MHT CET 2021 24th September Evening Shift (Mathematics) Question Paper with Solutions

We have $\bar{a}=3 \hat{i}+\hat{j}-\hat{k}, \bar{b}=2 \hat{i}-\hat{j}+23 \hat{k}$ and $\bar{c}=7 \hat{i}-\hat{j}+23 \hat{k}$

$\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=6-1-23 \neq 0$ and $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=14+1+529 \neq 0$

Thus $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are not mutually perpendicular.

Also for $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}, \frac{3}{2} \neq-1 \neq \frac{-1}{23}$.

Thus $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are not collinear.

Now $$\left|\begin{array}{ccc}3 & 1 & -1 \\ 2 & -1 & 23 \\ 7 & -1 & 23\end{array}\right|=3(-23+23)-(46-161)-(-2+7) \neq 0$$.

Thus $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non coplanar.

$\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}=\lambda \quad \text{.... (say)}$

Hence coordinates of any point on the given line are $(\lambda-2,2 \lambda+1,-2 \lambda-1)$.

This point is at a distance of 12 units from $(-2,1,-1)$.

$$\begin{aligned} & \text { This point is at a distance of } 12 \text { units from }(-2,1,-1) . \\ & \therefore 12=\sqrt{(\lambda-2+2)^2+(2 \lambda+1-1)^2+(-2 \lambda-1+1)^2}=\sqrt{\lambda^2+4 \lambda^2+4 \lambda^2}=3 \lambda \\ & \therefore \lambda= \pm 4 \Rightarrow \text { Required point }=(2,9,-9) \text { or }(-6,-7,7) \end{aligned}$$

$\therefore \lambda= \pm 4 \Rightarrow$ Required point $=(2,9,-9)$ or $(-6,-7,7)$

To find the volume of the parallelepiped formed by the coterminus edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$, we need to evaluate the scalar triple product of these vectors. The scalar triple product for vectors $\bar{u}, \bar{v}, \bar{w}$ is given by the determinant:

$|\bar{u} \cdot (\bar{v} \times \bar{w})|$

For the given vectors $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$, let's denote them as follows:

$\bar{u} = \bar{a}+\bar{b}$

$\bar{v} = \bar{b}+\bar{c}$

$\bar{w} = \bar{c}+\bar{a}$

Now, we need to compute the scalar triple product $(\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}))$.

We start by finding the cross product $(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a})$:

$(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}) = (\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a})$

Since the cross product of any vector with itself is zero, $(\bar{c} \times \bar{c}) = 0$, we simplify this to:

$(\bar{b}+\bar{c}) \times (\bar{c}+\bar{a}) = (\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})$

Next, we take the dot product of $\bar{a}+\bar{b}$ with this result:

$(\bar{a}+\bar{b}) \cdot \left[ (\bar{b} \times \bar{c}) + (\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a}) \right]$

Using the distributive property of the dot product over addition, we have:

$(\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{c}) + (\bar{a}+\bar{b}) \cdot (\bar{b} \times \bar{a}) + (\bar{a}+\bar{b}) \cdot (\bar{c} \times \bar{a})$

We know that the dot product of a vector with the cross product of two other vectors follows the property that:

$\bar{x} \cdot (\bar{y} \times \bar{z}) = [\bar{x} \bar{y} \bar{z}]$

Applying this property, we get:

$[(\bar{a}+\bar{b}) \bar{b} \bar{c}] + [(\bar{a}+\bar{b}) \bar{b} \bar{a}] + [(\bar{a}+\bar{b}) \bar{c} \bar{a}]$

We evaluate each of these scalar triple products separately:

$[(\bar{a}+\bar{b}) \bar{b} \bar{c}] = [\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{b} \bar{c}] = [\bar{a} \bar{b} \bar{c}] + 0 = [\bar{a} \bar{b} \bar{c}]$

$[(\bar{a}+\bar{b}) \bar{b} \bar{a}] = [\bar{a} \bar{b} \bar{a}] + [\bar{b} \bar{b} \bar{a}] = 0 + 0 = 0$

$[(\bar{a}+\bar{b}) \bar{c} \bar{a}] = [\bar{a} \bar{c} \bar{a}] + [\bar{b} \bar{c} \bar{a}] = 0 + [\bar{b} \bar{c} \bar{a}] = [\bar{b} \bar{c} \bar{a}]$

Summing these, we have:

$[\bar{a} \bar{b} \bar{c}] + 0 + [\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}] + [\bar{b} \bar{c} \bar{a}] = 2[\bar{a} \bar{b} \bar{c}]$

