WBJEE 2024 Paper 1 (Mathematics) Question Paper with Solutions

In order to find the points of extremum of the given integral, we start by considering the integral:

\int_\limits0^{x^2} \frac{t^2-5 t+4}{2+e^t} d t

Let's denote this integral by $F(x)$.

We need to find the critical points of $F(x)$, which are the points where its derivative with respect to $x$ is zero. Since $F(x)$ is an integral with an upper limit that depends on $x$, we will use the Leibniz rule for differentiation under the integral sign:

$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t, x) \, dt = f(b(x), x) \frac{d}{dx} b(x) - f(a(x), x) \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt$

In our case, the lower limit is 0 and is independent of $x$, so its derivative is zero. The function inside the integral does not explicitly depend on $x$, so the partial derivative with respect to $x$ is zero. Hence, we simplify to:

$\frac{d}{dx} F(x) = \frac{d}{dx} \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt = \left. \frac{t^2 - 5t + 4}{2 + e^t} \right|_{0}^{x^2} \cdot \frac{d}{dx} (x^2)$

Since the value of the integral at the lower limit is zero, we get:

$\frac{d}{dx} F(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot 2x$

Simplifying the expression inside the fraction, we have:

$\frac{d}{dx} F(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x$

To find the critical points, we must set the derivative equal to zero:

$\frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x = 0$

The fraction is zero if and only if the numerator is zero. Thus, we solve:

$(x^4 - 5x^2 + 4) \cdot 2x = 0$

This simplifies to two cases:

$2x (x^4 - 5x^2 + 4) = 0$

From the above equation, set each factor to zero:

$2x = 0 \Rightarrow x = 0$

and

$x^4 - 5x^2 + 4 = 0$

Letting $y = x^2$, we convert the quartic equation to a quadratic one:

$y^2 - 5y + 4 = 0$

Solving this quadratic equation by factoring:

$(y - 1)(y - 4) = 0$

So,

$y = 1 \quad \text{or} \quad y = 4$

Since $y = x^2$, we have:

$x^2 = 1 \quad \Rightarrow \quad x = \pm 1$

$x^2 = 4 \quad \Rightarrow \quad x = \pm 2$

Thus, the points of extremum are:

Option A: $\pm 1$

Option B: $\pm 2$

Therefore, the correct answer is both Option A and Option B.

From $𝐴𝑋=𝐵$, the equations are $𝑥-𝑦=2$, $𝑦-𝑧=1$, $𝑥+𝑦+𝑧=7$.

From the first two, $𝑥=𝑦+2$ and $𝑧=𝑦-1$. Substituting in the third equation,

$(𝑦+2)+𝑦+(𝑦-1)=7\Rightarrow 3𝑦+1=7\Rightarrow 𝑦=2$

So $𝑥=4$, $𝑦=2$, $𝑧=1$, giving $𝑋=⎛⎝⎜\begin{array}{c}4\\ 2\\ 1\end{array}⎞⎠⎟$.

Hence, the correct option is (D).

Let's solve the given differential equation and find the limit $\lim_{x \rightarrow 1} y(x)$.

The differential equation given is:

$x y^{\prime} + y - e^x = 0$

This is a first-order linear differential equation. We can rewrite it in the form:

$y^{\prime} + \frac{1}{x} y = \frac{e^x}{x}$

We will solve this using the integrating factor method. The integrating factor (IF) is given by:

$\mu(x) = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x|$

Since we are not given a specific interval for x, we assume x is positive:

$\mu(x) = x$

Multiplying both sides of the differential equation by the integrating factor:

$x y^{\prime} + y = e^x \Rightarrow x y^{\prime} + y = e^x \Rightarrow x (y^{\prime} + \frac{1}{x} y) = e^x$

$x \cdot y^{\prime} + y = e^x$

Recognize that the left side is the derivative of $xy$:

$\frac{d}{dx} (xy) = e^x$

Integrate both sides with respect to x:

$xy = \int e^x dx = e^x + C$

So we have:

$y = \frac{e^x + C}{x}$

Apply the initial condition $y(a) = b$:

$b = \frac{e^a + C}{a}$

$C = ab - e^a$

Thus, the solution to the differential equation is:

$y = \frac{e^x + ab - e^a}{x}$

Now, we need to determine:

$\lim\limits_{x \rightarrow 1} y(x)$

Substituting $x = 1$ into the solution:

$y(1) = \frac{e^1 + ab - e^a}{1} = e + ab - e^a$

Therefore:

The correct answer is $\text{Option D}$

$e + ab - e^a$

The diagonals of a cube meet at the center of the cube. To determine the angle between two diagonals, consider the geometric properties of the cube.

A cube has 12 face diagonals and 4 space diagonals. We are interested in the space diagonals because they span from one vertex of the cube to the opposite vertex.

Let's place the cube in a 3-dimensional Cartesian coordinate system with vertices at:

(0,0,0), (1,0,0), (1,1,0), (0,1,0), (0,0,1), (1,0,1), (1,1,1), and (0,1,1).

Consider the space diagonals of the cube, such as the one joining (0,0,0) to (1,1,1) and another joining (1,0,0) to (0,1,1). The direction vectors of these diagonals can be written as:

$\vec{d_1} = \langle 1, 1, 1 \rangle$

$\vec{d_2} = \langle -1, 1, 1 \rangle$

We use the dot product formula for vectors to find the angle between them, $\theta$, which is given by:

$\vec{d_1} \cdot \vec{d_2} = |\vec{d_1}| |\vec{d_2}| \cos \theta$

The dot product of these vectors is:

$\vec{d_1} \cdot \vec{d_2} = (1)(-1) + (1)(1) + (1)(1) = -1 + 1 + 1 = 1$

The magnitudes of the vectors are:

$|\vec{d_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

$|\vec{d_2}| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$

Substitute these values into the dot product formula:

$\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$

Therefore, the angle $\theta$ is:

$\theta = \cos^{-1} \left(\frac{1}{3}\right)$

Hence, the correct option is:

Option A: $\cos^{-1}\left(\frac{1}{3}\right)$

Given $𝑧=\cos 𝜃+𝑖\sin 𝜃$, so $\left|𝑧\right|=1$ and $\overline{𝑧}=\frac{1}{𝑧}=\cos 𝜃-𝑖\sin 𝜃$.

Since $𝑧$ is a root of ${𝑎}_0{𝑥}^{𝑛}+{𝑎}_1{𝑥}^{𝑛-1}+\cdots +{𝑎}_{𝑛-1}𝑥+{𝑎}_{𝑛}=0$ with all ${𝑎}_{𝑘}\in ℝ$,

${𝑎}_0{𝑧}^{𝑛}+{𝑎}_1{𝑧}^{𝑛-1}+\cdots +{𝑎}_{𝑛-1}𝑧+{𝑎}_{𝑛}=0.$

Taking the conjugate of both sides and using that every ${𝑎}_{𝑘}$ is real (so $\overline{{𝑎}_{𝑘}}={𝑎}_{𝑘}$),

${𝑎}_0{\overline{𝑧}}^{𝑛}+{𝑎}_1{\overline{𝑧}}^{𝑛-1}+\cdots +{𝑎}_{𝑛}=0,$

which means $\overline{𝑧}=\frac{1}{𝑧}$ is also a root of the same equation, i.e. $𝑃\left(\overline{𝑧}\right)=0$ where $𝑃\left(𝑥\right)={𝑎}_0{𝑥}^{𝑛}+{𝑎}_1{𝑥}^{𝑛-1}+\cdots +{𝑎}_{𝑛}$.

