MHT CET 2010 Mathematics Question Paper with Answer and Solution

(C) Given the parametric equations of the curve are $x=3 \cos \theta$ and $y=3 \sin \theta$.
Squaring and adding these equations,we get $x^2+y^2 = 9(\cos^2 \theta + \sin^2 \theta) = 9$,which represents a circle with radius $r=3$.
The point of tangency at $\theta = \frac{\pi}{4}$ is given by $x_1 = 3 \cos(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$ and $y_1 = 3 \sin(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.
Substituting the values,we get $x(\frac{3}{\sqrt{2}}) + y(\frac{3}{\sqrt{2}}) = 9$.
Multiplying both sides by $\frac{\sqrt{2}}{3}$,we obtain $x+y = 3\sqrt{2}$.

(A) The given equation of the circle is $x^{2}+y^{2}-6x-4y+9=0$.
Comparing this with the general form $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=-2, c=9$.
The center of the circle is $(-g, -f) = (3, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-2)^{2}-9} = \sqrt{9+4-9} = \sqrt{4} = 2$.
To check if the line $4x+3y-8=0$ is a tangent,we calculate the perpendicular distance $d$ from the center $(3, 2)$ to the line:
$d = \left|\frac{4(3)+3(2)-8}{\sqrt{4^{2}+3^{2}}}\right| = \left|\frac{12+6-8}{\sqrt{16+9}}\right| = \left|\frac{10}{5}\right| = 2$.
Since the perpendicular distance $d$ is equal to the radius $r$ $(d=r=2)$,the line is a tangent to the circle.

(C) The equation of the circle is $x^{2}+y^{2}-6x+4y-12=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=2, c=-12$.
The center is $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Any line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$.
The distance from the center $(3, -2)$ to the tangent line is equal to the radius:
$\frac{|4(3)+3(-2)+k|}{\sqrt{4^{2}+3^{2}}} = 5$
$\frac{|12-6+k|}{5} = 5$
$|6+k| = 25$
$6+k = 25$ or $6+k = -25$
$k = 19$ or $k = -31$.
Thus,the equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

(A) The equation of the circle is $x^{2} + y^{2} = r^{2}$.
Any point $P$ on the circle can be represented as $(r \cos \theta, r \sin \theta)$.
The slope of the tangent at $P$ is given by differentiating $x^{2} + y^{2} = r^{2}$ with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.
At point $P(r \cos \theta, r \sin \theta)$,the slope of the tangent $m_{t} = -\frac{r \cos \theta}{r \sin \theta} = -\cot \theta$.
The slope of the normal $m_{n}$ is $-\frac{1}{m_{t}} = \frac{1}{\cot \theta} = \tan \theta = \frac{\sin \theta}{\cos \theta}$.
The equation of the normal at $(r \cos \theta, r \sin \theta)$ is:
$y - r \sin \theta = \frac{\sin \theta}{\cos \theta} (x - r \cos \theta)$.
$y \cos \theta - r \sin \theta \cos \theta = x \sin \theta - r \sin \theta \cos \theta$.
$x \sin \theta - y \cos \theta = 0$.

(C) The forward difference operator $\Delta$ is defined as $\Delta u_{n} = u_{n+1} - u_{n}$.
For the third-order forward difference,we use the formula $\Delta^{3} u_{0} = (E-1)^{3} u_{0}$,where $E$ is the shift operator such that $E u_{n} = u_{n+1}$.
Expanding the operator,we get $\Delta^{3} u_{0} = (E^{3} - 3E^{2} + 3E - 1) u_{0}$.
This simplifies to $\Delta^{3} u_{0} = u_{3} - 3u_{2} + 3u_{1} - u_{0}$.
Substituting the given values $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$:
$\Delta^{3} u_{0} = 51 - 3(12) + 3(3) - 8$.
$\Delta^{3} u_{0} = 51 - 36 + 9 - 8$.
$\Delta^{3} u_{0} = 15 + 9 - 8 = 24 - 8 = 16$.

(B) The given equation of the curve is $9x^{2} + 25y^{2} = 225$.
Dividing both sides by $225$,we get $\frac{9x^{2}}{225} + \frac{25y^{2}}{225} = 1$,which simplifies to $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.
This is the equation of an ellipse of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a^{2} = 25$ and $b^{2} = 9$.
Thus,$a = 5$ and $b = 3$.
For an ellipse,the sum of the focal radii of any point on the curve is equal to the length of the major axis,which is $2a$.
Therefore,the sum of focal radii $= 2 \times 5 = 10$.