Therefore, the volume of the parallelepiped is:

$2[\bar{a} \bar{b} \bar{c}]$

Hence, the correct answer is Option D:

Option D: $2[\bar{a} \bar{b} \bar{c}]$

Here given plane passes through the point $(2,0,1)$ and let $\bar{b}=\hat{i}$ and $\bar{c}=\hat{i}+2 \hat{j}-3 \hat{k}$

Normal vector $$\overline{\mathrm{n}}=\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}\hat{i} & \hat{\mathrm{j}} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 2 & -3\end{array}\right|=3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$$

The equation of plane in scalar product from is $\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}$

Here $\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=2$

$\therefore \alpha=2$

Let

\begin{aligned} I & =\int_\limits2^5 2[x] d x \\ & =\int_\limits2^3 2(2) d x+\int_\limits3^4 2(3) d x+\int_\limits4^5 2(4) d x \\ & =4[x]_2^3+6[x]_3^4+8[x]_4^5=4+6+8=18 \end{aligned}

$$\begin{aligned} & \sqrt{\frac{d y}{d x}}-4 \frac{d y}{d x}-7 x=0 \\ & \therefore \sqrt{\frac{d y}{d x}}=4 \frac{d y}{d x}+7 x \text { and squaring both sides, we get } \\ & \frac{d y}{d x}=16\left(\frac{d y}{d x}\right)^2+56 x\left(\frac{d y}{d x}\right)+49 x^2 \\ & \therefore \text { Order }=1 \text {, degree }=2. \end{aligned}$$

$$\begin{aligned} & {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right]} \\ & R_3 \rightarrow R_3-R_1 \text { and } R_2 \rightarrow R_2-2 R_1 \\ & {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ -8 \\ -2 \end{array}\right]} \end{aligned}$$

$$\begin{aligned} \therefore x-y+z & =4 \quad \text{.... (1)}\\ 3 y-5 z & =-8 \quad \text{.... (2)}\\ 2 y & =-2 \quad \text{.... (3)} \end{aligned}$$

$$\begin{aligned} & \therefore y=-1 \quad \text {.....[from (3)]}\\ & \therefore-3-5 z=-8 \Rightarrow z=1 \quad \text {.....[from (2)]}\\ & \therefore x-(-1)+1=4 \Rightarrow x=2 \quad \text {.....[from (1)]}\\ & \therefore 2 x+y-z=2(2)-1-1=2 \end{aligned}$$

Let's start by breaking down the given statement and then simplify it.

The given statement is: "It is not true, if it is raining then weather is not pleasant."

In logical terms, we can write this statement as:

$\neg(p \rightarrow \neg q)$

To simplify this, we need to understand the implication operator ($\rightarrow$) in terms of logical operators. The implication $p \rightarrow \neg q$ is equivalent to $\neg p \lor \neg q$.

Therefore, the given statement can be rewritten as:

$\neg(\neg p \lor \neg q)$

Next, we apply De Morgan's laws to simplify $\neg(\neg p \lor \neg q)$, which gives us:

$p \land q$

So, the simplified form of the given statement is:

$p \land q$, which means "It is raining and the weather is pleasant."

Therefore, the correct option is:

Option D: It is raining and the weather is pleasant.

$$\begin{aligned} & x=e^t(\sin t-\cos t) \text { and } y=e^t(\sin t+\cos t) \\ & \therefore \quad \frac{d x}{d t}=e^t(\sin t-\cos t)+e^t(\cos t+\sin t)=2 e^t \sin t \\ & \quad \frac{d y}{d t}=e^t(\sin t+\cos t)+e^t(\cos t-\sin t)=2 e^t \cos t \\ & \therefore \quad \frac{d y}{d x}=\frac{2 e^t \cos t}{2 e^t \sin t} \Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}$$

$x^2-10x+y^2=0$

$x-10 x+25+y^2=25 \quad \Rightarrow$ centre $=(5,0)$ and $r=5$

$y=2 x$ is a chord of given circle.

Point of intersection of chord and circle is

$$\begin{aligned} & \text { Point of intersection of } \\ & x^2-10 x+25+4 x^2=25 \Rightarrow 5 x^2-10 x=0 \end{aligned} \quad \Rightarrow x^2-2 x=0$$

i.e. $x(x-2)=0 \Rightarrow x=0,2 \Rightarrow y=0,4$

Thus end points of the chord are $(0,0)$ and $(2,4)$

Mid point of the chord $=\left(\frac{2}{2}, \frac{4}{2}\right)=(1,2)$ and length of chord $=\sqrt{(2)^2+(4)^2}=\sqrt{20}$ is the diameter of required circle.

Hence equation of required circle is $(x-1)+(y-2)=\left(\frac{\sqrt{20}}{2}\right)^2$

i.e. $x^2+y^2-2 x-4 y=0$

To convert polar coordinates to Cartesian coordinates, we use the following formulas:

$x = r \cos(\theta)$

$y = r \sin(\theta)$

Given the polar coordinates $\left(2, \frac{\pi}{4}\right)$, we have:

$r = 2$

$\theta = \frac{\pi}{4}$

Now, let's calculate the Cartesian coordinates:

$x = 2 \cos\left(\frac{\pi}{4}\right)$

$y = 2 \sin\left(\frac{\pi}{4}\right)$

We know that:

$\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

So, substituting these values, we get:

$x = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

$y = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

Therefore, the Cartesian coordinates are $\left(\sqrt{2}, \sqrt{2}\right)$, which corresponds to Option A.

Refer figure.

Required area is shaded.

Area is unbounded.

Probability of getting $6=\frac{1}{6}$

$\mathrm{X}$ denotes number of times of getting 6. So $\mathrm{x}$ can take values $0,1,2$.

$$\begin{aligned} & P(x=0)=\frac{5}{6} \times \frac{5}{6}=\frac{25}{36} \\ & P(x=1) \quad=\left(\frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6}\right)=\frac{10}{36} \\ & P(x=2) \quad=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36} \\ & E(x)=\Sigma p_i x_i \\ & =\left(\frac{25}{36}\right)(0)+\left(\frac{10}{36}\right)(1)+\left(\frac{1}{36}\right)(2)=\frac{10+2}{36}=\frac{1}{3} \end{aligned}$$

$\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+8 \cot 8 \mathrm{~A}$

$=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+\frac{8}{\left(\frac{2 \tan 4 \mathrm{~A}}{1-\tan ^2 4 \mathrm{~A}}\right)}=\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+\frac{8\left(1-\tan ^2 4 \mathrm{~A}\right)}{2 \tan 4 \mathrm{~A}}$

$$\begin{aligned} & =\tan A+2 \tan 2 A+\frac{(4 \tan 4 A)(2 \tan 4 A)+8\left(1-\tan ^2 4 A^{\circ}\right)}{2 \tan 4 A} \\ & =\tan A+2 \tan 2 A+\frac{8}{2 \tan 4 A} \\ & =\tan A+2 \tan 2 A+\frac{8}{2\left(\frac{2 \tan 2 A}{1-\tan ^2 2 A}\right)}=\tan A+2 \tan 2 A+\frac{8\left(1-\tan ^2 2 A\right)}{4 \tan 2 A} \end{aligned}$$

$$\begin{aligned} & =\tan \mathrm{A}+\frac{(2 \tan 2 \mathrm{~A})(4 \tan 2 \mathrm{~A})+8\left(1-\tan ^2 2 \mathrm{~A}\right)}{4 \tan 2 \mathrm{~A}}=\tan \mathrm{A}+\frac{8}{4 \tan 2 \mathrm{~A}} \\ & =\tan \mathrm{A}+\frac{8}{4\left(\frac{2 \tan \mathrm{A}}{1-\tan ^2 \mathrm{~A}}\right)}=\tan \mathrm{A}+\frac{8\left(1-\tan ^2 \mathrm{~A}\right)}{8 \tan \mathrm{A}} \\ & =\frac{\tan \mathrm{A}(8 \tan \mathrm{A})+8\left(1-\tan ^2 \mathrm{~A}\right)}{8 \tan \mathrm{A}}=\frac{8}{8 \tan \mathrm{A}}=\cot \mathrm{A} \end{aligned}$$

We have $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$

$\bar{a} \cdot \bar{b}=|\bar{a}| \cdot|\bar{b}| \cdot \cos \theta$, where $\theta$ is angle between $\bar{a}$ and $\bar{b}$.

$\therefore \cos \theta=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}|}<0$, as $\theta$ is an obtuse angle.

$\therefore|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}|<0$

$\therefore\left(2 \lambda^2\right)(7)+(4 \lambda)(-2)+(1)(\lambda)<0$

$\therefore 14 \lambda^2-7 \lambda<0 \Rightarrow 7 \lambda(2 \lambda-1)<0 \Rightarrow \lambda(2 \lambda-1)<0$

$\therefore 0<\lambda<\frac{1}{2}$ i.e. $\lambda \in\left(0, \frac{1}{2}\right)$

$$\begin{aligned} & y=1+x e^y \\ & \therefore \frac{d y}{d x}=0+x e^y \frac{d y}{d x}+e^y \\ & \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1} \\ & \therefore \frac{d y}{d x}=\frac{-e^y}{\left(1+x e^y\right)-2}=\frac{-e^y}{y-2}=\frac{e^y}{2-y} \end{aligned}$$

$A(7,-8,1) ; B(p, q, 5)$ and $C(q+1,5 p, 0)$ are vertices of $\triangle A B C$ having centroid $G(3,-5, r)$

$$\begin{aligned} & \therefore 3=\frac{7+p+q+1}{3}, \quad-5=\frac{-8+q+5 p}{3}, r=\frac{1+5+0}{3} \\ & \therefore p+q=1 \quad \ldots(1), 5 p+q=-7 \quad \ldots(2), r=2 \end{aligned}$$

Solving (1) and (2), we get $\mathrm{p}=-2, \mathrm{q}=3$

$$\begin{aligned} & f(x)=x, \quad \text { if } x \leq 0 \\ &=0, \quad \text { if } x>0 \\ & \therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{n \rightarrow 0} x=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=0 \\ & f(0)=0 \end{aligned}$$

Thus $f(x)$ is continuous at $x=0$.

$$\begin{aligned} \mathrm{f}^{\prime}(\mathrm{x})=1, & \text { if } \mathrm{x} \leq 0 \\ =0, & \text { if } \mathrm{x}>0 \end{aligned}$$

Thus $f(x)$ is not differentiable at $x=0$.

We have $\mathrm{n}=50, \Sigma \mathrm{x}_{\mathrm{i}}^2=3050, \overline{\mathrm{x}}=6$

$$\begin{aligned} \text { S.D. } & =\sqrt{\frac{1}{n} \sum x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{3050}{50}-(6)^2}=\sqrt{61-36}=\sqrt{25}=5 \end{aligned}$$

We need to find the number of ways to form a committee of 5 with at most 2 ladies. This means we have three possible scenarios:

1. Scenario 1: No ladies

We need to choose 5 men out of 6. This can be done in $^6C_5 = \frac{6!}{5!1!} = 6$ ways.

2. Scenario 2: One lady

We need to choose 1 lady out of 4 and 4 men out of 6. This can be done in $^4C_1 \times ^6C_4 = \frac{4!}{1!3!} \times \frac{6!}{4!2!} = 4 \times 15 = 60$ ways.

3. Scenario 3: Two ladies

We need to choose 2 ladies out of 4 and 3 men out of 6. This can be done in $^4C_2 \times ^6C_3 = \frac{4!}{2!2!} \times \frac{6!}{3!3!} = 6 \times 20 = 120$ ways.

Therefore, the total number of ways to form the committee is 6 + 60 + 120 = 186.

So the correct answer is Option B: 186

$$\begin{aligned} & f(x)=x \log x \\ & \therefore f^{\prime}(x)=\frac{x}{x}+\log x=1+\log x \end{aligned}$$

When $1+\log x=0 \Rightarrow x=\frac{1}{e}$

$f^{\prime \prime}(x)=\frac{1}{x} \Rightarrow\left[f^{\prime \prime}(x)\right]_{x=\frac{1}{e}}=e>0$

Thus $f(x)$ is minimum at $x=\frac{1}{e}$ and

$f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}$

Let

$$\begin{aligned} I & =\int \frac{d x}{e^x+e^{-x}+2} \\ & =\int \frac{d x}{e^x+\frac{1}{e^x}+2}=\int \frac{e^x d x}{e^{2 x}+2 e^x+1}=\int \frac{e^x}{\left(e^x+1\right)^2} d x \end{aligned}$$

Put $e^x+1=t \Rightarrow e^x d x=d t$

$I=\int \frac{d t}{t^2}=\int t^{-2} d t=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c=\frac{-1}{e^x+1}+c$

We have $\left(x^2+y^2\right) \sin \theta+2 x y=0$

$\therefore \mathrm{a}=\sin \theta, \mathrm{b}=\sin \theta$ and $\mathrm{h}=1$ and let $\alpha$ be the acute angle between the given lines.

$$\begin{aligned} & \text { Then } \tan \alpha=\frac{\mid 2 \sqrt{(1)^2-(\sin \theta)(\sin \theta)}\mid}{\sin \theta+\sin \theta} \\ & =\frac{\left|2 \sqrt{1-\sin ^2 \theta}\right|}{2 \sin \theta}=\frac{2 \cos \theta}{2 \sin \theta}=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right) \\ & \therefore \alpha=\frac{\pi}{2}-\theta \\ & \end{aligned}$$