MHT CET 2019 2nd May Morning Shift (Mathematics) Question Paper with Solutions

Given, equation of hyperbola

${𝑥}^2-{𝑦}^2={𝑎}^2\text{... (i)}$

If $𝑃({𝑥}_1,{𝑦}_1)$ is a point on Eq. (i), then

${𝑥}_2^1-{𝑦}_2^1={𝑎}^2\text{... (ii)}$

$\begin{array}{cc}& \text{ Now, }𝑆𝑃=𝑒{𝑥}_1-𝑎\\ & \text{ and }𝑆{𝑃}^{\prime }=𝑒{𝑥}_1+𝑎\\ & \therefore 𝑆𝑃\cdot 𝑆{𝑃}^{\prime }=(𝑒{𝑥}_1-𝑎)(𝑒{𝑥}_1+𝑎)\\ & ={𝑒}^2\cdot {𝑥}_2^1-{𝑎}^2\end{array}$

$\begin{array}{cc}& =2{𝑥}_2^1-{𝑎}^2(\text{ since },\text{ for }{𝑥}^2-{𝑦}^2={𝑎}^2,𝜃=\sqrt{\surd 2})\\ & =2{𝑥}_2^1-({𝑥}_2^1-{𝑦}_2^1)\text{ [using Eq. (i)] }\\ & ={𝑥}_2^1+{𝑦}_2^1\end{array}$

$\begin{array}{cc}& \text{ We have }\\ & \begin{array}{cc}& 𝑓\left(𝑥\right)={\cos }^{-1}\left[\frac{1-(\log 𝑥{)}^2}{1+(\log 𝑥{)}^2}\right]\\ & 𝑓\left(𝑥\right)=2{\tan }^{-1}(\log 𝑥)[\because {\cos }^{-1}\left(\frac{1-{𝑥}^2}{1+{𝑥}^2}\right)=2{\tan }^{-1}𝑥]\\ & {𝑓}^{\prime }\left(𝑥\right)=\frac{2}{1+(\log 𝑥{)}^2}\times \frac{1}{𝑥}\\ & {𝑓}^{\prime }\left(𝑒\right)=\frac{2}{1+1}\times \frac{1}{𝑒}=\frac{1}{𝑒}\end{array}\end{array}$

Let equation of all circles having centre $(𝑎,𝑏)$ and given radius is 4 , is

$(𝑥-𝑎{)}^2+(𝑦-𝑏{)}^2=4$

Here, number of arbitrary constant in two $(𝑎,𝑏)$ so its order is 2.

We have, $𝐴=\left[\begin{array}{cc}𝑥& 1\\ 1& 0\end{array}\right]$

$\begin{array}{cc}\therefore \left|𝐴\right|& =(𝑥\times 0-1)=-1\\ \mathrm{adj}𝐴& =\left[\begin{array}{cc}0& -1\\ -1& 𝑥\end{array}\right]\\ {𝐴}^{-1}& =\frac{\mathrm{adj}𝐴}{\left|𝐴\right|}=(-1)\left[\begin{array}{cc}0& -1\\ -1& 𝑥\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -𝑥\end{array}\right]\end{array}$

Since, It is given, $𝐴={𝐴}^{-1}$

$\Rightarrow \left[\begin{array}{cc}𝑥& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -𝑥\end{array}\right]$

$\Rightarrow 𝑥=0$ (By equality of matrices)

Option (a), Here, $𝑓\left(0\right)={𝑒}^2$

and $\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0}{lim}(1+2𝑥{)}^{1/𝑥}={𝑒}^2$

$[\because \underset{𝑥\to 0}{lim}(1+𝜆𝑥{)}^{1/𝑥}={𝑒}^{𝜆}]$

$𝑓\left(0\right)=\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)={𝑒}^2$

$\Rightarrow 𝑓\left(𝑥\right)$ is continuous at $𝑥=0$.

