MHT CET 2025 26th April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \begin{array}{cc}& \frac{d𝑦}{d𝑥}=\cot 𝑥\cot 𝑦\\ & \Rightarrow \cot 𝑥d𝑥-\tan 𝑦d𝑦=0\end{array}\\ & \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \log (\sin 𝑥)-\log (\sec 𝑦)=\log 𝑐\\ & \Rightarrow \log \left(\frac{\sin 𝑥}{\sec 𝑦}\right)=\log 𝑐\Rightarrow \sin 𝑥=𝑐\sec 𝑦\end{array}\end{array}$

• Probability that a person is not a sportsperson = $\frac{1}{6}$.

• Thus, probability that a person is a sportsperson = $1-\frac{1}{6}=\frac{5}{6}$.

We need the probability that out of 6 family members, exactly 5 are sportspersons.

Step 1: Identify the distribution

This is a binomial probability problem:

$𝑃(𝑋=𝑘)=(\frac{𝑛}{𝑘}){𝑝}^{𝑘}(1-𝑝{)}^{𝑛-𝑘}$

where

• $𝑛=6$ (number of people),

• $𝑝=\frac{5}{6}$ (probability of a sportsperson),

• $𝑘=5$ (number of sportspersons).

Step 2: Apply the formula

$𝑃(𝑋=5)=(\frac{6}{5}){\left(\frac{5}{6}\right)}^5{\left(\frac{1}{6}\right)}^1$

$=6\cdot {\left(\frac{5}{6}\right)}^5\cdot \frac{1}{6}$

$=[6\cdot \frac{1}{6}]\cdot {\left(\frac{5}{6}\right)}^5$

$={\left(\frac{5}{6}\right)}^5$

Step 3: Match with options

The correct answer is:

$\overline{){\left(\frac{5}{6}\right)}^5}$

✅ Therefore, the answer is Option A.

$\begin{array}{cc}& \tan (𝜋\cos 𝜃)=\cot (𝜋\sin 𝜃)\\ & \Rightarrow \frac{\sin (𝜋\cos 𝜃)}{\cos (𝜋\cos 𝜃)}=\frac{\cos (𝜋\sin 𝜃)}{\sin (𝜋\sin 𝜃)}\\ & \Rightarrow \cos (𝜋\cos 𝜃)\cos (𝜋\sin 𝜃)-\sin (𝜋\cos 𝜃)\sin (𝜋\sin 𝜃)=0\end{array}$

$\begin{array}{cc}& \Rightarrow \cos (𝜋\cos 𝜃+𝜋\sin 𝜃)=0\\ & \Rightarrow 𝜋\cos 𝜃+𝜋\sin 𝜃=\frac{𝜋}{2}\\ & \Rightarrow \cos 𝜃+\sin 𝜃=\frac{1}{2}\\ & \Rightarrow \sqrt{\surd 2}(\frac{1}{\sqrt{\surd 2}}\cos 𝜃+\frac{1}{\sqrt{\surd 2}}\sin 𝜃)=\frac{1}{2}\\ & \Rightarrow \sin (\frac{𝜋}{4}+𝜃)=\frac{1}{2\sqrt{\surd 2}}\end{array}$

Given equation of line is $𝑥-𝑦-2=0$

Coordinates of $𝑀\equiv (2,0)$

Slope of $\mathrm{MN}=\frac{2-0}{4-2}=\frac{2}{2}=1$

$\begin{array}{cc}& \Rightarrow \tan 𝜃=1\\ & \Rightarrow 𝜃={45}^{\circ }\end{array}$

Since the line passes through $(2,0)$.

$\therefore$ Its equation is $𝑥=2$.

