MHT CET 2022 11th August Evening Shift (Mathematics) Question Paper with Solutions

Let's analyze each of the propositions given, starting with $p$ which states that for every natural number $n$, the expression $n^2 + n$ is even.

The expression $n^2 + n$ can be factored into $n(n + 1)$. Since any integer $n$ is either even or odd, if $n$ is even, $n(n + 1)$ will certainly be even because an even number times any other number is even. If $n$ is odd, then $n + 1$ is even, and again, an odd number times an even number is also even. Thus, in either case, the expression will be even, and $p$ is true.

Now let's look at $q$, which posits that for every natural number $n$, the expression $n^2 - n$ is odd.

Similarly, we can factor the expression into $n(n - 1)$, which is the product of two consecutive numbers. Just as in the first case, if $n$ is even, $n - 1$ is odd, resulting in an even number. If $n$ is odd, then $n - 1$ is even, which again results in an even number. Thus, $q$ is false because the product of two consecutive integers is always even.

Therefore, the truth value of $p$ is True (T), and the truth value of $q$ is False (F).

Next, we'll find the truth values of the compound statements:

• $p \wedge q$ is the conjunction of $p$ and $q$, and it is true if and only if both $p$ and $q$ are true. Since $q$ is false, $p \wedge q$ is false (F).


• $p \vee q$ is the disjunction of $p$ and $q$, and it is true if at least one of $p$ or $q$ is true. Since $p$ is true, $p \vee q$ is true (T).


• $p \rightarrow q$ is the conditional statement "if $p$ then $q$" and is only false if $p$ is true and $q$ is false (since the only way for an implication to be false is if a true premise leads to a false conclusion). Since this is precisely the case here, $p \rightarrow q$ is false (F).

So the correct truth values for $p \wedge q$, $p \vee q$, and $p \rightarrow q$ are respectively False (F), True (T), and False (F).

The right option based on the above reasoning is:

Option A $F, T, F$.

$\because f$ is $\uparrow$ in $(0,1]$ and $\downarrow$ in $[1,5)$

$\therefore f$ is both $\uparrow$ and $\uparrow$ at $x=1$

$\therefore f(1)=$ constant $\quad \therefore f^{\prime}(1)=0$

$$\begin{aligned} & \because f(x)=x^3-3(a-2) x^2+3 a x+7 \\ & \therefore f^{\prime}(x)=3 x^2-6(a-2) x+3 a \\ & \therefore f^{\prime}(1)=3-6(a-2)+3 a=15-3 a=0 \\ & \therefore a=5 \\ & \therefore f(x)=x^3-9 x^2+15 x+7 \\ & \therefore f(x)-14=x^3-9 x^2+15 x-7 \\ & \quad=(x-7)(x-1)^2 \quad \text{... factorising} \end{aligned}$$

$\therefore \frac{f(x)-14}{(x-1)^2}=x-7=0 \quad \therefore x=7$

$$\begin{aligned} I & =\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \mathrm{~d} x \\ & =\int \frac{\sqrt{\tan x}}{\left(\frac{\sin x}{\cos x}\right) \cdot \cos ^2 x} \mathrm{~d} x \\ & =\int \frac{\sqrt{\tan x}}{\tan x} \cdot \frac{1}{\cos ^2 x} \mathrm{~d} x \\ & =\int \frac{1}{\sqrt{\tan x}} \cdot \sec ^2 x \mathrm{~d} x \\ & =\int \frac{1}{\sqrt{t}} \mathrm{~d} t, \ldots t=\tan x \\ & =2 \cdot \sqrt{t}+C \\ & =2 \cdot \sqrt{\tan x}+C \end{aligned}$$

$$\begin{aligned} & \because \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \cdot \tan ^{-1} x \\ & \therefore 2\left[\tan ^{-1}\left(\frac{1-x}{1+1 \cdot x}\right)\right]=\tan ^{-1} x \\ & \therefore 2\left[\tan ^{-1} 1-\tan ^{-1} x\right]=\tan ^{-1} x \\ & \therefore 2\left(\frac{\pi}{4}-\tan ^{-1} x\right)=\tan ^{-1} x \end{aligned}$$

