MHT CET 2007 Mathematics Question Paper with Answer and Solution

(B) Given the function $f(x) = x^{2} - 78.8$.
The derivative is $f^{\prime}(x) = 2x$.
The initial approximation is $x_{0} = 14$.
According to the Newton-Raphson formula,the first approximation $x_{1}$ is given by:
$x_{1} = x_{0} - \frac{f(x_{0})}{f^{\prime}(x_{0})}$
$x_{1} = 14 - \frac{(14)^{2} - 78.8}{2 \times 14}$
$x_{1} = 14 - \frac{196 - 78.8}{28}$
$x_{1} = 14 - \frac{117.2}{28}$
$x_{1} = 14 - 4.1857...$
$x_{1} \approx 9.814$.

(B) Let the equation of the circle be $x^{2} + y^{2} + 2gx + 2fy + k = 0$.
Since it passes through $(2,0)$,we have $4 + 4g + k = 0 \Rightarrow k = -4 - 4g$.
Since it passes through $(0,1)$,we have $1 + 2f + k = 0 \Rightarrow k = -1 - 2f$.
Since it passes through $(4,5)$,we have $16 + 25 + 8g + 10f + k = 0 \Rightarrow 41 + 8g + 10f + k = 0$.
Solving these equations,we get $g = -\frac{13}{6}$,$f = -\frac{17}{6}$,and $k = \frac{14}{3}$.
The equation of the circle is $x^{2} + y^{2} - \frac{13}{3}x - \frac{17}{3}y + \frac{14}{3} = 0$,which simplifies to $3(x^{2} + y^{2}) - 13x - 17y + 14 = 0$.
Since $(0, c)$ lies on this circle,we substitute $x=0$ and $y=c$:
$3(0^{2} + c^{2}) - 13(0) - 17(c) + 14 = 0$.
$3c^{2} - 17c + 14 = 0$.
$(3c - 14)(c - 1) = 0$.
Thus,$c = 1$ or $c = \frac{14}{3}$.
Since $(0,1)$ is already a given point,the other point is $(0, \frac{14}{3})$,so $c = \frac{14}{3}$.

(C) The radius $r$ of the circle is the distance between the center $(1, 2)$ and the point $(4, 6)$ on the circle.
Using the distance formula: $r = \sqrt{(4-1)^2 + (6-2)^2}$
$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.
The area of the circle is given by the formula $A = \pi r^2$.
$A = \pi (5)^2 = 25 \pi$ sq unit.

(B) The equation of any line passing through the point $P(\alpha, \beta)$ is given by $\frac{x-\alpha}{\cos \theta} = \frac{y-\beta}{\sin \theta} = k$,where $k$ represents the distance from $P$ to any point on the line.
Any point on this line is $(\alpha + k \cos \theta, \beta + k \sin \theta)$.
Since this point lies on the circle $x^{2} + y^{2} = r^{2}$,we have:
$(\alpha + k \cos \theta)^{2} + (\beta + k \sin \theta)^{2} = r^{2}$
Expanding this,we get:
$\alpha^{2} + 2k\alpha \cos \theta + k^{2} \cos^{2} \theta + \beta^{2} + 2k\beta \sin \theta + k^{2} \sin^{2} \theta = r^{2}$
$k^{2}(\cos^{2} \theta + \sin^{2} \theta) + 2k(\alpha \cos \theta + \beta \sin \theta) + (\alpha^{2} + \beta^{2} - r^{2}) = 0$
$k^{2} + 2k(\alpha \cos \theta + \beta \sin \theta) + (\alpha^{2} + \beta^{2} - r^{2}) = 0$
This is a quadratic equation in $k$. Let the roots be $k_{1}$ and $k_{2}$,which represent the lengths $PA$ and $PB$.
The product of the roots $PA \cdot PB = k_{1}k_{2}$ is given by the constant term of the quadratic equation:
$PA \cdot PB = \alpha^{2} + \beta^{2} - r^{2}$.

(A) The distance between the two foci is $2ae = 7 - (-1) = 8$.
Since $e = 1/2$,we have $2a(1/2) = 8$,which implies $a = 8$.
The center of the ellipse is the midpoint of the foci: $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.
Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 64(1 - 1/4) = 64(3/4) = 48$.
Thus,$b = \sqrt{48} = 4\sqrt{3}$.
The equation of the ellipse is $\frac{(x-3)^2}{8^2} + \frac{y^2}{(4\sqrt{3})^2} = 1$.
The parametric coordinates are given by $(x, y) = (h + a \cos \theta, k + b \sin \theta)$,where $(h, k)$ is the center.
Substituting the values,we get $(3 + 8 \cos \theta, 4\sqrt{3} \sin \theta)$.

