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$.
For $x = 1$: $1 + 1 = A(1 - 2)(1 - 3) \implies 2 = A(-1)(-2) \implies 2 = 2A \implies A = 1$.
For $x = 2$: $2 + 1 = B(2 - 1)(2 - 3) \implies 3 = B(1)(-1) \implies 3 = -B \implies B = -3$.
For $x = 3$: $3 + 1 = C(3 - 1)(3 - 2) \implies 4 = C(2)(1) \implies 4 = 2C \implies C = 2$.
Thus,the partial fraction is $\frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3}$.

(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b \times c}{|b \times c|}$ form an orthonormal basis for the space $\mathbb{R}^3$.
Any vector $a$ can be expressed as a linear combination of these basis vectors:
$a = (a \cdot b)b + (a \cdot c)c + (a \cdot k)k$
Substituting $k = \frac{b \times c}{|b \times c|}$ into the expression,we get:
$a = (a \cdot b)b + (a \cdot c)c + \left(a \cdot \frac{b \times c}{|b \times c|}\right) \frac{b \times c}{|b \times c|}$
However,the term in the question is $\frac{a \cdot (b \times c)}{|b \times c|} (b \times c)$.
Note that $\frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = (a \cdot k) (|b \times c| k) = (a \cdot k) (\sin \alpha) k$,where $\alpha$ is the angle between $b$ and $c$.
Since $|b \times c| = \sin \alpha$,the expression simplifies to:
$(a \cdot b)b + (a \cdot c)c + (a \cdot k)k = a$.

(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5).$
Given $f(x) = f\left( \frac{x + 1}{x + 2} \right).$
Since $f(x) = f(-x)$,the equation $f(x) = f\left( \frac{x + 1}{x + 2} \right)$ is satisfied if $x = \frac{x + 1}{x + 2}$ or if $-x = \frac{x + 1}{x + 2}.$
Case $1$: $x = \frac{x + 1}{x + 2} \Rightarrow x^2 + 2x = x + 1 \Rightarrow x^2 + x - 1 = 0.$
Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}.$
Case $2$: $-x = \frac{x + 1}{x + 2} \Rightarrow -x^2 - 2x = x + 1 \Rightarrow x^2 + 3x + 1 = 0.$
Using the quadratic formula,$x = \frac{-3 \pm \sqrt{9 - 4(1)(1)}}{2} = \frac{-3 \pm \sqrt{5}}{2}.$
Thus,the four values are $\frac{-1 \pm \sqrt{5}}{2}$ and $\frac{-3 \pm \sqrt{5}}{2}.$ Note: The provided options suggest the intended roots are $\frac{3 \pm \sqrt{5}}{2}$ and $\frac{-3 \pm \sqrt{5}}{2}$ based on the equation $f(x) = f\left( \frac{-x+1}{-x+2} \right)$ logic.

(D) Given $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get:
$f(x) = \frac{1 - \cos 2x}{2} + \frac{1 - \cos(2x + \frac{2\pi}{3})}{2} + \cos x \cos(x + \frac{\pi}{3})$
$f(x) = 1 - \frac{1}{2} [\cos 2x + \cos(2x + \frac{2\pi}{3})] + \cos x \cos(x + \frac{\pi}{3})$
Using $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$f(x) = 1 - \frac{1}{2} [2 \cos(2x + \frac{\pi}{3}) \cos(-\frac{\pi}{3})] + \cos x \cos(x + \frac{\pi}{3})$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have:
$f(x) = 1 - \frac{1}{2} \cos(2x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$f(x) = 1 - \frac{1}{2} \cos(2x + \frac{\pi}{3}) + \frac{1}{2} [\cos(2x + \frac{\pi}{3}) + \cos(-\frac{\pi}{3})]$
$f(x) = 1 + \frac{1}{2} \cos(\frac{\pi}{3}) = 1 + \frac{1}{2} \cdot \frac{1}{2} = 1 + \frac{1}{4} = \frac{5}{4}$.
Since $f(x) = \frac{5}{4}$ for all $x \in R$,then $(g \circ f)(x) = g(f(x)) = g(\frac{5}{4})$.
Given $g(\frac{5}{4}) = 1$,therefore $(g \circ f)(x) = 1$.

(C) Given the equation $x{e^{xy}} = y + {\sin ^2}x$.
When $x = 0$,substituting into the equation gives $0 \cdot {e^0} = y + {\sin ^2}(0)$,which implies $y = 0$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{d}{dx}(x{e^{xy}}) = \frac{d}{dx}(y + {\sin ^2}x)$
${e^{xy}} + x{e^{xy}} \cdot \frac{d}{dx}(xy) = \frac{dy}{dx} + 2\sin x \cos x$
${e^{xy}} + x{e^{xy}} \left( y + x \frac{dy}{dx} \right) = \frac{dy}{dx} + \sin(2x)$
Substituting $x = 0$ and $y = 0$ into the differentiated equation:
${e^0} + 0 \cdot {e^0} (0 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + \sin(0)$
$1 + 0 = \frac{dy}{dx} + 0$
Therefore,$\frac{dy}{dx} = 1$.