Let $𝑅\left(𝑥\right)={𝑥}^{𝑛}𝑃\left(\frac{1}{𝑥}\right)={𝑎}_0+{𝑎}_1𝑥+{𝑎}_2{𝑥}^2+\cdots +{𝑎}_{𝑛}{𝑥}^{𝑛}$. Then

$𝑅\left(𝑧\right)={𝑧}^{𝑛}𝑃\left(\frac{1}{𝑧}\right)={𝑧}^{𝑛}𝑃\left(\overline{𝑧}\right)={𝑧}^{𝑛}\cdot 0=0,$

so ${𝑎}_0+{𝑎}_1𝑧+{𝑎}_2{𝑧}^2+\cdots +{𝑎}_{𝑛}{𝑧}^{𝑛}=0$.

Writing ${𝑧}^{𝑘}=\cos 𝑘𝜃+𝑖\sin 𝑘𝜃$ for $𝑘=0,1,\dots ,𝑛$ and equating the imaginary part of $𝑅\left(𝑧\right)=0$ to zero, with $\mathrm{Im}\left({𝑎}_0\right)=0$ since ${𝑎}_0$ is real,

${𝑎}_1\sin 𝜃+{𝑎}_2\sin 2𝜃+\cdots +{𝑎}_{𝑛}\sin 𝑛𝜃=0.$

(This can be checked directly: for $𝑛=2$ with real coefficients, $𝑃\left(𝑥\right)={𝑎}_0({𝑥}^2-2\cos 𝜃𝑥+1)$, giving ${𝑎}_1=-2{𝑎}_0\cos 𝜃$, ${𝑎}_2={𝑎}_0$, and ${𝑎}_1\sin 𝜃+{𝑎}_2\sin 2𝜃=-2{𝑎}_0\cos 𝜃\sin 𝜃+{𝑎}_0\sin 2𝜃=0$.)

Hence, the correct option is (C).

Since columns 2 and 3 of rows 1 and 3 are identical $(𝑥,1)$, apply ${𝑅}_3\to {𝑅}_3-{𝑅}_1$:

$𝑓\left(𝑥\right)=\mid \begin{array}{ccc}\cos 𝑥& 𝑥& 1\\ 2\sin 𝑥& {𝑥}^3& 2𝑥\\ \tan 𝑥-\cos 𝑥& 0& 0\end{array}\mid$

Expanding along the new third row,

$𝑓\left(𝑥\right)=(\tan 𝑥-\cos 𝑥)\mid \begin{array}{cc}𝑥& 1\\ {𝑥}^3& 2𝑥\end{array}\mid =(\tan 𝑥-\cos 𝑥)(2{𝑥}^2-{𝑥}^3)$

Therefore,

$\frac{𝑓\left(𝑥\right)}{{𝑥}^2}=(\tan 𝑥-\cos 𝑥)(2-𝑥)$

Taking the limit as $𝑥\to 0$,

$\underset{𝑥\to 0}{lim}\frac{𝑓\left(𝑥\right)}{{𝑥}^2}=(\tan 0-\cos 0)(2-0)=(0-1)\left(2\right)=-2$

Hence, the correct option is (B).

PC must be perpendicular to the tangent at P.

Let the square have vertices $𝑂=(0,0)$, $𝐴=(𝑎\cos 𝛼,𝑎\sin 𝛼)$ along the side making angle $𝛼$ with the $𝑥$-axis, and $𝐷=(-𝑎\sin 𝛼,𝑎\cos 𝛼)$ along the perpendicular side (angle $𝛼+\frac{𝜋}{2}$), so that the square lies above the $𝑥$-axis. The fourth vertex is $𝐵=𝐴+𝐷=(𝑎\cos 𝛼-𝑎\sin 𝛼,𝑎\sin 𝛼+𝑎\cos 𝛼)$.

Diagonal $𝑂𝐵$ passes through the origin, so its equation is $𝑦=\frac{𝑎(\sin 𝛼+\cos 𝛼)}{𝑎(\cos 𝛼-\sin 𝛼)}𝑥$, i.e.

$𝑦(\cos 𝛼-\sin 𝛼)=𝑥(\sin 𝛼+\cos 𝛼)$

which matches option (A).