$\begin{array}{cc}& \text{ Here, Amount }\left(𝐴\right)=𝑅𝑠800\\ & \text{ Principal }\left(𝑃\right)=𝑅𝑠400\\ & 𝑁=6\text{ years }\\ & 𝐴=𝑃{(1+\frac{𝑅}{100})}^{𝑁}\\ & \Rightarrow 800=400{(1+\frac{𝑅}{100})}^6\\ & \Rightarrow 1+\frac{𝑅}{100}=2^{\frac{1}{6}}\\ & \begin{array}{cc}𝐴& =𝑃{(1+\frac{𝑅}{100})}^{𝑁}=400{(1+\frac{𝑅}{100})}^{30}\\ & =400{\left(2^{\frac{1}{6}}\right)}^{30}\\ & =400\times 2^5\\ & =𝑅𝑠12800\end{array}\end{array}$

Given equation of circle is

${𝑥}^2+{𝑦}^2-4𝑥-2𝑦-20=0$

Here, $g=-2,f=-1,c=-20$

Radius $\left(r\right)=\sqrt{\surd g^2+f^2-c}$

$\begin{array}{cc}& =\sqrt{\surd 4+1+20}\\ & =5\text{ units }\end{array}$

Centre $\left(C\right)=(2,1)$

$\begin{array}{cc}𝑑\left(𝐴𝐶\right)=\sqrt{\surd (10-2{)}^2+(7-1{)}^2}& =\sqrt{\surd 8^2+6^2}=\sqrt{\surd 100}\\ & =10\text{ units }\end{array}$

Shortest distance of point A from circle

= Distance of point A from centre - radius

$=10-5$

$=5$ units

$\begin{array}{cc}& \Rightarrow 𝐴𝑀=5\text{ units }\\ & 𝐴{𝑀}^{\prime }=5+\text{ diameter }=15\text{ units }\end{array}$

$P(-1,-1,2)$ is the foot of perpendicular from origin to the plane.

$\therefore$ d.r.s. of normal are $-1-0,-1-0,2-0$, i.e., $-1,-1,2$

$\therefore$ Equation of required plane is

$\begin{array}{cc}& -1(𝑥+1)-1(𝑦+1)+2(𝑧-2)=0\\ & \Rightarrow 𝑥+𝑦-2𝑧+6=0\end{array}$

We know that,

$\begin{array}{cc}& \frac{𝑎}{\sin 𝐴}=\frac{𝑏}{\sin 𝐵}=\frac{𝑐}{\sin 𝐶}\\ & \Rightarrow \frac{5}{\left(\frac{3}{4}\right)}=\frac{7}{\sin 𝐵}\\ & \Rightarrow \sin 𝐵=\frac{21}{20}>1\end{array}$

which is not possible since $-1\le \sin 𝑥\le 1$

$\therefore$ No triangles can be formed.

$\begin{array}{cc}& \text{ Let }{\sin }^{-1}\left(4𝑥\right)=𝛼\text{ and }{\sin }^{-1}\left(4\sqrt{\surd 3}𝑥\right)=𝛽\\ & \begin{array}{cc}& \Rightarrow \sin 𝛼=4𝑥\text{ and }\sin 𝛽=4\sqrt{\surd 3}𝑥\\ & 𝛼+𝛽=\frac{-𝜋}{2}\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝛽=\frac{-𝜋}{2}-𝛼\\ & \Rightarrow \sin 𝛽=\sin (\frac{-𝜋}{2}-𝛼)=-\sin (\frac{𝜋}{2}+𝛼)\\ & \Rightarrow \sin 𝛽=-\cos 𝛼\\ & \Rightarrow \cos 𝛼=-4\sqrt{\surd 3}𝑥\\ & \text{ Now, }\cos 𝛼=\sqrt{\surd 1-{\sin }^2𝛼}\\ & \Rightarrow -4\sqrt{\surd 3}𝑥=\sqrt{\surd 1-16{𝑥}^2}\end{array}$

Squaring both sides, we get

$\begin{array}{cc}& 48{𝑥}^2=1-16{𝑥}^2\\ & \Rightarrow 64{𝑥}^2=1\\ & \Rightarrow {𝑥}^2=\frac{1}{64}\Rightarrow 𝑥=\pm \frac{1}{8}\end{array}$

For $𝑥=\frac{1}{8}$,

${\sin }^{-1}\left(\frac{1}{2}\right)+{\sin }^{-1}\left(\frac{\sqrt{\surd 3}}{2}\right)=\frac{𝜋}{6}+\frac{𝜋}{3}=\frac{𝜋}{2}\ne \frac{-𝜋}{2}$

For $𝑥=\frac{-1}{8}$,

${\sin }^{-1}\left(\frac{-1}{2}\right)+{\sin }^{-1}\left(\frac{-\sqrt{\surd 3}}{2}\right)=\frac{-𝜋}{6}-\frac{𝜋}{3}=\frac{-𝜋}{2}$

$\therefore 𝑥=\frac{-1}{8}$ is the required solution.