(A) The given curve is $9x^{2} + 16y^{2} = 144$.
Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.
Here,$a^{2} = 16$ and $b^{2} = 9$.
Let the equation of the tangent be $x + y = k$,which can be written as $y = -x + k$.
Comparing this with $y = mx + c$,we have $m = -1$ and $c = k$.
The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values,we get $k^{2} = 16(-1)^{2} + 9$.
$k^{2} = 16 + 9 = 25$.
$k = \pm 5$.
Thus,the equations of the tangents are $x + y = 5$ and $x + y = -5$.

(C) Given the values at $x = 0, 1, 2, 4$,we assume the function $f(x)$ is a polynomial of degree $3$ or less,implying the $4^{th}$ order forward difference is zero: $\Delta^{4} f(0) = 0$.
Using the operator $E$,where $Ef(x) = f(x+1)$,we have $(E-1)^{4} f(0) = 0$.
Expanding this,we get $(E^{4}-4E^{3}+6E^{2}-4E+1) f(0) = 0$.
This simplifies to $f(4) - 4f(3) + 6f(2) - 4f(1) + f(0) = 0$.
Substituting the known values $f(0)=1, f(1)=3, f(2)=9, f(4)=81$:
$81 - 4f(3) + 6(9) - 4(3) + 1 = 0$.
$81 - 4f(3) + 54 - 12 + 1 = 0$.
$124 - 4f(3) = 0$.
$4f(3) = 124$.
$f(3) = 31$.

(B) The given equations of the curves are $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$.
Dividing both by $144$,we get $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$.
For a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,the eccentricity $e$ is given by $e = \sqrt{1+\frac{b^{2}}{a^{2}}}$.
For the first curve,$a^{2}=9$ and $b^{2}=16$,so $e_{1} = \sqrt{1+\frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.
For the second curve,$a^{2}=16$ and $b^{2}=9$,so $e_{2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
Now,calculating $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}} = \frac{1}{(5/3)^{2}} + \frac{1}{(5/4)^{2}} = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$.

(B) The sum of the first $n$ natural numbers is given by the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
Substituting this into the limit expression:
$\lim _{n \rightarrow \infty} \frac{n(n+1)}{2n^{2}}$
$= \lim _{n \rightarrow \infty} \frac{n^2+n}{2n^2}$
$= \lim _{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2}$
As $n \rightarrow \infty$,$\frac{1}{n} \rightarrow 0$.
Therefore,the limit is $\frac{1+0}{2} = \frac{1}{2}$.

(A) Given limit: $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$
Factorizing the numerator: $15^{x}-5^{x}-3^{x}+1 = 5^{x}(3^{x}-1) - 1(3^{x}-1) = (3^{x}-1)(5^{x}-1)$
Using the identity $1-\cos 2x = 2\sin^{2}x$:
$\lim _{x \rightarrow 0} \frac{(3^{x}-1)(5^{x}-1)}{2\sin^{2}x}$
Dividing numerator and denominator by $x^{2}$:
$\lim _{x \rightarrow 0} \frac{(\frac{3^{x}-1}{x})(\frac{5^{x}-1}{x})}{2(\frac{\sin x}{x})^{2}}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} = \log a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$= \frac{(\log 3)(\log 5)}{2(1)^{2}} = \frac{(\log 3)(\log 5)}{2}$

(B) Given expression: $(x \cdot y)+[(x+y') \cdot y]'$
Using De Morgan's Law $(a \cdot b)' = a' + b'$:
$= (x \cdot y) + [(x+y')' + y']$
Using De Morgan's Law $(a+b)' = a' \cdot b'$ and involution law $(y')' = y$:
$= (x \cdot y) + [x' \cdot (y')' + y']$
$= (x \cdot y) + [x' \cdot y + y']$
$= x \cdot y + x' \cdot y + y'$
Factor out $y$ from the first two terms:
$= y \cdot (x + x') + y'$
Since $x + x' = 1$:
$= y \cdot 1 + y'$
$= y + y'$
Since $y + y' = 1$:
$= 1$

