MHT CET 2026 20th April Morning Shift (Mathematics) Question Paper with Solutions

Since $𝑤$ is a cube root of unity, we have

${𝑤}^3=1$

So powers of $𝑤$ repeat after every $3$ powers.

Now reduce each power using ${𝑤}^3=1$:

${𝑤}^{10}={𝑤}^9\cdot 𝑤=({𝑤}^3{)}^3\cdot 𝑤=𝑤$

${𝑤}^7={𝑤}^6\cdot 𝑤=({𝑤}^3{)}^2\cdot 𝑤=𝑤$

${𝑤}^5={𝑤}^3\cdot {𝑤}^2={𝑤}^2$

So the given expression becomes

${𝑤}^{10}-{𝑤}^7+{𝑤}^5-{𝑤}^2+1$

$=𝑤-𝑤+{𝑤}^2-{𝑤}^2+1$

$=1$

So the value is

$\overline{)1}$

Correct option: C.

Since ${𝑎}_1,{𝑎}_2,{𝑎}_3,\dots ,{𝑎}_{𝑛}$ are in arithmetic progression with common difference $𝑑$,

${𝑎}_{𝑟+1}-{𝑎}_{𝑟}=𝑑$

Now look at one term:

${\tan }^{-1}\left(\frac{𝑑}{1+{𝑎}_{𝑟}{𝑎}_{𝑟+1}}\right)$

Using $𝑑={𝑎}_{𝑟+1}-{𝑎}_{𝑟}$, we get

${\tan }^{-1}\left(\frac{{𝑎}_{𝑟+1}-{𝑎}_{𝑟}}{1+{𝑎}_{𝑟}{𝑎}_{𝑟+1}}\right)$

We know the formula

$\tan (𝐴-𝐵)=\frac{\tan 𝐴-\tan 𝐵}{1+\tan 𝐴\tan 𝐵}$

So,

$\frac{{𝑎}_{𝑟+1}-{𝑎}_{𝑟}}{1+{𝑎}_{𝑟}{𝑎}_{𝑟+1}}=\tan ({\tan }^{-1}{𝑎}_{𝑟+1}-{\tan }^{-1}{𝑎}_{𝑟})$

Hence each term behaves like

${\tan }^{-1}{𝑎}_{𝑟+1}-{\tan }^{-1}{𝑎}_{𝑟}$

So the whole sum becomes telescopic:

$({\tan }^{-1}{𝑎}_2-{\tan }^{-1}{𝑎}_1)+({\tan }^{-1}{𝑎}_3-{\tan }^{-1}{𝑎}_2)+\cdots +({\tan }^{-1}{𝑎}_{𝑛}-{\tan }^{-1}{𝑎}_{𝑛-1})$

All middle terms cancel, leaving

${\tan }^{-1}{𝑎}_{𝑛}-{\tan }^{-1}{𝑎}_1$

Now taking tangent,

$\tan ({\tan }^{-1}{𝑎}_{𝑛}-{\tan }^{-1}{𝑎}_1)=\frac{{𝑎}_{𝑛}-{𝑎}_1}{1+{𝑎}_1{𝑎}_{𝑛}}$

So the required value is

$\overline{)\frac{{𝑎}_{𝑛}-{𝑎}_1}{1+{𝑎}_1{𝑎}_{𝑛}}}$

Correct option: Option C.

In a right-angled triangle, one angle is ${90}^{\circ }$.

The three angles are in A.P. So we can write them as

$𝑎-𝑑,𝑎,𝑎+𝑑$

Their sum is ${180}^{\circ }$:

$(𝑎-𝑑)+𝑎+(𝑎+𝑑)={180}^{\circ }$

$3𝑎={180}^{\circ }$

$𝑎={60}^{\circ }$

So the middle angle is ${60}^{\circ }$.

Since the triangle is right-angled, one of the angles is ${90}^{\circ }$. Hence the three angles must be

${30}^{\circ },{60}^{\circ },{90}^{\circ }$

So $△𝐴𝐵𝐶$ is a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle.

In such a triangle, the sides are in the ratio

$1:\sqrt{\surd 3}:2$

The smallest side is opposite ${30}^{\circ }$. Given smallest side $=4$ units.