Given that $\overline{𝑎}=𝛼\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘},\overline{𝑏}=3\widehat{𝑖}-\widehat{𝑗}+𝛽\widehat{𝑘}$ and

$\overline{𝑐}=\widehat{𝑖}+2\widehat{𝑗}-2\widehat{𝑘}$

Projection of $\bar{a}$ on $\bar{c}=\frac{\bar{a}\cdot \bar{c}}{\left|\bar{c}\right|}$

$\begin{array}{cc}& \Rightarrow \frac{𝛼+6+2}{\sqrt{\surd 1+4+4}}=\frac{10}{3}\Rightarrow 𝛼+8=10\\ & \Rightarrow 𝛼=2\\ & \bar{b}\times \bar{c}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 𝛽\\ 1& 2& -2\end{array}\mid \\ & =\hat{i}(2-2𝛽)-\hat{j}(-6-𝛽)+\hat{k}(6+1)\end{array}$

On comparing with $-6\widehat{𝑖}+10\widehat{𝑗}+7\widehat{𝑘}$, we get

$\begin{array}{cc}& 2-2𝛽=-6\\ & \Rightarrow 𝛽=4\\ \therefore & 𝛼+𝛽=6\end{array}$

Given equation of parabola is

$\begin{array}{cc}& {𝑦}^2+4𝑦+4𝑥+2=0\\ & \Rightarrow {𝑦}^2+4𝑦+4=-4𝑥-2+4\\ & \Rightarrow (𝑦+2{)}^2=-4(𝑥-\frac{1}{2})\\ & \text{ Here }a=1,X=𝑥-\frac{1}{2}\end{array}$

Equation of directrix is $X=a$

$\begin{array}{cc}& \Rightarrow 𝑥-\frac{1}{2}=1\\ & \Rightarrow 𝑥=\frac{3}{2}\end{array}$

$\begin{array}{cc}& f\left(𝑥\right)=\sqrt{\surd 1+{\cos }^2\left({𝑥}^2\right)}\\ \therefore & f^{\prime }\left(𝑥\right)=\frac{1}{2\sqrt{\surd 1+{\cos }^2\left({𝑥}^2\right)}}\cdot (2\cos {𝑥}^2)\cdot (-\sin {𝑥}^2)\cdot \left(2𝑥\right)\\ \therefore & f^{\prime }\left(𝑥\right)=\frac{-𝑥\sin 2{𝑥}^2}{\sqrt{\surd 1+{\cos }^2\left({𝑥}^2\right)}}\\ \therefore & f^{\prime }\left(\frac{\sqrt{\surd 𝜋}}{2}\right)=\frac{-\frac{\sqrt{\surd 𝜋}}{2}\cdot \sin \frac{2𝜋}{4}}{\sqrt{\surd 1+{\cos }^2\frac{𝜋}{4}}}=-\frac{\frac{\sqrt{\surd 𝜋}}{2}\cdot 1}{\sqrt{\surd \frac{3}{2}}}=-\sqrt{\surd \frac{𝜋}{6}}\end{array}$

$\begin{array}{cc}& \text{ Let }\sqrt{\surd 6+8i}=𝑥+i𝑦\\ & \Rightarrow 6+8i=(𝑥+i𝑦{)}^2\\ & \Rightarrow 6+8i={𝑥}^2+2𝑥𝑦i-{𝑦}^2\end{array}$