(A) To find the dual of a Boolean expression,we replace the $OR$ operator $(+)$ with the $AND$ operator $(\cdot)$ and vice versa,and replace $0$ with $1$ and $1$ with $0$.
Given expression: $(x+y) \cdot (x^{\prime} \cdot 1)$.
Replacing $+$ with $\cdot$ and $\cdot$ with $+$,and replacing $1$ with $0$ (though $0$ is not present,we focus on the operators):
The dual is $(x \cdot y) + (x^{\prime} + 0)$.
Since $x^{\prime} + 0 = x^{\prime}$,the expression simplifies to $(x \cdot y) + x^{\prime}$.
However,looking at the provided options,the transformation of the operator $1$ is often treated as a constant in dual logic. Replacing the operators in $(x+y) \cdot (x^{\prime} \cdot 1)$ gives $(x \cdot y) + (x^{\prime} + 1)$.

(B) $(p \wedge q) \vee \sim p = \sim p \vee (p \wedge q)$ (By commutative law)
$= (\sim p \vee p) \wedge (\sim p \vee q)$ (By distributive law)
$= (p \vee \sim p) \wedge (\sim p \vee q)$ (By commutative law)
$= t \wedge (\sim p \vee q)$ (By complement law,where $t$ is a tautology)
$= \sim p \vee q$ (By identity law)

(B) Given truth values are $p = T, q = F, r = F$.
First,evaluate the components:
$\sim q = \sim F = T$
$\sim p = \sim T = F$
Now,evaluate the expressions in the brackets:
$p \wedge \sim q = T \wedge T = T$
$\sim p \vee r = F \vee F = F$
Finally,evaluate the implication:
$(p \wedge \sim q)$ $\rightarrow (\sim p \vee r) = T$ $\rightarrow F = F$.
Thus,the truth value is $F$.

(B) The circuit consists of two switches $p$ and $\sim p$ connected in parallel,which are then connected in series with a switch $q$.
The logical expression for the parallel combination of $p$ and $\sim p$ is $(p \lor \sim p)$.
Since $(p \lor \sim p) = T$ (a tautology,represented as $1$ in switching circuits),the expression becomes $1 \land q$.
Thus,the output is $1 \cdot q = q$.

(C) The equations of the lines passing through the origin with slopes $m_{1}=3$ and $m_{2}=-\frac{1}{3}$ are $y=3x$ and $y=-\frac{1}{3}x$.
These can be written as $(y-3x)=0$ and $(3y+x)=0$.
The combined equation of the pair of lines is given by the product of the individual equations:
$(y-3x)(3y+x)=0$
$3y^{2}+xy-9xy-3x^{2}=0$
$3y^{2}-8xy-3x^{2}=0$.

(A) The given equation is $x^{2}-xy-6y^{2}-7x+31y-18=0$.
Comparing this with the general equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$b=-6$,and $2h=-1$,which implies $h=-\frac{1}{2}$.
The angle $\theta$ between the pair of lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{(-\frac{1}{2})^{2} - (1)(-6)}}{1+(-6)} \right|$.
$\tan \theta = \left| \frac{2\sqrt{\frac{1}{4}+6}}{-5} \right| = \left| \frac{2\sqrt{\frac{25}{4}}}{-5} \right|$.
$\tan \theta = \left| \frac{2 \times \frac{5}{2}}{-5} \right| = \left| \frac{5}{-5} \right| = |-1| = 1$.
Therefore,$\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(C) The standard equation of a parabola with vertex at $(0,0)$ opening along the positive $x$-axis is $y^{2} = 4ax$.
The length of the latus rectum is given by $4a = \frac{16}{3}$.
Substituting $4a = \frac{16}{3}$ into the standard equation $y^{2} = 4ax$,we get $y^{2} = \frac{16}{3}x$.
Multiplying both sides by $3$,we obtain $3y^{2} = 16x$.

(D) The given parabola is $y^{2}=8x$. Comparing with $y^{2}=4ax$,we get $4a=8$,so $a=2$.
Any tangent to the parabola is given by $y=mx+\frac{a}{m}$,which can be written as $mx-y+\frac{2}{m}=0$.
For this line to be a tangent to the circle $x^{2}+y^{2}=2$ (with center $(0,0)$ and radius $r=\sqrt{2}$),the perpendicular distance from the center to the line must equal the radius.
Therefore,$\frac{|\frac{2}{m}|}{\sqrt{m^{2}+1}}=\sqrt{2}$.
Squaring both sides,$\frac{4}{m^{2}}=2(m^{2}+1)$ $\Rightarrow 2=m^{2}(m^{2}+1)$ $\Rightarrow m^{4}+m^{2}-2=0$.
Let $t=m^{2}$,then $t^{2}+t-2=0 \Rightarrow (t+2)(t-1)=0$. Since $m^{2} \geq 0$,we have $m^{2}=1$,so $m=\pm 1$.
If $m=1$,the tangent is $x-y+2=0 \Rightarrow y=x+2$.
If $m=-1$,the tangent is $-x-y-2=0 \Rightarrow x+y=-2$.
Comparing $x+y=-2$ with the given form $x+y=k$,we get $k=-2$.