So the sides are

$4,4\sqrt{\surd 3},8$

Here $8$ is the hypotenuse, and the perpendicular sides are $4$ and $4\sqrt{\surd 3}$.

Area of the right triangle is

$\frac{1}{2}\times \text{base}\times \text{height}$

$=\frac{1}{2}\times 4\times 4\sqrt{\surd 3}$

$=8\sqrt{\surd 3}$

So the area of $△𝐴𝐵𝐶$ is

$\overline{)8\sqrt{\surd 3}\text{ sq. units}}$

Correct option: B.

Given,

${\sec }^{-1}\left(\frac{{𝑥}^2+{𝑦}^2}{{𝑥}^2-{𝑦}^2}\right)=2𝑎$

Taking $\sec$ on both sides,

$\frac{{𝑥}^2+{𝑦}^2}{{𝑥}^2-{𝑦}^2}=\sec 2𝑎$

Now divide numerator and denominator by ${𝑥}^2$:

$\frac{1+\frac{{𝑦}^2}{{𝑥}^2}}{1-\frac{{𝑦}^2}{{𝑥}^2}}=\sec 2𝑎$

Let

$\frac{{𝑦}^2}{{𝑥}^2}=𝑡$

Then,

$\frac{1+𝑡}{1-𝑡}=\sec 2𝑎$

Cross-multiplying,

$1+𝑡=\sec 2𝑎(1-𝑡)$

$1+𝑡=\sec 2𝑎-𝑡\sec 2𝑎$

$𝑡(1+\sec 2𝑎)=\sec 2𝑎-1$

So,

$𝑡=\frac{\sec 2𝑎-1}{\sec 2𝑎+1}$

Now use $\sec 2𝑎=\frac{1}{\cos 2𝑎}$:

$𝑡=\frac{\frac{1}{\cos 2𝑎}-1}{\frac{1}{\cos 2𝑎}+1}$

$𝑡=\frac{1-\cos 2𝑎}{1+\cos 2𝑎}$

Using the identity,

$\frac{1-\cos 2𝑎}{1+\cos 2𝑎}={\tan }^2𝑎$

So,

$\frac{{𝑦}^2}{{𝑥}^2}={\tan }^2𝑎$

Hence,

${𝑦}^2={𝑥}^2{\tan }^2𝑎$

Since $𝑎$ is constant, differentiating both sides with respect to $𝑥$:

$2𝑦\frac{𝑑𝑦}{𝑑𝑥}=2𝑥{\tan }^2𝑎$

$𝑦\frac{𝑑𝑦}{𝑑𝑥}=𝑥{\tan }^2𝑎$

Comparing with

$𝑦\frac{𝑑𝑦}{𝑑𝑥}=𝑥\cdot 𝑓\left(𝑎\right)$

we get

$𝑓\left(𝑎\right)={\tan }^2𝑎$

Now,

$𝑓\left(\frac{2𝜋}{3}\right)={\tan }^2\left(\frac{2𝜋}{3}\right)$

$\tan \left(\frac{2𝜋}{3}\right)=-\sqrt{\surd 3}$

So,

${\tan }^2\left(\frac{2𝜋}{3}\right)=3$

Hence,

$𝑓\left(\frac{2𝜋}{3}\right)=3$

Correct option: C.

We need to find

$\cos {15}^{\circ }-\sin {15}^{\circ }$

Use the identity

$\cos 𝑥-\sin 𝑥=\sqrt{\surd 2}\cos (𝑥+{45}^{\circ })$

Putting $𝑥={15}^{\circ }$,

$\cos {15}^{\circ }-\sin {15}^{\circ }=\sqrt{\surd 2}\cos ({15}^{\circ }+{45}^{\circ })$

$=\sqrt{\surd 2}\cos {60}^{\circ }$

Since

$\cos {60}^{\circ }=\frac{1}{2}$

we get

$\sqrt{\surd 2}\times \frac{1}{2}=\frac{\sqrt{\surd 2}}{2}=\frac{1}{\sqrt{\surd 2}}$

