KCET 2017 Mathematics Question Paper with Answer and Solution

(A) The given series is $3+5+7+\ldots$ up to $n$ terms.
This is an Arithmetic Progression $(A.P.)$ with the first term $a = 3$ and common difference $d = 5 - 3 = 2$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula:
$S_n = \frac{n}{2}[2a + (n-1)d]$
Substituting the values $a = 3$ and $d = 2$:
$S_n = \frac{n}{2}[2(3) + (n-1)2]$
$S_n = \frac{n}{2}[6 + 2n - 2]$
$S_n = \frac{n}{2}[2n + 4]$
$S_n = n(n + 2)$

(A) The given equation of the ellipse is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=36$ and $b^{2}=16$.
The eccentricity $e$ of an ellipse is given by the formula $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$.
Substituting the values,we get $e=\sqrt{1-\frac{16}{36}}$.
$e=\sqrt{\frac{36-16}{36}}=\sqrt{\frac{20}{36}}$.
$e=\frac{\sqrt{20}}{6}=\frac{2 \sqrt{5}}{6}$.

(B) We use the trigonometric identity: $\cos ^{2} A - \sin ^{2} B = \cos(A+B) \cdot \cos(A-B)$.
Here,$A = 45^{\circ}$ and $B = 15^{\circ}$.
Substituting these values into the identity:
$\cos ^{2} 45^{\circ} - \sin ^{2} 15^{\circ} = \cos(45^{\circ} + 15^{\circ}) \cdot \cos(45^{\circ} - 15^{\circ})$
$= \cos(60^{\circ}) \cdot \cos(30^{\circ})$
$= \frac{1}{2} \times \frac{\sqrt{3}}{2}$
$= \frac{\sqrt{3}}{4}$.

(A) Given that,$ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $.
Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,we can rewrite the expression as:
$ \lim _{\theta \rightarrow 0} \frac{2 \sin^2(2 \theta)}{2 \sin^2(3 \theta)} = \lim _{\theta \rightarrow 0} \frac{\sin^2(2 \theta)}{\sin^2(3 \theta)} $.
Multiplying and dividing by $ (2 \theta)^2 $ and $ (3 \theta)^2 $ respectively:
$ \lim _{\theta \rightarrow 0} \left( \frac{\sin(2 \theta)}{2 \theta} \right)^2 \times \left( \frac{3 \theta}{\sin(3 \theta)} \right)^2 \times \frac{(2 \theta)^2}{(3 \theta)^2} $.
Since $ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $,the expression simplifies to:
$ 1^2 \times 1^2 \times \frac{4 \theta^2}{9 \theta^2} = \frac{4}{9} $.

(D) The given statement is of the form "If $P$,then $Q$",where $P$ is "$x$ is a prime number" and $Q$ is "$x$ is odd".
The contrapositive of the statement "If $P$,then $Q$" is defined as "If $\neg Q$,then $\neg P$".
Here,$\neg Q$ is "$x$ is not odd" and $\neg P$ is "$x$ is not a prime number".
Therefore,the contrapositive statement is "If $x$ is not odd,then $x$ is not a prime number".

(A) $1$. The line passes through the points $(0, 5)$ and $(4, 0)$. The equation of this line is given by the intercept form: $\frac{x}{4} + \frac{y}{5} = 1$,which simplifies to $5x + 4y = 20$. Since the shaded region is above the line,the inequality is $5x + 4y \geq 20$.
$2$. The region is bounded by the vertical line $x = 6$ on the right,and since the shaded area is to the left of this line,we have $x \leq 6$.
$3$. The region is bounded by the horizontal line $y = 3$ on the top,and since the shaded area is below this line,we have $y \leq 3$.
$4$. The region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.
$5$. Combining all these,the system of inequations is $5x + 4y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$.

(D) Given that ${ }^{n} C_{12}={ }^{n} C_{8}$.
We know the property of combinations that ${ }^{n} C_{r}={ }^{n} C_{k}$ implies either $r=k$ or $n=r+k$.
Here,$r=12$ and $k=8$.
Since $12 \neq 8$,we must have $n=12+8$.
Therefore,$n=20$.

(C) Given that,$\left(\frac{1+i}{1-i}\right)^{m} = 1$.
First,simplify the expression inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^{2}}{1^{2}-i^{2}} = \frac{1+i^{2}+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$.
Now,the equation becomes $i^{m} = 1$.
We know that the powers of $i$ follow a cycle:
$i^{1} = i$
$i^{2} = -1$
$i^{3} = -i$
$i^{4} = 1$
Thus,the least positive integral value of $m$ for which $i^{m} = 1$ is $m = 4$.

