MHT CET 2024 3rd May Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \text{ Let }I=\int \frac{d𝑥}{{\cos }^3𝑥\sqrt{\surd 2\sin 2𝑥}}\\ & \text{ Put }\sin 2𝑥=\frac{2\tan 𝑥}{1+{\tan }^2𝑥}\\ \therefore & I=\int \frac{{\sec }^2𝑥\cdot {\sec }^2𝑥}{2\sqrt{\surd 𝑥\tan 𝑥}}d𝑥\end{array}$

Let $\tan 𝑥=t$

$\begin{array}{cc}\therefore & {\sec }^2𝑑𝑥=𝑑𝑡\\ \therefore 𝐼& =\frac{1}{2}\int \frac{1+{𝑡}^2}{\sqrt{\surd 𝑡}}𝑑𝑡=\frac{1}{2}\int (\frac{1}{\sqrt{\surd 𝑡}}+{𝑡}^{\frac{3}{2}})𝑑𝑡\\ & =\frac{1}{2}[\frac{{𝑡}^{\frac{1}{2}}}{\frac{1}{2}}+\frac{{𝑡}^{\frac{5}{2}}}{\frac{5}{2}}]+𝑘=(\tan 𝜃{)}^{\frac{1}{2}}+\frac{1}{5}(\tan 𝜃{)}^{\frac{5}{2}}+𝑘\end{array}$

Comparing with $(\tan 𝑥{)}^A+C(\tan 𝑥{)}^B+k$, we get

$\begin{array}{cc}& 𝐴=\frac{1}{2},𝐵=\frac{5}{2},𝐶=\frac{1}{5}\\ \therefore & 5(𝐴+𝐵+𝐶)=5(\frac{1}{2}+\frac{5}{2}+\frac{1}{5})=16\end{array}$

$\begin{array}{cc}& \text{ Let }I=\int \frac{2{𝑥}^2-1}{({𝑥}^2+4)({𝑥}^2-3)}d𝑥\\ & \frac{2{𝑥}^2-1}{({𝑥}^2+4)({𝑥}^2-3)}=\frac{𝐴}{{𝑥}^2+4}+\frac{𝐵}{{𝑥}^2-3}\\ \therefore & 2{𝑥}^2-1=A({𝑥}^2-3)+B({𝑥}^2+4)\\ \therefore & 2{𝑥}^2-1=(A+B){𝑥}^2-(3A-4B)\end{array}$

$\begin{array}{cc}& \Rightarrow A+B=2\text{ and }3A-4B=1\\ & \Rightarrow A=\frac{9}{7}\text{ and }B=\frac{5}{7}\\ \therefore I& =\frac{9}{7}\int \frac{1}{{𝑥}^2+(2{)}^2}d𝑥+\frac{5}{7}\int \frac{1}{{𝑥}^2-(\sqrt{\surd 3}{)}^2}d𝑥\\ & =\frac{9}{7}\times \frac{1}{2}{\tan }^{-1}\left(\frac{𝑥}{2}\right)+\frac{5}{7}\times \frac{1}{2\sqrt{\surd 3}}\log \left(\frac{𝑥-\sqrt{\surd 3}}{𝑥+\sqrt{\surd 3}}\right)+c\\ & =\frac{9}{14}{\tan }^{-1}\left(\frac{𝑥}{2}\right)+\frac{5}{14\sqrt{\surd 3}}\log \left(\frac{𝑥-\sqrt{\surd 3}}{𝑥+\sqrt{\surd 3}}\right)+c\end{array}$

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{1-\tan 𝑥}{4𝑥-𝜋}\\ & \text{ Let }𝑥=\frac{𝜋}{4}+t\\ & \begin{array}{cc}& \therefore \text{ when }𝑥\to \frac{𝜋}{4},𝑡\to 0\\ & \therefore 𝑓\left(𝑥\right)=\frac{1-\tan (\frac{𝜋}{4}+𝑡)}{4(\frac{𝜋}{4}+𝑡)-𝜋}\\ & \therefore f\left(t\right)=\frac{1-\tan (\frac{𝜋}{4}+t)}{4t}\\ & =\frac{1-\left(\frac{1+\tan 𝑡}{1-\tan 𝑡}\right)}{4𝑡}\\ & =\frac{-2\tan 𝑡}{(1-\tan 𝑡)4𝑡}=\frac{-1}{2}\times \frac{\tan 𝑡}{𝑡}\times \frac{1}{1-\tan 𝑡}\\ & \therefore 𝑓\left(\frac{𝜋}{4}\right)=\underset{𝑡\to 0}{lim}𝑓\left(𝑡\right)=\frac{-1}{2}\end{array}\end{array}$

