MHT CET 2026 19th April Morning Shift (Mathematics) Question Paper with Solutions

For $𝑎\ne 0$, the equation

$𝑎{𝑥}^2-𝑎𝑥+1=0$

has real roots when its discriminant is non-negative.

Here,

$𝐴=𝑎,𝐵=-𝑎,𝐶=1$

So,

$𝐷={𝐵}^2-4𝐴𝐶={𝑎}^2-4𝑎=𝑎(𝑎-4)$

For $𝑎\in \{1,2,\dots ,9\}$, we need

$𝑎(𝑎-4)\ge 0$

This is true for

$𝑎=4,5,6,7,8,9$

Thus, there are $6$ favourable values.

For $𝑎=0$, the equation becomes $1=0$, which has no solution. Since there are $10$ possible values of $𝑎$,

$𝑃=\frac{6}{10}=\frac{3}{5}$

Hence, the correct answer is Option A: $\overline{)\frac{3}{5}}$.

Write the complex numbers in polar form:

$1+𝑖=\sqrt{\surd 2}(\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4})$

and

$1-𝑖=\sqrt{\surd 2}(\cos (-\frac{𝜋}{4})+𝑖\sin (-\frac{𝜋}{4})).$

Hence,

$\frac{(1+𝑖{)}^{𝑛}}{(1-𝑖{)}^{𝑛-2}}=2(\cos \frac{(𝑛-1)𝜋}{2}+𝑖\sin \frac{(𝑛-1)𝜋}{2}).$

This number is real when

$\sin \frac{(𝑛-1)𝜋}{2}=0.$

So $𝑛-1$ must be even. The smallest positive value is

$𝑛=1.$

Indeed, for $𝑛=1$,

$\frac{1+𝑖}{(1-𝑖{)}^{-1}}=(1+𝑖)(1-𝑖)=2,$

which is real.

$\overline{)\text{Option A: }1}$

First arrange the four men around the round table.

The number of circular arrangements of four men is

$(4-1)!=3!$

There are four gaps between the men. To ensure that no two women sit together, the three women must occupy three different gaps.

The three distinct women can be placed in these four gaps in

${}^4{𝑃}_3=\frac{4!}{(4-3)!}=4!$

ways.

Hence, the required number of arrangements is

$3!\times 4!$

So, the correct answer is Option B: $\overline{)3!\times 4!}$.

Let

$𝜃=\frac{𝜋}{5}={36}^{\circ }$

Then $3𝜃={108}^{\circ }$ and $2𝜃={72}^{\circ }$. Since

$\cos {108}^{\circ }=-\cos {72}^{\circ },$

we get

$\cos 3𝜃=-\cos 2𝜃.$

Put $𝑥=\cos 𝜃$. Using

$\cos 3𝜃=4{𝑥}^3-3𝑥\text{and}\cos 2𝜃=2{𝑥}^2-1,$

we have

$4{𝑥}^3-3𝑥=-(2{𝑥}^2-1).$

So,

$4{𝑥}^3+2{𝑥}^2-3𝑥-1=0,$

which factorises as

$(𝑥+1)(4{𝑥}^2-2𝑥-1)=0.$

Thus,

$𝑥=-1\text{or}𝑥=\frac{1\pm \sqrt{\surd 5}}{4}.$

Since $\cos {36}^{\circ }$ is positive,

$\cos \left(\frac{𝜋}{5}\right)=\frac{1+\sqrt{\surd 5}}{4}.$

Hence, the correct answer is Option C.

Given

$\cos 𝑥-\sin 𝑥=-1,$

we write it as

$1+\cos 𝑥-\sin 𝑥=0.$

Using

$1+\cos 𝑥=2{\cos }^2\frac{𝑥}{2}\text{and}\sin 𝑥=2\sin \frac{𝑥}{2}\cos \frac{𝑥}{2},$

we get

$2\cos \frac{𝑥}{2}(\cos \frac{𝑥}{2}-\sin \frac{𝑥}{2})=0.$

Hence,

$\cos \frac{𝑥}{2}=0$

or

$\cos \frac{𝑥}{2}=\sin \frac{𝑥}{2}.$

The first equation gives

$𝑥=(2𝑛+1)𝜋.$

In $[0,6𝜋]$, these values are

$𝑥=𝜋,3𝜋,5𝜋.$

Since $𝑥=𝑘\left(\frac{𝜋}{3}\right)$, the corresponding values of $𝑘$ are

$𝑘=3,9,15.$

The second equation gives $𝑥=\frac{𝜋}{2}+2𝑛𝜋$, which does not produce a natural-number value of $𝑘$.