(C) The total number of ways to form a four-digit number using $7$ distinct digits is given by $P(7, 4) = 7 \times 6 \times 5 \times 4 = 840$.
For the number to be $> 4000$,the first digit (thousands place) must be $4, 5, 6,$ or $7$.
There are $4$ choices for the first digit.
After choosing the first digit,the remaining $3$ positions can be filled by the remaining $6$ digits in $P(6, 3) = 6 \times 5 \times 4 = 120$ ways.
Thus,the number of favourable cases is $4 \times 120 = 480$.
Therefore,the required probability is $\frac{480}{840} = \frac{4}{7}$.

(C) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
The possible sums range from $2$ to $12$.
The prime numbers in this range are $2, 3, 5, 7, 11$.
Favourable outcomes for each sum:
Sum $= 2$: $(1, 1)$ - $1$ case
Sum $= 3$: $(1, 2), (2, 1)$ - $2$ cases
Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ - $4$ cases
Sum $= 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ - $6$ cases
Sum $= 11$: $(5, 6), (6, 5)$ - $2$ cases
Total favourable cases $= 1 + 2 + 4 + 6 + 2 = 15$.
Required probability $= \frac{15}{36} = \frac{5}{12}$.

(D) Given the curve equation: $y^{2} = ax^{2} + b$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 2ax$
$\frac{dy}{dx} = \frac{ax}{y}$.
The slope of the tangent at the point $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{a(2)}{3} = \frac{2a}{3}$.
The given equation of the tangent is $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$.
Equating the slopes: $\frac{2a}{3} = 4 \Rightarrow 2a = 12 \Rightarrow a = 6$.
Since the point $(2, 3)$ lies on the curve,it must satisfy the equation $y^{2} = ax^{2} + b$:
$(3)^{2} = 6(2)^{2} + b$
$9 = 6(4) + b$
$9 = 24 + b$
$b = 9 - 24 = -15$.
Thus,the values are $a = 6$ and $b = -15$.

(A) The volume $V$ of a solid generated by rotating a curve $y = f(x)$ about the $x$-axis between $x = a$ and $x = b$ is given by the formula: $V = \int_{a}^{b} \pi y^{2} dx$.
Given the curve $y^{2} = 4x$ and the limits $x = 4$ to $x = 5$.
Substituting the values into the formula:
$V = \int_{4}^{5} \pi (4x) dx$
$V = 4\pi \int_{4}^{5} x dx$
$V = 4\pi \left[ \frac{x^{2}}{2} \right]_{4}^{5}$
$V = 2\pi [x^{2}]_{4}^{5}$
$V = 2\pi (5^{2} - 4^{2})$
$V = 2\pi (25 - 16)$
$V = 2\pi (9) = 18\pi$ cubic units.

(D) For the function $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
At $x = -\frac{\pi}{2}$:
$\lim_{x \rightarrow -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.
$\lim_{x \rightarrow -\frac{\pi}{2}^+} (a \sin x + b) = a \sin(-\frac{\pi}{2}) + b = -a + b$.
Equating these,we get $-a + b = 2$ (Equation $1$).
At $x = \frac{\pi}{2}$:
$\lim_{x \rightarrow \frac{\pi}{2}^-} (a \sin x + b) = a \sin(\frac{\pi}{2}) + b = a + b$.
$\lim_{x \rightarrow \frac{\pi}{2}^+} \cos x = \cos(\frac{\pi}{2}) = 0$.
Equating these,we get $a + b = 0$ (Equation $2$).
Adding Equation $1$ and Equation $2$:
$(-a + b) + (a + b) = 2 + 0 \Rightarrow 2b = 2 \Rightarrow b = 1$.
Substituting $b = 1$ into Equation $2$:
$a + 1 = 0 \Rightarrow a = -1$.
Thus,the values are $a = -1$ and $b = 1$.