Let the equation of the line be

$\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}=1$

Since line passes through the point $(2,3)$, we get

$\begin{array}{cc}& \frac{2}{𝑎}+\frac{3}{𝑏}=1\\ \therefore & 2𝑏+3𝑎=𝑎𝑏\text{.... (i)}\end{array}$

As per the given condition, Area of the triangle $=12$ sq. units

$\begin{array}{cc}& \therefore \frac{1}{2}\left|𝑎𝑏\right|=12\\ & \therefore 𝑎𝑏=\pm 24\end{array}$

Case I :

$𝑎𝑏=24\text{.... (ii)}$

$\begin{array}{cc}\therefore & 2𝑏+3𝑎=24\text{....[from (i)]}\\ \therefore & 2{\left(\frac{24}{𝑎}\right)}^2+3𝑎=24\text{....[from (ii)]}\\ \therefore & 16+{𝑎}^2=8𝑎\\ \therefore & {𝑎}^2-8𝑎+16=0\\ \therefore & 𝑎=4\\ \therefore & 𝑏=6\end{array}$

$\begin{array}{cc}& \text{ Case II: }\\ & 𝑎𝑏=-24\\ & \therefore 2\left(\frac{-24}{𝑎}\right)+3𝑎=-24\text{.... [from (i) and (ii)]}\\ & \therefore {𝑎}^2+8𝑎-16=0\\ & \therefore 𝑎=\frac{-8\pm \sqrt{\surd 64+64}}{2}=-4\pm 4\sqrt{\surd 2}\end{array}$

$\therefore b$ will also have 2 values.

$\therefore$ The required number of lines is 3.

$\begin{array}{cc}& {\left[\frac{1+{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2}{\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}\right]}^{\frac{3}{2}}=𝑘𝑥\\ \therefore & {\left[\frac{1+{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2}{\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}\right]}^3=(𝑘𝑥{)}^2\\ \therefore & {[1+{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2]}^3=(𝑘𝑥{)}^2{\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}^3\\ \therefore & \text{ Order }=2,\text{ Degree }=3\end{array}$

Let $△\mathrm{ABC}$ be on isosceles triangle such that $\mathrm{AB}=\mathrm{AC}=𝑥$

$\therefore \angle \mathrm{ABC}=\angle \mathrm{ACB}=𝜃$

Draw seg $\mathrm{AD}⟂$ side BC at point D.

$\therefore △\mathrm{ABD}$ is a right angled triangle such that

$\mathrm{AD}=𝑥\sin 𝜃\text{ and }\mathrm{BD}=𝑥\cos 𝜃$

Similarly, in $△\mathrm{ACD}$,

$\mathrm{DC}=𝑥\cos 𝜃$

$\therefore$ In $△\mathrm{ABC}$,

Height $=\mathrm{AD}=𝑥\sin 𝜃$

Base $=\mathrm{BC}=𝑥\cos 𝜃+𝑥\cos 𝜃=2𝑥\cos 𝜃$

$\begin{array}{cc}\therefore A\left(△\mathrm{ABC}\right)& =\frac{1}{2}\times 𝑥\sin 𝜃\times 2𝑥\cos 𝜃\\ & =\frac{{𝑥}^2}{2}(2\sin 𝜃\cos 𝜃)\\ & =\frac{{𝑥}^2}{2}\sin 2𝜃\end{array}$

Since, $-1\le \sin 2𝜃\le 1$, for maximum value of $\sin 2𝜃$, Maximum area $=\frac{{𝑥}^2}{2}$ sq. units.

A contingency is a proposition (compound statement) that is neither a tautology (always true) nor a contradiction (always false). Let us check each option:

$p\wedge \neg p$

Always false (contradiction).

$p\vee \neg p$

Always true (tautology).

$p\vee q$

Can be true or false depending on the truth values of $p$ and $q$.

If both $p$ and $q$ are false, then $p\vee q$ is false.

Otherwise, it is true.

Hence it is not always true and not always false; so it is a contingency.