Thus, the number of possible values of $𝑘$ is

$\overline{)3}.$

Option A.

A line making equal intercepts of length $𝑎$ on the coordinate axes has the equation

$\frac{𝑥}{𝑎}+\frac{𝑦}{𝑎}=1$

or

$𝑥+𝑦-𝑎=0.$

Since the line is tangent to the circle ${𝑥}^2+{𝑦}^2=4$, its distance from the centre $(0,0)$ is equal to the radius $2$.

$\frac{|0+0-𝑎|}{\sqrt{\surd 1^2+1^2}}=2$

$\frac{\left|𝑎\right|}{\sqrt{\surd 2}}=2$

$\left|𝑎\right|=2\sqrt{\surd 2}$

Thus, the length of each intercept is

$\overline{)2\sqrt{\surd 2}}$

So, Option C is correct.

The two given lines intersect at a point lying on $𝐿$.

From

$𝑥-𝑦+1=0,$

we get

$𝑦=𝑥+1.$

Substituting in $2𝑥+𝑦-7=0$,

$2𝑥+𝑥+1-7=0$

$3𝑥=6\Rightarrow 𝑥=2.$

Thus,

$𝑦=2+1=3.$

So, $𝐿$ passes through $(2,3)$.

Since $𝐿$ cuts the positive coordinate axes at $𝐴(𝑎,0)$ and $𝐵(0,𝑏)$, its intercept form is

$\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}=1.$

As $(2,3)$ lies on this line,

$\frac{2}{𝑎}+\frac{3}{𝑏}=1.$

Using AM-GM inequality,

$\frac{2}{𝑎}+\frac{3}{𝑏}\ge 2\sqrt{\surd \frac{6}{𝑎𝑏}}.$

Hence,

$1\ge 2\sqrt{\surd \frac{6}{𝑎𝑏}},$

which gives

$𝑎𝑏\ge 24.$

Equality holds when

$\frac{2}{𝑎}=\frac{3}{𝑏}=\frac{1}{2},$

giving $𝑎=4$ and $𝑏=6$.

The area of $△𝑂𝐴𝐵$ is

$\frac{1}{2}𝑎𝑏.$

Therefore, the minimum area is

$\frac{1}{2}\times 24=\overline{)12\text{ square units}}.$

Hence, the correct answer is Option B.

For a pair of lines

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0,$

the combined equation of their angle bisectors is

$ℎ({𝑥}^2-{𝑦}^2)-(𝑎-𝑏)𝑥𝑦=0.$

Here,

$𝑎=1,ℎ=-𝑝,𝑏=-1.$

Substituting,

$-𝑝({𝑥}^2-{𝑦}^2)-2𝑥𝑦=0.$

Dividing by $-𝑝$,

${𝑥}^2+\frac{2}{𝑝}𝑥𝑦-{𝑦}^2=0.$

Comparing this with

${𝑥}^2-2𝑞𝑥𝑦-{𝑦}^2=0,$

we get

$-2𝑞=\frac{2}{𝑝}.$

Hence,

$𝑝𝑞=-1,$

or

$𝑝𝑞+1=0.$

So, the correct answer is Option D.

