KCET 2025 Mathematics Question Paper with Answer and Solution

(A) For the first inequality: $3x + 8 < 17$ $\Rightarrow 3x < 9$ $\Rightarrow x < 3$.
For the second inequality: $2x + 8 \geq 12$ $\Rightarrow 2x \geq 4$ $\Rightarrow x \geq 2$.
Thus,Statement $(I)$ is true.
The common set of solutions is the intersection of $x < 3$ and $x \geq 2$,which is the interval $[2, 3)$.
Since Statement $(II)$ claims the set is $(2, 3)$,which excludes $2$,Statement $(II)$ is false.

(D) The triangle inequality for complex numbers states that for any two complex numbers $Z_1$ and $Z_2$,the modulus of their sum is less than or equal to the sum of their moduli,i.e.,$\left|Z_1+Z_2\right| \leq \left|Z_1\right|+\left|Z_2\right|$.
Therefore,the statement $\left|Z_1+Z_2\right| \geq \left|Z_1\right|+\left|Z_2\right|$ is generally false.

(B) four-digit number is even if its unit digit is $0$ or $2$. Since repetition is not allowed,we consider two cases:
Case $1$: The unit digit is $0$.
The remaining $3$ positions can be filled by the remaining $3$ digits $(1, 2, 3)$ in $3! = 3 \times 2 \times 1 = 6$ ways.
Case $2$: The unit digit is $2$.
The thousands place cannot be $0$ or $2$. Thus,the thousands place can be filled by $1$ or $3$ ($2$ ways).
The remaining $2$ positions can be filled by the remaining $2$ digits in $2! = 2$ ways.
Total ways for Case $2 = 2 \times 2 = 4$ ways.
Total even numbers = $6 + 4 = 10$.

(B) An octagon has $n = 8$ sides.
The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Substituting $n = 8$ into the formula:
$\text{Number of diagonals} = \frac{8(8-3)}{2} = \frac{8 \times 5}{2} = \frac{40}{2} = 20$.
Thus,the number of diagonals is $20$.

(B) Let the first term of the $G$.$P$. be $a$ and the common ratio be $r$.
The $n^{\text{th}}$ term of a $G$.$P$. is given by $T_n = ar^{n-1}$.
Given:
$x = T_4 = ar^3$
$y = T_{10} = ar^9$
$z = T_{16} = ar^{15}$
Now,consider the product $xz$:
$xz = (ar^3)(ar^{15}) = a^2r^{18} = (ar^9)^2$
Since $y = ar^9$,we have $xz = y^2$.
Therefore,$y = \sqrt{xz}$.

(C) The number of terms in the binomial expansion of $(a + b)^m$ is given by $m + 1$.
Given the expression $(2x + 3)^{3n}$,the number of terms is $(3n + 1)$.
According to the problem,the number of terms is $22$.
So,$3n + 1 = 22$.
$3n = 21$.
$n = 7$.

(A) We evaluate each option:
$A$: $\cos 5 \pi = -1$ and $\cos 4 \pi = 1$. Since $-1 \neq 1$,this statement is incorrect.
$B$: $\sin 2 \pi = 0$ and $\sin (-2 \pi) = 0$. This is correct.
$C$: $\sin 4 \pi = 0$ and $\sin 6 \pi = 0$. This is correct.
$D$: $\tan 45^{\circ} = 1$ and $\tan (-315^{\circ}) = \tan (-315^{\circ} + 360^{\circ}) = \tan 45^{\circ} = 1$. This is correct.
Therefore,the incorrect statement is $A$.

(B) Given that $\cos x + \cos^2 x = 1$.
From this,we have $\cos x = 1 - \cos^2 x$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = \cos x$.
Now,we need to find the value of $\sin^2 x + \sin^4 x$.
Substituting $\sin^2 x = \cos x$,we get $\sin^2 x + \sin^4 x = \cos x + (\cos x)^2$.
Since $\cos x + \cos^2 x = 1$,the value is $1$.

(A) We know that the range of the sine function is $-1 \leq \sin x \leq 1$.
To find the minimum value of $f(x) = 1 - \sin x$,we need to subtract the maximum possible value of $\sin x$ from $1$.
Since the maximum value of $\sin x$ is $1$,we have:
$f_{\min} = 1 - (\sin x)_{\max} = 1 - 1 = 0$.

(B) The given line is $x + 6y = 5$,which can be written as $y = -\frac{1}{6}x + \frac{5}{6}$.
The slope of this line is $m_1 = -\frac{1}{6}$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-1/6} = 6$.
The equation of the line passing through $(-1, -3)$ with slope $m = 6$ is given by $y - y_1 = m(x - x_1)$.
$y - (-3) = 6(x - (-1))$
$y + 3 = 6(x + 1)$
$y + 3 = 6x + 6$
$6x - y = -3$.
To find the $x$-intercept,we set $y = 0$ in the equation:
$6x - 0 = -3$
$6x = -3$
$x = -\frac{3}{6} = -\frac{1}{2}$.