Option (b), Here, $𝑓\left(0\right)=-1$

and $\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0}{lim}(\sin 𝑥-\cos 𝑥)=\sin 0-\cos 0$

$\begin{array}{cc}& =-1\\ 𝑓\left(0\right)=\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)& =-1\end{array}$

$\Rightarrow 𝑓\left(𝑥\right)$ is continuous at $𝑥=0$ Option (c),

Here, $𝑓\left(0\right)=-1$

and $\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0}{lim}\frac{{𝑒}^{1/𝑥}-1}{{𝑒}^{1/𝑥}+1}=\underset{𝑥\to 0}{lim}\frac{1-{𝑒}^{-1/𝑥}}{1+{𝑒}^{-1/𝑥}}=1$

$\therefore 𝑓\left(0\right)\ne \underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)\Rightarrow 𝑓\left(𝑥\right)$ is not continuous option (d), Here $𝑓\left(0\right)=1$

$\begin{array}{cc}& \text{ and }\\ & \begin{array}{cc}& \underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0}{lim}\frac{{𝑒}^{5𝑥}-{𝑒}^{2𝑥}}{\sin 3𝑥}\times \frac{3𝑥}{3𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{{𝑒}^{5𝑥}-{𝑒}^{2𝑥}}{3𝑥}\times \underset{𝑥\to 0}{lim}\frac{3𝑥}{\sin 3𝑥}\\ & (1+\frac{5𝑥}{1!}+\frac{(5𝑥{)}^2}{2!}+\dots )-\\ & \frac{(1+\frac{2𝑥}{1!}+\frac{(2𝑥{)}^2}{2!}+\dots )}{3𝑥}\\ & (\because \underset{𝑥\to 0}{lim}\frac{3𝑥}{\sin 3𝑥}=1)\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =\underset{𝑥\to 0}{lim}\frac{(\frac{3𝑥}{1!}+\frac{21{𝑥}^2}{2!}+\dots .)}{3𝑥}=1\\ \therefore 𝑓\left(0\right)& =\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=1\end{array}\\ & \Rightarrow 𝑓\left(𝑥\right)\text{ is continuous at }𝑥=0\text{. }\end{array}$

Here, $𝑛=6$

$𝑝=25\%=\frac{1}{4},𝑞=1-𝑝=3/4$

Let $𝑋$ be the random variable, which denotes the number of cases that police station are solved.

$\therefore$ According to question,

$\begin{array}{cc}𝑃(𝑋& =5)+𝑃(𝑋=6)\\ & ={}^6{𝐶}_5\left(𝑝{)}^5\right(𝑞{)}^{6-5}+{}^6{𝐶}_6\left(𝑝{)}^6\right(𝑞{)}^{6-6}\\ & ={}^6{𝐶}_1{\left(\frac{1}{4}\right)}^5{\left(\frac{3}{4}\right)}^1+{}^6{𝐶}_0{\left(\frac{1}{4}\right)}^6{\left(\frac{3}{4}\right)}^0\\ & =\frac{1}{(4{)}^6}[6\times 3+1]=\frac{19}{4096}\end{array}$

We have, ${𝑆}_{𝑛}=\frac{4^{𝑛}-3^{𝑛}}{3^{𝑛}}$

For $𝑛=1,{𝑆}_1={𝑡}_1=\frac{4-3}{3}=\frac{1}{3}$

For $𝑛=2,{𝑆}_2={𝑡}_1+{𝑡}_2=\frac{4^2-3^2}{3^2}=\frac{7}{9}$

$\therefore {𝑡}_2=\frac{7}{9}-{𝑡}_1=\frac{7}{9}-\frac{1}{3}=\frac{7-3}{9}=\frac{4}{9}$

The intersection points of given curves are $(0,0)$ and $(1,1)$.