So the value is

$\overline{)\frac{1}{\sqrt{\surd 2}}}$

Correct option: C

Given,

$\cos 𝑥+\sin 𝑥=\cos 2𝑥+\sin 2𝑥$

Use the identity

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

So,

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

and

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

Thus,

$\sqrt{\surd 2}\sin (𝑥+\frac{𝜋}{4})=\sqrt{\surd 2}\sin (2𝑥+\frac{𝜋}{4})$

Cancel $\sqrt{\surd 2}$:

$\sin (𝑥+\frac{𝜋}{4})=\sin (2𝑥+\frac{𝜋}{4})$

For $\sin 𝐴=\sin 𝐵$,

$𝐴=𝐵+2𝑛𝜋$

or

$𝐴=𝜋-𝐵+2𝑛𝜋$

So first,

$𝑥+\frac{𝜋}{4}=2𝑥+\frac{𝜋}{4}+2𝑛𝜋$

$𝑥=-2𝑛𝜋$

Since $𝑛\in 𝑍$, this can be written as

$𝑥=2𝑛𝜋$

So,

$𝑝=2$

Now second,

$𝑥+\frac{𝜋}{4}=𝜋-(2𝑥+\frac{𝜋}{4})+2𝑛𝜋$

$𝑥+\frac{𝜋}{4}=\frac{3𝜋}{4}-2𝑥+2𝑛𝜋$

$3𝑥=\frac{𝜋}{2}+2𝑛𝜋$

$𝑥=\frac{𝜋}{6}+\frac{2𝑛𝜋}{3}$

This matches the form

$𝑥=\frac{𝑛𝑞𝜋}{3}+\frac{𝜋}{6}$

So,

$𝑞=2$

Hence,

$𝑝:𝑞=2:2=1:1$

So the correct option is

$\overline{)1:1}$

Option A.

Put $\sin 𝜃=𝑥$.

First equation:

$2{\sin }^2𝜃-2{\cos }^2𝜃+1=0$

Using ${\cos }^2𝜃=1-{\sin }^2𝜃$,

$2{\sin }^2𝜃-2(1-{\sin }^2𝜃)+1=0$

$2{\sin }^2𝜃-2+2{\sin }^2𝜃+1=0$

$4{\sin }^2𝜃-1=0$

${\sin }^2𝜃=\frac{1}{4}$

So,

$\sin 𝜃=\pm \frac{1}{2}$

Now solve the second equation:

$2{\sin }^2𝜃+3\sin 𝜃-2=0$

Put $\sin 𝜃=𝑥$:

$2{𝑥}^2+3𝑥-2=0$

$2{𝑥}^2+4𝑥-𝑥-2=0$

$(2𝑥-1)(𝑥+2)=0$

So,

$𝑥=\frac{1}{2}\text{or}𝑥=-2$

But $\sin 𝜃$ cannot be $-2$, because sine lies between $-1$ and $1$.

So the second equation gives:

$\sin 𝜃=\frac{1}{2}$

For common solutions, we need:

$\sin 𝜃=\frac{1}{2}$

In the interval $[0,2𝜋]$, this happens at:

$𝜃=\frac{𝜋}{6},\frac{5𝜋}{6}$

So the number of common solutions is:

$2$

Hence, the correct option is Option B.

The line is given in normal form as

$𝑥\cos 𝛼+𝑦\sin 𝛼=𝑝$

where $𝑝$ is the perpendicular distance of the line from the origin, and $𝛼$ is the angle made by this perpendicular with the positive $𝑋$-axis.

Here,

$𝛼={45}^{\circ }$

So,

$𝑥\cos {45}^{\circ }+𝑦\sin {45}^{\circ }=𝑝$

Since

$\cos {45}^{\circ }=\sin {45}^{\circ }=\frac{1}{\sqrt{\surd 2}}$

we get

$\frac{𝑥}{\sqrt{\surd 2}}+\frac{𝑦}{\sqrt{\surd 2}}=𝑝$

$𝑥+𝑦=𝑝\sqrt{\surd 2}$

Now this line cuts the coordinate axes at equal intercepts, because its equation is of the form

$𝑥+𝑦=𝑝\sqrt{\surd 2}$

So the intercepts on both axes are $𝑝\sqrt{\surd 2}$.