$$\begin{aligned} & \therefore \frac{\pi}{2}=3 \cdot \tan ^{-1} x \\ & \therefore \frac{\pi}{6}=\tan ^{-1} x \\ & \therefore x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \end{aligned}$$

$$\begin{aligned} & \bar{a}=\hat{\boldsymbol{i}}-\hat{\boldsymbol{k}} \equiv(1,0,-1) \\ & \bar{b}=x \hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+(1-x) \hat{\boldsymbol{k}} \equiv(x, 1,1-x) \\ & \bar{c}=y \hat{\boldsymbol{i}}+x \hat{\boldsymbol{j}}+(1+x-y) \hat{\boldsymbol{k}} \equiv(y, x, 1+x-y) \\ & \therefore\left[\begin{array}{ccc} \bar{a} & \bar{b} & \bar{c} \end{array}\right]=\left|\begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{array}\right| \\ & =\left|\begin{array}{ccc} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \end{array}\right| \quad \ldots \text { by } C_1+C_3 \\ & =1[(1)(1+x)-(x)(1)] \\ & =1+x-x \\ & =1 \\ & =\text { contains neither } x \text { nor } y \end{aligned}$$

$$\begin{aligned} & Z=5 x+2 y \\ & 2 x-y \geq 2, x+2 y \leq 8, x, y \geq 0 \\ & L_1: 2 x-y=2, L_2: x+2 y=8 \\ & \therefore A(1,0), B(0,-2) \text { lie on } L_1 \\ & \quad C(8,0), D(0,4) \text { lie on } L_2 \\ & L_1 \cap L_2 \equiv\left(\frac{12}{5}, \frac{14}{5}\right) \equiv E \end{aligned}$$

$\therefore$ Critical points are $A(1,0), C(8,0), E \equiv\left(\frac{12}{5}, \frac{14}{5}\right)$

$$\begin{aligned} \therefore Z_A & =5(1)+2(0)=5 \\ Z_C & =5(8)+2(0)=40 \\ Z_E & =5\left(\frac{12}{5}\right)+2\left(\frac{14}{5}\right)=\frac{88}{5}=17.6 \end{aligned}$$

$\therefore Z_{\text {max }}=40$

I=\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}=\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{2 \cdot \cos ^2 \frac{x}{2}}

=\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\left(\frac{1}{2} \sec ^2 \frac{x}{2}\right) \mathrm{d} x=\left[\tan \frac{x}{2}\right]_{\pi / 4}^{3 \pi / 4}

$=\tan \frac{3 \pi}{8}-\tan \frac{\pi}{8}=\frac{\sin (3 \pi / 8)}{\sin (3 \pi / 8)}-\frac{\sin (\pi / 8)}{\cos (\pi / 8)}$

$=\frac{\sin (3 \pi / 8) \cdot \cos (\pi / 8)-\cos (3 \pi / 8) \cdot \sin (\pi / 8)}{\cos (3 \pi / 8) \cdot \cos (\pi / 8)}$

$=\frac{\sin \left(\frac{3 \pi}{8}-\frac{\pi}{8}\right)}{\frac{1}{2}\left[\cos \left(\frac{3 \pi}{8}+\frac{\pi}{8}\right)+\cos \left(\frac{3 \pi}{8}-\frac{\pi}{8}\right)\right]}$

$=\frac{2 \times \sin \frac{\pi}{4}}{\cos \frac{\pi}{2}+\cos \frac{\pi}{4}}=\frac{2 \times \frac{1}{\sqrt{2}}}{0+\frac{1}{\sqrt{2}}}=2$

$$\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1} \\ & L_2=\frac{x-0}{2}=\frac{y-0}{-2}=\frac{z-0}{1} \end{aligned}$$