(B) Let $y=mx+c$ be a common tangent to the hyperbola $9x^{2}-16y^{2}=144$ and the circle $x^{2}+y^{2}=9$.
Rewriting the hyperbola as $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$,we have $a^{2}=16$ and $b^{2}=9$.
The condition for $y=mx+c$ to be a tangent to the hyperbola is $c^{2}=a^{2}m^{2}-b^{2}$,so $c^{2}=16m^{2}-9$ $(i)$.
The condition for $y=mx+c$ to be a tangent to the circle $x^{2}+y^{2}=3^{2}$ is $c^{2}=r^{2}(1+m^{2})$,so $c^{2}=9(1+m^{2})$ (ii).
Equating $(i)$ and (ii): $16m^{2}-9=9+9m^{2}$.
$7m^{2}=18$ $\Rightarrow m^{2}=\frac{18}{7}$ $\Rightarrow m=\pm 3\sqrt{\frac{2}{7}}$.
Substituting $m^{2}=\frac{18}{7}$ into (ii): $c^{2}=9(1+\frac{18}{7})=9(\frac{25}{7})=\frac{225}{7}$.
Thus,$c=\pm \frac{15}{\sqrt{7}}$.
The common tangent is $y=3\sqrt{\frac{2}{7}}x+\frac{15}{\sqrt{7}}$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.
Using the Taylor series expansion for $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - O(\theta^6)$,we have:
$\sin x = x - \frac{x^3}{6} + O(x^5)$
$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24})}{x^4}$.
$L = \lim _{x \rightarrow 0} \frac{\frac{5x^4}{24} - \frac{x^4}{24}}{x^4} = \lim _{x \rightarrow 0} \frac{4x^4/24}{x^4} = \frac{4}{24} = \frac{1}{6}$.

(A) The given limit is of the form $1^{\infty}$.
We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.
$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} = e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2} - 1\right)}$
$= e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1 - (x^{2}-4 x+2)}{x^{2}-4 x+2}\right)}$
$= e^{\lim _{x \rightarrow \infty} x \left(\frac{2 x-1}{x^{2}-4 x+2}\right)}$
$= e^{\lim _{x \rightarrow \infty} \frac{2 x^{2}-x}{x^{2}-4 x+2}}$
Dividing numerator and denominator by $x^{2}$,we get:
$= e^{\lim _{x \rightarrow \infty} \frac{2 - 1/x}{1 - 4/x + 2/x^{2}}} = e^{2/1} = e^{2}$.

(B) To find the dual of a logical statement,we replace $\vee$ with $\wedge$,$\wedge$ with $\vee$,$T$ with $F$,and $F$ with $T$.
Given statement: $[p \vee (\sim q)] \wedge (\sim p)$.
Replacing $\vee$ with $\wedge$ and $\wedge$ with $\vee$,we get:
$[p \wedge (\sim q)] \vee (\sim p)$.

(A) quadratic equation $ax^2 + bx + c = 0$ can have imaginary roots if the discriminant $D = b^2 - 4ac < 0$.
Therefore,the statement '$A$ quadratic equation always has a real root' is false,meaning its truth value is '$F$'.
Option $B$ is true as it represents the permutation formula.
Option $C$ is true because the cube roots of unity are $1, \omega, \omega^2$,which form a $GP$ with common ratio $\omega$.

(C) The equation of the bisectors of the angles between the lines $x^{2}-2 p x y-y^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
Here,$a=1, b=-1, h=-p$.
So,$\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$.
$\Rightarrow \frac{x^{2}-y^{2}}{2} = \frac{-xy}{p}$.
$\Rightarrow px^{2} + 2xy - py^{2} = 0$.
Since this equation represents the same pair of lines as $x^{2}-2qxy-y^{2}=0$,we compare the coefficients:
$\frac{p}{1} = \frac{2}{-2q} = \frac{-p}{-1}$.
From $\frac{p}{1} = \frac{2}{-2q}$,we get $p = -\frac{1}{q}$,which implies $pq = -1$ or $pq+1=0$.

(B) The total number of ways to place $3$ letters in $3$ envelopes is $3! = 6$.
There is only $1$ way to place all letters in their correct envelopes.
The probability that all letters are placed in the correct envelopes is $\frac{1}{3!} = \frac{1}{6}$.
The probability that all letters are not placed in the right envelopes is $1 - \frac{1}{6} = \frac{5}{6}$.

(D) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let the outcome of the first die be $x$ and the second die be $y$. We want the probability that $x < y$.
The possible pairs $(x, y)$ such that $x < y$ are:
For $x=1$,$y \in \{2, 3, 4, 5, 6\}$ ($5$ outcomes).
For $x=2$,$y \in \{3, 4, 5, 6\}$ ($4$ outcomes).
For $x=3$,$y \in \{4, 5, 6\}$ ($3$ outcomes).
For $x=4$,$y \in \{5, 6\}$ ($2$ outcomes).
For $x=5$,$y \in \{6\}$ ($1$ outcome).
For $x=6$,there are no possible values for $y$.
Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.

(B) Let $A$ and $B$ be two independent events.
The odds against event $A$ are $5:2$,so the probability of $A$ occurring is $P(A) = \frac{2}{5+2} = \frac{2}{7}$.
The probability of $A$ not occurring is $P(\bar{A}) = 1 - \frac{2}{7} = \frac{5}{7}$.
The odds in favour of event $B$ are $6:5$,so the probability of $B$ occurring is $P(B) = \frac{6}{6+5} = \frac{6}{11}$.
The probability of $B$ not occurring is $P(\bar{B}) = 1 - \frac{6}{11} = \frac{5}{11}$.
Since the events are independent,the probability that at least one event happens is $P(A \cup B) = 1 - P(\bar{A} \cap \bar{B}) = 1 - P(\bar{A})P(\bar{B})$.
$P(A \cup B) = 1 - (\frac{5}{7} \times \frac{5}{11}) = 1 - \frac{25}{77} = \frac{77-25}{77} = \frac{52}{77}$.