$\begin{array}{cc}\therefore \text{ Required area }& ={\int }_1^0((2𝑥-{𝑥}^2)-𝑥]𝑑𝑥\\ & ={\int }_1^0(𝑥-{𝑥}^2)𝑑𝑥={[\frac{{𝑥}^2}{2}-\frac{{𝑥}^3}{3}]}_1^0\end{array}$

$=\frac{1}{6}$ sq. unit

$\begin{array}{cc}& \text{ We have, }𝑥\frac{𝑑𝑦}{𝑑𝑥}=𝑦-𝑥\tan \left(\frac{𝑦}{𝑥}\right)\\ & \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑦}{𝑥}-\tan \left(\frac{𝑦}{𝑥}\right)\\ & \text{ Put }\frac{𝑦}{𝑥}=𝑣\Rightarrow 𝑦=𝑣𝑥\\ & \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=𝑣+𝑥\frac{𝑑𝑣}{𝑑𝑥}\\ & \therefore 𝑣+𝑥\frac{𝑑𝑣}{𝑑𝑥}=𝑣-\tan 𝑣\\ & \Rightarrow 𝑥\frac{𝑑𝑣}{𝑑𝑥}=-\tan 𝑣\\ & \Rightarrow \cot 𝑣𝑑𝑣=-\frac{𝑑𝑥}{𝑥}\\ & \Rightarrow \log (\sin 𝑣)=-\log \left(𝑥\right)+\log 𝑐\\ & \Rightarrow \log (\sin 𝑣\cdot 𝑥)=\log 𝑐\\ & \Rightarrow \sin \left(\frac{𝑦}{𝑥}\right)\cdot 𝑥=𝑐\\ & \Rightarrow 𝑥\sin \left(\frac{𝑦}{𝑥}\right)=𝑐\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& (𝑝\wedge 𝑞)\wedge [\sim 𝑟\vee (𝑝\wedge 𝑞\left)\right]\vee (\sim 𝑝\wedge 𝑞)\\ & \equiv (𝑝\wedge 𝑞)\vee (\sim 𝑝\wedge 𝑞)\\ & \equiv (𝑝\vee \sim 𝑝)\wedge 𝑞\equiv 𝑇\wedge 𝑞=𝑞\end{array}\end{array}$

$\begin{array}{cc}& \text{ Total number of balls }=10\\ & \begin{array}{cc}\text{ Required probability }& =\frac{{}^6{𝐶}_2+{}^4{𝐶}_2}{{}^{10}{𝐶}_2}\\ & =\frac{15+6}{45}=\frac{21}{45}=\frac{7}{15}\end{array}\end{array}$

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

To determine the rate at which the area of the circular ripple increases, start by considering the formula for the area of a circle:

$𝐴=𝜋{𝑟}^2$

where $𝐴$ is the area and $𝑟$ is the radius of the circle.

Given that the radius of the ripple increases at the rate of $\frac{𝑑𝑟}{𝑑𝑡}=5\text{cm/sec}$, we want to find the rate of change of the area, $\frac{𝑑𝐴}{𝑑𝑡}$, after 2 seconds.

First, apply the chain rule to differentiate the area with respect to time $𝑡$:

$\frac{𝑑𝐴}{𝑑𝑡}=\frac{𝑑}{𝑑𝑡}\left(𝜋{𝑟}^2\right)=2𝜋𝑟\frac{𝑑𝑟}{𝑑𝑡}$

After differentiating, substitute the known rate of change of the radius:

$\frac{𝑑𝐴}{𝑑𝑡}=2𝜋𝑟\cdot 5$

Now, determine the radius after 2 seconds. Since the radius increases at a rate of $5\text{cm/sec}$, after 2 seconds, the radius will be:

$𝑟=5\times 2=10\text{cm}$

Substitute $𝑟=10\text{cm}$ into the expression for $\frac{𝑑𝐴}{𝑑𝑡}$:

$\frac{𝑑𝐴}{𝑑𝑡}=2𝜋\cdot 10\cdot 5=100𝜋{\text{cm}}^2/\text{sec}$

Thus, the area increases at a rate of:

Option A: $100𝜋{\text{cm}}^2/\text{sec}$

We have, $𝑓\left(𝑥\right)=3𝑥-2,𝑔\left(𝑥\right)={𝑥}^2$

Then

$\begin{array}{cc}(𝑓\circ 𝑔)\left(𝑥\right)& =𝑓\left(𝑔\right(𝑥\left)\right)=𝑓\left({𝑥}^2\right)\\ & =3{𝑥}^2-2\end{array}$

$𝑝\to 𝑞$ means,

(i) $𝑝$ is sufficient for $𝑞$

(ii) $𝑝$ only if $𝑞$

(iii) $𝑞$ is necessary condition for $𝑝$

(iv) $\sim 𝑞<text>\to </text><text>\sim </text>𝑝$

$\begin{array}{cc}& \text{ Let }𝑓\left(𝑥\right)=𝑎{𝑥}^5+𝑏{𝑥}^3+𝑐𝑥\\ & \Rightarrow 𝑓(-𝑥)=-𝑎{𝑥}^5-𝑏{𝑥}^3-𝑐𝑥=-𝑓\left(𝑥\right)\\ & \Rightarrow 𝑓\left(𝑥\right)\text{ is an odd function. }\\ & \therefore {\int }_3^{-3}(𝑎{𝑥}^5+𝑏{𝑥}^3+𝑐𝑥)𝑑𝑥=0\\ & \therefore {\int }_3^{-3}(𝑎{𝑥}^5+𝑏{𝑥}^3+𝑐𝑥+𝑘)𝑑𝑥\text{ is depends only }𝑘\end{array}$

General equation of a circle with centre $(-𝑔,-𝑓)$ is

$\begin{array}{c}{𝑥}^2+{𝑦}^2+2𝑔𝑥+2𝑓𝑦+𝑐=0\\ \text{If}(-𝑔,-𝑓)=𝐴(-1,2)\end{array}$

Then equation of circle,

${𝑥}^2+{𝑦}^2+2𝑥-4𝑦+𝑐=0$

We have, $\left|𝐴\right|\ne 0$

$\Rightarrow {𝐴}^{-1}$ is exists.

$\begin{array}{cc}& \text{ Since, }(𝐴-2𝐼)(𝐴-4𝐼)=0\\ & \Rightarrow {𝐴}^2-𝐴\left(4𝐼\right)-2𝐼\left(𝐴\right)+8𝐼\cdot 𝐼=0\\ & \Rightarrow {𝐴}^2-4𝐴𝐼-2𝐴𝐼+8𝐼=0\\ & \Rightarrow {𝐴}^2-6𝐴𝐼+8𝐼=0\\ & \Rightarrow {𝐴}^2-6𝐴+8𝐼=0\end{array}$

$\begin{array}{cc}& \text{ On pre multiply both sides by }{𝐴}^{-1}\text{, we get }\\ & \begin{array}{cc}& {𝐴}^{-1}{𝐴}^2-6{𝐴}^{-1}𝐴+8{𝐴}^{-1}𝐼=0\\ \Rightarrow & 𝐼𝐴-6𝐼+8{𝐴}^{-1}=0\\ \Rightarrow & 𝐴-6𝐼+8{𝐴}^{-1}=0\\ \Rightarrow & 𝐴+8{𝐴}^{-1}=6𝐼\end{array}\end{array}$

$\begin{array}{cc}& \text{ Here, }(\frac{7+𝑝+𝑞+1}{3},\frac{-8+𝑞+5𝑝}{3},\frac{1+5+0}{3})\\ & =(3,-5,𝑟)\\ & \Rightarrow (\frac{8+𝑝+𝑞}{3},\frac{-8+5𝑝+𝑞}{3},2)=(3,-5,𝑟)\\ & \Rightarrow \frac{8+𝑝+𝑞}{3}=3;\frac{-8+5𝑝+𝑞}{3}=-5;2=𝑟\\ & \Rightarrow 8+𝑝+𝑞=9;-8+5𝑝+𝑞=-15;𝑟=2\\ & \Rightarrow 𝑝+𝑞=1,5𝑝+𝑞=-7,𝑟=2\\ & \Rightarrow 𝑝=-2,𝑞=3,𝑟=2\end{array}$