On comparing real and imaginary parts, we get

$\begin{array}{cc}& {𝑥}^2-{𝑦}^2=6\text{ and }2𝑥𝑦=8\\ & \Rightarrow {𝑥}^2-{𝑦}^2=6\text{ and }𝑥𝑦=4\\ & \Rightarrow {𝑥}^2-{𝑦}^2=6\text{ and }𝑦=\frac{4}{𝑥}\\ & \Rightarrow {𝑥}^2-\frac{16}{{𝑥}^2}=6\\ & \Rightarrow {𝑥}^4-6{𝑥}^2-16=0\end{array}$

$\Rightarrow {𝑥}^2=8\dots$ [considering only real roots]

$\Rightarrow 𝑥=\pm 2\sqrt{\surd 2}$

When $𝑥=2\sqrt{\surd 2},𝑦=\sqrt{\surd 2}$

When $𝑥=-2\sqrt{\surd 2},𝑦=-\sqrt{\surd 2}$

$\begin{array}{cc}\therefore \sqrt{\surd 6+8𝑖}& =\pm (2\sqrt{\surd 2}+\sqrt{\surd 2𝑖})\\ |\pm (2\sqrt{\surd 2}+\sqrt{\surd 2𝑖}\left)\right|& =\sqrt{\surd (2\sqrt{\surd 2}{)}^2+(\sqrt{\surd 2}{)}^2}\\ & =\sqrt{\surd 10}\\ & =\sqrt{\surd 2}\cdot \sqrt{\surd 5}\end{array}$

$\begin{array}{cc}& [𝑥{]}^2-5[𝑥]+6=0\\ & \Rightarrow [𝑥{]}^2-3[𝑥]-2[𝑥]+6=0\\ & \Rightarrow \left(\right[𝑥]-3)\left(\right[𝑥]-2)=0\\ & \Rightarrow \left[𝑥\right]=3\text{ or }\left[𝑥\right]=2\\ & \Rightarrow 𝑥\in [3,4)\text{ or }𝑥\in [2,3)\\ & \Rightarrow 𝑥\in [2,4)\end{array}$

$\begin{array}{cc}& \text{ Area of smaller part }=2{\int }_2^1\sqrt{\surd 4-{𝑥}^2}𝑑𝑥\\ & =2{[\frac{𝑥}{2}\sqrt{\surd 4-{𝑥}^2}+2{\sin }^{-1}\frac{𝑥}{2}]}_2^1\\ & =2[2\cdot \frac{𝜋}{2}-(\frac{\sqrt{\surd 3}}{2}+2\cdot \frac{𝜋}{6})]\\ & =\frac{4𝜋}{3}-\sqrt{\surd 3}\end{array}$

$\begin{array}{cc}& \text{ Since the function is continuous at }𝑥=0\text{. }\\ & \begin{array}{cc}\therefore & \underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=𝑓\left(0\right)\\ & \Rightarrow \underset{𝑥\to 0}{lim}\frac{\cos 𝑎𝑥-\cos 𝑏𝑥}{\cos 𝑐𝑥-\cos 𝑏𝑥}=-1\\ & \Rightarrow \underset{𝑥\to 0}{lim}\left[\frac{-2\sin \left(\frac{𝑎+𝑏}{2}\right)𝑥\sin \left(\frac{𝑎-𝑏}{2}\right)𝑥}{-2\sin \left(\frac{𝑐+𝑏}{2}\right)𝑥\sin \left(\frac{𝑐-𝑏}{2}\right)𝑥}\right]=-1\\ & \Rightarrow \frac{\left(\frac{𝑎+𝑏}{2}\right)\left(\frac{𝑎-𝑏}{2}\right)\underset{𝑥\to 0}{lim}\frac{\sin \left(\frac{𝑎+𝑏}{2}\right)𝑥}{\frac{𝑎+𝑏}{2}}\underset{𝑥\to 0}{lim}\frac{\sin \left(\frac{𝑎-𝑏}{2}\right)𝑥}{\frac{𝑎-𝑏}{2}}}{\left(\frac{𝑐+𝑏}{2}\right)\left(\frac{𝑐-𝑏}{2}\right)\underset{𝑥\to 0}{lim}\frac{\left(\frac{𝑐+𝑏}{2}\right)𝑥}{\frac{𝑐+𝑏}{2}}\underset{𝑥\to 0}{lim}\frac{\left(\frac{𝑐-𝑏}{2}\right)𝑥}{\frac{𝑐-𝑏}{2}}}=-1\end{array}\end{array}$