(C) Let $I = \int_{4}^{7} \frac{(11-x)^{2}}{x^{2}+(11-x)^{2}} d x$ $(i)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we get:
$I = \int_{4}^{7} \frac{(11-(11-x))^{2}}{(11-x)^{2}+(11-(11-x))^{2}} d x$
$I = \int_{4}^{7} \frac{x^{2}}{(11-x)^{2}+x^{2}} d x$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{4}^{7} \frac{(11-x)^{2} + x^{2}}{x^{2}+(11-x)^{2}} d x$
$2I = \int_{4}^{7} 1 d x$
$2I = [x]_{4}^{7} = 7 - 4 = 3$
$I = \frac{3}{2}$

(A) Let $I = \int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$.
Since $\operatorname{cosec} x = \frac{1}{\sin x}$,we have $\log (\operatorname{cosec} x) = \log (\sin x)^{-1} = -\log \sin x$.
Therefore,$I = -\int_{0}^{\pi / 2} \log \sin x d x$.
Using the standard definite integral result $\int_{0}^{\pi / 2} \log \sin x d x = -\frac{\pi}{2} \log 2$,we get:
$I = -(-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2$.

(B) Let $I = \int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$.
Using the trigonometric identity $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$,we can write $\tan ^{-1}\left(\frac{1-x}{1+x}\right) = \tan ^{-1}(1) - \tan ^{-1}(x)$.
Since $\tan ^{-1}(1) = \frac{\pi}{4}$,the integral becomes:
$I = \int_{0}^{1} \left( \frac{\pi}{4} - \tan ^{-1}(x) \right) \, dx$.
$I = \int_{0}^{1} \frac{\pi}{4} \, dx - \int_{0}^{1} \tan ^{-1}(x) \, dx$.
$I = \left[ \frac{\pi}{4} x \right]_{0}^{1} - p$.
$I = \frac{\pi}{4}(1 - 0) - p$.
$I = \frac{\pi}{4} - p$.

(D) The given equation is $a e^{x} + b e^{2x} + c e^{3x} + d = 0$.
This equation contains $4$ arbitrary constants,namely $a, b, c,$ and $d$.
By definition,the order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $4$ arbitrary constants,the order of the differential equation is $4$.

(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 = a^2$,or $x^2 + y^2 - 2ax = 0$.
Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $a = x + y \frac{dy}{dx}$.
Substituting the value of $a$ back into the original equation: $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.
$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \Rightarrow y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.
However,for a general family of circles with center $(a, 0)$ and radius $r$,the equation is $(x-a)^2 + y^2 = r^2$. Differentiating twice eliminates two constants $a$ and $r$. The standard form for circles with center on the $x$-axis is $y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

(B) Given differential equation is $y(1+\log x) \frac{dx}{dy} = x \log x$.
Rearranging the terms to separate the variables,we get:
$\frac{1+\log x}{x \log x} dx = \frac{1}{y} dy$.
Integrating both sides:
$\int \frac{1+\log x}{x \log x} dx = \int \frac{1}{y} dy$.
Split the integral on the left side:
$\int \frac{1}{x \log x} dx + \int \frac{1}{x} dx = \int \frac{1}{y} dy$.
Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes:
$\int \frac{1}{u} du + \int \frac{1}{x} dx = \int \frac{1}{y} dy$.
$\log |u| + \log |x| = \log |y| + \log |c|$.
Substituting $u = \log x$ back:
$\log |\log x| + \log |x| = \log |y| + \log |c|$.
Using the property $\log a + \log b = \log(ab)$:
$\log |x \log x| = \log |yc|$.
Taking the exponential of both sides:
$x \log x = yc$.

(D) Given,$x^{2} y^{5}=(x+y)^{7}$.
Taking the natural logarithm on both sides,we get:
$2 \ln x + 5 \ln y = 7 \ln (x+y)$.
Differentiating both sides with respect to $x$:
$\frac{2}{x} + \frac{5}{y} \frac{dy}{dx} = \frac{7}{x+y} \left(1 + \frac{dy}{dx}\right)$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} \left(\frac{5}{y} - \frac{7}{x+y}\right) = \frac{7}{x+y} - \frac{2}{x}$.
$\frac{dy}{dx} \left(\frac{5x + 5y - 7y}{y(x+y)}\right) = \frac{7x - 2x - 2y}{x(x+y)}$.
$\frac{dy}{dx} \left(\frac{5x - 2y}{y(x+y)}\right) = \frac{5x - 2y}{x(x+y)}$.
Thus,$\frac{dy}{dx} = \frac{y}{x}$.
Now,differentiating again with respect to $x$ using the quotient rule:
$\frac{d^{2}y}{dx^{2}} = \frac{x \frac{dy}{dx} - y(1)}{x^{2}}$.
Substituting $\frac{dy}{dx} = \frac{y}{x}$:
$\frac{d^{2}y}{dx^{2}} = \frac{x(y/x) - y}{x^{2}} = \frac{y - y}{x^{2}} = 0$.