(C) Given equation: $x^2+3y^2=12$
Divide by $12$: $\frac{x^2}{12} + \frac{y^2}{4} = 1$
This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 12$ and $b^2 = 4$.
Thus,$a = \sqrt{12} = 2\sqrt{3}$ and $b = 2$.
The length of the latus rectum is given by $\frac{2b^2}{a}$.
Length $= \frac{2 \times 4}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$ units.

(B) Let $L = \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(x^4-\sqrt{x})}{\frac{d}{dx}(\sqrt{x}-1)}$
$L = \lim _{x \rightarrow 1} \frac{4x^3 - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$
$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{4x^3}{\frac{1}{2\sqrt{x}}} - \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} \right)$
$L = \lim _{x \rightarrow 1} (8x^3\sqrt{x} - 1)$
$L = 8(1)^3\sqrt{1} - 1 = 8 - 1 = 7$.

(B) Step $1$: Calculate the mean $(\mu)$ of the data.
$\mu = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$.
Step $2$: Calculate the mean deviation about the mean using the formula $\frac{1}{N} \sum |x_i - \mu|$.
$|4-10| = 6$
$|7-10| = 3$
$|8-10| = 2$
$|9-10| = 1$
$|10-10| = 0$
$|12-10| = 2$
$|13-10| = 3$
$|17-10| = 7$
Sum of absolute deviations $= 6+3+2+1+0+2+3+7 = 24$.
Mean deviation $= \frac{24}{8} = 3$.

(C) Given $|A| = 3$ and $|B| = 6$.
For $A \cup B$,the formula is $|A \cup B| = |A| + |B| - |A \cap B|$.
To minimize $|A \cup B|$,we must maximize $|A \cap B|$. The maximum value of $|A \cap B|$ is $\min(|A|, |B|) = 3$.
Thus,$\min |A \cup B| = 3 + 6 - 3 = 6$. So,Statement $(I)$ is true.
For $A \cap B$,the maximum value is $\min(|A|, |B|) = 3$.
Thus,$\max |A \cap B| = 3$. So,Statement $(II)$ is true.

(A) First,we determine the elements of each set:
$A = \{ x : x \text{ is an integer and } x^2 - 9 = 0 \} = \{ -3, 3 \}$.
$B = \{ x : x \text{ is a natural number and } 2 \leq x < 5 \} = \{ 2, 3, 4 \}$.
$C = \{ x : x \text{ is a prime number } \leq 4 \} = \{ 2, 3 \}$.
Next,calculate $(B - C)$:
$B - C = \{ 2, 3, 4 \} - \{ 2, 3 \} = \{ 4 \}$.
Finally,calculate $(B - C) \cup A$:
$(B - C) \cup A = \{ 4 \} \cup \{ -3, 3 \} = \{ -3, 3, 4 \}$.

(B) Given equations are:
$1. 4x + 6y = 5$
$2. 8x + 12y = 10$
Step-by-step check:
Multiply Equation $(1)$ by $2$:
$2(4x + 6y) = 2(5) \Rightarrow 8x + 12y = 10$
Comparing this with Equation $(2)$,we see that both equations are identical.
Conclusion:
Since both equations represent the same line,the system has infinitely many solutions.

(D) The distance of a point $P(x, y, z)$ from the $yz$-plane is given by the absolute value of its $x$-coordinate,which is $|x|$.
Given the point $P(-3, 4, 5)$,the $x$-coordinate is $-3$.
Therefore,the distance from the $yz$-plane is $|-3| = 3 \text{ units}$.
Thus,the correct option is $D$.

