MHT CET 2021 22nd September Morning Shift (Mathematics) Question Paper with Solutions

The given parametric equations are:

$x = \frac{1 - t^2}{1 + t^2}$

and

$y = \frac{2 a t}{1 + t^2}$

To find $\frac{d y}{d x}$, we need to compute $\frac{d y}{d t}$ and $\frac{d x}{d t}$ first, and then form their ratio.

First, let's differentiate $x$ with respect to $t$:

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

Using the quotient rule, $\frac{d}{d t} \left( \frac{u}{v} \right) = \frac{u' v - u v'}{v^2}$, where $u = 1 - t^2$ and $v = 1 + t^2$:

$u' = \frac{d}{d t} (1 - t^2) = -2 t$

$v' = \frac{d}{d t} (1 + t^2) = 2 t$

Therefore,

$\frac{d x}{d t} = \frac{(-2 t) (1 + t^2) - (1 - t^2) (2 t)}{(1 + t^2)^2}$

Simplify the numerator:

$\frac{d x}{d t} = \frac{-2 t - 2 t^3 - 2 t + 2 t^3}{(1 + t^2)^2} = \frac{-4 t}{(1 + t^2)^2}$

Now, let's differentiate $y$ with respect to $t$:

$\frac{d y}{d t} = \frac{d}{d t} \left( \frac{2 a t}{1 + t^2} \right)$

Using the quotient rule again, where $u = 2 a t$ and $v = 1 + t^2$:

$u' = \frac{d}{d t} (2 a t) = 2 a$

$v' = \frac{d}{d t} (1 + t^2) = 2 t$

Therefore,

$\frac{d y}{d t} = \frac{(2 a) (1 + t^2) - (2 a t) (2 t)}{(1 + t^2)^2}$

Simplify the numerator:

$\frac{d y}{d t} = \frac{2 a + 2 a t^2 - 4 a t^2}{(1 + t^2)^2} = \frac{2 a - 2 a t^2}{(1 + t^2)^2} = \frac{2 a (1 - t^2)}{(1 + t^2)^2}$

Now, we find $\frac{d y}{d x}$ by dividing $\frac{d y}{d t}$ by $\frac{d x}{d t}$:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{2 a (1 - t^2)}{(1 + t^2)^2}}{\frac{-4 t}{(1 + t^2)^2}} = \frac{2 a (1 - t^2)}{-4 t} = \frac{a (1 - t^2)}{-2 t} = \frac{a (t^2 - 1)}{2 t}$

The correct answer is:

Option D

$\frac{a\left(t^2 - 1\right)}{2 t}$

Let the numbers be x, y, z.

x + y + z = 6 ....... (1)

x + 3z = 7 ....... (2)

3x + y + z = 12 ...... (3)

Eq. (3) $-$ (1), gives

2x = 6 $\Rightarrow$ x = 3

From eq. (2), we get

3 + 3z = 7 $\Rightarrow$ z = $\frac{4}{3}$

From eq. (1), we get

$$\begin{aligned} & 3+y+\frac{4}{3}=6 \Rightarrow y=\frac{5}{3} \\ & \therefore x y z=(3)\left(\frac{5}{3}\right)\left(\frac{4}{3}\right)=\frac{20}{3} \end{aligned}$$

Note : This problem can be alternatively solved as follows.

$$\begin{aligned} & {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 3 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 6 \\ 7 \\ 12 \end{array}\right]} \\ & R_3 \rightarrow R_3-R_1 \\ & {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 2 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 6 \\ 7 \\ 6 \end{array}\right]} \\ & \therefore x+y+z=6 \quad \text{.....(1)}\\ & x+3 z=7 \quad \text{.....(2)}\\ & 2 x=6 \quad \text{.....(3)}\\ & \therefore x=3 \quad \text{.....[From (3)]}\\ & \therefore 3+3 z=7 \quad \Rightarrow \quad z=\frac{4}{3} \quad \ldots[\text { From }(2)] \end{aligned}$$