$(p\wedge (p\to q))\to q$

This is actually a tautology (always true). One way to see this is that

$p\wedge (p\to q)$ essentially simplifies to $p\wedge (\neg p\vee q)$,

which is the same as $(p\wedge \neg p)\vee (p\wedge q)$ = $\text{False}\vee (p\wedge q)$ = $p\wedge q$. Thus the whole statement is $(p\wedge q)\to q$, which is always true.

Therefore, the contingency is:

(C) $p\vee q$.

Let $𝐼=\underset{0}{\overset{\frac{𝜋}{3}}{\int }}\frac{\tan 𝜃}{\sqrt{\surd 2\mathrm{ksec}𝜃}}𝑑𝜃$

$\begin{array}{cc}& =\frac{1}{\sqrt{\surd 2k}}{\int }_{\frac{𝜋}{3}}^0\frac{\sin 𝜃}{\cos 𝜃}\times \sqrt{\surd \cos 𝜃}d𝜃\\ & =\frac{1}{\sqrt{\surd 2k}}{\int }_{\frac{𝜋}{3}}^0\frac{\sin 𝜃}{\sqrt{\surd \cos 𝜃}}d𝜃\end{array}$

Let $\cos 𝜃=t$

$\therefore \sin 𝜃d𝜃=-\mathrm{dt}$

$\therefore$ when $𝜃=\frac{𝜋}{3},t=\frac{1}{2}$

when $𝜃=0,𝑡=1$

$\begin{array}{cc}& \therefore 𝐼=\frac{-1}{\sqrt{\surd 2𝑘}}{\int }_{\frac{1}{2}}^1\frac{1}{\sqrt{\surd 𝑡}}\mathrm{dt}\\ & =\frac{-1}{\sqrt{\surd 2𝑘}}\frac{{\left[{𝑡}^{\frac{1}{2}}\right]}_{\frac{1}{2}}^1}{\frac{1}{2}}\\ & =\frac{-\sqrt{\surd 2}}{\sqrt{\surd 𝑘}}(\frac{1}{\sqrt{\surd 2}}-1)=\frac{\sqrt{\surd 2}}{\sqrt{\surd 𝑘}}(1-\frac{1}{\sqrt{\surd 2}})\\ & \text{ Given that }𝐼=(1-\frac{1}{\sqrt{\surd 2}})\\ & \Rightarrow \frac{\sqrt{\surd 2}}{\sqrt{\surd 𝑘}}=1\Rightarrow 𝑘=2\end{array}$

$\begin{array}{cc}& \text{ Since }\sum _{𝑥=0}^{8}P(X=𝑥)=1\\ & \begin{array}{cc}\therefore k+2k+3k& +4k+4k+3k+2k+k+k=1\\ \Rightarrow & k=\frac{1}{21}\\ \therefore P(3<𝑥\le 6)& =P(X=4)+P(X=5)+P(X=6)\\ & =\frac{4}{21}+\frac{3}{21}+\frac{2}{21}\\ & =\frac{9}{21}=\frac{3}{7}\end{array}\end{array}$

$\begin{array}{cc}𝑦& =\log \left[e^{5𝑥}{\left(\frac{3𝑥-4}{𝑥+5}\right)}^{\frac{4}{3}}\right]\\ \therefore 𝑦& =5𝑥\log e+\frac{4}{3}\log (3𝑥-4)-\frac{4}{3}\log (𝑥+5)\\ \therefore \frac{d𝑦}{d𝑥}& =5+\frac{4}{3(3𝑥-4)}\times 3-\frac{4}{3(𝑥+5)}\times 1\\ & =5+\frac{4}{(3𝑥-4)}-\frac{4}{3(𝑥+5)}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}h\left(𝑥\right)& =\left[f\right(𝑥){]}^2+[g\left(𝑥\right){]}^2\\ \therefore h^{\prime }\left(𝑥\right)& =2\left[f\right(𝑥\left)\right]f^{\prime }\left(𝑥\right)+2\left[g\right(𝑥\left)\right]g^{\prime }\left(𝑥\right)\\ & =2[-f^{\prime \prime }(𝑥\left)\right]f^{\prime }\left(𝑥\right)+2\left[f^{\prime }\right(𝑥\left)\right]\cdot f^{\prime \prime }\left(𝑥\right)\\ & =0\end{array}\\ & \therefore h\left(𝑥\right)\text{ is a constant function. }\\ & \therefore ℎ\left(5\right)=1\Rightarrow ℎ\left(10\right)=1\end{array}$