(C) The sum of probabilities of all outcomes in a random experiment is $1$.
$p(w_1) + p(w_2) + p(w_3) + p(w_4) + p(w_5) = 1$.
Given $p(w_1) = \frac{1}{6}$,$p(w_2) = a$,$p(w_3) = b$,$p(w_4) = c$,and $p(w_5) = \frac{1}{12}$.
So,$\frac{1}{6} + a + b + c + \frac{1}{12} = 1$.
$\frac{2+1}{12} + a + b + c = 1$ $\Rightarrow \frac{3}{12} + a + b + c = 1$ $\Rightarrow a + b + c = 1 - \frac{1}{4} = \frac{3}{4}$.
Given $12a + 12b - 1 = 0$,we have $a + b = \frac{1}{12}$.
Substituting this into the sum: $\frac{1}{12} + c = \frac{3}{4} \Rightarrow c = \frac{3}{4} - \frac{1}{12} = \frac{9-1}{12} = \frac{8}{12} = \frac{2}{3}$.
Assuming the outcomes $w_2, w_3$ are equally likely,$a = b$.
Since $a + b = \frac{1}{12}$,then $2b = \frac{1}{12} \Rightarrow b = \frac{1}{24}$.
However,based on the provided options and standard interpretation of such problems where $a=b=c$ is not implied but specific constraints are given,the intended answer is $p(w_3) = \frac{1}{24}$ which is not listed. Re-evaluating the prompt's constraint $12a+12b=1$,if $a=b$,then $b = \frac{1}{24}$. If the question implies $p(w_3) = b = \frac{1}{24}$,none match. Given the structure,if $a=b=c$,then $3b = 3/4 \Rightarrow b = 1/4$. If $12a+12b=1$ and $a=b$,$24b=1 \Rightarrow b=1/24$. Given the options,$b = 1/24$ is not present. Assuming a typo in the question where $12a+12b=1$ was meant to be $a+b=...$,we select the most logical fit.

(B) The total number of faces on the die is $2 + 3 + 1 = 6$.
The probability of getting a $1$ is $P(1) = \frac{2}{6}$.
The probability of getting a $3$ is $P(3) = \frac{1}{6}$.
Since the events are mutually exclusive,$P(1 \text{ or } 3) = P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(C) Given that $A \subset B$,it implies that $A \cap B = A$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $A \cap B = A$,we get $P(A \mid B) = \frac{P(A)}{P(B)}$.
Since $A \subset B$,we have $P(A) \leq P(B)$,which implies $\frac{1}{P(B)} \leq \frac{1}{P(A)}$.
Multiplying both sides by $P(A)$ (assuming $P(A) > 0$),we get $\frac{P(A)}{P(B)} \leq 1$.
Also,since $P(B) \leq 1$,we have $\frac{P(A)}{P(B)} \geq P(A)$.
Thus,$P(A \mid B) \geq P(A)$.

(C) An equivalence relation $R$ on $A = \{a, b, c\}$ must be reflexive,symmetric,and transitive. Since $(b, c) \in R$ and $R$ is symmetric,$(c, b) \in R$. Since $R$ is reflexive,$(a, a), (b, b), (c, c) \in R$.
For transitivity,since $(b, c) \in R$ and $(c, b) \in R$,we must have $(b, b) \in R$ (which is true) and $(c, c) \in R$ (which is true).
Case $1$: If $a$ is related only to itself,$R_1 = \{(a, a), (b, b), (c, c), (b, c), (c, b)\}$. This is an equivalence relation.
Case $2$: If $a$ is related to $b$ and $c$,then by symmetry and transitivity,$a$ must be related to all elements. $R_2 = \{(a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a), (b, c), (c, b)\}$. This is the universal relation,which is also an equivalence relation.
Thus,there are $2$ such equivalence relations.

(A) Given that $A^2=A$.
We need to evaluate $(I-A)^3$.
Using the binomial expansion for matrices,$(I-A)^3 = I^3 - 3I^2A + 3IA^2 - A^3$.
Since $I^n = I$ and $A^2 = A$,we have $A^3 = A^2 \times A = A \times A = A^2 = A$.
Substituting these values:
$(I-A)^3 = I - 3A + 3A - A$.
$(I-A)^3 = I - A$.

(C) For the product $AB$ to be an identity matrix $I$,the number of columns in $A$ must be equal to the number of rows in $B$,and the resulting matrix $AB$ must be a square matrix.
Given that the order of matrix $B$ is $3 \times 4$,let the order of matrix $A$ be $m \times n$.
For the product $AB$ to be defined,the number of columns in $A$ $(n)$ must equal the number of rows in $B$ $(3)$. Thus,$n = 3$.
The resulting matrix $AB$ will have the order $m \times 4$.
Since $AB$ is an identity matrix,it must be a square matrix,meaning the number of rows must equal the number of columns. Therefore,$m = 4$.
Thus,the order of matrix $A$ is $4 \times 3$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
Calculate powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2^1 A$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} = 2^2 A$.
By induction,$A^n = 2^{n-1} A$.
For $n = 6$,$A^6 = 2^{6-1} A = 2^5 A = 32 A$.
Comparing $A^6 = k A$ with $A^6 = 32 A$,we get $k = 32$.

(B) diagonal matrix is a square matrix in which all elements except those in the main diagonal are zero. The diagonal elements themselves can be any value,including zero,but they are not required to be zero. Therefore,the statement that a diagonal matrix has all diagonal elements equal to zero is incorrect.

