MHT CET 2025 22nd April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \left(\frac{2+\sin 𝑥}{𝑦+1}\right)\frac{d𝑦}{d𝑥}=-\cos 𝑥\\ \therefore & \frac{1}{𝑦+1}d𝑦=\frac{-\cos 𝑥}{2+\sin 𝑥}d𝑥\\ & \text{ Integrating on both sides, we get }\\ & \log (𝑦+1)=-\log (2+\sin 𝑥)+c...\left(𝑖\right)\\ & \text{ When }𝑥=0,𝑦=1\\ \Rightarrow & c=2\log 2=\log 4\\ \therefore & \begin{array}{c}\text{ (i) }\Rightarrow \log (𝑦+1)=-\log (2+\sin 𝑥)+\log 4\\ \Rightarrow \log (𝑦+1)=\log \left(\frac{4}{2+\sin 𝑥}\right)\end{array}\\ \Rightarrow & 𝑦+1=\frac{4}{2+\sin 𝑥}\\ \Rightarrow & 𝑦=\frac{4}{2+\sin 𝑥}-1\\ \therefore & 𝑦\left(\frac{𝜋}{2}\right)=\frac{4}{3}-1=\frac{1}{3}\end{array}$

$\begin{array}{cc}& \left|𝐴\right|=1+{\cot }^2\frac{𝜃}{2}={\mathrm{cosec}}^2\frac{𝜃}{2}\ne 0\\ & \text{ Now, }\mathrm{adj}\left(𝐴\right)=\left[\begin{array}{cc}1& -\cot \frac{𝜃}{2}\\ \cot \frac{𝜃}{2}& 1\end{array}\right]\\ \therefore & {𝐴}^{-1}=\frac{1}{\left|𝐴\right|}(\mathrm{adj}(𝐴\left)\right)=\frac{1}{{\mathrm{cosec}}^2\frac{𝜃}{2}}\left[\begin{array}{cc}1& -\cot \frac{𝜃}{2}\\ \cot \frac{𝜃}{2}& 1\end{array}\right]\\ & ={\sin }^2\frac{𝜃}{2}{𝐴}^{𝑇}=\left(\frac{1-\cos 𝜃}{2}\right){𝐴}^{𝑇}\end{array}$

$\begin{array}{cc}& \bar{a},\bar{b},\bar{c}\text{ are unit vectors. }\\ & \begin{array}{cc}& |\bar{a}+\bar{b}{|}^2+|\bar{a}+\bar{c}{|}^2=8\\ & \Rightarrow |\bar{a}{|}^2+|\bar{b}{|}^2+2\bar{a}\cdot \bar{b}+|\bar{a}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{c}=8\\ & \Rightarrow 1+1+2\bar{a}\cdot \bar{b}+1+1+2\bar{a}\cdot \bar{c}=8\\ & \Rightarrow 2(\bar{a}\bar{b}+\bar{a}\bar{c})=4\\ & \Rightarrow \bar{a}\cdot \bar{b}+\bar{a}\cdot \bar{c}=2...\left(𝑖\right)\end{array}\end{array}$

Now,

$\begin{array}{cc}& |\overrightarrow{𝑎}+3\overrightarrow{𝑏}{|}^2+|\overrightarrow{𝑎}+3\overrightarrow{𝑐}{|}^2\\ & =|\overrightarrow{𝑎}{|}^2+|3\overrightarrow{𝑏}{|}^2+2\overrightarrow{𝑎}\cdot 3\overline{𝑏}+|\overrightarrow{𝑎}{|}^2+|3\overline{𝑐}{|}^2+2\overline{𝑎}\cdot 3\overline{𝑐}\\ & =1+9+6\overline{𝑎}\cdot \overline{𝑏}+1+9+6\overline{𝑎}\cdot \overline{𝑐}\\ & =20+6(\overline{𝑎}\cdot \overline{𝑏}+\overline{𝑎}\cdot \overline{𝑐})\\ & =20+6\left(2\right)\dots \text{ [From (i)] }\\ & =20+12\\ & =32\end{array}$

Step 1: Define the Probability

The chance that one component will survive is given as $𝑝=\frac{2}{3}$.

The chance that one component will not survive is $𝑞=1-𝑝=1-\frac{2}{3}=\frac{1}{3}$.