The ellipse is

$3{𝑥}^2+4{𝑦}^2=12$

Dividing by $12$,

$\frac{{𝑥}^2}{4}+\frac{{𝑦}^2}{3}=1$

For the ellipse $\frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1$, a tangent with slope $𝑚$ is

$𝑦=𝑚𝑥\pm \sqrt{\surd {𝑎}^2{𝑚}^2+{𝑏}^2}$

Here,

${𝑎}^2=4,{𝑏}^2=3,𝑚=-\frac{1}{2}$

Thus,

$𝑦=-\frac{𝑥}{2}\pm \sqrt{\surd 4\left(\frac{1}{4}\right)+3}$

$𝑦=-\frac{𝑥}{2}\pm 2$

Taking $𝑦=-\frac{𝑥}{2}+2$:

• At the $𝑋$-axis, $𝑦=0$, so $𝑥=4$. Hence, $𝐴=(4,0)$.

• At the $𝑌$-axis, $𝑥=0$, so $𝑦=2$. Hence, $𝐵=(0,2)$.

Therefore,

$\text{Area of }△𝐴𝑂𝐵=\frac{1}{2}\times 𝑂𝐴\times 𝑂𝐵$

$=\frac{1}{2}\times 4\times 2=4\text{ sq. units}$

The other parallel tangent gives the same area.

$\overline{)4\text{ sq. units}}$

Hence, Option A is correct.

For continuity at $𝑥=\frac{𝜋}{4}$,

$𝑓\left(\frac{𝜋}{4}\right)=\underset{𝑥\to \frac{𝜋}{4}}{lim}\frac{2\sqrt{\surd 2}-(\cos 𝑥+\sin 𝑥{)}^3}{1-\sin 2𝑥}.$

Let

$𝑡=\cos 𝑥+\sin 𝑥.$

Since

${𝑡}^2=(\cos 𝑥+\sin 𝑥{)}^2=1+\sin 2𝑥,$

we get

$1-\sin 2𝑥=2-{𝑡}^2=(\sqrt{\surd 2}-𝑡)(\sqrt{\surd 2}+𝑡).$

Also,

$2\sqrt{\surd 2}-{𝑡}^3=(\sqrt{\surd 2}{)}^3-{𝑡}^3=(\sqrt{\surd 2}-𝑡\left)\right(2+\sqrt{\surd 2}𝑡+{𝑡}^2).$

Thus,

$\underset{𝑥\to \frac{𝜋}{4}}{lim}𝑓\left(𝑥\right)=\underset{𝑡\to \sqrt{\surd 2}}{lim}\frac{2+\sqrt{\surd 2}𝑡+{𝑡}^2}{\sqrt{\surd 2}+𝑡}.$

Substituting $𝑡=\sqrt{\surd 2}$,

$𝑓\left(\frac{𝜋}{4}\right)=\frac{2+2+2}{2\sqrt{\surd 2}}=\frac{6}{2\sqrt{\surd 2}}=\frac{3\sqrt{\surd 2}}{2}.$

Hence, the correct answer is

$\overline{)\frac{3\sqrt{\surd 2}}{2}}$

Option A.

The expression inside the bracket can be factorised as

$1-\cos \frac{{𝑥}^2}{2}-\cos \frac{{𝑥}^2}{4}+\cos \frac{{𝑥}^2}{2}\cos \frac{{𝑥}^2}{4}=(1-\cos \frac{{𝑥}^2}{2})(1-\cos \frac{{𝑥}^2}{4}).$

Thus, the limit becomes

$\underset{𝑥\to 0}{lim}\frac{8}{{𝑥}^8}(1-\cos \frac{{𝑥}^2}{2})(1-\cos \frac{{𝑥}^2}{4}).$

Using the standard result

$\underset{𝑡\to 0}{lim}\frac{1-\cos 𝑡}{{𝑡}^2}=\frac{1}{2},$

we write

$1-\cos \frac{{𝑥}^2}{2}\sim \frac{1}{2}{\left(\frac{{𝑥}^2}{2}\right)}^2=\frac{{𝑥}^4}{8},$

and

$1-\cos \frac{{𝑥}^2}{4}\sim \frac{1}{2}{\left(\frac{{𝑥}^2}{4}\right)}^2=\frac{{𝑥}^4}{32}.$

Hence,

$\underset{𝑥\to 0}{lim}\frac{8}{{𝑥}^8}\left(\frac{{𝑥}^4}{8}\right)\left(\frac{{𝑥}^4}{32}\right)=\frac{8}{256}=\frac{1}{32}.$

So, the correct answer is

$\overline{)\frac{1}{32}}$

Option B.