(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n=2$ and $|A|=2$.
So,$|\operatorname{adj}(A)| = |A|^{2-1} = |A|^1 = 2$.
Given $\operatorname{adj}(A) = B = \left[\begin{array}{ll}1 & 3 \\ 1 & \alpha\end{array}\right]$,we calculate the determinant of $B$:
$|B| = (1)(\alpha) - (3)(1) = \alpha - 3$.
Since $|\operatorname{adj}(A)| = |B|$,we have $\alpha - 3 = 2$.
Therefore,$\alpha = 2 + 3 = 5$.

(A) Given the equation: $A^2 - 5A + 7I = 0$.
Since $A^2 - 5A + 7I = 0$,we have $7I = 5A - A^2$.
Multiplying both sides by $A^{-1}$ (assuming $|A| \neq 0$):
$A^{-1}(A^2 - 5A + 7I) = A^{-1}(0)$.
$A^{-1}A^2 - 5A^{-1}A + 7A^{-1}I = 0$.
$A - 5I + 7A^{-1} = 0$.
$7A^{-1} = 5I - A$.
$A^{-1} = \frac{1}{7}(5I - A)$.

(B) Given $A = \begin{bmatrix} k & 2 \\ 2 & k \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = (k \times k) - (2 \times 2) = k^2 - 4$.
We know the property of determinants that $|A^n| = |A|^n$.
Given $|A^3| = 125$,we can write this as $|A|^3 = 125$.
Substituting the value of $|A|$:
$(k^2 - 4)^3 = 125$.
Taking the cube root on both sides:
$k^2 - 4 = 5$.
$k^2 = 9$.
$k = \pm 3$.

(D) Given that $A$ is a square matrix of order $n = 3$ and $\operatorname{det}(A) = 3$.
We know that for a scalar $k$ and an $n \times n$ matrix $A$,$\operatorname{det}(kA) = k^n \operatorname{det}(A)$.
Also,$\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}$.
Therefore,$\operatorname{det}(3A^{-1}) = 3^3 \operatorname{det}(A^{-1})$.
Substituting the values,we get $\operatorname{det}(3A^{-1}) = 27 \times \frac{1}{\operatorname{det}(A)} = 27 \times \frac{1}{3} = 9$.

(D) Let $x = \cos \theta$,where $\theta = \cos^{-1} x$. Since $x \in [-1, 1]$,$\theta \in [0, \pi]$.
Substituting $x = \cos \theta$ into the right-hand side:
$\sin^{-1} (2 \cos \theta \sqrt{1 - \cos^2 \theta}) = \sin^{-1} (2 \cos \theta \sin \theta) = \sin^{-1} (\sin 2 \theta)$.
For this to equal $2 \cos^{-1} x = 2 \theta$,we require $\sin^{-1} (\sin 2 \theta) = 2 \theta$.
This holds when $2 \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,which implies $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.
Given $\theta \in [0, \pi]$,the intersection is $\theta \in [0, \frac{\pi}{4}]$.
Since $\theta = \cos^{-1} x$,we have $0 \leq \cos^{-1} x \leq \frac{\pi}{4}$.
Taking the cosine of all sides (noting that $\cos$ is a decreasing function in this interval):
$\cos(0) \geq x \geq \cos(\frac{\pi}{4})$,
which gives $1 \geq x \geq \frac{1}{\sqrt{2}}$ or $\frac{1}{\sqrt{2}} \leq x \leq 1$.

(C) We use the trigonometric identities $\sec ^2 \theta = 1 + \tan ^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot ^2 \theta$.
Substituting $\theta = \tan ^{-1} 2$ and $\theta = \cot ^{-1} 3$ respectively:
$= (1 + \tan ^2(\tan ^{-1} 2)) + (1 + \cot ^2(\cot ^{-1} 3))$
$= (1 + (\tan(\tan ^{-1} 2))^2) + (1 + (\cot(\cot ^{-1} 3))^2)$
$= (1 + 2^2) + (1 + 3^2)$
$= (1 + 4) + (1 + 9)$
$= 5 + 10 = 15$.

(B) For the function $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ to be defined,the expression inside the square root must be strictly greater than zero.
$\Rightarrow (x - 2)(x - 5) > 0$
The critical points are $x = 2$ and $x = 5$.
Testing the intervals $(-\infty, 2)$,$(2, 5)$,and $(5, \infty)$:
For $x < 2$,$(x - 2)(x - 5) > 0$.
For $2 < x < 5$,$(x - 2)(x - 5) < 0$.
For $x > 5$,$(x - 2)(x - 5) > 0$.
Thus,the domain is $x \in (-\infty, 2) \cup (5, \infty)$.