Area of the triangle formed with the coordinate axes is

$\frac{1}{2}\times 𝑝\sqrt{\surd 2}\times 𝑝\sqrt{\surd 2}$

$=\frac{1}{2}\times 2{𝑝}^2$

$={𝑝}^2$

Given area is $49$ sq. units, so

${𝑝}^2=49$

$𝑝=7$

Hence,

$𝑥+𝑦=7\sqrt{\surd 2}$

So the correct option is

$\overline{)𝑥+𝑦=7\sqrt{\surd 2}}$

Option B.

The given equation is

$16{𝑥}^2-24𝑥𝑦+9{𝑦}^2-8𝑥+6𝑦-35=0$

Notice that

$16{𝑥}^2-24𝑥𝑦+9{𝑦}^2=(4𝑥-3𝑦{)}^2$

and

$-8𝑥+6𝑦=-2(4𝑥-3𝑦)$

So the equation becomes

$(4𝑥-3𝑦{)}^2-2(4𝑥-3𝑦)-35=0$

Put

$4𝑥-3𝑦=𝑢$

Then

${𝑢}^2-2𝑢-35=0$

Factorising,

$(𝑢-7)(𝑢+5)=0$

So,

$𝑢=7\text{or}𝑢=-5$

Hence the two straight lines are

$4𝑥-3𝑦-7=0$

and

$4𝑥-3𝑦+5=0$

These are parallel lines. The locus of points equidistant from two parallel lines is the line exactly midway between them.

For the lines

$4𝑥-3𝑦=7$

and

$4𝑥-3𝑦=-5$

the middle value is

$\frac{7+(-5)}{2}=1$

So the required locus is

$4𝑥-3𝑦=1$

or

$4𝑥-3𝑦-1=0$

Hence the correct option is

$\overline{)4𝑥-3𝑦-1=0}$

Option A.

The centre of the circle must be equidistant from the points $(2,3)$ and $(4,5)$.

So, the centre lies on the perpendicular bisector of the line segment joining these two points.

Let the points be

$𝐴(2,3),𝐵(4,5)$

Midpoint of $𝐴𝐵$ is

$(\frac{2+4}{2},\frac{3+5}{2})=(3,4)$

Slope of $𝐴𝐵$ is

$\frac{5-3}{4-2}=1$

So, the slope of the perpendicular bisector is $-1$.

Equation of the perpendicular bisector passing through $(3,4)$ is

$𝑦-4=-1(𝑥-3)$

$𝑦-4=-𝑥+3$

$𝑥+𝑦=7$

The centre also lies on the given line

$4𝑥-𝑦-3=0$

So,

$𝑦=4𝑥-3$

Using $𝑥+𝑦=7$,

$𝑥+4𝑥-3=7$

$5𝑥=10$

$𝑥=2$

Then

$𝑦=4\left(2\right)-3=5$

So, the centre is

$(2,5)$

Radius is the distance from $(2,5)$ to $(2,3)$:

$𝑟=\sqrt{\surd (2-2{)}^2+(5-3{)}^2}$

$𝑟=2$

Thus, the circle is

$(𝑥-2{)}^2+(𝑦-5{)}^2=4$

So, the correct option is

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

For the ellipse

$\frac{{𝑥}^2}{25}+\frac{{𝑦}^2}{9}=1$

we have

${𝑎}^2=25,{𝑏}^2=9$

so

$𝑎=5,𝑏=3$

If $𝜃$ is the eccentric angle, then the point on the ellipse is written as

$𝑥=𝑎\cos 𝜃,𝑦=𝑏\sin 𝜃$

So here,

$𝑥=5\cos 𝜃,𝑦=3\sin 𝜃$

The distance of this point from the center $(0,0)$ is given as $5$.