Let $𝑝,𝑞,𝑟$ represent the switches ${𝑆}_1,{𝑆}_2,{𝑆}_3$ respectively.

In the upper branch, switch ${𝑆}_1$ is in series with the parallel combination of ${𝑆}_2$ and ${𝑆}_3$. So its Boolean expression is $𝑝\wedge (𝑞\vee 𝑟)$.

In the lower branch, the switches are ${𝑆}_1$, not ${𝑆}_2$, and not ${𝑆}_3$ in series. So its Boolean expression is $\sim 𝑟\wedge \sim 𝑞\wedge 𝑝$.

We can write this as $𝑝\wedge (\sim 𝑞\wedge \sim 𝑟)$.

The two branches are connected in parallel, so we join their expressions using $\vee$:

$[𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [𝑝\wedge (\sim 𝑞\wedge \sim 𝑟\left)\right]$

Using De Morgan's Law, $\sim 𝑞\wedge \sim 𝑟<text>\equiv </text><text>\sim </text>(𝑞\vee 𝑟)$.

So the expression becomes:

$[𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [𝑝\wedge \sim (𝑞\vee 𝑟\left)\right]$

Now $𝑝$ is common in both parts, so we take it outside:

$𝑝\wedge \left[\right(𝑞\vee 𝑟)\vee \sim (𝑞\vee 𝑟\left)\right]$

Here, $(𝑞\vee 𝑟)$ and $\sim (𝑞\vee 𝑟)$ are opposites. One of them will always be true.

So, $𝐴\vee \sim 𝐴\equiv 𝑇$.

Hence,

$𝑝\wedge 𝑇\equiv 𝑝$

This means the whole circuit works exactly like a circuit with only switch ${𝑆}_1$.

Since the statement contains $\sim 𝑞$ as a conjunction, it can be true only when $𝑞$ is false.

For $𝑞=\text{false}$,

$𝑝\vee 𝑞=𝑝$

and

$(\sim 𝑝\wedge 𝑞)\vee (𝑝\wedge \sim 𝑞)=(\sim 𝑝\wedge \text{false})\vee (𝑝\wedge \text{true})=𝑝.$

Thus, the given statement becomes

$𝑝\wedge 𝑝\wedge \sim 𝑞=𝑝\wedge \sim 𝑞.$

Hence, the correct answer is Option A:

$\overline{)𝑝\wedge \sim 𝑞}$

The tautology is Option A:

$\left[\right(𝑝\to 𝑞)\wedge \sim 𝑞]<text>\to </text><text>\sim </text>𝑝$

This represents the valid rule called modus tollens:

• If $𝑝\to 𝑞$ is true,

• and $𝑞$ is false,

• then $𝑝$ must also be false.

Using $𝑝\to 𝑞<text>\equiv </text><text>\sim </text>𝑝\vee 𝑞$,

$\left[\right(\sim 𝑝\vee 𝑞)\wedge \sim 𝑞]<text>\to </text><text>\sim </text>𝑝$

Whenever the antecedent is true, $\sim 𝑝$ is necessarily true. Hence, the whole implication is always true.

The other options are not tautologies:

• B and C are contradictions, so they are always false.

• In D, the expression inside the first bracket is a tautology, but the whole expression becomes $𝑟$. Since $𝑟$ may be false, it is not a tautology.

$\overline{)\text{Option A}}$

Using the half-angle relation in a triangle,

$\tan \frac{𝐴}{2}=\frac{𝑟}{𝑠-𝑎}\Rightarrow \cot \frac{𝐴}{2}=\frac{𝑠-𝑎}{𝑟},$

where $𝑠$ is the semiperimeter and $𝑟$ is the inradius. Similarly,

$\cot \frac{𝐵}{2}=\frac{𝑠-𝑏}{𝑟},\cot \frac{𝐶}{2}=\frac{𝑠-𝑐}{𝑟}.$

Hence,

$\cot \frac{𝐴}{2}:\cot \frac{𝐵}{2}:\cot \frac{𝐶}{2}=(𝑠-𝑎):(𝑠-𝑏):(𝑠-𝑐).$

Given

$𝑎:𝑏:𝑐=18:17:7,$

take $𝑎=18$, $𝑏=17$, and $𝑐=7$. Then

$𝑠=\frac{18+17+7}{2}=21.$

Thus,

$(𝑠-𝑎):(𝑠-𝑏):(𝑠-𝑐)=(21-18):(21-17):(21-7)=3:4:14.$

So the correct answer is Option C: $3:4:14$.