Step 2: Number of Components Tested

We are testing 4 components, so $𝑛=4$.

Step 3: Binomial Distribution

This situation follows a binomial distribution: $𝑋\sim 𝐵(𝑛=4,𝑝=\frac{2}{3})$.

Step 4: Probability That At Most 2 Survive

We want to find the probability that at most 2 components survive, or $𝑃(𝑋\le 2)$.

This means we need to add up the probabilities for exactly 0, 1, and 2 components surviving:

$𝑃(𝑋\le 2)=𝑃(𝑋=0)+𝑃(𝑋=1)+𝑃(𝑋=2)$

Step 5: Calculate Each Probability

$𝑃(𝑋=0)={}^4{𝐶}_0{\left(\frac{2}{3}\right)}^0{\left(\frac{1}{3}\right)}^4$ $𝑃(𝑋=1)={}^4{𝐶}_1{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^3$ $𝑃(𝑋=2)={}^4{𝐶}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2$

Step 6: Substitute Values and Add

$𝑃(𝑋\le 2)=\frac{1}{81}+\frac{8}{81}+\frac{24}{81}$

$=\frac{33}{81}$

$=\frac{33}{3^4}$

Distance from origin to the line $a𝑥+b𝑦+c=0$ is given by,

$\therefore \text{ Distance }=\frac{\left|𝑐\right|}{\sqrt{\surd {𝑎}^2+{𝑏}^2}}$

Consider option (D).

$\therefore$ Distance of a line $5𝑥-2𝑦=3$ from the origin

$\begin{array}{cc}& =\frac{\left|3\right|}{\sqrt{\surd 5^2+(-2{)}^2}}\\ & =\frac{3}{\sqrt{\surd 25+4}}=\frac{3}{\sqrt{\surd 29}}\approx 0.557\end{array}$

$\therefore$ Option (D): $5𝑥-2𝑦=3$ is the nearest line to the origin

$\begin{array}{cc}& \begin{array}{cc}& [𝑝\wedge \sim 𝑟]<text>\to </text><text>\sim </text>𝑟\wedge 𝑞\equiv 𝐹\\ & \Rightarrow 𝑝\wedge \sim 𝑟\equiv 𝑇\text{ and }\sim 𝑟\wedge 𝑞\equiv 𝐹\\ & \Rightarrow 𝑝\equiv 𝑇,\sim 𝑟\equiv 𝑇\\ & \Rightarrow 𝑝\equiv 𝑇,𝑟=𝐹\end{array}\\ & \text{ Consider }\sim 𝑟\wedge 𝑞\equiv 𝐹\end{array}$

$\Rightarrow 𝑞\equiv 𝐹\dots (<text>\because </text><text>\sim </text>𝑟\equiv 𝑇)$

We get,

$𝑝\equiv 𝑇,𝑞\equiv 𝐹,𝑟\equiv 𝐹$

Now consider option (D)

$\begin{array}{cc}\sim (𝑟\vee 𝑞)<text>\to </text><text>\sim </text>𝑝& <text>\equiv </text><text>\sim </text>(𝐹\vee 𝐹)<text>\to </text><text>\sim </text>𝑇\\ & <text>\equiv </text><text>\sim </text>\left(𝐹\right)\to 𝐹\\ & \equiv 𝑇\to 𝐹\\ & \equiv 𝐹\end{array}$

Let $\overline{𝑑}=𝑝\widehat{𝑖}+𝑞\widehat{𝑗}+𝑟\widehat{𝑘}$, where $𝑝,𝑞,𝑟\in 𝑅$

As $\overline{𝑏},\overline{𝑐},\overline{𝑑}$ are coplanar, we get

$\begin{array}{cc}& \mid \begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ 𝑝& 𝑞& 𝑟\end{array}\mid =0\\ \therefore & -1(-𝑟-𝑝)-1(-𝑞)=0\\ \therefore & 𝑝+𝑞+𝑟=0...\left(𝑖\right)\end{array}$

Also, given that $\bar{a}$ and $\bar{d}$ are perpendiculars.