$\Rightarrow \frac{\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right)}{\left(\frac{c+b}{2}\right)\left(\frac{c-b}{2}\right)}=-1\dots [\because \underset{𝑥\to 0}{lim}\frac{\sin 𝑥}{𝑥}=1]$

$\begin{array}{cc}& \Rightarrow \frac{{𝑎}^2-{𝑏}^2}{{𝑐}^2-{𝑏}^2}=-1\\ & \Rightarrow {𝑎}^2-{𝑏}^2={𝑏}^2-{𝑐}^2\\ & \Rightarrow {𝑎}^2+{𝑐}^2=2{𝑏}^2\\ & \Rightarrow {𝑎}^2,{𝑏}^2,{𝑐}^2\text{ are in A.P. }\end{array}$

$\begin{array}{cc}{\int }_1^0{\tan }^{-1}𝑥d𝑥& ={\left[({\tan }^{-1}𝑥)𝑥\right]}_1^0-{\int }_1^0\frac{1}{1+{𝑥}^2}𝑥d𝑥\\ & ={[𝑥{\tan }^{-1}𝑥-\frac{1}{2}\log \mid 1+{𝑥}^2\mid ]}_1^0\\ & =\frac{𝜋}{4}-\frac{1}{2}\log 2\\ & =\frac{𝜋}{4}-\log \sqrt{\surd 2}\end{array}$

$\begin{array}{cc}& \left[\right(𝑝\vee 𝑞)\wedge (𝑞\to 𝑟)\wedge (\sim 𝑟\left)\right]\to (𝑝\wedge 𝑞)\equiv 𝐹\\ & \Rightarrow (𝑝\vee 𝑞)\wedge (𝑞\to 𝑟)\wedge (\sim 𝑟)\equiv 𝑇\\ & \Rightarrow 𝑝\vee 𝑞\equiv 𝑇,𝑞\to 𝑟\equiv 𝑇,\sim 𝑟\equiv 𝑇\\ & \text{ Now, }\sim 𝑟\equiv 𝑇\Rightarrow 𝑟\equiv 𝐹\\ & 𝑞\to 𝑟\equiv 𝑇\Rightarrow 𝑞\to 𝐹\equiv 𝑇\Rightarrow 𝑞\equiv 𝐹\\ & 𝑝\vee 𝑞\equiv 𝑇\Rightarrow 𝑝\vee 𝐹\equiv 𝑇\Rightarrow 𝑝\equiv 𝑇\\ & \therefore 𝑝,𝑞,𝑟\equiv 𝑇,𝐹,𝐹\end{array}$

The symbolic form of the given logical circuit is

$\begin{array}{cc}& \left[\right(\sim 𝑝\wedge \sim 𝑞)\vee (𝑝)\vee (𝑞\left)\right]\wedge 𝑟\wedge \left[\right(𝑝\wedge 𝑞)\vee (\sim 𝑝\wedge 𝑞\left)\right]\\ & \equiv \left[\right(\sim 𝑝\vee 𝑝)\wedge (\sim 𝑞\vee 𝑝)\vee 𝑞]\wedge 𝑟\wedge \left[\right(𝑝\vee \sim 𝑝)\wedge 𝑞]\dots \left[𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒𝑙𝑎𝑤\right]\end{array}$

$\begin{array}{cc}& \equiv [𝑇\wedge (\sim 𝑞\vee 𝑝)\vee 𝑞]\wedge 𝑟\wedge (𝑇\wedge 𝑞)\text{ ...[Complement law] }\end{array}$

$\equiv [\sim 𝑞\vee 𝑝\vee 𝑞]\wedge 𝑟\wedge 𝑞\dots \left[\text{ Identity law }\right]$