Diagonal $𝐴𝐷$ joins $𝐴=(𝑎\cos 𝛼,𝑎\sin 𝛼)$ and $𝐷=(-𝑎\sin 𝛼,𝑎\cos 𝛼)$. Its slope is

$𝑚=\frac{𝑎\cos 𝛼-𝑎\sin 𝛼}{-𝑎\sin 𝛼-𝑎\cos 𝛼}=-\frac{\cos 𝛼-\sin 𝛼}{\sin 𝛼+\cos 𝛼}$

Using point $𝐴$: $𝑦-𝑎\sin 𝛼=-\frac{\cos 𝛼-\sin 𝛼}{\sin 𝛼+\cos 𝛼}(𝑥-𝑎\cos 𝛼)$.

Multiplying by $(\sin 𝛼+\cos 𝛼)$ and simplifying,

$𝑦(\sin 𝛼+\cos 𝛼)+𝑥(\cos 𝛼-\sin 𝛼)=𝑎[{\sin }^2𝛼+\sin 𝛼\cos 𝛼+{\cos }^2𝛼-\sin 𝛼\cos 𝛼]=𝑎$

which matches option (C).

Hence, the correct options are (A) and (C).

Since columns 2 and 3 of rows 1 and 3 are identical $(𝑥,1)$, apply ${𝑅}_3\to {𝑅}_3-{𝑅}_1$:

$𝑓\left(𝑥\right)=\mid \begin{array}{ccc}\cos 𝑥& 𝑥& 1\\ 2\sin 𝑥& {𝑥}^3& 2𝑥\\ \tan 𝑥-\cos 𝑥& 0& 0\end{array}\mid$

Expanding along the new third row,

$𝑓\left(𝑥\right)=(\tan 𝑥-\cos 𝑥)\mid \begin{array}{cc}𝑥& 1\\ {𝑥}^3& 2𝑥\end{array}\mid =(\tan 𝑥-\cos 𝑥)(2{𝑥}^2-{𝑥}^3)$

Therefore,

$\frac{𝑓\left(𝑥\right)}{{𝑥}^2}=(\tan 𝑥-\cos 𝑥)(2-𝑥)$

Taking the limit as $𝑥\to 0$,

$\underset{𝑥\to 0}{lim}\frac{𝑓\left(𝑥\right)}{{𝑥}^2}=(\tan 0-\cos 0)(2-0)=(0-1)\left(2\right)=-2$

Hence, the correct option is (B).

Expanding the given determinant using column operations (C1 → C1 − xC2), the determinant factorizes as:

$$\left|\begin{array}{lll}a_1+b_1 x & a_1 x+b_1 & c_1 \\ a_2+b_2 x & a_2 x+b_2 & c_2 \\ a_3+b_3 x & a_3 x+b_3 & c_3\end{array}\right| = -(x-1)(x+1)\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$$

Since this expression is given to be zero for all $a_i, b_i, c_i$, the equation is satisfied whenever $x=1$ or $x=-1$, regardless of the value of the third determinant. Hence $x=1$ and $x=-1$ are both always valid solutions.

Correct options: A, B

Let the required line meet $5𝑥-𝑦-4=0$ at $𝐴(𝑎,5𝑎-4)$ and $3𝑥+4𝑦-4=0$ at $𝐵(𝑏,{𝑦}_2)$, with $3𝑏+4{𝑦}_2-4=0\Rightarrow {𝑦}_2=\frac{4-3𝑏}{4}$.