$\begin{array}{cc}\therefore & \overline{𝑎}\cdot \overline{𝑑}=0\\ \therefore & 𝑝-𝑞=0\\ \therefore & 𝑝=𝑞...\left(𝑖𝑖\right)\end{array}$

Among the given options only option (D) satisfies equations (i) and (ii)

$\begin{array}{cc}& \text{ Given, }{𝑎}^4+{𝑏}^4+{𝑐}^4-2{𝑎}^2{𝑐}^2-2{𝑐}^2{𝑏}^2=0\\ & \Rightarrow {𝑎}^4+{𝑏}^4+{𝑐}^4=2{𝑎}^2{𝑐}^2+2{𝑐}^2{𝑏}^2...\left(𝑖\right)\end{array}$

Using cosine rule,

$\begin{array}{cc}& \cos 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}\\ & \Rightarrow {𝑎}^2+{𝑏}^2-{𝑐}^2=2𝑎𝑏\cos 𝐶\end{array}$

Squaring on both sides, we get

$\begin{array}{cc}{𝑎}^4+{𝑏}^4+{𝑐}^4+2{𝑎}^2{𝑏}^2-2{𝑏}^2{𝑐}^2-2{𝑎}^2{𝑐}^2& =4{𝑎}^2{𝑏}^2{\cos }^2𝐶\\ \Rightarrow 2{𝑎}^2{𝑐}^2+2{𝑐}^2{𝑏}^2+2{𝑎}^2{𝑏}^2-2{𝑏}^2{𝑐}^2-2{𝑎}^2{𝑐}^2=& 4{𝑎}^2{𝑏}^2{\cos }^2𝐶\dots \left[\text{ From (i) }\right]\end{array}$

$\begin{array}{cc}& \Rightarrow 2{𝑎}^2{𝑏}^2=4{𝑎}^2{𝑏}^2{\cos }^2𝐶\\ & \Rightarrow {\cos }^2𝐶=\frac{1}{2}\\ & \Rightarrow \cos 𝐶=\frac{1}{\sqrt{\surd 2}}\\ & \Rightarrow 𝐶={135}^{\circ }\end{array}$

Given planes are parallel

Then,

$\begin{array}{cc}& \frac{{𝑎}_1}{{𝑎}_2}=\frac{{𝑏}_1}{{𝑏}_2}=\frac{{𝑐}_1}{{𝑐}_2}\\ & \Rightarrow \frac{2}{4}=\frac{-𝜆}{-1}=\frac{1}{𝜇}\\ & \Rightarrow 𝜆=\frac{1}{2}\text{ and }𝜇=2\\ & \Rightarrow 𝜆+𝜇=\frac{1}{2}+2=\frac{5}{2}\end{array}$

$\begin{array}{cc}& \text{ Let }𝑦=\frac{{𝑥}^2-𝑥+1}{{𝑥}^2+𝑥+1}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{({𝑥}^2+𝑥+1)(2𝑥-1)-({𝑥}^2-𝑥+1)(2𝑥+1)}{{({𝑥}^2+𝑥+1)}^2}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{2{𝑥}^2-2}{{({𝑥}^2+𝑥+1)}^2}\end{array}$

For maximum or minimum,

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}=0\\ & \Rightarrow 2{𝑥}^2-2=0\\ & \Rightarrow 𝑥=-1,1\\ & \frac{d^2𝑦}{d{𝑥}^2}=\frac{4(-{𝑥}^3+3𝑥+1)}{{𝑥}^2+𝑥+1}\end{array}$

At $𝑥=-1,\frac{d^2𝑦}{d{𝑥}^2}<0$, the function will occupy maximum value.

$\therefore 𝑓(-1)=3$

At $𝑥=1,\frac{d^2𝑦}{d{𝑥}^2}>0$, the function will occupy minimum value.

$\therefore 𝑓\left(1\right)=\frac{1}{3}$

Max. value - Min. value $=3-\frac{1}{3}=\frac{8}{3}$

l: $\frac{𝑥-1}{2}=\frac{𝑦+2}{3}=\frac{𝑧-3}{4}$ and

l2: $\frac{𝑥}{2}=\frac{𝑦-1}{3}=\frac{𝑧+1}{4}$

Let ${𝑟}_1$ and ${𝑟}_2$ be the position vectors of point on lines ${𝑙}_1$ and ${𝑙}_2$ respectively.