So,

$\sqrt{\surd {𝑥}^2+{𝑦}^2}=5$

Squaring both sides,

${𝑥}^2+{𝑦}^2=25$

Substitute $𝑥=5\cos 𝜃$ and $𝑦=3\sin 𝜃$:

$(5\cos 𝜃{)}^2+(3\sin 𝜃{)}^2=25$

$25{\cos }^2𝜃+9{\sin }^2𝜃=25$

Now use ${\cos }^2𝜃=1-{\sin }^2𝜃$:

$25(1-{\sin }^2𝜃)+9{\sin }^2𝜃=25$

$25-25{\sin }^2𝜃+9{\sin }^2𝜃=25$

$25-16{\sin }^2𝜃=25$

${\sin }^2𝜃=0$

So,

$\sin 𝜃=0$

Hence,

$𝜃=0$

among the given options.

So the correct option is

$\overline{)0}$

Option A.

We use the standard limits:

$\underset{𝑡\to 0}{lim}\frac{\log (1+𝑡)}{𝑡}=1$

and

$\underset{𝑡\to 0}{lim}\frac{{𝑒}^{𝑡}-1}{𝑡}=1$

Given,

$\underset{𝑥\to 0}{lim}\frac{𝑥\log (1+4𝑥)}{({𝑒}^{4𝑥}-1{)}^2}$

Rewrite it in a form where these standard limits can be used:

$\frac{𝑥\log (1+4𝑥)}{({𝑒}^{4𝑥}-1{)}^2}=\frac{𝑥\left(4𝑥\right)\left(\frac{\log (1+4𝑥)}{4𝑥}\right)}{(4𝑥{)}^2{\left(\frac{{𝑒}^{4𝑥}-1}{4𝑥}\right)}^2}$

Now simplify:

$=\frac{4{𝑥}^2}{16{𝑥}^2}\cdot \frac{\frac{\log (1+4𝑥)}{4𝑥}}{{\left(\frac{{𝑒}^{4𝑥}-1}{4𝑥}\right)}^2}$

$=\frac{1}{4}\cdot \frac{\frac{\log (1+4𝑥)}{4𝑥}}{{\left(\frac{{𝑒}^{4𝑥}-1}{4𝑥}\right)}^2}$

As $𝑥\to 0$,

$\frac{\log (1+4𝑥)}{4𝑥}\to 1$

and

$\frac{{𝑒}^{4𝑥}-1}{4𝑥}\to 1$

So the limit is

$\frac{1}{4}\cdot \frac{1}{1^2}=\frac{1}{4}$

Hence, the correct option is

$\overline{)\frac{1}{4}}$

Option A.

First find the truth values of $𝑝$, $𝑞$, and $𝑟$.

For statement $𝑝$:

$𝑝:\text{ If }{3}^4>4^3,\text{ then }{3}^3>4^4$

Now,

$3^4=81,4^3=64$

So $3^4>4^3$ is true.

Also,

$3^3=27,4^4=256$

So $3^3>4^4$ is false.

Thus $𝑝$ is of the form

$𝑇\Rightarrow 𝐹$

which is false.

So,

$𝑝=𝐹$

Now consider statement $𝑞$:

$𝑞:\text{ The roots of }{𝑥}^2-2𝑥+2=0\text{ are real if and only if Mumbai is in Maharashtra.}$

For the quadratic equation,

${𝑥}^2-2𝑥+2=0$

the discriminant is

$𝐷={𝑏}^2-4𝑎𝑐=(-2{)}^2-4(1\left)\right(2)=4-8=-4$

Since $𝐷<0$, the roots are not real.

So the first part is false.

Mumbai is in Maharashtra, so the second part is true.

Thus $𝑞$ is of the form

$𝐹\iff 𝑇$

which is false.

So,

$𝑞=𝐹$

Now statement $𝑟$ is:

$𝑟:\text{ Statement }𝑝\text{ is true or statement }𝑞\text{ is false.}$

Since $𝑝$ is false, “statement $𝑝$ is true” is false.

Since $𝑞$ is false, “statement $𝑞$ is false” is true.

So,

$𝑟=𝐹\vee 𝑇=𝑇$

Hence,

$𝑝=𝐹,𝑞=𝐹,𝑟=𝑇$

Now check the options.