(B) Given the equation: $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5$ ... $(i)$
Replacing $x$ with $\frac{1}{x}$ in $(i)$,we get: $af\left( {\frac{1}{x}} \right) + bf(x) = x - 5$ ... $(ii)$
To eliminate $f\left( {\frac{1}{x}} \right)$,multiply $(i)$ by $a$ and $(ii)$ by $b$:
$a^2 f(x) + ab f\left( {\frac{1}{x}} \right) = \frac{a}{x} - 5a$
$b^2 f(x) + ab f\left( {\frac{1}{x}} \right) = bx - 5b$
Subtracting the second from the first: $(a^2 - b^2) f(x) = \frac{a}{x} - bx - 5a + 5b$
Integrating both sides from $1$ to $2$:
$(a^2 - b^2) \int_1^2 f(x) dx = \int_1^2 \left( \frac{a}{x} - bx - 5(a - b) \right) dx$
$= \left[ a \log |x| - \frac{b x^2}{2} - 5(a - b)x \right]_1^2$
$= \left( a \log 2 - \frac{b(4)}{2} - 5(a - b)(2) \right) - \left( a \log 1 - \frac{b(1)}{2} - 5(a - b)(1) \right)$
$= a \log 2 - 2b - 10a + 10b - 0 + \frac{b}{2} + 5a - 5b$
$= a \log 2 - 5a + \frac{7}{2}b$
Thus,$\int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[ a \log 2 - 5a + \frac{7}{2}b \right]$.

(C) Given the function $L(x) = \int_1^x \frac{1}{t} dt$.
Evaluating the integral,we get $L(x) = [\ln |t|]_1^x$.
$L(x) = \ln |x| - \ln |1| = \ln |x|$.
Now,consider $L(xy) = \ln |xy| = \ln |x| + \ln |y|$.
Since $L(x) = \ln |x|$ and $L(y) = \ln |y|$,we have $L(xy) = L(x) + L(y)$.
Thus,the correct option is $C$.

(A) The plane containing vectors $i$ and $i + j$ has a normal vector $n_1 = i \times (i + j) = i \times i + i \times j = 0 + k = k.$ The equation of this plane is $r \cdot k = 0,$ which implies $z = 0.$
The plane containing vectors $i - j$ and $i + k$ has a normal vector $n_2 = (i - j) \times (i + k) = i \times i + i \times k - j \times i - j \times k = 0 - j + k - i = -i - j + k.$ The equation of this plane is $r \cdot (-i - j + k) = 0,$ which implies $-x - y + z = 0$ or $x + y - z = 0.$
The vector $a$ is parallel to the line of intersection of these two planes,so $a$ must be parallel to the cross product of the normals $n_1$ and $n_2$: $v = n_1 \times n_2 = k \times (-i - j + k) = -(k \times i) - (k \times j) + (k \times k) = -j + i + 0 = i - j.$
Thus,$a$ is parallel to $i - j.$
Let $\theta$ be the angle between $a = i - j$ and $b = i - 2j + 2k.$
$\cos \theta = \frac{a \cdot b}{|a| |b|} = \frac{(1)(1) + (-1)(-2) + (0)(2)}{\sqrt{1^2 + (-1)^2} \sqrt{1^2 + (-2)^2 + 2^2}} = \frac{1 + 2}{\sqrt{2} \sqrt{1 + 4 + 4}} = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}.$
Since $a$ can be in the direction of $i - j$ or $-(i - j)$,the angle $\theta$ satisfies $\cos \theta = \pm \frac{1}{\sqrt{2}}.$
Therefore,$\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$.

(C) The function $f(x) = [x]\sin \left( \frac{\pi}{[x + 1]} \right)$ is defined if $[x + 1] \neq 0$.
Since $[x + 1] = 0$ for $0 \le x + 1 < 1$,which implies $-1 \le x < 0$,the domain of $f$ is $R - [-1, 0)$.
Thus,the domain is $\left\{ x \in R \mid x \notin [-1, 0) \right\}$.
The function $[x]$ is discontinuous at all integers $I$.
For $x \in I$,$f(x)$ is discontinuous unless the product $[x]\sin \left( \frac{\pi}{[x + 1]} \right)$ is continuous.
At $x = 0$,$f(0) = [0]\sin \left( \frac{\pi}{[1]} \right) = 0$.
For $x \to 0^+$,$[x] = 0$,so $f(x) = 0 \cdot \sin \left( \frac{\pi}{1} \right) = 0$.
Thus,$f$ is continuous at $x = 0$.
For other integers $n \in I \setminus \{0\}$,the function remains discontinuous.
Therefore,the set of points of discontinuity is $I - \{0\}$.

(D) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.
Using the property $\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} }$,we have:
$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}(2\pi - x)}}{{{{\sin }^{2n}}(2\pi - x) + {{\cos }^{2n}}(2\pi - x)}}dx} $
$I = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Adding the two expressions for $I$:
$2I = \int_0^{2\pi } {\frac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$I = \pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Using the property $\int_0^{2a} {f(x)dx = 2\int_0^a {f(x)dx} }$ if $f(2a-x) = f(x)$:
$I = 2\pi \int_0^{\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $i$)
Similarly,$I = 4\pi \int_0^{\pi /2} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ (Equation $ii$)
Adding $(i)$ and $(ii)$:
$2I = 4\pi \int_0^{\pi /2} {1 dx} = 4\pi \left( \frac{\pi }{2} \right) = 2{\pi ^2}$
Therefore,$I = {\pi ^2}$.

(A) We know that $P$ (exactly one of $A$ or $B$ occurs) = $P(A) + P(B) - 2P(A \cap B) = p$ ..... $(i)$
Similarly,$P(B) + P(C) - 2P(B \cap C) = p$ ..... $(ii)$
and $P(C) + P(A) - 2P(C \cap A) = p$ ..... $(iii)$
Adding $(i), (ii)$ and $(iii)$,we get $2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = 3p$
Therefore,$P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3p}{2}$ ..... $(iv)$
We are given $P(A \cap B \cap C) = p^2$ ..... $(v)$
The probability of at least one of the three events $A, B$ and $C$ occurring is given by the inclusion-exclusion principle:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$
Substituting $(iv)$ and $(v)$ into the expression:
$P(A \cup B \cup C) = \frac{3p}{2} + p^2 = \frac{3p + 2p^2}{2}$