$\therefore {\overline{𝑟}}_1=(1,-2,3)\text{ and }{\overline{𝑟}}_2=(0,1,-1)$

and direction ratios of line ${𝑙}_1$ and ${𝑙}_2$ are

$\bar{d}=(2,3,4)$

$\therefore$ Lines are parallel

Distance between parallel lines ${𝑙}_1$ and ${𝑙}_2$ is given by,

$\begin{array}{cc}& D=\frac{\mid ({\overline{𝑟}}_2-{\overline{𝑟}}_1)\times \overline{𝑑}\mid }{\left|\overline{𝑑}\right|}\\ & {\overline{𝑟}}_2-{\overline{𝑟}}_1=(-1,3,-4)\\ \therefore & ({\overline{𝑟}}_2-{\overline{𝑟}}_1)\times \overline{𝑑}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ -1& 3& -4\\ 2& 3& 4\end{array}\mid \\ & =\widehat{𝑖}(12+12)-\widehat{𝑗}(-4+8)+\widehat{𝑘}(-3-6)\\ & =24\widehat{𝑖}-4\widehat{𝑗}-9\widehat{𝑘}\\ \Rightarrow & \mid ({\overline{𝑟}}_2-{\overline{𝑟}}_1)\times \overrightarrow{𝑑}\mid =\sqrt{\surd (24{)}^2+(-4{)}^2+(-9{)}^2}\\ & =\sqrt{\surd 673}\end{array}$

$\begin{array}{cc}\therefore & \text{ And }\left|\bar{d}\right|=\sqrt{\surd (2{)}^2+(3{)}^2+(4{)}^2}=\sqrt{\surd 29}\\ \therefore & \text{ Distance }\left(D\right)=\frac{\sqrt{\surd 673}}{\sqrt{\surd 29}}\\ & \text{ Perimeter of square }=4\times \text{ Distance between }\\ & \text{ lines }\\ & =4\times \frac{\sqrt{\surd 673}}{\sqrt{\surd 29}}\\ & =\frac{4\sqrt{\surd 673}}{\sqrt{\surd 29}}\text{ units }\end{array}$

$\tan \left(\frac{B-C}{2}\right)=𝑥\cot \frac{A}{2}$

We have,

$\tan \left(\frac{B-C}{2}\right)=\left(\frac{b-c}{b+c}\right)\cot \frac{A}{2}$

                                                    …[Napier's Analogies]

$\begin{array}{cc}\therefore & \left(\frac{b-c}{b+c}\right)\cot \frac{A}{2}=𝑥\cot \frac{A}{2}\\ & \Rightarrow 𝑥=\frac{b-c}{b+c}\end{array}$

A coin is tossed until: Head $\left(H\right)$ appears or 4 consecutive tails (TTTT) occur

Let $X=$ number of tosses until the process stops.

$\begin{array}{cc}\therefore & 𝑃(𝑋=1)=𝑃\left(𝐻\right)=\frac{1}{2}\\ \therefore & 𝑃(𝑋=2)=𝑃\left(𝑇\right)𝑃\left(𝐻\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\\ \therefore & 𝑃(𝑋=3)=𝑃\left(𝑇\right)𝑃\left(𝑇\right)𝑃\left(𝐻\right)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}\end{array}$

$\begin{array}{cc}& P(X=4)=P\left(T\right)P\left(T\right)P\left(T\right)P\left(H\right)={\left(\frac{1}{2}\right)}^4=\frac{1}{16}\\ & \text{ or }P\left(T\right)P\left(T\right)P\left(T\right)P\left(T\right)={\left(\frac{1}{2}\right)}^4=\frac{1}{16}\\ \therefore & P(X=4)=\frac{1}{16}+\frac{1}{16}=\frac{2}{16}=\frac{1}{8}\end{array}$

We are tasked with finding the degree of the differential equation:

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}+3{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2={𝑥}^2\log \left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)$

Step 1: Recall the concept of degree of a differential equation

• The degree of a differential equation is the power of the highest order derivative after the equation is made polynomial in all derivatives.

• That means:

1. There should be no transcendental (logarithmic, exponential, trigonometric, etc.) functions of derivatives.

2. No fractional powers of derivatives.

If such functions appear in the differential equation and cannot be removed, the degree is not defined.

Step 2: Examine our equation

Highest order derivative: $\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}$.