(A) The expansion of $(x+a)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.
For $(x+a)^n - (x-a)^n$,the terms with even powers of $a$ cancel out,leaving only the terms with odd powers of $a$.
If $n$ is odd,the number of terms in the expansion of $(x+a)^n - (x-a)^n$ is given by $\frac{n+1}{2}$.
Here,$n = 47$,which is an odd number.
Therefore,the number of terms is $\frac{47+1}{2} = \frac{48}{2} = 24$.

(A) The coefficient of variation $( CV )$ is defined by the formula:
$ CV = \frac{\sigma}{\bar{x}} \times 100 $
Where $ \sigma $ is the standard deviation and $ \bar{x} $ is the arithmetic mean.
Given $ CV = 60 $ and $ \sigma = 24 $.
Substituting these values into the formula:
$ 60 = \frac{24}{\bar{x}} \times 100 $
$ 60 = \frac{2400}{\bar{x}} $
$ \bar{x} = \frac{2400}{60} $
$ \bar{x} = 40 $
Therefore,the arithmetic mean is $ 40 $.

(A) Given the line equation: $y = 3x - 1$.
Comparing this with the slope-intercept form $y = mx + c$,the slope $m_1 = 3$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$.
Therefore,$3 \cdot m_2 = -1$,which gives $m_2 = -\frac{1}{3}$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $(y - y_1) = m(x - x_1)$.
Substituting $(x_1, y_1) = (1, 2)$ and $m = -\frac{1}{3}$:
$(y - 2) = -\frac{1}{3}(x - 1)$.
Multiplying both sides by $3$:
$3(y - 2) = -(x - 1) \Rightarrow 3y - 6 = -x + 1$.
Rearranging the terms: $x + 3y - 7 = 0$.

(C) Given that $A$ is a subset of $B$,denoted as $A \subset B$.
By the definition of subset,every element of $A$ is also an element of $B$.
Therefore,the union of $A$ and $B$ is simply $B$,i.e.,$A \cup B = B$.
Taking the number of elements on both sides,we get $n(A \cup B) = n(B)$.
Similarly,the intersection of $A$ and $B$ is $A$,i.e.,$A \cap B = A$,which implies $n(A \cap B) = n(A)$.

(A) Given the inequality $|x-2| \leq 1$.
We know that the property $|x| \leq a$ is equivalent to $-a \leq x \leq a$.
Applying this property to the given inequality,we get $-1 \leq x-2 \leq 1$.
Adding $2$ to all parts of the inequality,we get $-1 + 2 \leq x \leq 1 + 2$.
This simplifies to $1 \leq x \leq 3$.
Therefore,the solution set is $x \in [1, 3]$.

(D) Given that,$y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$.
Dividing the numerator and denominator by $\cos x$,we get:
$y = \tan^{-1}\left(\frac{\tan x + 1}{1 - \tan x}\right)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = \pi/4$ and $B = x$:
$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + x\right)\right) = \frac{\pi}{4} + x$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + x\right) = 0 + 1 = 1$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $(K, 0), (4, 0), (0, 2)$ and $\text{Area} = 4$.
Substituting the values:
$4 = \frac{1}{2} |K(0 - 2) + 4(2 - 0) + 0(0 - 0)|$
$4 = \frac{1}{2} |-2K + 8|$
$8 = |-2K + 8|$
This implies:
$-2K + 8 = 8$ or $-2K + 8 = -8$
Case $1$: $-2K = 0 \Rightarrow K = 0$.
Case $2$: $-2K = -16 \Rightarrow K = 8$.
Thus,the value of $K$ is $0$ or $8$.

(C) Given the binary operation $a * b = \frac{a}{b+1}$.
Check for commutativity:
$a * b = \frac{a}{b+1}$
$b * a = \frac{b}{a+1}$
Since $\frac{a}{b+1} \neq \frac{b}{a+1}$ for all $a, b \in R - \{-1\}$,the operation is not commutative.
Check for associativity:
$(a * b) * c = \left(\frac{a}{b+1}\right) * c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$
$a * (b * c) = a * \left(\frac{b}{c+1}\right) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a(c+1)}{b+c+1}$
Since $\frac{a}{(b+1)(c+1)} \neq \frac{a(c+1)}{b+c+1}$,the operation is not associative.
Therefore,the operation $*$ is neither commutative nor associative.