(D) Given $f(x) = \sin([\pi^2]x) - \sin([-\pi^2]x)$.
Since $\pi^2 \approx 9.869$,we have $[\pi^2] = 9$.
Since $-\pi^2 \approx -9.869$,we have $[-\pi^2] = -10$.
Thus,$f(x) = \sin(9x) - \sin(-10x) = \sin(9x) + \sin(10x)$.
Check options:
$A) f(0) = \sin(0) + \sin(0) = 0$ (True).
$B) f(\frac{\pi}{2}) = \sin(\frac{9\pi}{2}) + \sin(5\pi) = 1 + 0 = 1$ (True).
$C) f(\frac{\pi}{4}) = \sin(\frac{9\pi}{4}) + \sin(\frac{10\pi}{4}) = \sin(\frac{\pi}{4}) + \sin(\frac{5\pi}{2}) = \frac{1}{\sqrt{2}} + 1$ (True).
$D) f(\pi) = \sin(9\pi) + \sin(10\pi) = 0 + 0 = 0$. Since $0 \neq -1$,this statement is not true.

(A) For $f(x) = \sin x$ on $[0, \frac{\pi}{2}]$,the function is strictly increasing,so it is one-one.
For $g(x) = \cos x$ on $[0, \frac{\pi}{2}]$,the function is strictly decreasing,so it is one-one.
Thus,Statement $(I)$ is true.
Now,consider $(f+g)(x) = \sin x + \cos x$.
We evaluate the function at the endpoints of the interval:
$(f+g)(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.
$(f+g)(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1$.
Since $(f+g)(0) = (f+g)(\frac{\pi}{2})$ but $0 \neq \frac{\pi}{2}$,the function $f+g$ is not one-one.
Thus,Statement $(II)$ is false.

(C) $f(x) = x|x| = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases}$. The derivative $f'(x) = 2|x|$ exists for all $x$,so it is differentiable in $(-1, 1)$. Thus,$(a) \to (ii)$.
$(b)$ $f(x) = \sqrt{|x|}$. At $x=0$,the left-hand derivative is $\lim_{h \to 0^-} \frac{\sqrt{-h}-0}{h} = -\infty$ and right-hand derivative is $\infty$. Thus,it is not differentiable at $x=0$,which is in $(-1, 1)$. Thus,$(b) \to (iv)$.
$(c)$ $f(x) = x+[x]$. This function is strictly increasing in $(-1, 1)$ because $x$ is strictly increasing and $[x]$ is non-decreasing. Thus,$(c) \to (iii)$.
$(d)$ $f(x) = |x-1|+|x+1|$. In $(-1, 1)$,$f(x) = -(x-1) + (x+1) = 2$,which is a constant function and therefore continuous in $(-1, 1)$. Thus,$(d) \to (i)$.
The correct matching is $a-(ii), b-(iv), c-(iii), d-(i)$.

(B) To check the continuity at $x=0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
For $LHL$: $\lim_{x \to 0^-} f(x) = \lim_{h \to 0} \frac{e^{1/(0-h)}-1}{e^{1/(0-h)}+1} = \lim_{h \to 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$. Since $e^{-1/h} \to 0$ as $h \to 0^+$,$LHL$ $= \frac{0-1}{0+1} = -1$.
For $RHL$: $\lim_{x \to 0^+} f(x) = \lim_{h \to 0} \frac{e^{1/h}-1}{e^{1/h}+1} = \lim_{h \to 0} \frac{1-e^{-1/h}}{1+e^{-1/h}} = \frac{1-0}{1+0} = 1$.
Since $LHL$ $\neq$ $RHL$,the limit does not exist at $x=0$.
Therefore,the function is not continuous at $x=0$.

(C) For $f(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.
Left-hand limit $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (e^x + ax) = e^0 + a(0) = 1$.
Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} b(x - 1)^2 = b(0 - 1)^2 = b$.
Since $f(x)$ is continuous,$LHL = RHL$,so $b = 1$.
Now,for differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.
$LHD = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} (e^x + a) = e^0 + a = 1 + a$.
$RHD = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 2b(x - 1) = 2b(0 - 1) = -2b$.
Equating $LHD$ and $RHD$: $1 + a = -2b$.
Substituting $b = 1$: $1 + a = -2(1) \Rightarrow 1 + a = -2 \Rightarrow a = -3$.
Thus,$a = -3$ and $b = 1$.

(C) Given $y = \frac{\cos x}{1+\sin x}$.
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(-\sin x)(1+\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}$
$\frac{dy}{dx} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2} = \frac{-(\sin x + \sin^2 x + \cos^2 x)}{(1+\sin x)^2}$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$\frac{dy}{dx} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x}$. This matches option $(a)$.
Now,simplify $\frac{-1}{1+\sin x}$ using trigonometric identities:
$1+\sin x = 1+\cos\left(\frac{\pi}{2}-x\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$.
Therefore,$\frac{dy}{dx} = \frac{-1}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = -\frac{1}{2}\sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$. This matches option $(c)$.
Thus,both $(a)$ and $(c)$ are correct.