$\begin{array}{cc}& \text{ Let }\overline{𝐚}=8\widehat{𝐢}-6\widehat{𝐣},\bar{b}=3\hat{i}+4\hat{j}-12\hat{k}\\ & \begin{array}{cc}& \therefore \text{ Area of parallelogram }=\frac{1}{2}|\bar{a}\times \bar{b}|\\ & \bar{a}\times \bar{b}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 8& -6& 0\\ 3& 4& -12\end{array}\mid =72\widehat{𝑖}+96\widehat{𝑗}-50\widehat{𝑘}\\ & \therefore \frac{1}{2}|\bar{a}\times \bar{b}|=\sqrt{\surd 5184+9216+2500}\times \frac{1}{2}\\ & =\sqrt{\surd 16900}\times \frac{1}{2}=130\times \frac{1}{2}=65\end{array}\end{array}$

Line L passes through $(13,32)$.

$\begin{array}{cc}\therefore & \frac{13}{5}+\frac{32}{𝑏}=1\\ & \Rightarrow 𝑏=-20\end{array}$

So, equation of $𝐿$ is $\frac{𝑥}{5}-\frac{𝑦}{20}=1\Rightarrow 4𝑥-𝑦=20$

Slope of $𝐿$ is $m_1=4$.

Slope of $\frac{𝑥}{𝑐}+\frac{𝑦}{3}=1$ is $m_2=-\frac{3}{𝑐}$

$\begin{array}{cc}& \Rightarrow -\frac{3}{𝑐}=4\\ & \Rightarrow 𝑐=-\frac{3}{4}\end{array}$

Equation of line K is $-\frac{4𝑥}{3}+\frac{𝑦}{3}=1$

$\Rightarrow 4𝑥-𝑦=-3$

Distance between $𝐿$ and $𝐾$ is $\mid \frac{20+3}{\sqrt{\surd 16+1}}\mid =\frac{23}{\sqrt{\surd 17}}$

$\begin{array}{cc}& \left|\overrightarrow{a}\right|=\sqrt{\surd 27},\left|\bar{b}\right|=7\text{ and }|\overrightarrow{a}\times \bar{b}|=35\\ & \text{ We know that }\\ & \begin{array}{cc}& |\overrightarrow{a}\times \bar{b}|=\left|\overrightarrow{a}\right|\left|\bar{b}\right|\sin 𝜃\\ \therefore & \sin 𝜃=\frac{|\overrightarrow{a}\times \bar{b}|}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{35}{\sqrt{\surd 27}\times 7}=\frac{5}{\sqrt{\surd 27}}\\ \therefore & \cos 𝜃=\sqrt{\surd 1-\frac{25}{27}}=\sqrt{\surd \frac{2}{27}}\end{array}\end{array}$

$\begin{array}{cc}\text{ Now, }\bar{a}\cdot \bar{b}& =\left|\bar{a}\right|\left|\bar{b}\right|\cos 𝜃\\ & =\sqrt{\surd 27}\times 7\times \sqrt{\surd \frac{2}{27}}=7\sqrt{\surd 2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {\tan }^{-1}\sqrt{\surd 𝑥(𝑥+1)}+{\sin }^{-1}\sqrt{\surd {𝑥}^2+𝑥+1}=\frac{𝜋}{2}\\ & {\tan }^{-1}\sqrt{\surd 𝑥(𝑥+1)}\text{ is defined when }\\ & 𝑥(𝑥+1)\ge 0\text{.... (i)}\\ & {\sin }^{-1}\sqrt{\surd {𝑥}^2+𝑥+1}\text{ is defined when }\\ & 𝑥(𝑥+1)+1\le 1\text{ or }𝑥(𝑥+1)\le 0\text{.... (ii)}\end{array}\\ & \begin{array}{cc}& \text{ From (i) and (ii), }\\ & 𝑥(𝑥+1)=0\Rightarrow 𝑥=0\text{ or }-1.\end{array}\\ & \text{ Hence, number of solutions is }2\text{. }\end{array}$