(A) Given that,$\sin x = \frac{2t}{1+t^2}$ and $\tan y = \frac{2t}{1-t^2}$.
We know the standard trigonometric substitutions:
$x = \sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1}t$ (for $|t| \le 1$).
$y = \tan^{-1}\left(\frac{2t}{1-t^2}\right) = 2\tan^{-1}t$ (for $|t| < 1$).
Since $x = 2\tan^{-1}t$ and $y = 2\tan^{-1}t$,it follows that $x = y$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(A) We know that for a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X) = 1$.
Substituting the given values:
$0.3 + k + 2k + 2k = 1$
$0.3 + 5k = 1$
$5k = 1 - 0.3$
$5k = 0.7$
$k = \frac{0.7}{5}$
$k = 0.14$

(A) The area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi$ is given by the integral of the absolute value of the function:
$Area = \int_{0}^{\pi} |\cos x| dx$
Since $\cos x \geq 0$ for $x \in [0, \pi/2]$ and $\cos x \leq 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$Area = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx$
$Area = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$
$Area = (\sin(\pi/2) - \sin 0) - (\sin \pi - \sin(\pi/2))$
$Area = (1 - 0) - (0 - 1) = 1 + 1 = 2$ sq. units.

(A) The Mean Value Theorem states that for a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$,there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = x^{2}$,$a = 2$,and $b = 4$.
First,calculate $f(a) = f(2) = 2^{2} = 4$ and $f(b) = f(4) = 4^{2} = 16$.
Next,calculate the derivative $f'(x) = 2x$.
Using the formula $f'(c) = \frac{f(4) - f(2)}{4 - 2}$,we get $2c = \frac{16 - 4}{2}$.
$2c = \frac{12}{2} = 6$.
Therefore,$c = 3$.

(B) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $.
Divide numerator and denominator by $ \cos^{2} x $:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x $.
Let $ \tan x = t $,then $ \sec^{2} x d x = d t $.
When $ x = 0, t = 0 $ and when $ x = \frac{\pi}{2}, t \to \infty $.
$ I = \int_{0}^{\infty} \frac{d t}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d t}{t^{2} + (b/a)^{2}} $.
Using the formula $ \int \frac{d x}{x^{2} + k^{2}} = \frac{1}{k} \tan^{-1}(\frac{x}{k}) $:
$ I = \frac{1}{a^{2}} \cdot \frac{1}{(b/a)} \left[ \tan^{-1} \left( \frac{t}{b/a} \right) \right]_{0}^{\infty} $.
$ I = \frac{1}{ab} \left( \tan^{-1}(\infty) - \tan^{-1}(0) \right) = \frac{1}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2ab} $.

(C) The normal vector to the plane $2x - 3y + 6z - 11 = 0$ is $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.
The direction vector of the $X$-axis is $\vec{d} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between a line with direction vector $\vec{d}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \frac{|\vec{n} \cdot \vec{d}|}{|\vec{n}| |\vec{d}|}$.
Calculating the dot product: $\vec{n} \cdot \vec{d} = (2)(1) + (-3)(0) + (6)(0) = 2$.
Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ and $|\vec{d}| = 1$.
Thus,$\sin \theta = \frac{2}{7 \times 1} = \frac{2}{7}$.
Given that $\theta = \sin^{-1}(\alpha)$,we have $\sin^{-1}(\alpha) = \sin^{-1}(\frac{2}{7})$.
Therefore,$\alpha = \frac{2}{7}$.

(B) Let the radius of the sphere be $r = 4 \text{ cm}$.
The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.
Differentiating $V$ and $S$ with respect to $r$:
$\frac{dV}{dr} = 4 \pi r^2$
$\frac{dS}{dr} = 8 \pi r$
Using the chain rule,the rate of change of volume with respect to surface area is:
$\frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
At $r = 4 \text{ cm}$,we have:
$\frac{dV}{dS} = \frac{4}{2} = 2 \text{ cm}^3 \text{ cm}^{-2}$.

(C) Given that the function $f(x)$ is continuous at $x = 2$,the left-hand limit ($L$.$H$.$L$.) must equal the right-hand limit ($R$.$H$.$L$.).
$L$.$H$.$L$. = $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} Kx^2 = K(2)^2 = 4K$.
$R$.$H$.$L$. = $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} 3 = 3$.
Since the function is continuous,$4K = 3$.
Therefore,$K = \frac{3}{4}$.