(C) The given equation $xy+2x+2y+4=0$ can be factored as $(x+2)(y+2)=0$,which represents the lines $x=-2$ and $y=-2$.
Given the third line is $x+y+2=0$.
To find the vertices of the triangle,we solve the equations in pairs:
$1$. Intersection of $x=-2$ and $y=-2$ gives vertex $C(-2,-2)$.
$2$. Intersection of $x=-2$ and $x+y+2=0$ gives $-2+y+2=0$,so $y=0$. Vertex $A(-2,0)$.
$3$. Intersection of $y=-2$ and $x+y+2=0$ gives $x-2+2=0$,so $x=0$. Vertex $B(0,-2)$.
The vertices are $A(-2,0)$,$B(0,-2)$,and $C(-2,-2)$.
Since the lines $x=-2$ and $y=-2$ are perpendicular,$\Delta ABC$ is a right-angled triangle with the right angle at $C(-2,-2)$.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.
Midpoint of $AB = (\frac{-2+0}{2}, \frac{0-2}{2}) = (-1,-1)$.

(D) Let the equation of the tangent to the parabola $y^2=4x$ be $y=mx+\frac{1}{m}$.
For the parabola $x^2=-32y$,the equation of the tangent with slope $m$ is $y=mx-a m^2$,where $x^2=4ay$ gives $a=-8$.
Thus,the tangent is $y=mx-(-8)m^2$,which simplifies to $y=mx+8m^2$.
Comparing the two tangent equations,we have $\frac{1}{m}=8m^2$.
This implies $m^3=\frac{1}{8}$,so $m=\frac{1}{2}$.
Substituting $m=\frac{1}{2}$ into the first tangent equation: $y=\frac{1}{2}x+\frac{1}{1/2} = \frac{1}{2}x+2$.
Multiplying by $2$,we get $2y=x+4$,or $x-2y+4=0$.

(B) For the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$,the foci are $(\pm ae, 0)$. Here $a^2=16$ and $b^2$ is the square of the semi-minor axis. The eccentricity $e_1 = \sqrt{1-\frac{b^2}{16}}$. The focus is $(\pm 4 \times \sqrt{1-\frac{b^2}{16}}, 0) = (\pm \sqrt{16-b^2}, 0)$.
For the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$,we rewrite it as $\frac{x^2}{(12/5)^2}-\frac{y^2}{(9/5)^2}=1$. Here $a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.
The eccentricity $e_2 = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{81/25}{144/25}} = \sqrt{1+\frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.
The foci are $(\pm ae_2, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
Since the foci coincide,$\sqrt{16-b^2} = 3$.
Squaring both sides,$16-b^2 = 9$,which gives $b^2 = 7$.

(C) Given the differential equation for the slope of the tangent:
$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right)$
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation:
$v + x\frac{dv}{dx} = v - \cos^2(v)$
$x\frac{dv}{dx} = -\cos^2(v)$
Separating the variables:
$\sec^2(v) dv = -\frac{1}{x} dx$
Integrating both sides:
$\int \sec^2(v) dv = -\int \frac{1}{x} dx$
$\tan(v) = -\log|x| + C$
Substituting $v = \frac{y}{x}$ back:
$\tan\left(\frac{y}{x}\right) = -\log|x| + C$
The curve passes through $\left(1, \frac{\pi}{4}\right)$,so:
$\tan\left(\frac{\pi/4}{1}\right) = -\log(1) + C$
$\tan\left(\frac{\pi}{4}\right) = 0 + C \Rightarrow C = 1$
Thus,$\tan\left(\frac{y}{x}\right) = 1 - \log(x) = \log(e) - \log(x) = \log\left(\frac{e}{x}\right)$
Therefore,$y = x \tan^{-1}\left[\log\left(\frac{e}{x}\right)\right]$.

(D) Let the rectangle be inscribed in a circle of radius $r$. Let the sides of the rectangle be $2x$ and $2y$. Since the rectangle is inscribed in a circle,the diagonal of the rectangle is the diameter of the circle,so $(2x)^2 + (2y)^2 = (2r)^2$,which simplifies to $x^2 + y^2 = r^2$,or $y = \sqrt{r^2 - x^2}$.
The area $A$ of the rectangle is $A = (2x)(2y) = 4x\sqrt{r^2 - x^2}$.
To find the maximum area,differentiate $A$ with respect to $x$:
$\frac{dA}{dx} = 4 \left( \sqrt{r^2 - x^2} + x \cdot \frac{1}{2\sqrt{r^2 - x^2}} \cdot (-2x) \right) = 4 \left( \frac{r^2 - x^2 - x^2}{\sqrt{r^2 - x^2}} \right) = \frac{4(r^2 - 2x^2)}{\sqrt{r^2 - x^2}}$.
Setting $\frac{dA}{dx} = 0$ gives $r^2 - 2x^2 = 0$,so $x = \frac{r}{\sqrt{2}}$.
Substituting $x = \frac{r}{\sqrt{2}}$ into the area formula:
$A = 4 \left( \frac{r}{\sqrt{2}} \right) \sqrt{r^2 - \frac{r^2}{2}} = 4 \left( \frac{r}{\sqrt{2}} \right) \left( \frac{r}{\sqrt{2}} \right) = 4 \cdot \frac{r^2}{2} = 2r^2$.
Thus,the maximum area is $2r^2$.