$\begin{array}{cc}& \int \frac{1}{{𝑥}^2+1}𝑑𝑥=\frac{1}{{𝑥}^2+1}\\ & \int 1𝑑𝑥-\int (\frac{𝑑}{𝑑𝑥}\left(\frac{1}{{𝑥}^2+1}\right)\int 1𝑑𝑥)𝑑𝑥\text{(Integration by parts)}\\ & =\frac{𝑥}{{𝑥}^2+1}-\int \frac{-2𝑥}{{({𝑥}^2+1)}^2}𝑥𝑑𝑥\\ & =\frac{𝑥}{{𝑥}^2+1}+2\int \frac{{𝑥}^2+1-1}{{({𝑥}^2+1)}^2}𝑑𝑥\\ & =\frac{𝑥}{{𝑥}^2+1}+2\int \frac{𝑥}{({𝑥}^2+1)}𝑑𝑥-2\int \frac{𝑑𝑥}{{({𝑥}^2+1)}^2}\text{.... (i)}\end{array}$

And also, we know that

$\int \frac{𝑑𝑥}{{𝑥}^2+1}={\tan }^{-1}\left(𝑥\right)\text{.... (ii)}$

From Eqs. (i) and (ii), we get

$\begin{array}{cc}& {\tan }^{-1}𝑥=\frac{𝑥}{{𝑥}^2+1}+2{\tan }^{-1}𝑥-2\int \frac{𝑑𝑥}{{({𝑥}^2+1)}^2}\\ \Rightarrow & 2\int \frac{𝑑𝑥}{{({𝑥}^2+1)}^2}={\tan }^{-1}𝑥+\frac{𝑥}{{𝑥}^2+1}\\ \Rightarrow & \int \frac{𝑑𝑥}{{({𝑥}^2+1)}^2}=\frac{1}{2}{\tan }^{-1}𝑥+\frac{𝑥}{2({𝑥}^2+1)}+𝑐\end{array}$

$\begin{array}{cc}& \text{ We have, }𝜃=\frac{17𝜋}{3}\\ & \begin{array}{cc}\therefore \tan 𝜃-\cot 𝜃& =\tan \left(\frac{17𝜋}{3}\right)-\cot \left(\frac{17𝜋}{3}\right)\\ & =\tan (5𝜋+\frac{2𝜋}{3})-\cot (5𝜋+\frac{2𝜋}{3})\\ & =\tan \frac{2𝜋}{3}-\cot \frac{2𝜋}{3}\\ & =\tan (𝜋-\frac{𝜋}{3})-\cot (𝜋-\frac{𝜋}{3})\\ & =-\tan 𝜋/3+\cot 𝜋/3\\ & =-\sqrt{\surd 3}+\frac{1}{\sqrt{\surd 3}}=\frac{-3+1}{\sqrt{\surd 3}}=\frac{-2}{\sqrt{\surd 3}}\end{array}\end{array}$

Let $𝑦={\log }_{{𝑒}^2}(\log 𝑥)$

$\Rightarrow 𝑦=\frac{1}{2}{\log }_6(\log 𝑥)(\because {\log }_{{𝑎}^{𝑛}}(𝑥)=\frac{1}{𝑛}{\log }_{𝑎}𝑥)$