Since $(1,5)$ is the midpoint of $𝐴𝐵$,

$\frac{𝑎+𝑏}{2}=1\Rightarrow 𝑎+𝑏=2\text{...(i)}$

$\frac{(5𝑎-4)+{𝑦}_2}{2}=5\Rightarrow {𝑦}_2=14-5𝑎\text{...(ii)}$

Equating ${𝑦}_2$ from the two expressions,

$\frac{4-3𝑏}{4}=14-5𝑎\Rightarrow 4-3𝑏=56-20𝑎\Rightarrow 3𝑏=20𝑎-52$

From (i), $𝑏=2-𝑎$, so

$3(2-𝑎)=20𝑎-52\Rightarrow 6-3𝑎=20𝑎-52\Rightarrow 23𝑎=58\Rightarrow 𝑎=\frac{58}{23}$

Then $𝑏=2-𝑎=-\frac{12}{23}$, and

$𝐴=(\frac{58}{23},\frac{198}{23})$

The required line passes through $(1,5)$ and $𝐴$. Its direction ratios are

$(\frac{58}{23}-1,\frac{198}{23}-5)=(\frac{35}{23},\frac{83}{23})\propto (35,83)$

So the slope is $\frac{83}{35}$, and the line is

$𝑦-5=\frac{83}{35}(𝑥-1)\Rightarrow 35𝑦-175=83𝑥-83\Rightarrow 83𝑥-35𝑦+92=0$

Hence, the correct option is (B).

Since $𝛼,𝛽$ are roots of $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$, $𝑎{𝑥}^2+𝑏𝑥+𝑐=𝑎(𝑥-𝛼)(𝑥-𝛽)$.

As $𝑥\to 𝛽$, $𝑎(𝑥-𝛼)(𝑥-𝛽)\to 0$, so using $1-\cos 𝜃\sim \frac{{𝜃}^2}{2}$ for small $𝜃$,

$\underset{𝑥\to 𝛽}{lim}\frac{1-\cos (𝑎(𝑥-𝛼)(𝑥-𝛽))}{(𝑥-𝛽{)}^2}=\underset{𝑥\to 𝛽}{lim}\frac{\frac{1}{2}{𝑎}^2(𝑥-𝛼{)}^2(𝑥-𝛽{)}^2}{(𝑥-𝛽{)}^2}$

$=\frac{{𝑎}^2}{2}\underset{𝑥\to 𝛽}{lim}(𝑥-𝛼{)}^2=\frac{{𝑎}^2}{2}(𝛽-𝛼{)}^2=\frac{{𝑎}^2}{2}(𝛼-𝛽{)}^2$

Hence, the correct option is (D).

To find the value of $n$ in the expression $1000! = 3^n \times m$ where $m$ is an integer not divisible by 3, we need to determine the largest power of 3 that divides $1000!$. This can be accomplished by using Legendre's formula, which is also known as de Polignac's formula. The formula is used to determine the exponent of a prime number $p$ in the factorization of $n!$.

Legendre's formula is given by:

$e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \left\lfloor \frac{n}{p^4} \right\rfloor + \cdots$

where $\left\lfloor x \right\rfloor$ represents the floor function, which gives the greatest integer less than or equal to $x$, and $p$ is the prime number we are interested in (in this case, 3).

We need to find:

$e_3(1000!) = \left\lfloor \frac{1000}{3} \right\rfloor + \left\lfloor \frac{1000}{3^2} \right\rfloor + \left\lfloor \frac{1000}{3^3} \right\rfloor + \left\lfloor \frac{1000}{3^4} \right\rfloor + \cdots$

Let's calculate each term step by step:

$\left\lfloor \frac{1000}{3} \right\rfloor = \left\lfloor 333.3333 \right\rfloor = 333$

$\left\lfloor \frac{1000}{9} \right\rfloor = \left\lfloor 111.1111 \right\rfloor = 111$

$\left\lfloor \frac{1000}{27} \right\rfloor = \left\lfloor 37.037 \right\rfloor = 37$

$\left\lfloor \frac{1000}{81} \right\rfloor = \left\lfloor 12.3457 \right\rfloor = 12$

$\left\lfloor \frac{1000}{243} \right\rfloor = \left\lfloor 4.115 \right\rfloor = 4$

$\left\lfloor \frac{1000}{729} \right\rfloor = \left\lfloor 1.371 \right\rfloor = 1$

For powers higher than 3^6, the terms will be zero since 1000 divided by any higher powers of 3 will be less than 1.