Using the given condition, $𝑖+𝑗$ is even when $𝑖$ and $𝑗$ are both odd or both even. Hence,

$𝐴=⎡⎣⎢\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}⎤⎦⎥.$

The cofactor matrix of $𝐴$ is

$⎡⎣⎢\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ -1& 0& 1\end{array}⎤⎦⎥.$

Since $\mathrm{adj}\left(𝐴\right)$ is the transpose of the cofactor matrix, and this matrix is symmetric,

$\mathrm{adj}\left(𝐴\right)=⎡⎣⎢\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ -1& 0& 1\end{array}⎤⎦⎥.$

Hence, the correct answer is Option B.

Let

$𝐴=⎡⎣⎢\begin{array}{ccc}1+𝑝𝑞& 𝑝& 0\\ 𝑞& 1+𝑝𝑞& 𝑝\\ 0& 𝑞& 1\end{array}⎤⎦⎥.$

First, find its determinant:

$\begin{array}{cc}\left|𝐴\right|& =(1+𝑝𝑞)\mid \begin{array}{cc}1+𝑝𝑞& 𝑝\\ 𝑞& 1\end{array}\mid -𝑝\mid \begin{array}{cc}𝑞& 𝑝\\ 0& 1\end{array}\mid \\ & =(1+𝑝𝑞)((1+𝑝𝑞)-𝑝𝑞)-𝑝𝑞\\ & =1+𝑝𝑞-𝑝𝑞\\ & =1.\end{array}$

Using

${𝐴}^{-1}=\frac{1}{\left|𝐴\right|}\mathrm{adj}𝐴,$

the adjoint of $𝐴$ is

$\mathrm{adj}𝐴=⎡⎣⎢\begin{array}{ccc}1& -𝑝& {𝑝}^2\\ -𝑞& 1+𝑝𝑞& -(𝑝+{𝑝}^2𝑞)\\ {𝑞}^2& -(𝑞+𝑝{𝑞}^2)& 1+𝑝𝑞+{𝑝}^2{𝑞}^2\end{array}⎤⎦⎥.$

Since $\left|𝐴\right|=1$,

$\overline{){𝐴}^{-1}=⎡⎣⎢\begin{array}{ccc}1& -𝑝& {𝑝}^2\\ -𝑞& 1+𝑝𝑞& -(𝑝+{𝑝}^2𝑞)\\ {𝑞}^2& -(𝑞+𝑝{𝑞}^2)& 1+𝑝𝑞+{𝑝}^2{𝑞}^2\end{array}⎤⎦⎥}$

Hence, the correct answer is $\overline{)\text{Option C}}$.

The principal value range of ${\sin }^{-1}𝑥$ is

$[-\frac{𝜋}{2},\frac{𝜋}{2}].$

Since $\cos \frac{𝑘𝜋}{4}$ repeats after every $8$ values, consider one complete cycle:

$\begin{array}{ccccccccc}𝑘& 1& 2& 3& 4& 5& 6& 7& 8\\ {\sin }^{-1}(\cos \frac{𝑘𝜋}{4})& \frac{𝜋}{4}& 0& -\frac{𝜋}{4}& -\frac{𝜋}{2}& -\frac{𝜋}{4}& 0& \frac{𝜋}{4}& \frac{𝜋}{2}\end{array}$

The sum of these eight terms is $0$.