(D) Given $y=a \sin ^3 t$ and $x=a \cos ^3 t$.
First,differentiate $y$ with respect to $t$: $\frac{d y}{d t} = 3a \sin ^2 t \cos t$.
Next,differentiate $x$ with respect to $t$: $\frac{d x}{d t} = 3a \cos ^2 t (-\sin t) = -3a \cos ^2 t \sin t$.
Now,find $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3a \sin ^2 t \cos t}{-3a \cos ^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t$.
At $t = \frac{3 \pi}{4}$,$\frac{d y}{d x} = -\tan(\frac{3 \pi}{4}) = -(-1) = 1$.

(B) Let $u = \sin x$ and $v = \log x$.
We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$.
Using the chain rule,we have $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,find the derivative of $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$.
Next,find the derivative of $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$.
Now,substitute these into the formula: $\frac{du}{dv} = \frac{\cos x}{1/x} = x \cos x$.

(A) Given the function $f(x) = \tan x - x$.
To determine the nature of the function,we find its derivative with respect to $x$:
$f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1$.
Using the trigonometric identity $1 + \tan^2 x = \sec^2 x$,we have $\sec^2 x - 1 = \tan^2 x$.
Thus,$f'(x) = \tan^2 x$.
Since the square of any real number is always non-negative,$\tan^2 x \geq 0$ for all $x$ in the domain.
Therefore,$f'(x) \geq 0$,which implies that the function $f(x)$ is always increasing.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
Take $x^4$ common from the bracket: $I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.
Let $t = 1 + \frac{1}{x^4} = 1 + x^{-4}$.
Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int (1 + x^{-4})^{-3/4} (x^{-5} dx) = \int t^{-3/4} \left(-\frac{1}{4} dt\right) = -\frac{1}{4} \int t^{-3/4} dt$.
Integrating with respect to $t$:
$I = -\frac{1}{4} \left(\frac{t^{1/4}}{1/4}\right) + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:
$I = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c$.

(A) To evaluate the integral $\int \frac{dx}{(x+1)(x+2)}$,we use partial fractions.
We can write the integrand as:
$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$
$1 = A(x+2) + B(x+1)$
Setting $x = -1$,we get $A = 1$.
Setting $x = -2$,we get $B = -1$.
Thus,$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$.
Integrating both sides:
$\int \left( \frac{1}{x+1} - \frac{1}{x+2} \right) dx = \int \frac{1}{x+1} dx - \int \frac{1}{x+2} dx$
$= \log |x+1| - \log |x+2| + c$
$= \log \left| \frac{x+1}{x+2} \right| + c$

(D) Let $f(x) = \sin^5 x \cos^4 x$.
We check if the function is even or odd by evaluating $f(-x)$.
$f(-x) = \sin^5(-x) \cos^4(-x) = (-\sin x)^5 (\cos x)^4 = -\sin^5 x \cos^4 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) \, dx = 0$.
Therefore,$\int_{-1}^1 \sin^5 x \cos^4 x \, dx = 0$.

(A) We know that $1 + \sin \theta = \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2$.
Substituting $\theta = \frac{x}{2}$,we get $\sqrt{1 + \sin \frac{x}{2}} = |\cos \frac{x}{4} + \sin \frac{x}{4}|$.
Since $x \in [0, 2\pi]$,$\frac{x}{4} \in [0, \frac{\pi}{2}]$,where both $\sin \frac{x}{4}$ and $\cos \frac{x}{4}$ are positive.
Thus,the integral becomes $\int_0^{2\pi} (\cos \frac{x}{4} + \sin \frac{x}{4}) dx$.
Evaluating the integral: $[4 \sin \frac{x}{4} - 4 \cos \frac{x}{4}]_0^{2\pi}$.
$= (4 \sin \frac{\pi}{2} - 4 \cos \frac{\pi}{2}) - (4 \sin 0 - 4 \cos 0)$.
$= (4(1) - 4(0)) - (4(0) - 4(1)) = 4 - (-4) = 8$.

(B) Let $I = \int_0^1 \log \left(\frac{1-x}{x}\right) d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) d x$
$I = \int_0^1 \log \left(\frac{x}{1-x}\right) d x$
Adding the two expressions for $I$:
$2I = \int_0^1 \left[ \log \left(\frac{1-x}{x}\right) + \log \left(\frac{x}{1-x}\right) \right] d x$
$2I = \int_0^1 \log \left( \frac{1-x}{x} \times \frac{x}{1-x} \right) d x$
$2I = \int_0^1 \log(1) d x$
Since $\log(1) = 0$,we have $2I = 0$,which implies $I = 0$.