(A) function $f(x)$ is monotonically increasing if $f'(x) \geq 0$ for all $x \in R$.
Given $f(x) = kx - \sin x$.
Differentiating with respect to $x$,we get $f'(x) = k - \cos x$.
For $f(x)$ to be monotonically increasing,$f'(x) \geq 0$ for all $x \in R$.
$k - \cos x \geq 0 \implies k \geq \cos x$.
Since the maximum value of $\cos x$ is $1$,for $k \geq \cos x$ to hold for all $x$,$k$ must be greater than or equal to the maximum value of $\cos x$.
Therefore,$k \geq 1$.

(A) Given,$f(x) = \log(1+x) - \frac{2x}{2+x}$.
First,we find the derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4 + 2x - 2x}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$
$f'(x) = \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 \geq 0$ and $(2+x)^2 > 0$ for $x > -1$ (domain of $\log(1+x)$),the sign of $f'(x)$ depends on $(1+x)$.
Thus,$f'(x) > 0$ when $1+x > 0$,which means $x > -1$.
However,checking the options provided,$f'(x) > 0$ for all $x > 0$ is clearly satisfied.
Therefore,the function is increasing on $(0, \infty)$.

(B) Given,$f(x) = x^{25}(1-x)^{75}$.
To find the maximum value,we calculate the derivative $f'(x)$ using the product rule:
$f'(x) = 25x^{24}(1-x)^{75} - 75x^{25}(1-x)^{74}$.
Factor out the common terms:
$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$.
$f'(x) = 25x^{24}(1-x)^{74} (1-4x)$.
Setting $f'(x) = 0$ gives critical points $x = 0, 1, 1/4$.
For $x \in (0, 1/4)$,$f'(x) > 0$ (function is increasing).
For $x \in (1/4, 1)$,$f'(x) < 0$ (function is decreasing).
Since the derivative changes sign from positive to negative at $x = 1/4$,the function attains its maximum value at $x = 1/4$.

(C) Given,$f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$. The critical points are found by setting $f^{\prime}(x)=0$.
$f^{\prime}(x) = 6x^{2} - 18ax + 12a^{2} = 6(x^{2} - 3ax + 2a^{2}) = 6(x-a)(x-2a) = 0$.
So,the critical points are $x=a$ and $x=2a$.
We find the second derivative $f^{\prime \prime}(x) = 12x - 18a$.
For a maximum at $p$,$f^{\prime \prime}(p) < 0 \Rightarrow 12p - 18a < 0 \Rightarrow p < \frac{3}{2}a$.
For a minimum at $q$,$f^{\prime \prime}(q) > 0 \Rightarrow 12q - 18a > 0 \Rightarrow q > \frac{3}{2}a$.
Comparing the critical points $a$ and $2a$ with these conditions,we get $p=a$ and $q=2a$.
Given $p^{2}=q$,we substitute the values: $a^{2} = 2a$.
$a^{2} - 2a = 0 \Rightarrow a(a-2) = 0$.
Thus,$a=0$ or $a=2$.
If $a=0$,$f(x)=2x^{3}+1$,which is a strictly increasing function and has no local maxima or minima.
Therefore,$a=2$ is the only valid solution.

(A) Given,$f(x) = \begin{cases} e^{-x}, & x \geq 0 \\ e^{x}, & x < 0 \end{cases}$
$LHL = \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0} e^{x} = 1$
$RHL = \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0} e^{-x} = 1$
Also,$f(0) = e^{0} = 1$
Since $LHL = RHL = f(0)$,the function is continuous for every value of $x$.
Now,we check for differentiability at $x = 0$:
$LHD = \left(\frac{d}{dx} e^{x}\right)_{x = 0} = \left[e^{x}\right]_{x = 0} = 1$
$RHD = \left(\frac{d}{dx} e^{-x}\right)_{x = 0} = \left[-e^{-x}\right]_{x = 0} = -1$
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = 0$.
Therefore,$f(x) = e^{-|x|}$ is continuous everywhere but not differentiable at $x = 0$.

(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.
We know that $e^{x}$ is differentiable for all $x \in R$.
The function $g(x) = |x-1|$ is continuous everywhere but is not differentiable at $x = 1$ because the left-hand derivative and right-hand derivative at $x = 1$ are not equal.
Specifically,at $x = 1$,the left-hand derivative is $-e^{1} = -e$ and the right-hand derivative is $e^{1} = e$.
Since the product of a non-differentiable function and a non-zero differentiable function at a point is non-differentiable at that point,$f(x)$ is not differentiable at $x = 1$.
Therefore,the set of points where $f(x)$ is differentiable is $R - \{1\}$.