Summing these values, we get:

$e_3(1000!) = 333 + 111 + 37 + 12 + 4 + 1 = 498$

Therefore, the value of $n$ is:

$n = 498$

The correct option is:

Option A

For $𝑥\ge 0$, $\left|𝑥\right|=𝑥$, so $𝑓\left(𝑥\right)=\frac{{𝑒}^{𝑥}-{𝑒}^{-𝑥}}{{𝑒}^{𝑥}+{𝑒}^{-𝑥}}=\tanh 𝑥,$ which increases from $0$ to a value approaching $1$ as $𝑥\to \infty$, so $𝑓\left(𝑥\right)\in [0,1)$ for $𝑥\ge 0$.

For $𝑥<0$, $\left|𝑥\right|=-𝑥$, so $𝑓\left(𝑥\right)=\frac{{𝑒}^{-𝑥}-{𝑒}^{-𝑥}}{{𝑒}^{𝑥}+{𝑒}^{-𝑥}}=0.$

Thus $𝑓\left(𝑥\right)=0$ for every $𝑥<0$ as well as for $𝑥=0$, so distinct inputs give the same output: $𝑓$ is not one-one.

Also the range of $𝑓$ is $[0,1)$, which is not all of $ℝ$, so $𝑓$ is not onto.

Hence, the correct option is (D).

Given $𝑓\left(𝑥\right)=\frac{{𝑒}^{𝑥}}{1+{𝑒}^{𝑥}}$.

$𝑓\left(𝑎\right)+𝑓(-𝑎)=\frac{{𝑒}^{𝑎}}{1+{𝑒}^{𝑎}}+\frac{{𝑒}^{-𝑎}}{1+{𝑒}^{-𝑎}}=\frac{{𝑒}^{𝑎}}{1+{𝑒}^{𝑎}}+\frac{1}{1+{𝑒}^{𝑎}}=1$

Let $𝑝=𝑓(-𝑎)$ and $𝑞=𝑓\left(𝑎\right)$, so that $𝑝+𝑞=1$.

${𝐼}_1={\int }_{𝑞}^{𝑝}𝑥𝑔(𝑥(1-𝑥))𝑑𝑥,{𝐼}_2={\int }_{𝑞}^{𝑝}𝑔(𝑥(1-𝑥))𝑑𝑥$

In ${𝐼}_1$, put $𝑥=1-𝑢\Rightarrow 𝑑𝑥=-𝑑𝑢$. When $𝑥=𝑝$, $𝑢=1-𝑝=𝑞$; when $𝑥=𝑞$, $𝑢=1-𝑞=𝑝$.

${𝐼}_1={\int }_{𝑢=𝑝}^{𝑢=𝑞}(1-𝑢)𝑔((1-𝑢)𝑢)(-𝑑𝑢)={\int }_{𝑞}^{𝑝}(1-𝑢)𝑔(𝑢(1-𝑢))𝑑𝑢$

Since $𝑔(𝑢(1-𝑢))=𝑔((1-𝑢)𝑢)$ is the same function of $𝑥$ used in ${𝐼}_1,{𝐼}_2$,

${𝐼}_1={\int }_{𝑞}^{𝑝}𝑔(𝑢(1-𝑢))𝑑𝑢-{\int }_{𝑞}^{𝑝}𝑢𝑔(𝑢(1-𝑢))𝑑𝑢={𝐼}_2-{𝐼}_1$

$\Rightarrow 2{𝐼}_1={𝐼}_2\Rightarrow \frac{{𝐼}_2}{{𝐼}_1}=2$

Hence, the correct option is (C).

Let $a, b, c$ be distinct odd natural numbers. Suppose the equation $ax^2+bx+c=0$ has a rational root $\frac{p}{q}$ (in lowest terms). Then $p$ divides $c$ and $q$ divides $a$, so both $p$ and $q$ must be odd (since $a, c$ are odd).