(C) Given,$x = \sec \theta$ and $y = \tan \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta$
$\frac{dy}{d\theta} = \sec^2 \theta$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sec^2 \theta}{\sec \theta \tan \theta} = \frac{\sec \theta}{\tan \theta} = \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{1}{\sin \theta} = \csc \theta$.
Next,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\csc \theta) = \frac{d}{d\theta}(\csc \theta) \cdot \frac{d\theta}{dx}$
$= (-\csc \theta \cot \theta) \cdot \frac{1}{\sec \theta \tan \theta}$
$= -\frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{1} \cdot \frac{\cos \theta}{\sin \theta} = -\frac{\cos^3 \theta}{\sin^3 \theta} = -\cot^3 \theta$.
At $\theta = \frac{\pi}{4}$:
$\frac{d^2y}{dx^2} = -\cot^3(\frac{\pi}{4}) = -(1)^3 = -1$.

(C) Let $u = (\log x)^{x}$.
Taking logarithm on both sides,$\log u = x \log(\log x)$.
Differentiating with respect to $x$,we get $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1$.
So,$\frac{du}{dx} = (\log x)^{x} \left[ \frac{1}{\log x} + \log(\log x) \right]$.
Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^{x} [ \frac{1}{\log x} + \log(\log x) ]}{1/x}$.
Therefore,$\frac{du}{dv} = x(\log x)^{x} \left[ \frac{1}{\log x} + \log(\log x) \right]$.

(D) Given,$x=f(t)$ and $y=g(t)$.
First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{dt} = f'(t)$ and $\frac{dy}{dt} = g'(t)$.
Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$.
Now,we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g'(t)}{f'(t)} \right) = \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) \cdot \frac{dt}{dx}$.
Using the quotient rule for differentiation:
$\frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2}$.
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{f'(t)}$,we have:
$\frac{d^2y}{dx^2} = \left[ \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2} \right] \cdot \frac{1}{f'(t)} = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^3}$.

(C) We know that the shift operator $E$ is defined as $E = 1 + \Delta$,where $\Delta$ is the forward difference operator.
Thus,$(1+\Delta)^{n} f(a) = E^{n} f(a)$.
By the definition of the shift operator,$E^{n} f(a) = f(a+nh)$.
Therefore,$(1+\Delta)^{n} f(a) = f(a+nh)$.

(A) Using the definition of the forward difference operator $\Delta f(x) = f(x+1) - f(x)$,we have $f(4) - f(3) = \Delta f(3)$.
Since $\Delta f(x) = f(x+1) - f(x)$,we can write $f(3) = f(2) + \Delta f(2)$.
Thus,$\Delta f(3) = \Delta [f(2) + \Delta f(2)] = \Delta f(2) + \Delta^2 f(2)$.
Further,since $\Delta^2 f(x) = \Delta^2 f(x-1) + \Delta^3 f(x-1)$,we expand $\Delta^2 f(2)$ as $\Delta^2 [f(1) + \Delta f(1)] = \Delta^2 f(1) + \Delta^3 f(1)$.
Substituting this back,we get $f(4) - f(3) = \Delta f(2) + \Delta^2 f(1) + \Delta^3 f(1)$.

(A) Given that $g$ is the inverse of $f$,so $g(x) = f^{-1}(x)$.
This implies $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
Therefore,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.
Given $f^{\prime}(x) = \frac{1}{1+x^{2}}$,we substitute $g(x)$ for $x$:
$f^{\prime}(g(x)) = \frac{1}{1+[g(x)]^{2}}$.
Substituting this into the expression for $g^{\prime}(x)$:
$g^{\prime}(x) = \frac{1}{\frac{1}{1+[g(x)]^{2}}} = 1 + [g(x)]^{2}$.