Now,

$2026=8\times 253+2.$

Thus, the first $2024$ terms form $253$ complete cycles and have sum $0$. The remaining terms correspond to $𝑘\equiv 1,2\left(\mathrm{mod}8\right)$:

${\sin }^{-1}(\cos \frac{𝜋}{4})+{\sin }^{-1}(\cos \frac{2𝜋}{4})=\frac{𝜋}{4}+0.$

Hence,

$\overline{)\frac{𝜋}{4}}$

So, the correct choice is Option C.

Since

$\tan \frac{𝜋}{4}=1$

and the principal value of ${\sin }^{-1}\left(1\right)$ is $\frac{𝜋}{2}$, the given equation becomes

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

Hence,

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

Taking sine on both sides,

$\sqrt{\surd \frac{3}{𝑥}}=\sin \frac{𝜋}{3}=\frac{\sqrt{\surd 3}}{2}.$

Squaring,

$\frac{3}{𝑥}=\frac{3}{4}\Rightarrow 𝑥=4.$

Now check the options. For $𝑥=4$,

${𝑥}^2-𝑥-12=16-4-12=0.$

Thus, $𝑥$ is a root of

$\overline{){𝑥}^2-𝑥-12=0}.$

Hence, the correct answer is $\overline{)\text{Option B}}$.

Given

$𝑓\left(𝑥\right)={𝑥}^2,𝑔\left(𝑥\right)=\left[{𝑥}^2\right]$

First,

$𝑔\left(\frac{3}{2}\right)=\left[{\left(\frac{3}{2}\right)}^2\right]=\left[\frac{9}{4}\right]=2$

So,

$(𝑓\circ 𝑔)\left(\frac{3}{2}\right)=𝑓\left(2\right)=2^2=4$

Also,

$𝑓\left(\frac{3}{2}\right)=\frac{9}{4}$

Hence,

$(𝑔\circ 𝑓)\left(\frac{3}{2}\right)=𝑔\left(\frac{9}{4}\right)=\left[{\left(\frac{9}{4}\right)}^2\right]=\left[\frac{81}{16}\right]=5$

Thus,

$(𝑓\circ 𝑔)\left(\frac{3}{2}\right)+(𝑔\circ 𝑓)\left(\frac{3}{2}\right)=4+5=9$

So the correct answer is Option C: $9$.

Using the identity $\sin 2𝜃=2\sin 𝜃\cos 𝜃$, take $𝜃={\sin }^{-1}𝑥$.

Then $\sin 𝜃=𝑥$ and $\cos 𝜃=\sqrt{\surd 1-{𝑥}^2}$. Hence,

$𝑦=2𝑥\sqrt{\surd 1-{𝑥}^2}.$

Differentiating using the product rule,

$\frac{𝑑𝑦}{𝑑𝑥}=2\sqrt{\surd 1-{𝑥}^2}+2𝑥\left(\frac{-𝑥}{\sqrt{\surd 1-{𝑥}^2}}\right).$

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{2(1-{𝑥}^2)-2{𝑥}^2}{\sqrt{\surd 1-{𝑥}^2}}=\frac{2-4{𝑥}^2}{\sqrt{\surd 1-{𝑥}^2}}.$

Thus, the correct answer is Option A.

Given the parametric equations

$𝑥=𝑎{𝑡}^2-1,𝑦={𝑡}^3+1$

For parametric functions,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{\frac{𝑑𝑦}{𝑑𝑡}}{\frac{𝑑𝑥}{𝑑𝑡}}$

Now,

$\frac{𝑑𝑥}{𝑑𝑡}=2𝑎𝑡,\frac{𝑑𝑦}{𝑑𝑡}=3{𝑡}^2$

Thus,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{3{𝑡}^2}{2𝑎𝑡}=\frac{3𝑡}{2𝑎}$

Again differentiating with respect to $𝑥$,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{\frac{𝑑}{𝑑𝑡}\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}{\frac{𝑑𝑥}{𝑑𝑡}}=\frac{\frac{3}{2𝑎}}{2𝑎𝑡}=\frac{3}{4{𝑎}^2𝑡}$

At $𝑡=1$,

$\frac{3}{4{𝑎}^2}=\frac{3}{16}$

So,

$4{𝑎}^2=16$

${𝑎}^2=4$

Since $𝑎>0$,

$\overline{)𝑎=2}$

Hence, the correct answer is Option D.