(B) In a Linear Programming Problem $(LPP)$,if the objective function attains the same optimal value at two distinct vertices of the feasible region,then it also attains that same optimal value at every point on the line segment joining these two vertices. This is a fundamental property of the convex set of feasible solutions in an $LPP$.

(A) Given the matrix equation:
$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Multiplying the first matrix by $2$,we get:
$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Adding the two matrices on the left side:
$\begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
By comparing the corresponding elements of the two matrices:
$2+y = 5 \Rightarrow y = 5-2 = 3$
$2x+2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$
Thus,the values are $x=3$ and $y=3$.

(A) Given that $f(x) = 8x^3$ and $g(x) = x^{1/3}$.
By the definition of composite function,$(f \circ g)(x) = f(g(x))$.
Substitute $g(x)$ into $f(x)$:
$(f \circ g)(x) = f(x^{1/3}) = 8(x^{1/3})^3$.
Using the exponent rule $(a^m)^n = a^{m \times n}$,we get $(x^{1/3})^3 = x^{(1/3) \times 3} = x^1 = x$.
Therefore,$(f \circ g)(x) = 8x$.

(B) For a symmetric matrix,we know that: $A^{T} = A$ $(1)$
For a skew-symmetric matrix,we know that: $A^{T} = -A$ $(2)$
Equating $(1)$ and $(2)$,we get: $A = -A$
Adding $A$ to both sides: $2A = 0$
Therefore,$A = 0$,which means $A$ is a zero matrix.

(D) The function $f(x) = \sec^{-1} x$ is the inverse of the secant function restricted to the domain $[0, \pi] - \{\frac{\pi}{2}\}$.
By definition,the range of the principal value branch of $\sec^{-1} x$ is the set of all values in the interval $[0, \pi]$ excluding $\frac{\pi}{2}$,because $\sec x$ is undefined at $\frac{\pi}{2}$.
Therefore,the range is $[0, \pi] - \{\frac{\pi}{2}\}$.

(A) We need to evaluate the integral $ I = \int_{-5}^{5} |x+2| \, dx $.
Since $|x+2| = -(x+2)$ for $x < -2$ and $|x+2| = (x+2)$ for $x \ge -2$,we split the integral at $x = -2$:
$ I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx $
Evaluating the first part:
$ -\left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} = -\left( (\frac{4}{2} - 4) - (\frac{25}{2} - 10) \right) = -\left( -2 - 2.5 \right) = 4.5 $
Evaluating the second part:
$ \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5} = \left( (\frac{25}{2} + 10) - (\frac{4}{2} - 4) \right) = (12.5 + 10) - (-2) = 22.5 + 2 = 24.5 $
Adding both parts:
$ I = 4.5 + 24.5 = 29 $

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
We are given the equation $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.
Squaring both sides of the equation,we get:
$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{0}|^2$
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Substituting the values $|\vec{a}|^2 = 1, |\vec{b}|^2 = 1, |\vec{c}|^2 = 1$:
$1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

(A) We are given the integral $ I = \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx $.
Using the trigonometric identity $\cos 2A = 2\cos^2 A - 1$,we can rewrite the numerator:
$\cos 2x - \cos 2\theta = (2\cos^2 x - 1) - (2\cos^2 \theta - 1) = 2\cos^2 x - 2\cos^2 \theta = 2(\cos^2 x - \cos^2 \theta)$.
Now,substitute this into the integral:
$I = \int \frac{2(\cos^2 x - \cos^2 \theta)}{\cos x - \cos \theta} dx$.
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:
$I = \int \frac{2(\cos x - \cos \theta)(\cos x + \cos \theta)}{\cos x - \cos \theta} dx$.
Canceling the common term $(\cos x - \cos \theta)$:
$I = 2 \int (\cos x + \cos \theta) dx$.
Integrating with respect to $x$ (noting that $\cos \theta$ is a constant):
$I = 2(\sin x + x \cos \theta) + C$.

(A) Let $y = \cos ^{-1}(2x^2 - 1)$ and $z = \cos ^{-1} x$.
We know that $\cos ^{-1}(2x^2 - 1) = 2 \cos ^{-1} x$ for $x \in [0, 1]$.
Therefore,$y = 2z$.
Differentiating $y$ with respect to $z$,we get $\frac{dy}{dz} = \frac{d}{dz}(2z) = 2$.