$\therefore 3+y+\frac{4}{3}=6 \Rightarrow y=\frac{5}{3}$ .....[From (1)]

We know that in any triangle ABC, the sum of angles is $180^\circ$. Therefore, we have:

$A + B + C = 180^\circ$

This implies:

$A + B - C = 180^\circ - 2C$

Now, let's use the sine rule, which states:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

From this, we get:

$\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$

Now, we can use the formula for the sine of half an angle:

$\sin \frac{1}{2} (A + B - C) = \sin \frac{1}{2} (180^\circ - 2C) = \sin (90^\circ - C) = \cos C$

We can express $\cos C$ using the cosine rule:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Now, let's substitute all these values into the given expression:

$2ab \sin \frac{1}{2} (A + B - C) = 2ab \cos C = 2ab \cdot \frac{a^2 + b^2 - c^2}{2ab} = a^2 + b^2 - c^2$

Therefore, the correct answer is Option B: $a^2 + b^2 - c^2$.

The given equation represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$.

We know that the product of the slopes of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is given by: $m_1m_2 = \frac{a}{b}$

We are also given that $m_1 = 2m_2$. Substituting this into the product of slopes equation, we get:

$2m_2^2 = \frac{a}{b}$.

Now, we need to find the relationship between $h^2$ and $ab$. To do this, we'll use the fact that the sum of the slopes of the lines is given by: $m_1 + m_2 = -\frac{2h}{b}$.

Substituting $m_1 = 2m_2$, we get: $3m_2 = -\frac{2h}{b}$.

Squaring both sides of this equation, we get: $9m_2^2 = \frac{4h^2}{b^2}$.

Now, we can substitute the value of $m_2^2$ from the product of slopes equation into this equation: $9 \times \frac{a}{2b} = \frac{4h^2}{b^2}$.

Simplifying this, we get: $h^2 = \frac{9ab}{8}$.

Therefore, $h^2 : ab = \boxed{9 : 8}$.

So the correct answer is option C.

To determine the correct equation of the circle that passes through the origin and intercepts the x-axis at -2 and the y-axis at 3, let's start by understanding the general form of the equation of a circle in a Cartesian coordinate system. The general form is:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Here, the center of the circle is $(-g, -f)$ and the radius is given by $\sqrt{g^2 + f^2 - c}$.

Given that the circle passes through the origin (0,0), we can substitute these coordinates into the general equation to find $c$. Substituting $(0,0)$, we get:

$0^2 + 0^2 + 2g \cdot 0 + 2f \cdot 0 + c = 0$

This simplifies to:

$c = 0$

So the equation simplifies to:

$x^2 + y^2 + 2gx + 2fy = 0$

Next, let's use the intercepts on the x-axis and y-axis to determine $g$ and $f$. The intercept on the x-axis is -2, so the circle passes through the point (-2, 0). Substituting these coordinates into the equation:

$(-2)^2 + 0^2 + 2g(-2) + 2f(0) = 0$

Which simplifies to:

$4 - 4g = 0 \implies g = 1$

Similarly, the intercept on the y-axis is 3, so the circle passes through the point (0, 3). Substituting these coordinates into the equation:

$0^2 + 3^2 + 2g(0) + 2f(3) = 0$

Which simplifies to:

$9 + 6f = 0 \implies f = -\frac{3}{2}$

Thus, substituting $g = 1$ and $f = -\frac{3}{2}$ into the general equation of the circle, we get:

$x^2 + y^2 + 2(1)x + 2\left(-\frac{3}{2}\right)y = 0$

This further simplifies to:

$x^2 + y^2 + 2x - 3y = 0$

Therefore, the correct equation of the circle is option C:

$x^2 + y^2 + 2x - 3y = 0$

$$\begin{aligned} \text { Let } I & =\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x \\ & =\int\left(\frac{\sin x}{\cos x}\right)\left(\sec ^6 x\right)(\sec x) d x \\ & =\int\left(\sec ^6 x\right)(\sec x \tan x) d x \end{aligned}$$