(B) Given,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$.
On differentiating with respect to $x$,we get:
$f^{\prime}(x) = \frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)$
$f^{\prime}(x) = \frac{1+x^{2}}{\sqrt{(1-x^{2})^{2}}} \times \frac{2(1-x^{2})}{(1+x^{2})^{2}}$
$f^{\prime}(x) = \frac{2}{1+x^{2}} \times \frac{1-x^{2}}{|1-x^{2}|}$
This simplifies to:
$f^{\prime}(x) = \begin{cases} \frac{2}{1+x^{2}}, & \text{if } |x| < 1 \\ -\frac{2}{1+x^{2}}, & \text{if } |x| > 1 \end{cases}$
At $x = 1$ and $x = -1$,the derivative does not exist because the left-hand derivative and right-hand derivative are not equal.
Therefore,$f(x)$ is differentiable on $R - \{-1, 1\}$.

(B) Let $I = \int_{0}^{\pi / 2} \frac{\cos 3x + 1}{2 \cos x - 1} dx$.
Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos 3x + 1 = 4 \cos^3 x - 3 \cos x + 1$.
Since $\cos(\pi/3) = 1/2$,we can write $2 \cos x - 1 = 2(\cos x - \cos(\pi/3))$.
Also,$4 \cos^3(\pi/3) - 3 \cos(\pi/3) + 1 = 4(1/8) - 3(1/2) + 1 = 1/2 - 3/2 + 1 = 0$.
Thus,the numerator is $4 \cos^3 x - 3 \cos x + 1 = 4 \cos^3 x - 3 \cos x - (4 \cos^3(\pi/3) - 3 \cos(\pi/3))$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we simplify the integrand:
$\frac{4(\cos^3 x - \cos^3(\pi/3)) - 3(\cos x - \cos(\pi/3))}{2(\cos x - \cos(\pi/3))} = \frac{4(\cos^2 x + \cos x \cos(\pi/3) + \cos^2(\pi/3)) - 3}{2} = \frac{4 \cos^2 x + 2 \cos x + 1 - 3}{2} = 2 \cos^2 x + \cos x - 1$.
Using $\cos^2 x = \frac{1 + \cos 2x}{2}$,we get $2(\frac{1 + \cos 2x}{2}) + \cos x - 1 = 1 + \cos 2x + \cos x - 1 = \cos 2x + \cos x$.
Now,$I = \int_{0}^{\pi / 2} (\cos 2x + \cos x) dx = [\frac{1}{2} \sin 2x + \sin x]_{0}^{\pi / 2} = (\frac{1}{2} \sin \pi + \sin(\pi/2)) - (0 + 0) = 0 + 1 = 1$.

(C) Given the integral $\int_{0}^{1} \frac{1}{1+x} dx$ with $n=2$ subintervals.
Here,the interval is $[0, 1]$,so $h = \frac{1-0}{2} = \frac{1}{2}$.
The points are $x_0 = 0$,$x_1 = 0.5$,and $x_2 = 1$.
The corresponding values of $y = f(x) = \frac{1}{1+x}$ are:
$y_0 = f(0) = \frac{1}{1+0} = 1$
$y_1 = f(0.5) = \frac{1}{1+0.5} = \frac{1}{1.5} = \frac{2}{3}$
$y_2 = f(1) = \frac{1}{1+1} = \frac{1}{2}$
Using Simpson's one-third rule: $\int_{a}^{b} f(x) dx \approx \frac{h}{3} [y_0 + 4y_1 + y_2]$.
Substituting the values: $\int_{0}^{1} \frac{1}{1+x} dx \approx \frac{0.5}{3} [1 + 4(\frac{2}{3}) + 0.5] = \frac{1}{6} [1 + \frac{8}{3} + 0.5] = \frac{1}{6} [1.5 + \frac{8}{3}] = \frac{1}{6} [\frac{3}{2} + \frac{8}{3}] = \frac{1}{6} [\frac{9+16}{6}] = \frac{25}{36}$.

(B) Let $I = \int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$.
We can rewrite the argument of the $\tan^{-1}$ function as:
$\frac{2x-1}{1+x-x^2} = \frac{x + (x-1)}{1 - x(x-1)}$.
Using the identity $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$I = \int_{0}^{1} (\tan^{-1}(x) + \tan^{-1}(x-1)) d x$.
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,let $f(x) = \tan^{-1}(x-1)$.
Then $\int_{0}^{1} \tan^{-1}(x-1) d x = \int_{0}^{1} \tan^{-1}((1-x)-1) d x = \int_{0}^{1} \tan^{-1}(-x) d x = -\int_{0}^{1} \tan^{-1}(x) d x$.
Substituting this back into the integral:
$I = \int_{0}^{1} \tan^{-1}(x) d x - \int_{0}^{1} \tan^{-1}(x) d x = 0$.