$\begin{array}{cc}& \sqrt{\surd 3}\sec 𝑥+\mathrm{cosec}𝑥+2(\tan 𝑥-\cot 𝑥)=0\\ \therefore & \frac{\sqrt{\surd 3}}{2}\sec 𝑥+\frac{1}{2}\mathrm{cosec}𝑥=\cot 𝑥-\tan 𝑥\\ \therefore & \frac{\sqrt{\surd 3}}{2}\times \frac{1}{\cos 𝑥}+\frac{1}{2}\times \frac{1}{\sin 𝑥}=\frac{\cos 𝑥}{\sin 𝑥}-\frac{\sin 𝑥}{\cos 𝑥}\\ \therefore & \frac{\sqrt{\surd 3}}{2}\sin 𝑥+\frac{1}{2}\cos 𝑥={\cos }^2𝑥-{\sin }^2𝑥\\ \therefore & \cos (\frac{𝜋}{3}-𝑥)=\cos 2𝑥\\ \therefore & \frac{𝜋}{3}-𝑥=2𝑛𝜋\pm 2𝑥\end{array}$

for $𝑛=0:𝑥=\frac{𝜋}{9}\in (-𝜋,𝜋)$ or $\frac{-𝜋}{3}\in (-𝜋,𝜋)$

for $n=1:𝑥=\frac{-5𝜋}{9}\in (-𝜋,𝜋)$ or $𝑥=\frac{5𝜋}{3}\notin (-𝜋,𝜋)$

for $n=-1:𝑥=\frac{-7𝜋}{3}\notin (-𝜋,𝜋)$ or $𝑥=\frac{7𝜋}{9}\in (-𝜋,𝜋)$

for $n=2:𝑥=\frac{11𝜋}{3}\notin (-𝜋,𝜋)$ or $𝑥=\frac{-11𝜋}{9}\notin (-𝜋,𝜋)$

$\therefore$ Distinct solutions are : $\frac{𝜋}{9},\frac{-𝜋}{3},\frac{-5𝜋}{9},\frac{7𝜋}{9}$

$\therefore$ Required sum $=\frac{𝜋-3𝜋-5𝜋+7𝜋}{9}=0$

Word BARRACK has 7 letters in which ' $𝐴$ ' and ' $𝑅$ ' repeats twice.

Case I :

All four letters are different. (B, A, R, C, K)

$\therefore$ No. of letters $={}^5{C}_4\times 4!=120$

Case II :

' $𝑅$ ' repeats twice and remaining letter three letters are different (B, A, C, K)

$\therefore$ No. of letters $={}^4{C}_2\times \frac{4!}{2!}=72$

Case III :

' A ' repeats twice and remaining letter three letters are different (B, R, C, K)

$\therefore$ No. of letters $={}^4{C}_2\times \frac{4!}{2!}=72$

Case IV :

Both ' $𝐴$ ' and ' $𝑅$ ' repeat twice.

$\therefore$ No. of letters $=\frac{4!}{2\times 2!}=6$

$\therefore$ Total no. of letters form $=120+72+72+6$ $=270$

$\begin{array}{cc}& \text{ Given, the rate of increasing the radius }\\ & \begin{array}{cc}& =\frac{\mathrm{dr}}{\mathrm{dt}}=8\mathrm{cm}/\sec \\ & \text{ Area }=𝐴=𝜋{𝑟}^2\\ & \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=2𝜋r\frac{\mathrm{dr}}{\mathrm{dt}}\\ & \Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=2𝜋\left(12\right)\left(8\right)\dots [\because r=12\mathrm{cm}]\\ & =192𝜋{\mathrm{cm}}^2/\sec \end{array}\end{array}$

The points of intersection of ${𝑥}^2=4𝑦$ and $𝑥=4𝑦-2$ are $(2,1)$ and $(-1,\frac{1}{4})$.

$\begin{array}{cc}& \text{ Required area }\\ & ={\int }_2^{-1}\frac{1}{4}(𝑥+2)d𝑥-{\int }_2^{-1}\frac{1}{4}{𝑥}^{2}d𝑥\\ & =\frac{1}{4}{[\frac{{𝑥}^2}{2}+2𝑥]}_2^{-1}-\frac{1}{4}{\left[\frac{{𝑥}^3}{3}\right]}_2^{-1}\\ & =\frac{9}{8}\text{ sq. units }\end{array}$

Initial amount $=64\mathrm{gms}$

Half life period $=5$ years

$\therefore$ Amount of substance left after 5 years $=\frac{64}{2}=32\mathrm{gms}$

Amount of substance left after 10 years $=\frac{32}{2}=16\mathrm{gms}$

Amount of substance left after 15 years $=\frac{16}{2}=8\mathrm{gms}$

Equation of the plane containing $\frac{𝑥}{4}=\frac{𝑦}{3}=\frac{𝑧}{2}$ and $\frac{𝑥}{2}=\frac{𝑦}{-4}=\frac{𝑧}{3}$ is

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑥& 𝑦& 𝑧\\ 4& 3& 2\\ 2& -4& 3\end{array}\mid =0\\ & \Rightarrow 17𝑥-8𝑦-22𝑧=0\end{array}$

Now required plane is perpendicular to this plane.