(A) Given function is $f(x) = x^{2} + 2x - 5$.
To find the interval where the function is strictly increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 5) = 2x + 2$.
For a function to be strictly increasing,we must have $f'(x) > 0$.
$2x + 2 > 0$
$2x > -2$
$x > -1$.
Thus,the function is strictly increasing in the interval $(-1, \infty)$.

(A) Given that,$y = \log(\log x) \quad (1)$
Differentiating Eq. $(1)$ with respect to $x$,we get:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \quad (2)$
Now,differentiating Eq. $(2)$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}((x \log x)^{-1}) = -1(x \log x)^{-2} \cdot \frac{d}{dx}(x \log x)$
$\frac{d^2y}{dx^2} = -\frac{1}{(x \log x)^2} \cdot [x \cdot \frac{1}{x} + \log x \cdot 1]$
$\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}$

(A) Given differential equation is $\frac{dy}{dx} + y = 1$ $(y \neq 1)$.
Rearranging the terms,we get $\frac{dy}{dx} = 1 - y$.
Separating the variables,we have $\frac{dy}{1-y} = dx$.
Integrating both sides,we get $\int \frac{dy}{1-y} = \int dx$.
This gives $-\log |1-y| = x + C$.
Using the property of logarithms,$-\log |a| = \log |\frac{1}{a}|$,we get $\log \left|\frac{1}{1-y}\right| = x + C$.

(A) We need to evaluate the integral $ I = \int \sqrt{x^{2}+2 x+5} \, dx $.
First,complete the square for the quadratic expression: $ x^{2}+2x+5 = (x+1)^{2} + 4 = (x+1)^{2} + 2^{2} $.
Now the integral becomes $ I = \int \sqrt{(x+1)^{2} + 2^{2}} \, dx $.
Using the standard formula $ \int \sqrt{t^{2}+a^{2}} \, dt = \frac{t}{2} \sqrt{t^{2}+a^{2}} + \frac{a^{2}}{2} \log \left|t + \sqrt{t^{2}+a^{2}}\right| + C $,where $ t = x+1 $ and $ a = 2 $.
Substituting these values: $ I = \frac{x+1}{2} \sqrt{(x+1)^{2} + 2^{2}} + \frac{2^{2}}{2} \log \left|(x+1) + \sqrt{(x+1)^{2} + 2^{2}}\right| + C $.
Simplifying the expression: $ I = \frac{1}{2}(x+1) \sqrt{x^{2}+2x+5} + 2 \log \left|x+1 + \sqrt{x^{2}+2x+5}\right| + C $.
Thus,the correct option is $ A $.

(A) Given the differential equation:
$x \frac{dy}{dx} + 2y = x^2$
Dividing both sides by $x$ (since $x \neq 0$):
$\frac{dy}{dx} + \frac{2}{x} y = x$
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x$.
The integrating factor ($I$.$F$.) is given by:
$I.F. = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx}$
$= e^{2 \log |x|} = e^{\log |x^2|} = x^2$
Thus,the integrating factor is $x^2$.

(B) Two events $A$ and $B$ are independent if and only if the occurrence of one does not affect the probability of the other.
If $A$ and $B$ are independent,then their complements $A^{\prime}$ and $B^{\prime}$ are also independent.
By the definition of independent events,the probability of the intersection of two independent events is the product of their individual probabilities.
Therefore,$P(A^{\prime} \cap B^{\prime}) = P(A^{\prime}) \cdot P(B^{\prime})$.
Since $P(A^{\prime}) = 1 - P(A)$ and $P(B^{\prime}) = 1 - P(B)$,we have $P(A^{\prime} \cap B^{\prime}) = (1 - P(A))(1 - P(B))$.
Thus,option $B$ is the correct condition for independence.

(B) Given the function $f(x) = \sqrt{9 - x^2}$.
For the function to be defined,the expression inside the square root must be non-negative:
$9 - x^2 \geq 0 \Rightarrow x^2 \leq 9 \Rightarrow -3 \leq x \leq 3$.
Thus,the domain of the function is $x \in [-3, 3]$.
When $x = 0$,$f(0) = \sqrt{9 - 0} = 3$.
When $x = \pm 3$,$f(\pm 3) = \sqrt{9 - 9} = 0$.
Since the square root function always yields non-negative values,the minimum value is $0$ and the maximum value is $3$.
Therefore,the range of the function is $[0, 3]$.