On differentiating both sides w.r.t. $𝑥$, we get

$\begin{array}{cc}\frac{𝑑𝑦}{𝑑𝑥}& =\frac{1}{2}\frac{1}{\log 𝑥}\frac{𝑑}{𝑑𝑥}(\log 𝑥)\\ & =\frac{1}{\log {𝑥}^2}\cdot \frac{1}{𝑥}=\frac{1}{𝑥\log {𝑥}^2}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \cos 𝐴=\frac{\sin 𝐵}{\sin 𝐶}\\ \Rightarrow & \frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}=\frac{𝑏}{𝑐}(\because \frac{\sin 𝐴}{𝑎}=\frac{\sin 𝐵}{𝑏}=\frac{\sin 𝐶}{𝑐}=𝑘)\\ \Rightarrow & {𝑏}^2+{𝑐}^2-{𝑎}^2=2{𝑏}^2\\ \Rightarrow & {𝑐}^2-{𝑎}^2={𝑏}^2\\ \Rightarrow & {𝑐}^2={𝑎}^2+{𝑏}^2\\ \Rightarrow & △𝐴𝐵𝐶\text{ right angled triangle at }\angle 𝐶.\end{array}\end{array}$

In any GP if $(𝑚+𝑛{)}^{\text{th }}$ term i.e. ${𝑎}_{𝑚+𝑛}=𝑝$ and $(𝑚-𝑛{)}^{\text{th }}$ term i.e. ${𝑎}_{𝑚-𝑛}=𝑞$, then ${𝑎}_{𝑚}=\sqrt{\surd 𝑝𝑞}$

We know that, $\Sigma 𝑝𝑖=1$

$\begin{array}{cc}\Rightarrow & 𝑘+3𝑘+5𝑘+7𝑘+8𝑘+𝑘=1\\ \Rightarrow & 25𝑘=1\\ \Rightarrow & 𝑘=\frac{1}{25}\end{array}$

Now, $𝑃(2\le 𝑋<5)$

$\begin{array}{cc}& =𝑃(𝑋=2)+𝑃(𝑋=3)+𝑃(𝑋=4)\\ & =3𝑘+5𝑘+7𝑘=15𝑘\\ & =15\times \frac{1}{25}=\frac{3}{5}\end{array}$

We have, $𝑦={\log }_{𝑒}𝑥\Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{𝑥}$

$\therefore$ Slope of normal at $𝑃(1,0)$

$\begin{array}{cc}& =-\frac{1}{(𝑑𝑦/𝑑𝑥{)}_{(1,0)}}\\ & =\frac{-1}{1}=-1\end{array}$

$\therefore$ Equation of normal at $𝑃(1,0)$

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

$\begin{array}{cc}& \text{ We have, }\sin 𝑥\cos 𝑥=\frac{1}{4}\\ & \begin{array}{c}\Rightarrow 2\sin 𝑥\cos 𝑥=\frac{1}{2}\\ \Rightarrow \sin 2𝑥=\frac{1}{2}\\ \Rightarrow 2𝑥={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{𝜋}{6}\\ \Rightarrow 𝑥=\frac{𝜋}{12}\\ \text{ If }𝑥\in (0,\frac{𝜋}{2}),\text{ then }𝑥=\frac{𝜋}{12},\frac{𝜋}{2}-\frac{𝜋}{12}=\frac{𝜋}{12},\frac{5𝜋}{12}\end{array}\end{array}$

We have $𝐚+𝐛,𝐛+𝐜$ and $𝐜+𝐚$ are coterminous edge of a parallelopiped then its volume is $[𝐚+𝐛𝐛+𝐜𝐜+𝐚]$

$=2\left[𝐚𝑏𝐜\right]$

We have,

$\begin{array}{cc}& \text{ c.d.f as }𝐹\left(𝑥\right)=\frac{𝑥-25}{10}\\ & \begin{array}{cc}\therefore 𝑃(27\le 𝑥\le 33)& =𝐹\left(33\right)-𝐹\left(27\right)\\ & =\left(\frac{33-25}{10}\right)-\left(\frac{27-25}{10}\right)\\ & =\frac{8}{10}-\frac{2}{10}=\frac{6}{10}=\frac{3}{5}\end{array}\end{array}$