$\begin{array}{cc}& \equiv \left[\right(\sim 𝑞\vee 𝑞)\vee 𝑝]\wedge 𝑟\wedge 𝑞\text{ ...[Commutative law] }\end{array}$

$\equiv (𝑇\vee 𝑝)\wedge 𝑟\wedge 𝑞\dots \left[\text{ Complement law }\right]$

$\equiv 𝑇\wedge 𝑟\wedge 𝑞$…[Identity law]

$\equiv r\wedge q$…[Identity law]

$\begin{array}{cc}& \equiv q\wedge r\text{ ...[Commutative law] }\end{array}$

$\begin{array}{cc}& \text{ By Napier’s analogy, }\\ & \begin{array}{cc}& \tan \left(\frac{𝐴-𝐵}{2}\right)=\left(\frac{𝑎-𝑏}{𝑎+𝑏}\right)\cot \left(\frac{𝐶}{2}\right)\\ & \cot \left(\frac{𝐴+𝐵}{2}\right)\tan \left(\frac{𝐴-𝐵}{2}\right)\\ & =\left(\frac{𝑎-𝑏}{𝑎+𝑏}\right)\cot \left(\frac{𝐶}{2}\right)\cot (\frac{𝜋}{2}-\frac{𝐶}{2})\\ & =\left(\frac{𝑎-𝑏}{𝑎+𝑏}\right)\cot \left(\frac{𝐶}{2}\right)\tan \left(\frac{𝐶}{2}\right)=\frac{𝑎-𝑏}{𝑎+𝑏}\end{array}\end{array}$

$\begin{array}{cc}& (𝑎+𝑏+𝑐)(𝑎+𝑏-𝑐)=3𝑎𝑏\\ & \Rightarrow (𝑎+𝑏{)}^2-{𝑐}^2=3𝑎𝑏\\ & \Rightarrow {𝑎}^2+{𝑏}^2+2𝑎𝑏-{𝑐}^2=3𝑎𝑏\\ & \Rightarrow {𝑎}^2+{𝑏}^2-{𝑐}^2=𝑎𝑏\\ & \Rightarrow \frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{𝑎𝑏}=1\\ & \Rightarrow \frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}=\frac{1}{2}\\ & \Rightarrow \cos 𝐶=\frac{1}{2}\\ & \Rightarrow 𝐶=\frac{𝜋}{3}\end{array}$

$\bar{b}\cdot \bar{c}=8$

$\Rightarrow \left|\bar{b}\right|\left|\bar{c}\right|\cos 𝜃=8$

$\begin{array}{cc}& \Rightarrow 2\left|\overline{𝑐}\right|\cos {45}^{\circ }=8\\ & \Rightarrow \left|\overline{𝑐}\right|=4\sqrt{\surd 2}\\ & \begin{array}{cc}|\overline{𝑏}\times \overline{𝑐}|& =\left|\overline{𝑏}\right|\left|\overline{𝑐}\right|\sin 𝜃\\ & =8\sqrt{\surd 2}\sin {45}^{\circ }\\ & =8\end{array}\end{array}$

Since $\bar{a},\bar{b}$ and $\bar{c}$ are coplanar and $\bar{b}\times \bar{c}$ is a vector perpendicular to the plane containing $\bar{b}$ and $\bar{c}$.

$\therefore$ Angle between $\bar{a}$ and $\bar{b}\times \bar{c}=𝜙={90}^{\circ }$

$\begin{array}{cc}|\bar{a}\times (\bar{b}\times \bar{c}\left)\right|& =|\bar{a}\Vert |\bar{b}\times \bar{c}|\sin 𝜙\\ & =1\times 8\times \sin {90}^{\circ }\\ & =8\end{array}$

We are asked:

$\left|\overrightarrow{𝑎}\right|=2,\left|\overrightarrow{𝑏}\right|=3,\left|\overrightarrow{𝑐}\right|=4,$

with conditions:

$\overrightarrow{𝑎}⟂(\overrightarrow{𝑏}+\overrightarrow{𝑐}),\overrightarrow{𝑏}⟂(\overrightarrow{𝑐}+\overrightarrow{𝑎}),\overrightarrow{𝑐}⟂(\overrightarrow{𝑎}+\overrightarrow{𝑏}).$

We want $|\overrightarrow{𝑎}+\overrightarrow{𝑏}+\overrightarrow{𝑐}|$.