Consider option (C)

$\left(17\right)\left(6\right)+(-8)(-67)+(-22)\left(29\right)=0$

$\therefore$ Option (C) is correct.

$\begin{array}{cc}& (81{)}^{{\sin }^2𝑥}+(81{)}^{{\cos }^2𝑥}=30\text{.... (i)}\\ & \text{ Let }𝑦={81}^{{\sin }^2𝑥}\\ \therefore & {81}^{{\cos }^2𝑥}={81}^{(1-{\sin }^2𝑥)}=\frac{81}{{81}^{{\sin }^2𝑥}}=\frac{81}{𝑦}\end{array}$

$\therefore$ Equation (i) becomes

$\begin{array}{cc}& 𝑦+\frac{81}{𝑦}=30\\ \therefore & {𝑦}^2-30𝑦+81=0\\ \therefore & (𝑦-27)(𝑦-3)=0\end{array}$

$\therefore 𝑦=27\text{ or }3$

$\therefore {81}^{{\sin }^2𝑥}=27$ or ${81}^{{\sin }^2𝑥}=3$

$\therefore 3^{4{\sin }^2𝑥}=3^3$ or $3^{4{\sin }^2𝑥}=3^1$

$\therefore 4{\sin }^2𝑥=3$ or $4{\sin }^2𝑥=1$

$\therefore \sin 𝑥=\frac{\sqrt{\surd 3}}{2},\frac{1}{2}$ $\dots [\because 𝑥\in [0,𝜋\left]\right]$

$\therefore 𝑥=\frac{𝜋}{3},\frac{2𝜋}{3},\frac{𝜋}{6},\frac{5𝜋}{6}\dots [\because 𝑥\in [0,𝜋\left]\right]$

$\therefore$ Required number of roots are 4.

$\begin{array}{cc}& \text{ Let }𝐼=\int (2𝑥+4)\sqrt{\surd 𝑥-1}𝑑𝑥\\ & \text{ Let }𝑥-1=𝑡\\ \therefore 𝑥& =1+𝑡\\ \therefore & 𝑑𝑥=𝑑𝑡\\ \therefore 𝐼& =\int \left[2\right(1+𝑡)+4]\sqrt{\surd 𝑡}𝑑𝑡\\ & =\int (6\sqrt{\surd 𝑡}+2{𝑡}^{\frac{3}{2}})𝑑𝑡\\ & =\frac{6{𝑡}^{\frac{3}{2}}}{\frac{3}{2}}+2\frac{{𝑡}^{\frac{5}{2}}}{\frac{5}{2}}+𝑐\\ & =4(𝑥-1{)}^{\frac{3}{2}}+\frac{4}{5}(𝑥-1{)}^{\frac{5}{2}}+𝑐\end{array}$

$\begin{array}{cc}& \text{ Comparing with }a(𝑥-1{)}^{\frac{5}{2}}+b(𝑥-1{)}^{\frac{3}{2}}+c\text{, we get }\\ & \begin{array}{cc}& 𝑎=\frac{4}{5},𝑏=4\\ \therefore & 2𝑎+𝑏=\frac{8}{5}+4=\frac{28}{5}\end{array}\end{array}$

$𝑋\sim 𝐵(6,\frac{1}{2})$

Here, $𝑛=6,𝑝=\frac{1}{2},𝑞=\frac{1}{2}$

Consider, $|𝑥-4|\le 2$

$\begin{array}{cc}\therefore & -2\le 𝑥-4\le 2\\ \therefore & 2\le 𝑥\le 6\\ \therefore & P\left[\right|𝑥-4|\le 2]=P(2\le 𝑥\le 6)\\ & =1-\left[P\right(X=0)+P(X=1\left)\right]\\ & =1-[{}^6{C}_0{p}^0{q}^6+{}^6{C}_1{p}^1{q}^5]\\ & =1-[\frac{1}{2^6}+\frac{6}{2^6}]\\ & =1-\frac{7}{64}=\frac{57}{64}\end{array}$