It appears in two places:

• Linearly on the LHS: $\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}$.

• Inside a logarithm: $\log \left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)$.

Because the second derivative $\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}$ appears inside $\log (\cdot )$, the equation is not a polynomial in the derivatives.

Step 3: Conclusion

Since we cannot remove the logarithmic involvement of the highest order derivative, the degree of the equation is not defined.

✅ Answer: Option D (Not defined)

$\begin{array}{cc}& \text{ According to given condition, }\\ & \begin{array}{cc}& {\left(\sqrt{\surd {𝑦}^2+{𝑧}^2}\right)}^2+{\left(\sqrt{\surd {𝑥}^2+{𝑧}^2}\right)}^2+{\left(\sqrt{\surd {𝑥}^2+{𝑦}^2}\right)}^2=324\\ & {𝑦}^2+{𝑧}^2+{𝑥}^2+{𝑧}^2+{𝑥}^2+{𝑦}^2=324\\ & 2{𝑥}^2+2{𝑦}^2+2{𝑧}^2=324\\ & {𝑥}^2+{𝑦}^2+{𝑧}^2=\frac{324}{2}\\ & {𝑥}^2+{𝑦}^2+{𝑧}^2=162\end{array}\\ & \begin{array}{cc}\therefore \text{ Distance of }\mathrm{OP}& =\sqrt{\surd {𝑥}^2+{𝑦}^2+{𝑧}^2}\\ & =\sqrt{\surd 162}\\ & =9\sqrt{\surd 2}\end{array}\end{array}$

Given expression

$\begin{array}{cc}& =\sum _{𝑖=0}^{99}[\frac{1}{2}+\frac{𝑖}{100}]\\ & =\sum _{𝑖=0}^{49}[\frac{1}{2}+\frac{𝑖}{100}]+\sum _{𝑖=50}^{99}[\frac{2}{3}+\frac{𝑖}{100}]\\ & =0+\sum _{𝑖=50}^{99}[\frac{1}{2}+\frac{𝑖}{100}]\dots \left[\begin{array}{c}\because \frac{1}{2}\le \frac{1}{2}+\frac{𝑖}{100}<1\\ \text{ For }𝑖=0,1,2,\dots 49\end{array}\right]\end{array}$

$=50$ [each term in the summation is one or more than 1 but less than 2 when $i=50,51,52$, ….99]

We are given the lines:

$3𝑥=2𝑦=-𝑧$

and

$-𝑥=6𝑦=-4𝑧$

We want the angle between them.

Step 1: Find direction vector of the first line

Equation: $3𝑥=2𝑦=-𝑧=𝑘.$

So we can parametrize with $𝑘$:

$𝑥=\frac{𝑘}{3},𝑦=\frac{𝑘}{2},𝑧=-𝑘$

Thus the direction vector is proportional to

$(\frac{1}{3},\frac{1}{2},-1).$

To clear denominators multiply by 6:

$(2,3,-6).$

Step 2: Find direction vector of the second line

Equation: $-𝑥=6𝑦=-4𝑧=𝑚.$

So

$𝑥=-𝑚,𝑦=\frac{𝑚}{6},𝑧=-\frac{𝑚}{4}.$

Direction vector:

$(-1,\frac{1}{6},-\frac{1}{4}).$

Multiply through by 12:

$(-12,2,-3).$

Step 3: Angle between vectors

Dot product:

$(2,3,-6)\cdot (-12,2,-3)=2(-12)+3\left(2\right)+(-6)(-3).$

Compute:

$=-24+6+18=0.$

So vectors are perpendicular.

Step 4: Conclusion

Thus the angle between the lines is

$\frac{𝜋}{2}.$

Answer:

$\text{Option C: }\frac{𝜋}{2}$

$\begin{array}{cc}& 𝑦+\frac{d}{d𝑥}\left(𝑥𝑦\right)=𝑥(\sin 𝑥+\log 𝑥)\\ & 𝑦+𝑥\frac{d𝑦}{d𝑥}+𝑦=𝑥(\sin 𝑥+\log 𝑥)\\ & 𝑥\frac{d𝑦}{d𝑥}+2𝑦=𝑥(\sin 𝑥+\log 𝑥)\\ & \frac{d𝑦}{d𝑥}+2\left(\frac{𝑦}{𝑥}\right)=\sin 𝑥+\log 𝑥\\ & \Rightarrow P=\frac{2}{𝑥},Q=\sin 𝑥+\log 𝑥\\ & \begin{array}{cc}\text{ I.F }& =e^{\int \mathrm{pd}𝑥}=e^{\int \frac{2}{𝑥}}d𝑥\\ & =e^{2\log 𝑥}\\ & ={𝑥}^2\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑦\times \text{ I.F }=\int \text{ Q }\times \text{ I.F }d𝑥+c\\ & 𝑦{𝑥}^2=\int (\sin 𝑥+\log 𝑥){𝑥}^{2}d𝑥+c\\ & 𝑦{𝑥}^2=\int ({𝑥}^2\sin 𝑥+{𝑥}^2\log 𝑥)d𝑥+c\\ & 𝑦{𝑥}^2=\int {𝑥}^2\sin 𝑥d𝑥+\int {𝑥}^2\log 𝑥d𝑥+c\\ & =-{𝑥}^2\cos 𝑥-\int \left(2𝑥\right(-\cos 𝑥\left)\right)d𝑥+\left(\frac{{𝑥}^3}{3}\right)\log 𝑥-\int (\frac{1}{𝑥}\times \frac{{𝑥}^3}{3})\mathrm{dx}\end{array}$

$\begin{array}{cc}& =-{𝑥}^2\cos 𝑥+2[\int 𝑥\cos 𝑥d𝑥]+\frac{{𝑥}^3}{3}\log 𝑥-\frac{1}{3}\int {𝑥}^{2}d𝑥+c\\ & =-{𝑥}^2\cos 𝑥+2[𝑥\sin 𝑥-\int (\sin 𝑥\left)d𝑥\right]+\frac{{𝑥}^3}{3}\log 𝑥-\frac{1}{3}\left(\frac{{𝑥}^3}{3}\right)+c\end{array}$

$\begin{array}{cc}& \text{ Dividing by }{𝑥}^2\text{, we get }\\ & 𝑦=-\cos 𝑥+\frac{2\sin 𝑥}{𝑥}+\frac{2}{{𝑥}^2}\cos 𝑥+\frac{𝑥}{3}\log 𝑥-\frac{𝑥}{9}+\frac{𝑐}{{𝑥}^2}\end{array}$

$\begin{array}{cc}& 𝑥𝑦-𝑥+𝑦-1=0\\ & \Rightarrow 𝑥(𝑦-1)+(𝑦-1)=0\\ & \Rightarrow (𝑥+1)(𝑦-1)=0\\ & \Rightarrow 𝑥=-1\text{ or }𝑦=1\end{array}$

$\therefore$ Point of intersection is $(-1,1)$

Substituting $𝑥=-1,𝑦=1$, in equation $𝑥+k𝑦-3=0$, we get

$\begin{array}{cc}& (-1)+k\left(1\right)-3=0\\ & \Rightarrow k=3+1=4\end{array}$

Since $𝑓\left(𝑥\right)$ is continuous in $[0,𝜋]$.

$\therefore$ it is continuous at $𝑥=\frac{𝜋}{4}$ and $𝑥=\frac{𝜋}{2}$.

$\begin{array}{cc}\therefore & \underset{𝑥\to {\left(\frac{𝑥}{4}\right)}^{-}}{lim}f\left(𝑥\right)=\underset{𝑥\to {\left(\frac{𝑥}{4}\right)}^{+}}{lim}f\left(𝑥\right)\\ \Rightarrow & \underset{𝑥\to {\left(\frac{𝑥}{4}\right)}^{-}}{lim}(𝑥+a\sqrt{\surd 2}\sin 𝑥)=\underset{𝑥\to {\left(\frac{𝑥}{4}\right)}^{+}}{lim}(2𝑥\cot 𝑥+b)\\ \Rightarrow & \frac{𝜋}{4}+a\sqrt{\surd 2}\left(\frac{1}{\sqrt{\surd 2}}\right)=2\left(\frac{𝜋}{4}\right)\left(1\right)+b\\ \Rightarrow & \frac{𝜋}{4}+a=\frac{𝜋}{2}+b\\ \Rightarrow & a-b=\frac{𝜋}{4}...\left(𝑖\right)\end{array}$