Put $\sec x=t \Rightarrow \sec x \tan x d x=d t$

$\therefore I=\int t^6 d t=\frac{t^7}{7}+c=\frac{\sec ^7 x}{7}+c$

Let's consider the given points where the plane meets the coordinate axes:

• The plane meets the X-axis at point $A(a, 0, 0)$
• The plane meets the Y-axis at point $B(0, b, 0)$
• The plane meets the Z-axis at point $C(0, 0, c)$

The centroid of triangle $\triangle \mathrm{ABC}$ is calculated as follows:

The coordinates of the centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by:

$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

For triangle $\mathrm{ABC}$, the centroid is given as $(1,2,3)$:

$\left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \right) = (1, 2, 3)$

This can be decomposed into three separate equations:

$\frac{a}{3} = 1 \implies a = 3$

$\frac{b}{3} = 2 \implies b = 6$

$\frac{c}{3} = 3 \implies c = 9$

Thus, the intercepts of the plane are $3$ on the X-axis, $6$ on the Y-axis, and $9$ on the Z-axis.

The general equation of a plane with intercepts $a$, $b$, and $c$ is:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Substituting the values of $a$, $b$, and $c$, we get:

$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

Therefore, the equation of the plane is:

Option C: $\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$

To find the equation of the perpendicular bisector of the line segment joining $A(-2, 3)$ and $B(6, -5)$, we need to follow these steps:

Step 1: Determine the midpoint of the line segment $AB$.

The midpoint $M$ of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Substituting the coordinates of points $A$ and $B$:

$M = \left( \frac{-2 + 6}{2}, \frac{3 - 5}{2} \right)$

$M = \left( \frac{4}{2}, \frac{-2}{2} \right)$

$M = (2, -1)$

So, the midpoint $M$ of line segment $AB$ is $(2, -1)$.

Step 2: Find the slope of the line segment $AB$.

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting the coordinates of points $A$ and $B$:

$m = \frac{-5 - 3}{6 - (-2)}$

$m = \frac{-8}{6 + 2}$

$m = \frac{-8}{8}$

$m = -1$

So, the slope of line segment $AB$ is $-1$.

Step 3: Determine the slope of the perpendicular bisector.

The slope of the perpendicular bisector of a line is the negative reciprocal of the slope of the line. Therefore, the slope $m_{\text{perpendicular}}$ is given by:

$m_{\text{perpendicular}} = -\frac{1}{m}$

Since $m = -1$:

$m_{\text{perpendicular}} = -\frac{1}{-1}$

$m_{\text{perpendicular}} = 1$

So, the slope of the perpendicular bisector is $1$.

Step 4: Use the point-slope form to write the equation of the perpendicular bisector.

The point-slope form of the equation of a line with slope $m$ passing through point $(x_1, y_1)$ is:

$y - y_1 = m(x - x_1)$

Using the slope $m_{\text{perpendicular}} = 1$ and the midpoint $(2, -1)$:

$y - (-1) = 1(x - 2)$

$y + 1 = x - 2$

Rearranging to standard form:

$x - y = 3$

Therefore, the equation of the perpendicular bisector is:

Option D: $x - y = 3$

$$\begin{aligned} & \sin ^{-1}\left(\frac{d y}{d x}\right)=x+y \\ & \therefore \frac{d y}{d x}=\sin (x+y) \\ & \text { Put } x+y=t \quad \Rightarrow \quad 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}-1=\sin t \quad \Rightarrow \quad \frac{d t}{d x}=1+\sin t \\ & \therefore \int \frac{d t}{1+\sin t}=\int d x\end{aligned}$$

$$\begin{aligned} & \therefore \int \frac{(1-\sin t) d t}{1-\sin ^2 t}=\int d x \Rightarrow \int \frac{1-\sin t}{\cos ^2 t} d t=\int d x \Rightarrow \int\left(\sec ^2 t-\sec t \tan t\right) d t=\int d x \\ & \therefore \tan t-\sec t=x+c \\ & \therefore \tan (x+y)-\sec (x+y)=x+c\end{aligned}$$

y = mx + $\frac{4}{m}$ ..... (1)