For a rational root to exist, the discriminant $D=b^2-4ac$ must be a perfect square. Since $a, c$ are odd, $4ac \equiv 4 \pmod 8$ is even but $b^2$ is odd (as $b$ is odd), so $D = b^2 - 4ac$ is odd. If $x=\frac{p}{q}$ is a root with $p, q$ odd integers, substituting back gives $ap^2+bpq+cq^2=0$. Since $a,b,c,p,q$ are all odd, $ap^2+bpq+cq^2$ is a sum of three odd terms, which is odd, and cannot equal zero.

This contradiction shows no rational root can exist. Hence the number of rational roots must be 0.

Correct option: A

Let's solve the given differential equation and find the limit $\underset{𝑥\to 1}{lim}𝑦\left(𝑥\right)$.

The differential equation given is:

$𝑥{𝑦}^{\prime }+𝑦-{𝑒}^{𝑥}=0$

This is a first-order linear differential equation. We can rewrite it in the form:

${𝑦}^{\prime }+\frac{1}{𝑥}𝑦=\frac{{𝑒}^{𝑥}}{𝑥}$

We will solve this using the integrating factor method. The integrating factor (IF) is given by:

$𝜇\left(𝑥\right)={𝑒}^{\int \frac{1}{𝑥}𝑑𝑥}={𝑒}^{\ln \left|𝑥\right|}=\left|𝑥\right|$

Since we are not given a specific interval for x, we assume x is positive:

$𝜇\left(𝑥\right)=𝑥$

Multiplying both sides of the differential equation by the integrating factor:

$𝑥{𝑦}^{\prime }+𝑦={𝑒}^{𝑥}\Rightarrow 𝑥{𝑦}^{\prime }+𝑦={𝑒}^{𝑥}\Rightarrow 𝑥({𝑦}^{\prime }+\frac{1}{𝑥}𝑦)={𝑒}^{𝑥}$

$𝑥\cdot {𝑦}^{\prime }+𝑦={𝑒}^{𝑥}$

Recognize that the left side is the derivative of $𝑥𝑦$:

$\frac{𝑑}{𝑑𝑥}\left(𝑥𝑦\right)={𝑒}^{𝑥}$

Integrate both sides with respect to x:

$𝑥𝑦=\int {𝑒}^{𝑥}𝑑𝑥={𝑒}^{𝑥}+𝐶$

So we have:

$𝑦=\frac{{𝑒}^{𝑥}+𝐶}{𝑥}$

Apply the initial condition $𝑦\left(𝑎\right)=𝑏$:

$𝑏=\frac{{𝑒}^{𝑎}+𝐶}{𝑎}$

$𝐶=𝑎𝑏-{𝑒}^{𝑎}$

Thus, the solution to the differential equation is:

$𝑦=\frac{{𝑒}^{𝑥}+𝑎𝑏-{𝑒}^{𝑎}}{𝑥}$

Now, we need to determine:

$\underset{𝑥\to 1}{lim}𝑦\left(𝑥\right)$

Substituting $𝑥=1$ into the solution:

$𝑦\left(1\right)=\frac{{𝑒}^1+𝑎𝑏-{𝑒}^{𝑎}}{1}=𝑒+𝑎𝑏-{𝑒}^{𝑎}$

Therefore:

The correct answer is $\text{Option D}$

$𝑒+𝑎𝑏-{𝑒}^{𝑎}$

Let's solve the given differential equation and find the limit $\underset{𝑥\to 1}{lim}𝑦\left(𝑥\right)$.