$\therefore$ their vector equations are

$$\begin{aligned} & \bar{r}=\bar{a}_1+m \bar{b} \text { and } \bar{r}=\bar{a}_2+n \bar{b}, \text { where } \\ & \bar{a}_1 \equiv(1,2,3), \bar{a}_2 \equiv(0,0,0), \bar{b} \equiv(2,-2,1) \end{aligned}$$

$$\begin{aligned} & \therefore\left(\bar{a}_1-\bar{a}_2\right) \times \bar{b}=\left|\begin{array}{rrr} \hat{\boldsymbol{i}} & \hat{\boldsymbol{j}} & \hat{\boldsymbol{k}} \\ -1 & -2 & -3 \\ 2 & -2 & 1 \end{array}\right| \equiv(-8,-5,6) \\ & \therefore\left|\left(\bar{a}_2-\bar{a}_1\right) \times \bar{b}\right|=\sqrt{64+25+36}=5 \sqrt{5} \\ & \quad|\bar{b}|=\sqrt{4+4+1}=3 \\ & \therefore \mathrm{d}\left(L_1, L_2\right)=\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \times \bar{b}}{b}\right|=\frac{5 \sqrt{5}}{3} \end{aligned}$$

The cotangent of an angle being equal to 0 implies that the angle in question is an odd multiple of $\frac{\pi}{2}$. This is because the cotangent function, defined as $\cot(x) = \frac{1}{\tan(x)}$, becomes undefined when $\tan(x)$ is 0, which occurs at multiples of $\pi$. And $\cot(x) = 0$ when $\tan(x)$ goes to infinity, which happens at odd multiples of $\frac{\pi}{2}$ since the tangent function has its asymptotes there.

Since $\cot(A + B) = 0$, the angle $A + B$ must be an odd multiple of $\frac{\pi}{2}$. For simplicity, let's assume $A + B = \frac{\pi}{2}$ as the result would be the same for any odd multiple:

$A + B = \frac{\pi}{2} \Rightarrow B = \frac{\pi}{2} - A$

We can use this to determine $\sin(A + 2B)$:

$\sin(A + 2B) = \sin(A + 2(\frac{\pi}{2} - A))$

$\sin(A + 2B) = \sin(A + \pi - 2A)$

$\sin(A + 2B) = \sin(\pi - A)$

Now, since sine is a function with symmetry in an odd fashion around the y-axis (odd function), we have:

$\sin(\pi - A) = \sin A$

Therefore, the answer is Option A, which is $\sin A$. This result holds true for any odd multiple of $\frac{\pi}{2}$ since the sine function is periodic with period $2\pi$. Even if $A + B$ were, say, $\frac{3\pi}{2}$ (another odd multiple of $\frac{\pi}{2}$), the result would be the same due to the periodic nature of the sine function.

$$\begin{aligned} & \because \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x+y}{x-y} \quad \therefore \text { put } y=v x \\ & \therefore v+x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{3+v}{1-v} \quad \therefore x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{3+v}{1-v}-v \\ & \therefore x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{3+v-v+v^2}{1-v} \quad \therefore x \cdot \frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{3+v^2}{1-v} \\ & \therefore \int \frac{1-v}{3+v^2} \mathrm{~d} v=\int \frac{1}{x} \mathrm{~d} v \\ & \therefore \int \frac{1}{\sqrt{3}^2+v^2} \mathrm{~d} v-\frac{1}{2} \int \frac{2 v}{3+v^2} \mathrm{~d} v=\log x \\ & \therefore \frac{1}{\sqrt{3}} \cdot \tan ^{-1}\left(\frac{v}{\sqrt{3}}\right)-\frac{1}{2} \cdot \log \left(3+v^2\right)=\log x+C \\ & \therefore \frac{1}{\sqrt{3}} \cdot \tan ^{-1}\left(\frac{y}{x \sqrt{3}}\right)-\log \left(\frac{y^2+3 x^2}{x^2}\right)^{1 / 2} \\ & \quad=\log x+C \end{aligned}$$

Note : $$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$$

$$A \cdot B=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 3\end{array}\right] \cdot\left[\begin{array}{ll}2 & 3 \\ 1 & 2 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}5 & 9 \\ 10 & 17\end{array}\right]$$

$\therefore|A B|=85-90=-5$

$$\therefore(A B)^{-1}=\frac{1}{-5}\left[\begin{array}{rr} 17 & -9 \\ -10 & 5 \end{array}\right]=\left[\begin{array}{rr} -17 / 5 & 9 / 5 \\ 2 & -1 \end{array}\right]$$

Joint equation of a pair of lines, through the origin, making an equilateral triangle with the line $y = b$, is

$\sqrt{3} x^2-y^2=0$

To find the vector perpendicular to the plane containing the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}}$, we can use the cross product of these two vectors.