$\therefore \frac{d y}{d x}=m$

Substituting value of $m$ in equation (1), we get

$$\begin{aligned} & y=\left(\frac{d y}{d x}\right) x+\frac{4}{\left(\frac{d y}{d x}\right)} \\ & \therefore y\left(\frac{d y}{d x}\right)=\left(\frac{d y}{d x}\right)^2 x+4 \end{aligned}$$

$$\begin{aligned} & y^2-9 x y+18 x^2=0 \\ & \therefore(y-3 x)(y-6 x)=0 \end{aligned}$$

Thus three lines forming triangle are

$y=3 x, y=6 x, y=9$

Their point of intersections are $(0,0),(3,9),\left(\frac{3}{2}, 9\right)$

The area of the triangle formed by the points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substituting the points $(0, 0)$, $(3, 9)$, and $\left(\frac{3}{2}, 9\right)$ into the formula:

$\text{Area} = \frac{1}{2} \left| 0 (9 - 9) + 3 \left(9 - 0\right) + \frac{3}{2} (0 - 9) \right|$

$\text{Area} = \frac{1}{2} \left| 0 + 27 - \frac{27}{2} \right|$

$\text{Area} = \frac{1}{2} \left| \frac{54}{2} - \frac{27}{2} \right|$

$\text{Area} = \frac{1}{2} \left| \frac{27}{2} \right|$

$\text{Area} = \frac{27}{4}$

Therefore, the area of the triangle is $\frac{27}{4}$ square units. Hence, the correct answer is:

Option C $\frac{27}{4}$ sq. units

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

$$\begin{aligned} & \therefore \int \frac{d v}{3}=-\int \frac{d x}{x} \\ & \therefore \frac{-1}{2 v^2}=-\log x-c \quad \Rightarrow \frac{1}{2 v^2}=\log x+c \\ & \therefore \frac{y^2}{2 x^2}=\log x+c \quad \Rightarrow y^2=x^2 \log x^2+2 x^2 c \end{aligned}$$

We have $y(1)=-2$

$$\begin{aligned} & \therefore 4=0+2 c \Rightarrow c=2 \\ & \therefore y^2=x^2 \log x^2+4 x^2 \end{aligned}$$

$y^2=a x^2+b x+c$

Differentiatig w.r.t. $x$, we get

$$\begin{aligned} & 2 y \frac{d y}{d x}=2 a x+b \quad \Rightarrow \quad 2 y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2=2 a \\ & \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=a \\ & \therefore y^3 \frac{d^2 y}{d x^2}=-\left(a x^2+b x+c\right)\left[\left(\frac{2 a x+b}{2}\right)^2-a\right] \quad \text{.....(1)} \end{aligned}$$

R.H.S. of eq. (1) is a function of '$x$' only.

Angles A, B, C of $\triangle \mathrm{ABC}$ are in A.P.

$$\begin{aligned} & \therefore \angle B=60^{\circ} \Rightarrow \angle A+\angle C=120^{\circ} \\ & \text { Also } \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \quad \ldots \text { (let) }\\ & \therefore a=k \sin A \quad \& \quad c=k\sin C \\ & \frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A \\ & =\frac{k \sin A}{k \sin C}(2 \sin C \cos C)+\frac{k \sin C}{k \sin A}(2 \sin A \cos A) \\ & =2 \sin A \cos C+2 \cos A \sin C \quad=2(\sin A \cos C+\cos A \sin C) \\ & =2 \sin (A+C)=2 \sin \left(120^{\circ}\right)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \end{aligned}$$

$$\begin{aligned} & y(1+\log x)=\left(\log x^x\right) \frac{d y}{d x} \\ & \therefore \int \frac{(1+\log x)}{\left(\log x^x\right)}=\int \frac{d y}{y} \end{aligned}$$