(C) The area $A$ is given by the definite integral of the function $y = \sin \left(\frac{x}{3}\right)$ from $x = 0$ to $x = 3\pi$.
$A = \int_0^{3 \pi} \sin \left(\frac{x}{3}\right) dx$
We know that the integral of $\sin(kx)$ is $-\frac{1}{k} \cos(kx)$. Here $k = \frac{1}{3}$,so the integral is $-3 \cos \left(\frac{x}{3}\right)$.
$A = \left[ -3 \cos \left(\frac{x}{3}\right) \right]_0^{3 \pi}$
$A = -3 \left[ \cos \left(\frac{3 \pi}{3}\right) - \cos \left(\frac{0}{3}\right) \right]$
$A = -3 [ \cos(\pi) - \cos(0) ]$
Since $\cos(\pi) = -1$ and $\cos(0) = 1$:
$A = -3 [ -1 - 1 ] = -3 [ -2 ] = 6 \text{ sq. units}$.

(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:
$A = 2 \int_0^{16} x \, dy = 2 \int_0^{16} \sqrt{y} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_0^{16}$
$A = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_0^{16}$
$A = \frac{4}{3} \left( 16^{3/2} - 0^{3/2} \right)$
$A = \frac{4}{3} \times (4^2)^{3/2} = \frac{4}{3} \times 4^3$
$A = \frac{4}{3} \times 64 = \frac{256}{3} \text{ sq. units}$

(D) The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2 y}{d x^2}$,so the order $a = 2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of the highest derivative $\frac{d^2 y}{d x^2}$ is $2$,so the degree $b = 2$.
Therefore,$a - b = 2 - 2 = 0$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
First,we find the Integrating Factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values:
$y \cdot \sec x = \int \sec x \cdot \sec x dx + c$
$y \sec x = \int \sec^2 x dx + c$
$y \sec x = \tan x + c$.

(C) Given vectors are $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 4\hat{k}$,and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.
Since $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$,their dot product must be zero:
$(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$
First,calculate $\vec{a} + \lambda\vec{b}$:
$\vec{a} + \lambda\vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k}) = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + 4\lambda)\hat{k}$
Now,take the dot product with $\vec{c} = \hat{i} + \hat{j} + \hat{k}$:
$((1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + 4\lambda)\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(1 + \lambda)(1) + (2 - \lambda)(1) + (1 + 4\lambda)(1) = 0$
$1 + \lambda + 2 - \lambda + 1 + 4\lambda = 0$
$4 + 4\lambda = 0$
$4\lambda = -4$
$\lambda = -1$

(D) We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
Substituting the given values: $12 = 10 \times 2 \times \cos \theta$.
$12 = 20 \cos \theta \implies \cos \theta = \frac{12}{20} = \frac{3}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Now,$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
$|\vec{a} \times \vec{b}| = 10 \times 2 \times \frac{4}{5} = 20 \times \frac{4}{5} = 16$.

(A) Statement $(I)$ : The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. If $|\vec{a}|=0$ or $|\vec{b}|=0$,then $\vec{a} \cdot \vec{b} = 0 \times |\vec{b}| \cos \theta = 0$ or $\vec{a} \cdot \vec{b} = |\vec{a}| \times 0 \times \cos \theta = 0$. Thus,Statement $(I)$ is true.
Statement $(II)$ : The cross product of two vectors is defined as $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n}$. If $\vec{a} \times \vec{b} = \vec{0}$,then $\sin \theta = 0$,which implies $\theta = 0$ or $\theta = \pi$. This means the vectors are parallel or collinear,not perpendicular. Thus,Statement $(II)$ is false.

(A) The direction cosines of a line are given by $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha, \beta, \gamma$ are the angles made with the $x, y, z$ axes respectively.
Here,$\alpha = 90^{\circ}, \beta = 60^{\circ}, \gamma = \theta$.
The sum of the squares of the direction cosines is always $1$,so $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2 90^{\circ} + \cos^2 60^{\circ} + \cos^2 \theta = 1$.
$(0)^2 + (\frac{1}{2})^2 + \cos^2 \theta = 1$.
$0 + \frac{1}{4} + \cos^2 \theta = 1$.
$\cos^2 \theta = 1 - \frac{1}{4} = \frac{3}{4}$.
Since $\theta$ is acute,$\cos \theta = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$ or $\frac{\pi}{6}$ radians.

(B) Let the required line pass through the point $P(0,1,2)$ with direction ratios $(a, b, c)$.
The equation of the line is $\frac{x-0}{a} = \frac{y-1}{b} = \frac{z-2}{c}$.
The given line is $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$,which has direction ratios $(2, 3, -2)$.
Since the two lines are perpendicular,the dot product of their direction ratios must be zero:
$2a + 3b - 2c = 0$.
Checking the options:
For option $C$: The direction ratios are $(3, 4, 3)$.
$2(3) + 3(4) - 2(3) = 6 + 12 - 6 = 12 \neq 0$.
For option $B$: The direction ratios are $(-3, 4, 3)$.
$2(-3) + 3(4) - 2(3) = -6 + 12 - 6 = 0$.
Thus,the line $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$ is perpendicular to the given line.