(A) The system of equations can be represented as $AX = B$,where $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 5 & 5 & 9 \end{bmatrix}$.
To determine the nature of the solution,we calculate the determinant $\Delta = |A|$.
$\Delta = 1(9-15) - 2(18-15) + 3(10-5)$
$\Delta = 1(-6) - 2(3) + 3(5)$
$\Delta = -6 - 6 + 15 = 3$.
Since $\Delta \neq 0$,the system of equations has a unique solution (only one solution).

(D) Given that $\overrightarrow{a}, \overrightarrow{b}$,and $\overrightarrow{c}$ are non-coplanar vectors,their scalar triple product is non-zero:
$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = \left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right| \neq 0$.
Let $\Delta = \left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$.
We are given the determinant equation:
$\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right| = 0$.
Using the property of determinants,we can split this into two determinants:
$\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right| + \left|\begin{array}{lll}a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3\end{array}\right| = 0$.
In the second determinant,take $a, b, c$ common from rows $1, 2, 3$ respectively:
$\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right| + abc \left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right| = 0$.
Note that $\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right| = \left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right| = \Delta$ (by swapping columns twice).
Thus,$\Delta + abc \Delta = 0 \Rightarrow \Delta(1 + abc) = 0$.
Since $\Delta \neq 0$,we must have $1 + abc = 0$,which implies $abc = -1$.

(A) The given solution is $y=a \cos x+b \sin x+c e^{-x}$.
Since the solution contains $3$ arbitrary constants $(a, b, c)$,the order of the corresponding differential equation is equal to the number of arbitrary constants.
Therefore,the order of the differential equation is $3$.

(D) The general equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^{2} = 4a(y-k)$,where $h, k,$ and $a$ are arbitrary constants.
Since there are $3$ arbitrary constants,we need to differentiate the equation $3$ times.
Let $(x-h)^{2} = A(y-k)$,where $A = 4a$.
Differentiating with respect to $x$: $2(x-h) = A y_{1}$.
Differentiating again with respect to $x$: $2 = A y_{2}$.
Differentiating a third time with respect to $x$: $0 = A y_{3}$.
Since $A = 4a \neq 0$,we must have $y_{3} = 0$.
None of the given options match $y_{3} = 0$.

(B) Given $x=\phi(t)$ and $y=\psi(t)$.
First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi^{\prime}(t)}{\phi^{\prime}(t)}$.
Now,we differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) = \frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \cdot \frac{dt}{dx}$.
Using the quotient rule for $\frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)$:
$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{2}} \cdot \frac{1}{\phi^{\prime}}$.
$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{3}}$.

(A) Given,$y = \log_{\cos x} \sin x = \frac{\log \sin x}{\log \cos x}$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \log \sin x$ and $v = \log \cos x$.
Then $u' = \frac{1}{\sin x} \cdot \cos x = \cot x$ and $v' = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.
$\frac{dy}{dx} = \frac{(\log \cos x)(\cot x) - (\log \sin x)(-\tan x)}{(\log \cos x)^2}$.
$\frac{dy}{dx} = \frac{\cot x \log \cos x + \tan x \log \sin x}{(\log \cos x)^2}$.

(C) We have,$y = \log |x| = \begin{cases} \log x, & x > 0 \\ \log (-x), & x < 0 \end{cases}$
For $x > 0$,$\frac{dy}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$.
For $x < 0$,$\frac{dy}{dx} = \frac{d}{dx}(\log (-x)) = \frac{1}{-x} \times \frac{d}{dx}(-x) = \frac{1}{-x} \times (-1) = \frac{1}{x}$.
Combining both cases,we get $\frac{dy}{dx} = \frac{1}{x}$ for all $x \neq 0$.

(A) Given,$y^{2}=a x^{2}+b x+c$.
On differentiating with respect to $x$,we get $2 y \frac{d y}{d x}=2 a x+b$,which implies $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$.
Again differentiating with respect to $x$,we get $2(\frac{d y}{d x})^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 2 a$.
Dividing by $2$,we have $(\frac{d y}{d x})^{2} + y \frac{d^{2} y}{d x^{2}} = a$.
Substituting $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$,we get $y \frac{d^{2} y}{d x^{2}} = a - (\frac{2 a x + b}{2 y})^{2}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - (2 a x + b)^{2}}{4 y^{2}}$.
Substituting $y^{2} = a x^{2} + b x + c$,we get $y \frac{d^{2} y}{d x^{2}} = \frac{4 a (a x^{2} + b x + c) - (4 a^{2} x^{2} + 4 a b x + b^{2})}{4 y^{2}}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a^{2} x^{2} + 4 a b x + 4 a c - 4 a^{2} x^{2} - 4 a b x - b^{2}}{4 y^{2}}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4 y^{2}}$.
Multiplying both sides by $y^{2}$,we get $y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4}$,which is a constant.