We know that $\frac{d}{d x}\left(\log x^x\right)=(1+\log x) d x$

$\therefore \log \left|\log \mathrm{x}^{\mathrm{x}}\right| \quad=\log |\mathrm{y}|+\log \mathrm{c}$

We have $\mathrm{y}(\mathrm{e})=\mathrm{e}^2$

$$\begin{aligned} & \therefore \log \left|\log \mathrm{e}^{\mathrm{e}}\right|=\log \left|\mathrm{e}^2\right|+\log \mathrm{c} \\ & \therefore \log |\mathrm{e} \log \mathrm{e}| \quad=2 \log |\mathrm{e}|+\log \mathrm{c} \\ & \therefore 1 \quad=2+\log \mathrm{c} \Rightarrow \log \mathrm{c}=-1 \\ & \therefore \log \left|\log \mathrm{x}^{\mathrm{x}}\right|=\log |\mathrm{y}|-\log \mathrm{e} \\ & \therefore \log |\mathrm{x} \log \mathrm{x}|+\log \mathrm{e} \quad=\log |\mathrm{y}| \\ & \therefore \mathrm{e} \mathrm{x} \log \mathrm{x}-\mathrm{y}=0 \end{aligned}$$

To solve the integral $\int {{e^x}\left( {{{x - 1} \over {{x^2}}}} \right)dx}$, let's first simplify the integrand.

Rewrite the integrand as follows:

$\int {e^x \left( \frac{x - 1}{x^2} \right) dx} = \int {e^x \left( \frac{x}{x^2} - \frac{1}{x^2} \right) dx} = \int {e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx}$

So, the integral becomes:

$\int {e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx} = \int {e^x \frac{1}{x} dx} - \int {e^x \frac{1}{x^2} dx}$

We need to evaluate each term separately. Let's start with the first term:

$\int {e^x \frac{1}{x} dx}$

Applying integration by parts, let:

$u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$

$dv = e^x dx \Rightarrow v = e^x$

Using the integration by parts formula $\int u dv = uv - \int v du$:

$\int {e^x \frac{1}{x} dx} = \frac{e^x}{x} - \int {e^x \left(-\frac{1}{x^2} \right) dx}$

Note that the second term inside the integration is the same as the second term we need to evaluate originally. So we can write:

$\int {e^x \frac{1}{x} dx} = \frac{e^x}{x} + \int {e^x \frac{1}{x^2} dx}$

Now, notice that:

$\int {e^x \frac{1}{x^2} dx}$

can cancel out the $\int {e^x \frac{1}{x^2} dx}$ from earlier, leaving us with:

$\int {e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx} = \frac{e^x}{x}$

Thus, the solution to the integral is:

$\int {{e^x}\left( {{{x - 1} \over {{x^2}}}} \right)dx = \frac{{e^x}}{x} + c }$

The correct answer is:

Option D: $\frac{{e^x}}{x} + c$

We have A = \left[ {\matrix{ 1 & 0 & 1 \cr 0 & 2 & 3 \cr 1 & 2 & 1 \cr } } \right]

$\therefore$ |A| = (2 $-$ 6) + (0 $-$ 2) = $-$6

adj A = {\left[ {\matrix{ { - 4} & 3 & { - 2} \cr 2 & 0 & { - 2} \cr { - 2} & { - 3} & 2 \cr } } \right]^T}

$\therefore$ {A^{ - 1}} = {{\left[ {\matrix{ { - 4} & 2 & { - 2} \cr 3 & 0 & { - 3} \cr { - 2} & { - 2} & 2 \cr } } \right]} \over { - 6}}

$|{A^{ - 1}}| = {1 \over {{{( - 6)}^3}}}[( - 4)( - 6) - 2(0) - 2( - 6)] = {{36} \over { - 216}} = {{ - 1} \over 6}$

This problem can be alternately solved as follows :

$$|A|=\left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right|=(-4)+(-2)=-6$$

We know that $\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}$

$\therefore\left|A^{-1}\right|=\frac{1}{-6}$