(B) Given $ \Delta = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} $.
Taking $ x, y, z $ common from $ R_1, R_2, R_3 $ respectively:
$ \Delta = xyz \begin{vmatrix} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{vmatrix} $.
Multiply $ C_3 $ by $ xyz $:
$ \Delta = \begin{vmatrix} A & x & yz \\ B & y & zx \\ C & z & xy \end{vmatrix} $.
Taking the transpose of the determinant (rows become columns):
$ \Delta = \begin{vmatrix} A & B & C \\ x & y & z \\ yz & zx & xy \end{vmatrix} = \Delta_1 $.
Thus,$ \Delta_1 = \Delta $.

(A) We are given that $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$.
We know the identity $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ for any real $x$.
This implies $\tan ^{-1} x=\frac{\pi}{2}-\cot ^{-1} x$ and $\tan ^{-1} y=\frac{\pi}{2}-\cot ^{-1} y$.
Substituting these into the given equation:
$(\frac{\pi}{2}-\cot ^{-1} x) + (\frac{\pi}{2}-\cot ^{-1} y) = \frac{4 \pi}{5}$.
$\pi - (\cot ^{-1} x + \cot ^{-1} y) = \frac{4 \pi}{5}$.
$\cot ^{-1} x + \cot ^{-1} y = \pi - \frac{4 \pi}{5}$.
$\cot ^{-1} x + \cot ^{-1} y = \frac{\pi}{5}$.

(A) The coordinates of the given point are $P(x, y, z) = (6, 7, 8)$.
The perpendicular distance of any point $(x, y, z)$ from the $XY$-plane is given by the absolute value of its $z$-coordinate,which is $|z|$.
Here,the $z$-coordinate is $8$.
Therefore,the perpendicular distance from the $XY$-plane is $|8| = 8$.

(B) Let $ I = \int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x $.
Using the property $ \int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x $,we get:
$ I = \int_{0}^{\pi / 2} \frac{\tan ^{7}(\frac{\pi}{2}-x)}{\cot ^{7}(\frac{\pi}{2}-x)+\tan ^{7}(\frac{\pi}{2}-x)} d x $.
Since $ \tan(\frac{\pi}{2}-x) = \cot x $ and $ \cot(\frac{\pi}{2}-x) = \tan x $,we have:
$ I = \int_{0}^{\pi / 2} \frac{\cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x $.
Adding the two expressions for $ I $:
$ 2I = \int_{0}^{\pi / 2} \frac{\tan ^{7} x + \cot ^{7} x}{\tan ^{7} x+\cot ^{7} x} d x = \int_{0}^{\pi / 2} 1 d x $.
$ 2I = [x]_{0}^{\pi / 2} = \frac{\pi}{2} $.
Therefore,$ I = \frac{\pi}{4} $.

(D) Two vectors $\vec{a}$ and $\vec{b}$ are orthogonal if their dot product is zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
Given vectors are $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Calculating the dot product:
$\vec{a} \cdot \vec{b} = (2)(1) + (\lambda)(2) + (1)(3) = 0$
$2 + 2\lambda + 3 = 0$
$5 + 2\lambda = 0$
$2\lambda = -5$
$\lambda = -\frac{5}{2}$

(B) The curve is $y=x^2$ and the line is $y=16$. The intersection points are found by setting $x^2=16$,which gives $x = \pm 4$.
Since the region is symmetric about the $y$-axis,the area $A$ is given by:
$A = 2 \int_0^{16} x \, dy = 2 \int_0^{16} \sqrt{y} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_0^{16}$
$A = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_0^{16}$
$A = \frac{4}{3} \left( 16^{3/2} - 0^{3/2} \right)$
$A = \frac{4}{3} \times (4^2)^{3/2} = \frac{4}{3} \times 4^3$
$A = \frac{4}{3} \times 64 = \frac{256}{3} \text{ sq. units}$

(C) Given that,the probability of selecting a defective pen is $p = \frac{10}{100} = \frac{1}{10}$.
So,the probability of selecting a non-defective pen is $q = 1 - p = \frac{9}{10}$.
Since the pens are drawn with replacement,this follows a binomial distribution with $n = 5$ trials.
We need to find the probability that at most one pen is defective,i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$.
Using the binomial formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}\right)^{0} \left(\frac{9}{10}\right)^{5} = \left(\frac{9}{10}\right)^{5}$.
$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}\right)^{4} = 5 \times \frac{1}{10} \times \left(\frac{9}{10}\right)^{4} = \frac{1}{2} \left(\frac{9}{10}\right)^{4}$.
Therefore,$P(X \leq 1) = \left(\frac{9}{10}\right)^{5} + \frac{1}{2} \left(\frac{9}{10}\right)^{4}$.