Step 1: Convert perpendicular conditions into dot products

• $\overrightarrow{𝑎}\cdot (\overrightarrow{𝑏}+\overrightarrow{𝑐})=0\Rightarrow \overrightarrow{𝑎}\cdot \overrightarrow{𝑏}+\overrightarrow{𝑎}\cdot \overrightarrow{𝑐}=0$

• $\overrightarrow{𝑏}\cdot (\overrightarrow{𝑐}+\overrightarrow{𝑎})=0\Rightarrow \overrightarrow{𝑏}\cdot \overrightarrow{𝑐}+\overrightarrow{𝑏}\cdot \overrightarrow{𝑎}=0$

• $\overrightarrow{𝑐}\cdot (\overrightarrow{𝑎}+\overrightarrow{𝑏})=0\Rightarrow \overrightarrow{𝑐}\cdot \overrightarrow{𝑎}+\overrightarrow{𝑐}\cdot \overrightarrow{𝑏}=0$

So,

(1) $𝑎\cdot 𝑏+𝑎\cdot 𝑐=0$

(2) $𝑏\cdot 𝑐+𝑏\cdot 𝑎=0$

(3) $𝑐\cdot 𝑎+𝑐\cdot 𝑏=0$

Step 2: Solve the system

From (1): $𝑎\cdot 𝑏=-𝑎\cdot 𝑐$.

From (2): $𝑏\cdot 𝑐=-𝑏\cdot 𝑎$.

From (3): $𝑐\cdot 𝑎=-𝑐\cdot 𝑏$.

Let $𝑥=𝑎\cdot 𝑏,𝑦=𝑏\cdot 𝑐,𝑧=𝑐\cdot 𝑎.$

Then equations become:

1. $𝑥+𝑧=0⟹𝑧=-𝑥.$

2. $𝑥+𝑦=0⟹𝑦=-𝑥.$

3. $𝑧+𝑦=0.$

But if $𝑧=-𝑥,𝑦=-𝑥,$ then (3) says $(-𝑥)+(-𝑥)=-2𝑥=0⟹𝑥=0.$

So $𝑥=0,𝑦=0,𝑧=0.$

Thus all dot products vanish:

$\overrightarrow{𝑎},\overrightarrow{𝑏},\overrightarrow{𝑐}$ are mutually perpendicular.

Step 3: Compute magnitude of sum

$|\overrightarrow{𝑎}+\overrightarrow{𝑏}+\overrightarrow{𝑐}{|}^2=|\overrightarrow{𝑎}{|}^2+|\overrightarrow{𝑏}{|}^2+|\overrightarrow{𝑐}{|}^2\left(\text{since dot products vanish}\right)$

$=2^2+3^2+4^2=4+9+16=29.$

So

$|\overrightarrow{𝑎}+\overrightarrow{𝑏}+\overrightarrow{𝑐}|=\sqrt{\surd 29}.$

✅ Final Answer:

Option C: $\sqrt{\surd 29}$

$\begin{array}{cc}& \text{ Let }𝐼=\int \frac{{𝑥}^4+1}{𝑥{({𝑥}^2+1)}^2}𝑑𝑥\\ & =\int \frac{{𝑥}^4+1+2{𝑥}^2-2{𝑥}^2}{𝑥{({𝑥}^2+1)}^2}𝑑𝑥\\ & =\int \frac{{({𝑥}^2+1)}^2-2{𝑥}^2}{𝑥{({𝑥}^2+1)}^2}𝑑𝑥\\ & =\int \frac{1}{𝑥}𝑑𝑥-\int \frac{2𝑥}{{({𝑥}^2+1)}^2}𝑑𝑥\end{array}$

In second integral, put ${𝑥}^2+1=t\Rightarrow 2𝑥d𝑥=\mathrm{dt}$