Option A:

$(𝑝\vee 𝑞)\wedge 𝑟=(𝐹\vee 𝐹)\wedge 𝑇=𝐹\wedge 𝑇=𝐹$

Option B:

$𝑝\vee (𝑞\wedge 𝑟)=𝐹\vee (𝐹\wedge 𝑇)=𝐹\vee 𝐹=𝐹$

Option C:

$𝑝\wedge (𝑞\vee 𝑟)=𝐹\wedge (𝐹\vee 𝑇)=𝐹\wedge 𝑇=𝐹$

Option D:

$(𝑝\wedge 𝑞)\vee 𝑟=(𝐹\wedge 𝐹)\vee 𝑇=𝐹\vee 𝑇=𝑇$

So the statement with truth value true is

$\overline{)\text{Option D }(𝑝\wedge 𝑞)\vee 𝑟}$

Statement I:

“If $1$ is not a prime number, then $2$ is not a prime number.”

Here,

$1$ is not a prime number is true.

But $2$ is not a prime number is false, because $2$ is the smallest prime number.

An implication of the form “If $𝑝$, then $𝑞$” is false only when $𝑝$ is true and $𝑞$ is false.

So statement I is not true.

Statement II:

$𝑒$ is a vowel, and

$12\times 3=36$

Both parts are true.

So statement II is true.

Statement III:

“It is not true that $14$ is a composite number and $3$ is an even number.”

Now,

$14$ is a composite number is true.

$3$ is an even number is false.

So the statement “$14$ is composite and $3$ is even” is false.

Its negation becomes true.

So statement III is true.

Statement IV:

$\sqrt{\surd 5}$ is an irrational number is true.

Also, $3+\sqrt{\surd 5}$ is a real number. Since every real number can be written as a complex number with imaginary part $0$, we can write

$3+\sqrt{\surd 5}=(3+\sqrt{\surd 5})+0𝑖$

So $3+\sqrt{\surd 5}$ is also a complex number.

Thus statement IV is true.

Hence, only statement I is not true.

The correct option is:

$\overline{)\text{Option B: Only I}}$

The given statement is

$\sim 𝑝\vee \sim 𝑞$

We can write it in conditional form using

$𝑎\to 𝑏<text>\equiv </text><text>\sim </text>𝑎\vee 𝑏$

Here,

$\sim 𝑝\vee \sim 𝑞\equiv 𝑝\to \sim 𝑞$

Now, the inverse of $𝑝\to \sim 𝑞$ is formed by negating both parts:

$\sim 𝑝\to 𝑞$

Now we need the negation of this inverse:

$\sim (\sim 𝑝\to 𝑞)$

Since

$\sim 𝑝\to 𝑞\equiv 𝑝\vee 𝑞$

So,

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

Using De Morgan’s law,

$\sim (𝑝\vee 𝑞)<text>=</text><text>\sim </text>𝑝\wedge \sim 𝑞$

So the correct answer is

$\overline{)\sim 𝑝\wedge \sim 𝑞}$

Option C.

Given

$𝑋𝐴={𝐵}^{𝑇}$

and

${𝐴}^{-1}𝑌=𝐵$

From the first equation,

$𝑋={𝐵}^{𝑇}{𝐴}^{-1}$

From the second equation, multiply both sides by $𝐴$:

$𝐴\left({𝐴}^{-1}𝑌\right)=𝐴𝐵$

Since $𝐴{𝐴}^{-1}=𝐼$,

$𝑌=𝐴𝐵$

Now,

$𝑋𝑌=\left({𝐵}^{𝑇}{𝐴}^{-1}\right)\left(𝐴𝐵\right)$

Using associativity of matrix multiplication,

$𝑋𝑌={𝐵}^{𝑇}\left({𝐴}^{-1}𝐴\right)𝐵$

Since ${𝐴}^{-1}𝐴=𝐼$,

$𝑋𝑌={𝐵}^{𝑇}𝐼𝐵$

$𝑋𝑌={𝐵}^{𝑇}𝐵$

Now,

$𝐵=⎡⎣⎢\begin{array}{c}1\\ 0\\ 1\end{array}⎤⎦⎥$

So,

${𝐵}^{𝑇}=\left[\begin{array}{ccc}1& 0& 1\end{array}\right]$

Thus,

${𝐵}^{𝑇}𝐵=\left[\begin{array}{ccc}1& 0& 1\end{array}\right]⎡⎣⎢\begin{array}{c}1\\ 0\\ 1\end{array}⎤⎦⎥$

$=1^2+0^2+1^2$

$=2$

So,

$𝑋𝑌=\left[2\right]$

Correct option: Option D.