(C) We know that $\int \frac{1}{x^{2}+a^{2}} d x = \frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c$.
Given integral is $I = \int \frac{1}{16 x^{2}+9} d x$.
Taking $16$ as a common factor from the denominator:
$I = \frac{1}{16} \int \frac{1}{x^{2}+\frac{9}{16}} d x = \frac{1}{16} \int \frac{1}{x^{2}+\left(\frac{3}{4}\right)^{2}} d x$.
Using the formula with $a = \frac{3}{4}$:
$I = \frac{1}{16} \times \frac{1}{3/4} \tan ^{-1}\left(\frac{x}{3/4}\right)+c$.
$I = \frac{1}{16} \times \frac{4}{3} \tan ^{-1}\left(\frac{4 x}{3}\right)+c$.
$I = \frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c$.

(B) Given that $f(x) = x$ and $g(x) = \sin x$.
First,we find the composite function $f(g(x))$:
$f(g(x)) = f(\sin x) = \sin x$.
Now,we evaluate the integral:
$\int f(g(x)) \, dx = \int \sin x \, dx$.
Using the standard integral formula $\int \sin x \, dx = -\cos x + c$,we get:
$\int f(g(x)) \, dx = -\cos x + c$.

(B) Let $I = \int e^{\tan x}(\sec ^{2} x + \sec ^{3} x \sin x) d x$.
Since $\sec ^{3} x \sin x = \sec ^{2} x \cdot \sec x \sin x = \sec ^{2} x \tan x$,the integral becomes:
$I = \int e^{\tan x}(\sec ^{2} x + \sec ^{2} x \tan x) d x$.
$I = \int e^{\tan x}(1 + \tan x) \sec ^{2} x d x$.
Let $t = \tan x$,then $dt = \sec ^{2} x d x$.
Substituting these into the integral:
$I = \int e^{t}(1 + t) d t$.
$I = \int (e^{t} + t e^{t}) d t$.
Using the integration by parts formula $\int (f(t) + f'(t)) e^{t} d t = f(t) e^{t} + c$,where $f(t) = t$ and $f'(t) = 1$:
$I = t e^{t} + c$.
Substituting back $t = \tan x$:
$I = \tan x e^{\tan x} + c$.

(B) The feasible region is determined by the constraints $x + y \leq 7$,$2x + 3y \leq 16$,$x \geq 0$,and $y \geq 0$. The vertices of the feasible region are $O(0, 0)$,$A(0, 16/3)$,$B(5, 2)$,and $C(7, 0)$.
We evaluate the objective function $Z = 3x + 2y$ at each vertex:
At $O(0, 0): Z = 3(0) + 2(0) = 0$
At $A(0, 16/3): Z = 3(0) + 2(16/3) = 32/3 \approx 10.67$
At $B(5, 2): Z = 3(5) + 2(2) = 15 + 4 = 19$
At $C(7, 0): Z = 3(7) + 2(0) = 21$
Comparing these values,the maximum value of $Z$ is $21$ at point $C(7, 0)$.

(B) Given,$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
The determinant of matrix $A$ is denoted by $|A|$ or $\det(A)$.
$|A| = (a \times d) - (b \times c) = ad - bc$.
The expression $|A|^{-1}$ represents the multiplicative inverse of the determinant value $|A|$.
Therefore,$|A|^{-1} = \frac{1}{|A|} = \frac{1}{ad - bc}$.

(A) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (\cos \theta)(-\cos \theta) - (-\sin \theta)(-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -(\cos^2 \theta + \sin^2 \theta) = -1$.
Next,we find the adjoint of $A$,$\text{adj}(A)$,by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{-1} \begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$.
Thus,$A^{-1} = A$.

(C) When two coins are tossed simultaneously,the sample space is $S = \{HH, HT, TH, TT\}$.
Let $X$ be the random variable representing the number of heads.
The possible values for $X$ are $0, 1, 2$.
The probabilities are:
$P(X=0) = P(TT) = \frac{1}{4}$
$P(X=1) = P(HT, TH) = \frac{2}{4} = \frac{1}{2}$
$P(X=2) = P(HH) = \frac{1}{4}$
The expected value $E(X)$ is given by $\sum x_i P(x_i) = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{1}{2} = 1$.
Alternatively,since this follows a binomial distribution with $n=2$ and $p=\frac{1}{2}$,$E(X) = np = 2 \times \frac{1}{2} = 1$.