(D) The reflection of a point $(x, y, z)$ in the $XY$-plane results in the change of the sign of the $z$-coordinate,while the $x$ and $y$ coordinates remain unchanged.
Therefore,the reflection of the point $(\alpha, \beta, \gamma)$ in the $XY$-plane is $(\alpha, \beta, -\gamma)$.

(C) Given that,matrix $A$ is of order $3 \times 3$.
We know that for any square matrix $A$ of order $n \times n$,the property of the determinant is $|KA| = K^{n}|A|$.
Here,the order of the matrix is $n = 3$.
Substituting $n = 3$ into the formula,we get $|KA| = K^{3}|A|$.

(D) Given matrices are $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \tan^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} $ and $ B = \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix} $.
Subtracting $ B $ from $ A $:
$ A - B = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(\pi x) + \cos^{-1}(\pi x) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) - \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) + \tan^{-1}(\pi x) \end{bmatrix} $.
Assuming the standard identity $ \sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2} $ and $ \tan^{-1} \theta + \cot^{-1} \theta = \frac{\pi}{2} $,we simplify the terms.
For the diagonal elements:
$ \frac{1}{\pi} \sin^{-1}(\pi x) + \cos^{-1}(\pi x) $ and $ \frac{1}{\pi} \cot^{-1}(\pi x) + \tan^{-1}(\pi x) $.
Given the structure of the question,it simplifies to $ \frac{1}{2} I $ where $ I $ is the identity matrix.

(A) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{2}=\frac{d^{2}y}{dx^{2}}$.
To find the degree,we first identify the highest order derivative present in the equation.
The highest order derivative is $\frac{d^{2}y}{dx^{2}}$,which has an order of $2$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in terms of derivatives.
In the given equation,the highest order derivative $\frac{d^{2}y}{dx^{2}}$ has an exponent of $1$.
Therefore,the degree of the differential equation is $1$.

(B) Given the curve equation: $y^{2} = x$ $(1)$
Slope of the tangent $m$ is given by $\frac{dy}{dx}$.
We know that the slope $m = \tan(\theta)$,where $\theta$ is the angle with the $x$-axis.
Here,$\theta = \frac{\pi}{4}$,so $m = \tan(\frac{\pi}{4}) = 1$.
Differentiating equation $(1)$ with respect to $x$:
$2y \frac{dy}{dx} = 1$
Substituting $m = \frac{dy}{dx} = 1$:
$2y(1) = 1 \Rightarrow y = \frac{1}{2}$.
Substituting $y = \frac{1}{2}$ into equation $(1)$:
$(\frac{1}{2})^{2} = x \Rightarrow x = \frac{1}{4}$.
Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

(A) Given that,$y = \left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right|$.
The derivative of a determinant is the sum of determinants obtained by differentiating one row at a time while keeping other rows constant.
$\frac{dy}{dx} = \left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right| + \left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ 0 & 0 & 0 \\ a & b & c\end{array}\right| + \left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ 0 & 0 & 0\end{array}\right|$.
Since a determinant with a row of zeros is equal to $0$,the second and third determinants vanish.
Therefore,$\frac{dy}{dx} = \left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right|$.

(B) We are given the definite integral $\int_{0.2}^{3.5} [x] \, dx$,where $[x]$ denotes the greatest integer function.
We split the interval $[0.2, 3.5]$ based on the points where the greatest integer function changes its value:
$\int_{0.2}^{3.5} [x] \, dx = \int_{0.2}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx$
Since $[x] = 0$ for $x \in [0.2, 1)$,$[x] = 1$ for $x \in [1, 2)$,$[x] = 2$ for $x \in [2, 3)$,and $[x] = 3$ for $x \in [3, 3.5]$,we have:
$= \int_{0.2}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{3.5} 3 \, dx$
$= 0 + [x]_{1}^{2} + 2[x]_{2}^{3} + 3[x]_{3}^{3.5}$
$= 0 + (2 - 1) + 2(3 - 2) + 3(3.5 - 3)$
$= 0 + 1 + 2(1) + 3(0.5)$
$= 1 + 2 + 1.5 = 4.5$

(C) We are given the integral $ I = \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x $.
To solve this,we rewrite the numerator as $(x+4-1)$:
$ I = \int e^{x} \frac{(x+4-1)}{(x+4)^{2}} d x $.
Splitting the fraction,we get:
$ I = \int e^{x} \left[ \frac{x+4}{(x+4)^{2}} - \frac{1}{(x+4)^{2}} \right] d x $.
$ I = \int e^{x} \left[ \frac{1}{x+4} - \frac{1}{(x+4)^{2}} \right] d x $.
We know the standard integration formula:
$ \int e^{x} (f(x) + f'(x)) d x = e^{x} f(x) + C $.
Here,let $ f(x) = \frac{1}{x+4} $.
Then,$ f'(x) = -\frac{1}{(x+4)^{2}} $.
Substituting these into the formula,we get:
$ I = e^{x} \left( \frac{1}{x+4} \right) + C = \frac{e^{x}}{x+4} + C $.