Given

$𝑦={\cot }^{-1}\left(\frac{1+\sin 5𝑥}{\cos 5𝑥}\right)$

Let

$𝑢=\frac{1+\sin 5𝑥}{\cos 5𝑥}.$

Using

$\frac{𝑑}{𝑑𝑥}({\cot }^{-1}𝑢)=-\frac{{𝑢}^{\prime }}{1+{𝑢}^2},$

we get

${𝑢}^{\prime }=\frac{5{\cos }^{2}5𝑥+5\sin 5𝑥(1+\sin 5𝑥)}{{\cos }^{2}5𝑥}=\frac{5(1+\sin 5𝑥)}{{\cos }^{2}5𝑥}.$

Also,

$1+{𝑢}^2=1+\frac{(1+\sin 5𝑥{)}^2}{{\cos }^{2}5𝑥}=\frac{2(1+\sin 5𝑥)}{{\cos }^{2}5𝑥}.$

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=-\frac{\frac{5(1+\sin 5𝑥)}{{\cos }^{2}5𝑥}}{\frac{2(1+\sin 5𝑥)}{{\cos }^{2}5𝑥}}=-\frac{5}{2}.$

Thus, the correct answer is Option D, $\overline{)-\frac{5}{2}}$.

Using the change of base formula,

${\log }_{𝑥}10=\frac{1}{{\log }_{10}𝑥}.$

Let

$𝑦={\log }_{10}𝑥.$

Then

${\log }_{𝑥}10=\frac{1}{𝑦}.$

So,

$\frac{𝑑𝑦}{𝑑({\log }_{𝑥}10)}=\frac{𝑑𝑦}{𝑑(1/𝑦)}=\frac{1}{-\frac{1}{{𝑦}^2}}=-{𝑦}^2.$

Substituting $𝑦={\log }_{10}𝑥$,

$\frac{𝑑({\log }_{10}𝑥)}{𝑑({\log }_{𝑥}10)}=-({\log }_{10}𝑥{)}^2.$

Hence, the correct answer is Option B.

Let the length and breadth of the rectangular base be $𝑥$ m and $𝑦$ m.

The depth is $4$ m. Using the given volume,

$4𝑥𝑦=36$

Hence,

$𝑥𝑦=9$

The cost of the base is

$100𝑥𝑦=100\left(9\right)=900$

The total area of the four sides is

$2\left(4𝑥\right)+2\left(4𝑦\right)=8(𝑥+𝑦)$

So, the cost of the sides is

$50\times 8(𝑥+𝑦)=400(𝑥+𝑦)$

Thus, the total cost is

$𝐶=900+400(𝑥+𝑦)$

For a fixed product $𝑥𝑦=9$, the sum $𝑥+𝑦$ is minimum when $𝑥=𝑦$. By AM-GM inequality,

$𝑥+𝑦\ge 2\sqrt{\surd 𝑥𝑦}=2\sqrt{\surd 9}=6$

The minimum occurs at $𝑥=𝑦=3$ m.

Therefore, the minimum cost is

${𝐶}_{min}=900+400\left(6\right)=900+2400=\text{Rs. }3300$

Hence, the correct answer is Option C: Rs. $3300$.

The particle hits the ground when its height becomes zero.

$128𝑡-16{𝑡}^2=0$

$16𝑡(8-𝑡)=0$

So, $𝑡=0$ or $𝑡=8$ seconds. Since $𝑡=0$ is the starting time, it returns to the ground after $8$ seconds.

Velocity is the derivative of height:

$𝑣\left(𝑡\right)={𝑠}^{\prime }\left(𝑡\right)=128-32𝑡$

At $𝑡=8$:

$𝑣\left(8\right)=128-32\left(8\right)=128-256=-128\text{ ft/sec}$

The negative sign shows that the particle is moving downward.

Hence, the correct answer is Option A: $-128$ ft/sec.

For Rolle’s theorem, we need a point $𝑐\in (0,1)$ such that