(A) Both Statement $(I)$ and Statement $(II)$ are true in a Linear Programming Problem $(LPP)$:
Statement $(I)$:
The objective function in an $LPP$ is always linear,meaning it can be expressed as a linear equation with the variables raised to the power of $1$.
Statement $(II)$:
The linear inequalities that restrict the variables in an $LPP$ are called constraints.
Explanation:
In a linear programming problem,you are trying to optimize (maximize or minimize) an objective function (a linear equation) while adhering to certain constraints (linear inequalities) that limit the possible values of the variables.

(C) To find the maximum value of the objective function $z=3x+4y$,we identify the feasible region defined by the constraints:
$1$. $x+y \leq 40$
$2$. $x+2y \leq 60$
$3$. $x, y \geq 0$
The corner points of the feasible region are determined by the intersection of these lines and the axes:
- Intersection of $x+y=40$ and $x+2y=60$: Subtracting the first from the second gives $y=20$,which implies $x=20$. Point: $(20, 20)$.
- Intersection of $x+y=40$ with the $x$-axis $(y=0)$: Point $(40, 0)$.
- Intersection of $x+2y=60$ with the $y$-axis $(x=0)$: Point $(0, 30)$.
- The origin $(0, 0)$ is also a corner point.
Now,evaluate $z=3x+4y$ at each corner point:
- At $(0, 0)$: $z = 3(0) + 4(0) = 0$
- At $(40, 0)$: $z = 3(40) + 4(0) = 120$
- At $(0, 30)$: $z = 3(0) + 4(30) = 120$
- At $(20, 20)$: $z = 3(20) + 4(20) = 60 + 80 = 140$
The maximum value is $140$ at the point $(20, 20)$.

(D) Given that $A$ and $B$ are non-mutually exclusive events,we have $P(A \cap B) \neq 0$.
Given the condition $P(A \mid B) = P(B \mid A)$.
Using the definition of conditional probability,we have:
$\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$
Since $P(A \cap B) = P(B \cap A)$,we can divide both sides by $P(A \cap B)$ because $P(A \cap B) \neq 0$:
$\frac{1}{P(B)} = \frac{1}{P(A)}$
Therefore,$P(A) = P(B)$.

(C) Statement $(I)$: If $E$ and $F$ are independent,then $P(E \cap F) = P(E)P(F)$. We know that $P(E^{\prime} \cap F^{\prime}) = P((E \cup F)^{\prime}) = 1 - P(E \cup F) = 1 - [P(E) + P(F) - P(E \cap F)] = 1 - P(E) - P(F) + P(E)P(F) = (1 - P(E))(1 - P(F)) = P(E^{\prime})P(F^{\prime})$. Thus,$E^{\prime}$ and $F^{\prime}$ are independent. Statement $(I)$ is true.
Statement $(II)$: If $E$ and $F$ are mutually exclusive,then $P(E \cap F) = 0$. For them to be independent,we require $P(E \cap F) = P(E)P(F)$. Since $P(E) > 0$ and $P(F) > 0$,$P(E)P(F) > 0$. Thus,$P(E \cap F) \neq P(E)P(F)$,meaning they cannot be independent. Statement $(II)$ is true.

(D) Let $A$ be the event that Meera visits temple $A$ and $B$ be the event that she visits temple $B$. Let $F$ be the event that she meets her friend.
Given:
$P(A) = \frac{2}{5}$
$P(B) = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$
$P(F|A) = \frac{1}{3}$
$P(F|B) = \frac{2}{7}$
We need to find the probability that she met her friend at temple $B$,which is $P(B|F)$.
Using Bayes' Theorem:
$P(B|F) = \frac{P(B) \times P(F|B)}{P(A) \times P(F|A) + P(B) \times P(F|B)}$
$P(B|F) = \frac{\frac{3}{5} \times \frac{2}{7}}{(\frac{2}{5} \times \frac{1}{3}) + (\frac{3}{5} \times \frac{2}{7})}$
$P(B|F) = \frac{\frac{6}{35}}{\frac{2}{15} + \frac{6}{35}}$
Finding a common denominator for the denominator: $15 = 3 \times 5$ and $35 = 7 \times 5$,so $LCM$ is $105$.
$P(B|F) = \frac{\frac{6}{35}}{\frac{14}{105} + \frac{18}{105}} = \frac{\frac{18}{105}}{\frac{14+18}{105}} = \frac{18}{32} = \frac{9}{16}$