(A) Given $f(x) = \frac{ax + b}{cx + d}$.
We are given that $f \circ f(x) = x$.
This implies $f(f(x)) = x$.
Substituting $f(x)$ into the function:
$f\left(\frac{ax + b}{cx + d}\right) = x$
$\frac{a\left(\frac{ax + b}{cx + d}\right) + b}{c\left(\frac{ax + b}{cx + d}\right) + d} = x$
Multiplying numerator and denominator by $(cx + d)$:
$\frac{a(ax + b) + b(cx + d)}{c(ax + b) + d(cx + d)} = x$
$\frac{a^2x + ab + bcx + bd}{acx + bc + cdx + d^2} = x$
$\frac{(a^2 + bc)x + (ab + bd)}{(ac + cd)x + (bc + d^2)} = x$
$(a^2 + bc)x + (ab + bd) = x((ac + cd)x + (bc + d^2))$
$(a^2 + bc)x + b(a + d) = (ac + cd)x^2 + (bc + d^2)x$
For this to hold for all $x$,the coefficients must match. Comparing the constant term,we have $b(a + d) = 0$. If $b \neq 0$,then $a + d = 0$,which means $d = -a$. If $d = -a$,the expression becomes $\frac{ax + b}{cx - a}$,and $f(f(x)) = \frac{a(\frac{ax+b}{cx-a}) + b}{c(\frac{ax+b}{cx-a}) - a} = \frac{a^2x + ab + bcx - ab}{acx + bc - acx + a^2} = \frac{x(a^2 + bc)}{a^2 + bc} = x$. Thus,$d = -a$ is the required condition.

(B) Let $y = f(x) = x^{3} + 5$.
To find the inverse,we express $x$ in terms of $y$:
$y - 5 = x^{3}$
$x = (y - 5)^{1/3}$
Since $f^{-1}(y) = x$,we have $f^{-1}(y) = (y - 5)^{1/3}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = (x - 5)^{1/3}$.

(B) Let $I = \int \cos ^{3} x \cdot e^{\log (\sin x)} d x$.
Since $e^{\log (\sin x)} = \sin x$,the integral becomes $I = \int \cos ^{3} x \sin x d x$.
Let $u = \cos x$. Then $du = -\sin x d x$,which implies $\sin x d x = -du$.
Substituting these into the integral,we get $I = \int u^{3} (-du) = -\int u^{3} du$.
Integrating with respect to $u$,we get $I = -\frac{u^{4}}{4} + c$.
Substituting back $u = \cos x$,we obtain $I = -\frac{\cos ^{4} x}{4} + c$.

(A) Let $I = \int \frac{x + \sin x}{1 + \cos x} dx$.
Using the half-angle formulas,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} dx$
$I = \int \left( \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int \left( \frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$
Now,apply integration by parts to the first term $\int \frac{x}{2} \sec^2 \frac{x}{2} dx$ taking $u = \frac{x}{2}$ and $dv = \sec^2 \frac{x}{2} dx$:
$\int \frac{x}{2} \sec^2 \frac{x}{2} dx = \frac{x}{2} \cdot 2 \tan \frac{x}{2} - \int (1) \cdot \tan \frac{x}{2} dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx$.
Substituting this back into the expression for $I$:
$I = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx + \int \tan \frac{x}{2} dx + c$
$I = x \tan \frac{x}{2} + c$.

(B) The given constraints are $-x_{1} + x_{2} \leq 1$,$-x_{1} + 3x_{2} \leq 9$,and $x_{1}, x_{2} \geq 0$.
To determine the nature of the feasible region,we plot the lines $-x_{1} + x_{2} = 1$ and $-x_{1} + 3x_{2} = 9$.
For $-x_{1} + x_{2} = 1$,the intercepts are $(0, 1)$ and $(-1, 0)$.
For $-x_{1} + 3x_{2} = 9$,the intercepts are $(0, 3)$ and $(-9, 0)$.
Since the region is defined by $x_{1}, x_{2} \geq 0$ (first quadrant) and the inequalities allow the region to extend infinitely in the direction of increasing $x_{1}$ and $x_{2}$,the feasible region is unbounded.

(C) In a linear programming problem,we deal with a convex feasible region formed by the intersection of linear inequalities. The terms 'Optimal solution','Feasible solution',and 'Objective function' are standard components of linear programming. The term 'Concave region' is not used in this context.

(A) We know that the forward difference operator $\Delta$ is defined as $\Delta f(x) = f(x+h) - f(x)$ and the backward difference operator $\nabla$ is defined as $\nabla f(x) = f(x) - f(x-h)$.
Now,consider the expression $\Delta \nabla f(x)$:
$\Delta \nabla f(x) = \Delta [f(x) - f(x-h)]$
$= \Delta f(x) - \Delta f(x-h)$
$= [f(x+h) - f(x)] - [f(x) - f(x-h)]$
$= [f(x+h) - f(x)] - \nabla f(x)$
$= \Delta f(x) - \nabla f(x)$
Therefore,$\Delta \nabla = \Delta - \nabla$.

(C) Given matrices are $A = \begin{bmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{bmatrix}$.
Calculating the product $AB$:
$AB = \begin{bmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{bmatrix} \begin{bmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{bmatrix}$
$= \begin{bmatrix} \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta & \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta \\ \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta & \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta \end{bmatrix}$
$= \begin{bmatrix} \cos \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) & \cos \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \\ \sin \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) & \sin \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \end{bmatrix}$
$= \cos(\alpha - \beta) \begin{bmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{bmatrix}$.
For $AB$ to be a null matrix,we must have $\cos(\alpha - \beta) = 0$.
This implies $\alpha - \beta = (2n + 1) \frac{\pi}{2}$ for some integer $n$,which is an odd multiple of $\pi / 2$.