The cross product is defined as:

For our vectors, this becomes:

Now that we have the perpendicular vector, let us compute the magnitude of the projection of the vector $2\hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+ \hat{\mathbf{k}}$ onto $\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$.

The projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$ is given by the formula:

But we are only interested in the magnitude of the projection, which simplifies to:

Let's find the dot product of vectors $2\hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+ \hat{\mathbf{k}}$ and $\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$:

Next, let's find the magnitude of $\mathbf{b}$, which is $\|\mathbf{b}\|$:

Finally, the magnitude of the projection is:

$$\begin{aligned} & \because y=\cos \left(\sin x^2\right) \\ & \therefore \text { at } x=\sqrt{\frac{\pi}{2}}, \\ & \mathrm{y}=\cos \left(\sin \frac{\pi}{2}\right)=\cos 1 \end{aligned}$$

$$\begin{aligned} \text { (1) } \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} & =2 x \cdot \cos x^2 \cdot\left[-\sin \left(\sin x^2\right)\right] \\ & =-2 x \cdot \cos x^2 \cdot \sin \left(\sin x^2\right) \end{aligned}$$

$$\begin{aligned} \left.\therefore \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=\sqrt{\pi / 2}} & =-2 \cdot \sqrt{\frac{\pi}{2}} \cdot \cos \frac{\pi}{2} \cdot \sin \left(\sin \frac{\pi}{2}\right) \\ & =-2 \cdot \sqrt{\frac{\pi}{2}} \cdot(0) \cdot \sin 1 \\ & =0 \end{aligned}$$

To find the new level of production when 25 more workers are employed, we need to integrate the given rate of change of production with respect to the number of workers over the interval from the current number of workers to the current number plus 25.

The rate of change of production is given by the differential equation:

To find the production function $P(x)$, we integrate the differential equation with respect to $x$:

Integrating term by term, we get:

Where $C$ is the constant of integration.

We know that when the number of workers was 0, the firm was manufacturing 2000 items. We can use this information to find the constant $C$:

Now, the production function with the constant $C$ included is:

The new level of production after employing 25 more workers is $P(25)$, provided that the initial number of workers was 0. If there was an initial number of workers $x_0$ where $x_0 \ge 0$, we need to evaluate the increase in production from $x_0$ to $x_0 + 25$.

However, without loss of generality and because no initial number of workers $x_0$ is provided, let's proceed by assuming the firm is starting from 0 workers. We add 25 workers to find:

Thus, the correct answer is 3500, which corresponds to Option D.

First, let's write down the equation given:

To solve for |z|, we need to convert this equation into something that we can easily manipulate. Let's represent z as a complex number in the standard form:

where x and y are the real and imaginary parts of z, respectively. Thus, the modulus of z, denoted by |z|, is given by:

Now, let's substitute z with x + iy in the original equation:

Separating the real and imaginary parts, we get:

Since we know that $y = 1$, we can substitute this into the first equation and then solve for $x$:

Let's isolate $\sqrt{x^2 + 1}$ on one side:

Squaring both sides to eliminate the square root:

Cancelling out $x^2$ from both sides and simplifying:

Now we can find $|z|$ using the value of $x$ and the fact that $y = 1$:

Thus, the correct answer is Option B: $\frac{5}{3}$.

To find the matrix product $A \cdot (\text{adj}A)$, we first need to understand the properties of a matrix and its adjugate (adjoint) matrix. The adjugate of a matrix $A$, denoted as $\text{adj}A$, is the transpose of the cofactor matrix of $A$. One important property relating a matrix and its adjugate involves the determinant of the matrix, given by the equation:

where $I$ is the identity matrix of the same size as $A$, and $\text{det}A$ is the determinant of the matrix $A$. This relation tells us that the product of a matrix and its adjugate is a diagonal matrix, where each diagonal element is equal to the determinant of the original matrix.

Given the matrix $A$ and the equations $xyz = 60$ and $8x + 4y + 3z = 20$, we are asked to find $A \cdot (\text{adj}A)$. According to the property mentioned, this requires us to find the determinant of the given matrix $A$.

The determinant of $A$, $\text{det}A$, for a 3x3 matrix $$\left[\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]$$ is given by: