MHT CET 2020 16th October Morning Shift (Mathematics) Question Paper with Solutions

Here, $P(3,2,1)$ is the foot of the perpendicular drawn from the origin to the plane

$\therefore O P$ is normal to the plane

Thus, direction ratio of the normal to the plane are $(3,2,1)$

Now, since the plane passes through $(3,2,1)$

$\therefore$ It equation is

$$\begin{array}{r} 3(x-3)+2(y-2)+1(z-1)=0 \\ 3 x-9+2 y-4+z-1=0 \\ 3 x+2 y+z=14 \end{array}$$

We have,

$$\begin{aligned} & I=\int \frac{\sin \theta}{\sin 3 \theta} d \theta \\ & I=\int \frac{\sin \theta}{3 \sin \theta-4 \sin ^3 \theta} d \theta \end{aligned}$$

$$\begin{aligned} & I=\int \frac{d \theta}{3-4 \sin ^2 \theta} \\ & I=\int \frac{\sec ^2 \theta d \theta}{3 \sec ^2 \theta-4 \tan ^2 \theta}\left[\because \text { divide by } \cos ^2 \theta\right] \\ & I=\int \frac{\sec ^2 \theta d \theta}{3+3 \tan ^2 \theta-4 \tan ^2 \theta} \\ & I=\int \frac{\sec ^2 \theta d \theta}{3-\tan ^2 \theta} \end{aligned}$$

Put, $\tan \theta=t \Rightarrow \sec ^2 \theta d \theta=d t$

$$\begin{aligned} \quad I & =\int \frac{d t}{(\sqrt{3})^2-t^2}=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C \\ \quad I & =\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\tan \theta}{\sqrt{3}-\tan \theta}\right|+C \\ \therefore \quad K & =\sqrt{3} \end{aligned}$$

Area of region bounded by the curve $y=4 x^3-6 x^2+4 x+1$ and line $x=1, x=5$ and $X$ - axis, is

$$\begin{aligned} & \int_1^5\left(4 x^3-6 x^2+4 x+1\right) d x \\ & =\left[x^4-2 x^3+2 x^2+x\right]_1^5 \\ & =[(625-250+50+5)-(1-2+2+1)] \\ & =430-2=428 \text { sq unit } \end{aligned}$$

We have,

$$\begin{aligned} & \sin ^{-1}\left(\frac{-1}{2}\right)+\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \\ & -\sin ^{-1}\left(\frac{1}{2}\right)+\pi-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ & =-\frac{\pi}{6}+\pi-\frac{\pi}{6}=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}=\cos ^{-1}\left(\frac{-1}{2}\right) \end{aligned}$$

We have,

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

Put, $\quad x-\frac{1}{x}=t \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t$

$$\begin{aligned} & \therefore \quad I=\int \sqrt{t} ~d t \\ & \quad I=\frac{2}{3}(t)^{3 / 2}+C \\ & \quad I=\frac{2}{3}\left(x-\frac{1}{x}\right)^{3 / 2}+C \\ & \therefore \quad K=3 / 2 \end{aligned}$$

To convert cartesian coordinates $(x,y)$ to polar coordinates $(r,\theta)$, the formulas used are:

• $r = \sqrt{x^2 + y^2}$ for the distance from the origin,
• $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ for the angle with the positive x-axis.

Given the cartesian coordinates $(-2, -2)$, let's calculate the polar coordinates.

• First, calculate the radial distance $r$:

  $r = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.$

• Next, calculate the angle $\theta$:

  Since the point is in the third quadrant (both x and y are negative), the angle $\theta$ with the positive x-axis extends to the third quadrant. The basic angle (i.e., ignoring the quadrant) is $\tan^{-1}\left(\frac{-2}{-2}\right) = \tan^{-1}(1)$.

The basic angle is $\frac{\pi}{4}$ because $\tan^{-1}(1) = \frac{\pi}{4}$. However, since the point (-2,-2) lies in the third quadrant, the actual angle $\theta$ we're interested in is $\pi + \frac{\pi}{4} = \frac{5\pi}{4}$.

So, the correct polar coordinates of the point $(-2, -2)$ are $(2\sqrt{2}, \frac{5\pi}{4})$.

Therefore, the correct option is:

Option D $\left(2 \sqrt{2}, \frac{5 \pi}{4}\right)$.

We have,

$I=\int_0^a \sqrt{\frac{x}{a-x}} d x$

Put, $x=a \sin ^2 \theta \Rightarrow d x=2 a \sin \theta \cos \theta d \theta$

When, $x=Q, \theta=0$ and $x=a, \theta=\pi / 2$

$$\begin{aligned} \therefore \quad I & =\int_0^{\pi / 2} \sqrt{\frac{a \sin ^2 \theta}{a\left(1-\sin ^2 \theta\right)}} 2 a \sin \theta \cos \theta d \theta \\ \quad I & =2 a \int_0^{\pi / 2} \frac{\sin \theta}{\cos \theta}(\sin \theta \cos \theta) d \theta \\ \quad I & =a \int_0^{\pi / 2} 2 \sin ^2 \theta d \theta \\ I & =a \int_0^{\pi / 2}(1-\cos 2 \theta) d \theta \\ I &=a\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2} \\ I & =a\left[\frac{\pi}{2}-0\right]=\left(\frac{\pi}{2}\right) a \end{aligned}$$

Let $(A, 0)$ and $(0, B)$ are $x$ and $y$ intercept respectively,

$\therefore$ Equation of line is

$\frac{x}{A}+\frac{y}{B}=1$

Now, $(a,-2 a)$ is mid-point of intercept

$\therefore a=\frac{A}{2}$ and $-2 a=B / 2 \Rightarrow A=2 a$ and $B=-4 a$

$\therefore$ Required equation of line is

$\frac{x}{2 a}-\frac{y}{4 a}=1 \Rightarrow 2 x-y=4 a$

We have,

$$\begin{aligned} & & \frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}} & =0 \\ \Rightarrow & & x \sqrt{1+y}+y \sqrt{1+x} & =0 \\ \Rightarrow & & x \sqrt{1+y} & =-y \sqrt{1+x} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad x^2(1+y)=y^2(1+x) \\ & \Rightarrow \quad x^2+x^2 y-y^2-x y^2=0 \\ & \Rightarrow \quad\left(x^2-y^2\right)+x^2 y-x y^2=0 \\ & \Rightarrow(x+y)(x-y)+x y(x-y)=0 \\ & \Rightarrow \quad(x-y)(x+y+x y)=0 \\ & \Rightarrow \quad y(1+x)=-x \quad[\because x-y \neq 0] \\ & \Rightarrow \quad y=\frac{-x}{1+x} \\ & \Rightarrow \quad \frac{d y}{d x}=-\left[\frac{(1+x)(1)-(x)}{(1+x)^2}\right] \\ & \Rightarrow \quad(1+x)^2 \frac{d y}{d x}=-1 \\ & \end{aligned}$$

Given line are

$\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$

and $\mathbf{r}=(3 \hat{\mathbf{i}}+\hat{\mathbf{k}})+\lambda^{\prime}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$

Here, $\mathbf{b}_1=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}_2=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$

$\therefore$ Angle between lines is

$$\begin{aligned} \cos \theta & =\left|\frac{\mathbf{b}_1 \cdot \mathbf{b}_2}{\left|\mathbf{b}_1\right|\left|\mathbf{b}_2\right|}\right| \\ \therefore \quad \cos \theta & =\left|\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+1+4} \sqrt{4+1+1}}\right| \\ \cos \theta & =\frac{2+1-2}{6}=\frac{1}{6} \\ \therefore \quad & \theta=\cos ^{-1}\left(\frac{1}{6}\right) \end{aligned}$$

Given minimise $z=5 x+8 y$

Subject to constrants

$$x+y \geq 5,\begin{array}{ll} 0 \leq x \leq 4, y \geq 2, y, \geq 0 \\ \text { Corner point } & z=5 x+8 y \\ A(3,2) & 15+16=31 \\ B(4,2) & 20+16=36 \\ C(0,5) & 0+40=40 \end{array}$$

$\therefore$ Minimum value = 31

Given vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+m \hat{\mathbf{k}}$ are coplanar

$$\therefore \quad\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 3 & m \end{array}\right|=0$$

$$\begin{array}{rr} \Rightarrow & 1(-m-3)-1(m-2)+1(3+2)=0 \\ \Rightarrow & -m-3-m+2+5=0 \\ \Rightarrow & 2 m=4 \Rightarrow m=2 \end{array}$$

To find the probability that an arrangement of the letters in 'LOGARITHM' starts with a vowel and ends with a consonant, we must first consider the number of ways to accomplish this scenario and the total number of possible arrangements of the letters in 'LOGARITHM'.

The word 'LOGARITHM' consists of 9 letters, including 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).

The total number of arrangements of the 9 letters (without any restrictions) is given by the factorial of 9, denoted as 9!, which is:

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Now, let's consider the arrangements that start with a vowel and end with a consonant. There are 3 choices for the first letter (since there are 3 vowels) and 6 choices for the last letter (since there are 6 consonants).

After choosing the first and last letters, we are left with 7 letters to be arranged freely. The number of ways to arrange these 7 letters is 7!, which is:

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

So, the total number of ways to arrange the letters of 'LOGARITHM' starting with a vowel and ending with a consonant is the product of the choices for the first and last letters and the number of arrangements of the remaining letters:

$3 \times 6 \times 7! = 18 \times 7!$

Thus, the probability of this specific arrangement happening, given all possible arrangements, is the ratio of the favorable outcomes to the total number of outcomes:

$P(\text{{Start with vowel and end with consonant}}) = \frac{{18 \times 7!}}{{9!}}$

Since $7!$ is a part of both the numerator and the denominator, it can be simplified further. The $7!$ cancels out, leaving us with:

$P(\text{{Start with vowel and end with consonant}}) = \frac{18}{9 \times 8} = \frac{18}{72} = \frac{1}{4}$

Therefore, the correct answer is Option D:

$\frac{1}{4}$

Given, $$A=\left[\begin{array}{ccc}1 & 3 & 2 \\ -2 & 0 & 1 \\ 5 & 2 & 1\end{array}\right]$$

Cofactor of second row of matrix

$$\begin{aligned} & \text { i.e., } A_{21}=-\left|\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right|=-(3-4)=1 \\ & A_{22}=\left|\begin{array}{ll} 1 & 2 \\ 5 & 1 \end{array}\right|=1-10=-9 \\ & A_{23}=-\left|\begin{array}{ll} 1 & 3 \\ 5 & 2 \end{array}\right|=-(2-15)=13 \\ & \therefore A_{21}+A_{22}+A_{23}=1-9+13=5 \end{aligned}$$

Given,

$p \rightarrow(\sim p \vee q)$

Truth table of given statement

$p$	$q$	$\sim p$	$\sim p\vee q$	$p\rightarrow (\sim p\vee q)$
T	T	F	T	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

$\therefore p \rightarrow(\sim p \vee q)$ is false when $p=\mathrm{T}, q=\mathrm{F}$

$\therefore$ Truth table of $p$ and $q$ are $T, F$ respectively.

We have,

$I=\int_2^3 \frac{x}{x^2-1} d x$

Put, $x^2-1=t \Rightarrow 2 x d x=d t=x d x=\frac{d t}{2}$

When, $x=2, t=3$ and $x=3, t=8$

$$\begin{aligned} & \therefore \quad I=\frac{1}{2} \int_3^{8} \frac{dt}{t} \Rightarrow I=\frac{1}{2}[\log t]_3^8=\frac{1}{2}[\log 8-\log 3] \\ & \Rightarrow \quad I=\frac{1}{2} \log (8 / 3) \\ \end{aligned}$$

Given, $f(x)=\log (\sin x)$

$a=\frac{\pi}{6}, b=\frac{5 \pi}{6}$

By Lagrange's mean value theorem (L.M.V.T)

$\begin{gathered} f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ f^{\prime}(x)=\cot x \\ f(b)=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)=-\log 2 \\ f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)=-\log 2 \\ \therefore f^{\prime}(c)=\cot C=\frac{-\log 2+\log 2}{\frac{5 \pi}{6}-\frac{\pi}{\pi / 6}}=0 \\ =\cot C=0=C=\pi / 2 \end{gathered}$

Given equation of pairs of lines

$k x y+5 x+3 y+2=0$

We know that, if

$a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$

is a pair of straight line, then

$$\begin{aligned} & a b c+2 f g h-a f^2-b g^2-c h^2=0 \\ & \text { in equation, } k x y+5 x+3 y+2=0 \\ & \text { Here, } a=0, b=0, h=k / 2, g=5 / 2, f=\frac{3}{2}, c=2 \end{aligned}$$

Here, $a=0, b=0, h=k / 2, g=5 / 2, f=\frac{3}{2}, c=2$

$\begin{gathered} \therefore 0+2\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{k}{2}\right)-0-0-2\left(\frac{k}{2}\right)^2=0 \\ \Rightarrow \quad \frac{15 k}{4}-\frac{k^2}{2}=0 \Rightarrow \frac{k}{2}\left(\frac{15}{2}-k\right)=0 \\ =k=\frac{15}{2} k \neq 0 \text {, if } k=0 \end{gathered}$

Then only one straight line.

Here,

$$\begin{aligned} & \quad n=10, p=0.8, q=0.2, r=2 \\ & \therefore \text { Required probability }={ }^{10} C_2(0.8)^2(0.2)^8 \\ & \quad={ }^{10} C_2\left(\frac{4}{5}\right)^2\left(\frac{1}{5}\right)^8=\frac{45 \times 16}{5^{10}}=\frac{9 \times 16}{5^9}=\frac{144}{5^9} \end{aligned}$$

We have,

$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$

Squaring both sides, we get

$$\begin{aligned} & \frac{x}{y}+\frac{y}{x}+2=16 \\ & x^2+y^2=14 x y \end{aligned}$$

On differentiating w.r.t. '$x$', we get

$$\begin{aligned} 2 x+2 y \frac{d y}{d x} & =14\left(x \frac{d y}{d x}+y\right) \\ x+y \frac{d y}{d x} & =7 x \frac{d y}{d x}+7 y \\ \frac{d y}{d x}(y-7 x) & =7 y-x \\ \frac{d y}{d x} & =\frac{7 y-x}{y-7 x} \end{aligned}$$

Given, $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\quad \text{..... (i)}$

$$\begin{aligned} & \Rightarrow \frac{b^2+c^2-a^2}{2 a b c}=\frac{a^2+c^2-b^2}{2 a b c}=\frac{a^2+b^2-c^2}{2 a b c} \\ & \Rightarrow b^2+c^2-a^2=a^2+c^2-b^2=a^2+b^2-c^2 \\ & \Rightarrow a=b=c \end{aligned}$$

$\therefore \triangle A B C$ is an equilateral triangle

$$\begin{aligned} \therefore \text { Area of } \triangle A B C & =\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4}(\sqrt{6})^2 \\ & =\frac{\sqrt{3} \times 6}{4}=\frac{3 \sqrt{3}}{2} \text { sq units } \end{aligned}$$

We have, $\frac{\sin A-\sin C}{\cos C-\cos A}=\cot B$

$\Rightarrow \quad \frac{2 \cos \left(\frac{A+C}{2}\right) \frac{\sin (A-C)}{2}}{2 \sin \left(\frac{C+A}{2}\right) \sin \left(\frac{A-C}{2}\right)}=\cot B$

$$\begin{aligned} & \Rightarrow \quad \cot \left(\frac{A+C}{2}\right)=\cot B \\ & \Rightarrow \quad A+C=2 B \\ \end{aligned}$$

$\therefore A, B, C$ are in Arithmetic progression

Here, $p, q$ represent switches $S_1$ and $S_2$ closed respectively.

Here, $S_1, S_2,\left(S_1^{\prime}\right.$ and $\left.S_2^{\prime}\right)$ are in parallel and $S_1^{\prime}$ and $S_2^{\prime}$ are in series

$\therefore I \equiv(p \vee q) \vee(\sim p \wedge \sim q)$

The given equation of the line is $5x + 2y + 7 = 0$. The slope of any line perpendicular to this line can be found by taking the negative reciprocal of the slope of this line. The slope of the given line is $-\frac{5}{2}$ since in the form $y = mx + c$, where $m$ is the slope, comparing it with the given equation, we have $2y = -5x - 7$, or $y = -\frac{5}{2}x - \frac{7}{2}$, thus $m = -\frac{5}{2}$. The slope of the line perpendicular to it would be $\frac{2}{5}$, as we take the negative reciprocal of the original slope to find the perpendicular slope.

Now, when we express the differential form of the equation representing lines with this slope, we use the fact that the slope is also the derivative of $y$ with respect to $x$ (i.e., $\frac{dy}{dx}$ for the lines), which should equal the perpendicular slope found above ($\frac{2}{5}$). Thus, we set $\frac{dy}{dx} = \frac{2}{5}$ and rearranging this in terms of differential forms (multiplying both sides by $dx$) we get:

$5dy = 2dx$

Therefore, the correct differential equation representing all lines perpendicular to $5x + 2y + 7 = 0$ is given by Option B: $5dy-2dx=0$.

To find the value of $\tan (\theta - \phi)$, we first need to calculate $\sin \theta, \cos \theta, \sin \phi, \text{ and } \cos \phi$, then use the formula for $\tan (A - B)$.

Given $\sin \theta = -\frac{12}{13}$ and $\cos \phi = -\frac{4}{5}$, and both $\theta$ and $\phi$ are in the third quadrant, where sine and cosine are both negative, the given values are already fitting the quadrant conditions.

In the third quadrant:

• $\sin^2 \theta = 1 - \cos^2 \theta$
• $\cos^2 \phi = 1 - \sin^2 \phi$

To find $\cos \theta$:

We know that $\sin^2 \theta + \cos^2 \theta = 1$, substituting the given value of sin we get:

$\left(-\frac{12}{13}\right)^2 + \cos^2 \theta = 1$

$\frac{144}{169} + \cos^2 \theta = 1$

$\cos^2 \theta = 1 - \frac{144}{169}$

$\cos^2 \theta = \frac{169}{169} - \frac{144}{169}$

$\cos^2 \theta = \frac{25}{169}$

$\cos \theta = \pm\sqrt{\frac{25}{169}}$

Because $\theta$ is in the third quadrant where cosine is negative, we take the negative root:

$\cos \theta = -\frac{5}{13}$

To find $\sin \phi$:

We also know that $\cos^2 \phi + \sin^2 \phi = 1$, substituting the given value of cos we get:

$\left(-\frac{4}{5}\right)^2 + \sin^2 \phi = 1$

$\frac{16}{25} + \sin^2 \phi = 1$

$\sin^2 \phi = 1 - \frac{16}{25}$

$\sin^2 \phi = \frac{25}{25} - \frac{16}{25}$

$\sin^2 \phi = \frac{9}{25}$

$\sin \phi = \pm\sqrt{\frac{9}{25}}$

Because $\phi$ is in the third quadrant where sine is negative, we take the negative root:

$\sin \phi = -\frac{3}{5}$

Now, to find $\tan (\theta-\phi)$:

The formula for $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$, we use:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}$

$\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$

Substituting into the formula:

$\tan (\theta - \phi) = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}}$

$= \frac{\frac{48}{20} - \frac{15}{20}}{1 + \frac{36}{20}}$

$= \frac{\frac{33}{20}}{\frac{20}{20} + \frac{36}{20}}$

$= \frac{33}{20 + 36}$

$= \frac{33}{56}$

Therefore, the correct answer is Option B: $\frac{33}{56}$.