The differential equation given is:

$𝑥{𝑦}^{\prime }+𝑦-{𝑒}^{𝑥}=0$

This is a first-order linear differential equation. We can rewrite it in the form:

${𝑦}^{\prime }+\frac{1}{𝑥}𝑦=\frac{{𝑒}^{𝑥}}{𝑥}$

We will solve this using the integrating factor method. The integrating factor (IF) is given by:

$𝜇\left(𝑥\right)={𝑒}^{\int \frac{1}{𝑥}𝑑𝑥}={𝑒}^{\ln \left|𝑥\right|}=\left|𝑥\right|$

Since we are not given a specific interval for x, we assume x is positive:

$𝜇\left(𝑥\right)=𝑥$

Multiplying both sides of the differential equation by the integrating factor:

$𝑥{𝑦}^{\prime }+𝑦={𝑒}^{𝑥}\Rightarrow 𝑥{𝑦}^{\prime }+𝑦={𝑒}^{𝑥}\Rightarrow 𝑥({𝑦}^{\prime }+\frac{1}{𝑥}𝑦)={𝑒}^{𝑥}$

$𝑥\cdot {𝑦}^{\prime }+𝑦={𝑒}^{𝑥}$

Recognize that the left side is the derivative of $𝑥𝑦$:

$\frac{𝑑}{𝑑𝑥}\left(𝑥𝑦\right)={𝑒}^{𝑥}$

Integrate both sides with respect to x:

$𝑥𝑦=\int {𝑒}^{𝑥}𝑑𝑥={𝑒}^{𝑥}+𝐶$

So we have:

$𝑦=\frac{{𝑒}^{𝑥}+𝐶}{𝑥}$

Apply the initial condition $𝑦\left(𝑎\right)=𝑏$:

$𝑏=\frac{{𝑒}^{𝑎}+𝐶}{𝑎}$

$𝐶=𝑎𝑏-{𝑒}^{𝑎}$

Thus, the solution to the differential equation is:

$𝑦=\frac{{𝑒}^{𝑥}+𝑎𝑏-{𝑒}^{𝑎}}{𝑥}$

Now, we need to determine:

$\underset{𝑥\to 1}{lim}𝑦\left(𝑥\right)$

Substituting $𝑥=1$ into the solution:

$𝑦\left(1\right)=\frac{{𝑒}^1+𝑎𝑏-{𝑒}^{𝑎}}{1}=𝑒+𝑎𝑏-{𝑒}^{𝑎}$

Therefore:

The correct answer is $\text{Option D}$

$𝑒+𝑎𝑏-{𝑒}^{𝑎}$

$𝑥=4-{𝑦}^2$ and $𝑦$-axis

${𝑦}^2=4-𝑥\Rightarrow$ Parabola

$\begin{array}{cc}& \text{ Required area }=2{\int }_2^0(4-{𝑦}^2)𝑑𝑦=2{[4𝑦-\frac{{𝑦}^3}{3}]}_2^0\\ \\ & =2[(4.2-\frac{2^3}{3})-(0\left)\right]=\frac{32}{3}\text{ sq. unit }\end{array}$

To solve the given integral, let's consider the integral:

$\int_{0}^{\pi} e^{\cos^2 x} \cdot \cos^3((2n+1)x) \, dx$

First, observe the integrand carefully. It includes the term $\cos^3((2n+1)x)$ which can be expanded using the trigonometric identity for cosine:

$\cos^3(\theta) = \frac{1}{4} \left( \cos(3\theta) + 3\cos(\theta) \right)$

Applying this identity to the integrand, we have:

$\cos^3((2n+1)x) = \frac{1}{4} \left( \cos(3(2n+1)x) + 3\cos((2n+1)x) \right)$

Substitute this back into the integral:

$\int_{0}^{\pi} e^{\cos^2 x} \cdot \frac{1}{4} \left( \cos(3(2n+1)x) + 3\cos((2n+1)x) \right) \, dx$

This can be broken down into two separate integrals:

$\frac{1}{4} \int_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx + \frac{3}{4} \int_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx$

Now, consider the properties of definite integrals and the cosine function over the interval from 0 to $\pi$. For odd multiples of $x$, the integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.

Specifically, for any odd integer $k$:

$\int_{0}^{\pi} \cos(kx) \, dx = 0$

Therefore:

$\int_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx = 0$

and

$\int_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx = 0$

Thus, combining these results:

$\frac{1}{4} \times 0 + \frac{3}{4} \times 0 = 0$