(A) The direction ratios of the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ are given by $(2-3, -1-4, 5-(-1)) = (-1, -5, 6)$.
Since the plane is normal to this line,the normal vector to the plane is $\vec{n} = (-1, -5, 6)$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the point $(2, -3, 1)$ and the normal vector $(-1, -5, 6)$,we get:
$-1(x-2) - 5(y+3) + 6(z-1) = 0$
$-x + 2 - 5y - 15 + 6z - 6 = 0$
$-x - 5y + 6z - 19 = 0$
Multiplying by $-1$,we get $x + 5y - 6z + 19 = 0$.

(A) Let the point on the line be $(x, y, z)$. The equation of the line is $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4} = k$.
From this,we get $x = 2k+1$,$y = -3k+2$,and $z = 4k-3$.
Since this point lies on the plane $2x+4y-z=1$,we substitute these coordinates into the plane equation:
$2(2k+1) + 4(-3k+2) - (4k-3) = 1$
$4k + 2 - 12k + 8 - 4k + 3 = 1$
$-12k + 13 = 1$
$-12k = -12$
$k = 1$.
Substituting $k=1$ back into the coordinate expressions:
$x = 2(1)+1 = 3$
$y = -3(1)+2 = -1$
$z = 4(1)-3 = 1$.
Thus,the required point is $(3, -1, 1)$.

(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4\hat{i}-2\hat{j}$,$\vec{b} = \hat{i}+4\hat{j}-3\hat{k}$,and $\vec{c} = -\hat{i}+5\hat{j}+\hat{k}$.
We need to find $\angle ABC$,which is the angle between vectors $\vec{BA}$ and $\vec{BC}$.
First,calculate $\vec{BA} = \vec{a} - \vec{b} = (4-1)\hat{i} + (-2-4)\hat{j} + (0-(-3))\hat{k} = 3\hat{i} - 6\hat{j} + 3\hat{k}$.
Next,calculate $\vec{BC} = \vec{c} - \vec{b} = (-1-1)\hat{i} + (5-4)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} + \hat{j} + 4\hat{k}$.
Now,find the dot product $\vec{BA} \cdot \vec{BC} = (3)(-2) + (-6)(1) + (3)(4) = -6 - 6 + 12 = 0$.
Since the dot product of $\vec{BA}$ and $\vec{BC}$ is $0$,the vectors are perpendicular to each other.
Therefore,$\angle ABC = \frac{\pi}{2}$.

(C) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = (\vec{u} \times \vec{v}) \cdot \vec{w}$.
Given edges are $2\vec{a}, 2\vec{b}, 2\vec{c}$.
Volume $= [2\vec{a} \ 2\vec{b} \ 2\vec{c}]$
$= (2\vec{a} \times 2\vec{b}) \cdot 2\vec{c}$
$= 4(\vec{a} \times \vec{b}) \cdot 2\vec{c}$
$= 8(\vec{a} \times \vec{b}) \cdot \vec{c}$
$= 8[\vec{a} \vec{b} \vec{c}]$.

(D) Given that $\overrightarrow{p}=\frac{\overrightarrow{b} \times \overrightarrow{c}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}, \overrightarrow{q}=\frac{\overrightarrow{c} \times \overrightarrow{a}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}, \overrightarrow{r}=\frac{\overrightarrow{a} \times \overrightarrow{b}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}$.
We need to calculate $\overrightarrow{a} \cdot \overrightarrow{p}+\overrightarrow{b} \cdot \overrightarrow{q}+\overrightarrow{c} \cdot \overrightarrow{r}$.
Substituting the values,we get:
$\overrightarrow{a} \cdot \overrightarrow{p} = \overrightarrow{a} \cdot \frac{\overrightarrow{b} \times \overrightarrow{c}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
$\overrightarrow{b} \cdot \overrightarrow{q} = \overrightarrow{b} \cdot \frac{\overrightarrow{c} \times \overrightarrow{a}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
$\overrightarrow{c} \cdot \overrightarrow{r} = \overrightarrow{c} \cdot \frac{\overrightarrow{a} \times \overrightarrow{b}}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{c} \overrightarrow{a} \overrightarrow{b}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = \frac{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]}{[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]} = 1$.
Therefore,$\overrightarrow{a} \cdot \overrightarrow{p}+\overrightarrow{b} \cdot \overrightarrow{q}+\overrightarrow{c} \cdot \overrightarrow{r} = 1+1+1 = 3$.