(A) For any square matrix $B$,we have $B(\operatorname{adj} B) = |B| I_{n}$.
Taking $B = \operatorname{adj} A$,we get $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |\operatorname{adj} A| I_{n}$.
Since $|\operatorname{adj} A| = |A|^{n-1}$,we have $(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} I_{n}$.
Multiplying both sides by $A$ on the left,we get $A(\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} A$.
Since $A(\operatorname{adj} A) = |A| I_{n}$,we have $(|A| I_{n})[\operatorname{adj}(\operatorname{adj} A)] = |A|^{n-1} A$.
Dividing both sides by $|A|$ (assuming $|A| \neq 0$),we get $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.

(A) Any vector $\overrightarrow{r}$ in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ can be written as $\overrightarrow{r} = m\overrightarrow{b} + n\overrightarrow{c}$. For simplicity,consider $\overrightarrow{r} = m\overrightarrow{b} + \overrightarrow{c}$ (assuming the vector is not parallel to $\overrightarrow{c}$).
$\overrightarrow{r} = m(\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} + \hat{j} - 2\hat{k}) = (m+1)\hat{i} + (2m+1)\hat{j} + (-m-2)\hat{k}$.
The projection of $\overrightarrow{r}$ on $\overrightarrow{a}$ is given by $\frac{|\overrightarrow{r} \cdot \overrightarrow{a}|}{|\overrightarrow{a}|} = \sqrt{\frac{2}{3}}$.
$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$,so $|\overrightarrow{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
$\overrightarrow{r} \cdot \overrightarrow{a} = 2(m+1) - 1(2m+1) + 1(-m-2) = 2m + 2 - 2m - 1 - m - 2 = -m - 1$.
Thus,$\frac{|-m-1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{2}{\sqrt{6}}$.
$|-m-1| = 2$,which implies $-m-1 = 2$ or $-m-1 = -2$.
If $-m-1 = 2$,then $m = -3$. Substituting $m = -3$ into $\overrightarrow{r}$: $\overrightarrow{r} = -2\hat{i} - 5\hat{j} + \hat{k}$.
If $-m-1 = -2$,then $m = 1$. Substituting $m = 1$ into $\overrightarrow{r}$: $\overrightarrow{r} = 2\hat{i} + 3\hat{j} - 3\hat{k}$.
Comparing with the options,$2\hat{i} + 3\hat{j} - 3\hat{k}$ is present.

(D) Since the three given vectors are coplanar,their scalar triple product must be zero. Alternatively,one vector can be expressed as a linear combination of the other two. Let $\overrightarrow{u} = \overrightarrow{a}+\lambda \overrightarrow{b}+3 \overrightarrow{c}$,$\overrightarrow{v} = -2 \overrightarrow{a}+3 \overrightarrow{b}-4 \overrightarrow{c}$,and $\overrightarrow{w} = \overrightarrow{a}-3 \overrightarrow{b}+5 \overrightarrow{c}$.
For coplanar vectors,the determinant of their coefficients must be zero:
$\begin{vmatrix} 1 & \lambda & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$1(3 \times 5 - (-4) \times (-3)) - \lambda((-2) \times 5 - (-4) \times 1) + 3((-2) \times (-3) - 3 \times 1) = 0$
$1(15 - 12) - \lambda(-10 + 4) + 3(6 - 3) = 0$
$1(3) - \lambda(-6) + 3(3) = 0$
$3 + 6\lambda + 9 = 0$
$6\lambda + 12 = 0$
$6\lambda = -12$
$\lambda = -2$

(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces:
$\overrightarrow{F} = (2 \hat{i}-5 \hat{j}+6 \hat{k}) + (-\hat{i}+2 \hat{j}-\hat{k}) = \hat{i}-3 \hat{j}+5 \hat{k}$
The displacement vector $\overrightarrow{d}$ is given by $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$:
$\overrightarrow{d} = (6 \hat{i}+\hat{j}-3 \hat{k}) - (4 \hat{i}-3 \hat{j}-2 \hat{k}) = 2 \hat{i}+4 \hat{j}-\hat{k}$
The work done $W$ is the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{d} = (\hat{i}-3 \hat{j}+5 \hat{k}) \cdot (2 \hat{i}+4 \hat{j}-\hat{k})$
$W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 \text{ unit}$

(D) Given,$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
$\therefore \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b})$
Squaring both sides,we get:
$|\overrightarrow{c}|^2 = (\overrightarrow{a}+\overrightarrow{b}) \cdot (\overrightarrow{a}+\overrightarrow{b})$
$|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b})$
Since $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}| \cos \theta$,where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$:
$|\overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta$
Substituting the given values:
$7^2 = 3^2 + 5^2 + 2(3)(5) \cos \theta$
$49 = 9 + 25 + 30 \cos \theta$
$49 = 34 + 30 \cos \theta$
$15 = 30 \cos \theta$
$\cos \theta = \frac{15}{30} = \frac{1}{2}$
Therefore,$\theta = \frac{\pi}{3}$.