(D) Let $ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $.
Using the property $ \int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx $,we have:
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{1}{e^{\sin(-x)}+1} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{1}{e^{-\sin x}+1} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{e^{\sin x}+1} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+e^{\sin x}}{1+e^{\sin x}} \right) dx $
$ I = \int_{0}^{\frac{\pi}{2}} 1 dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $.

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
We are given that $\sqrt{3}\vec{a}-\vec{b}$ is also a unit vector,so $|\sqrt{3}\vec{a}-\vec{b}| = 1$.
Squaring both sides,we get $|\sqrt{3}\vec{a}-\vec{b}|^2 = 1^2$.
Using the property $|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have:
$3|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3}(\vec{a} \cdot \vec{b}) = 1$.
Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$:
$3(1)^2 + (1)^2 - 2\sqrt{3}(|\vec{a}||\vec{b}|\cos \theta) = 1$.
$3 + 1 - 2\sqrt{3}(1)(1)\cos \theta = 1$.
$4 - 2\sqrt{3}\cos \theta = 1$.
$2\sqrt{3}\cos \theta = 3$.
$\cos \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos \theta = \frac{\sqrt{3}}{2}$,the angle $\theta = 30^{\circ}$.

(D) Given that,$f: R \rightarrow R$ and $f(x)=x^{4}$.
For $f$ to be one-one,$f(x_{1}) = f(x_{2})$ must imply $x_{1} = x_{2}$.
Here,$x_{1}^{4} = x_{2}^{4} \Rightarrow x_{1} = \pm x_{2}$.
For example,$f(1) = 1^{4} = 1$ and $f(-1) = (-1)^{4} = 1$.
Since $f(1) = f(-1)$ but $1 \neq -1$,$f$ is not one-one.
For $f$ to be onto,the range must equal the codomain $R$.
Since $x^{4} \geq 0$ for all $x \in R$,the range is $[0, \infty)$.
Since the range $[0, \infty) \neq R$,$f$ is not onto.
Therefore,$f$ is neither one-one nor onto.

(B) Given line: $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$ $(1)$
Point $P(-2, 4, -5)$.
Any point $Q$ on the line is given by $Q(3\lambda - 3, 5\lambda + 4, 6\lambda - 8)$.
The direction ratios of the line $PQ$ are $(3\lambda - 3 - (-2), 5\lambda + 4 - 4, 6\lambda - 8 - (-5)) = (3\lambda - 1, 5\lambda, 6\lambda - 3)$.
Since $PQ$ is perpendicular to the given line with direction ratios $(3, 5, 6)$,their dot product is zero:
$3(3\lambda - 1) + 5(5\lambda) + 6(6\lambda - 3) = 0$
$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$
$70\lambda - 21 = 0 \Rightarrow \lambda = \frac{21}{70} = \frac{3}{10}$.
Substituting $\lambda = \frac{3}{10}$ in $Q$,we get $Q\left(3(\frac{3}{10}) - 3, 5(\frac{3}{10}) + 4, 6(\frac{3}{10}) - 8\right) = Q\left(-\frac{21}{10}, \frac{55}{10}, -\frac{62}{10}\right)$.
The distance $PQ = \sqrt{(-\frac{21}{10} + 2)^2 + (\frac{55}{10} - 4)^2 + (-\frac{62}{10} + 5)^2}$
$PQ = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1 + 225 + 144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}}$.

(D) Given the determinant equation: $\left|\begin{array}{cc}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
Expanding both sides:
$(3 \times 1) - (x \times x) = (3 \times 1) - (2 \times 4)$
$3 - x^2 = 3 - 8$
$3 - x^2 = -5$
Subtracting $3$ from both sides:
$-x^2 = -8$
$x^2 = 8$
Taking the square root on both sides:
$x = \pm \sqrt{8}$
$x = \pm 2 \sqrt{2}$