For the matrix

$𝐴=\left[\begin{array}{cc}𝑎& 1\\ 1& 𝑏\end{array}\right]$

where $𝑎$ and $𝑏$ are the roots of

${𝑥}^2-4𝑥+2=0$

Using the relation between roots and coefficients,

$𝑎+𝑏=4$

and

$𝑎𝑏=2$

Now,

$det\left(𝐴\right)=𝑎𝑏-1\cdot 1=𝑎𝑏-1=2-1=1$

So,

${𝐴}^{-1}=\left[\begin{array}{cc}𝑏& -1\\ -1& 𝑎\end{array}\right]$

Now add $𝐴$ and ${𝐴}^{-1}$:

$𝐴+{𝐴}^{-1}=\left[\begin{array}{cc}𝑎& 1\\ 1& 𝑏\end{array}\right]+\left[\begin{array}{cc}𝑏& -1\\ -1& 𝑎\end{array}\right]$

$=\left[\begin{array}{cc}𝑎+𝑏& 0\\ 0& 𝑎+𝑏\end{array}\right]$

Since $𝑎+𝑏=4$,

$𝐴+{𝐴}^{-1}=\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]=4{𝐼}_2$

So,

$𝑘=4$

Correct option: C.

For the function

$𝑓\left(𝑥\right)=\sqrt{\surd 𝑥-1}+\sqrt{\surd 6-𝑥}$

both square roots should be defined.

For $\sqrt{\surd 𝑥-1}$, the quantity inside the square root must be non-negative:

$𝑥-1\ge 0$

$𝑥\ge 1$

For $\sqrt{\surd 6-𝑥}$, we need:

$6-𝑥\ge 0$

$𝑥\le 6$

So $𝑥$ must satisfy both conditions together:

$𝑥\ge 1\text{and}𝑥\le 6$

Hence,

$1\le 𝑥\le 6$

So the domain is

$[1,6]$

The correct option is Option B: $[1,6]$.

For the combination ${}^{(10-𝑥)}{𝐶}_{(𝑥-6)}$ to be defined, we must have

${}^{𝑛}{𝐶}_{𝑟}$

defined only when:

$𝑛\text{ and }𝑟\text{ are non-negative integers, and }𝑟\le 𝑛$

Here,

$𝑛=10-𝑥$

and

$𝑟=𝑥-6$

So we need:

$10-𝑥\ge 0$

which gives

$𝑥\le 10$

Also,

$𝑥-6\ge 0$

which gives

$𝑥\ge 6$

Now the lower number of the combination must not be greater than the upper number:

$𝑥-6\le 10-𝑥$

$2𝑥\le 16$

$𝑥\le 8$

So combining all conditions:

$6\le 𝑥\le 8$

But $𝑥$ must be an integer because the values in ${}^{𝑛}{𝐶}_{𝑟}$ must be integers.

Hence,

$𝑥\in \{6,7,8\}$

So the correct option is:

$\overline{)\text{Option D }\{6,7,8\}}$

For $𝑓\left(𝑥\right)=[{\tan }^2𝑥]$, we first check what happens near $𝑥=0$.

We know

$\tan 0=0$

so

$𝑓\left(0\right)=[{\tan }^{2}0]=\left[0\right]=0$

Now, when $𝑥$ is very close to $0$, $\tan 𝑥$ is also very close to $0$. Hence ${\tan }^2𝑥$ is also close to $0$ and is non-negative.

For all $𝑥$ sufficiently close to $0$, we have

$0\le {\tan }^2𝑥<1$

So its greatest integer value is

$[{\tan }^2𝑥]=0$

Thus, near $𝑥=0$,

$𝑓\left(𝑥\right)=0$

So,

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

and

$𝑓\left(0\right)=0$

Hence,

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

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

Also, since $𝑓\left(𝑥\right)=0$ in a neighbourhood of $0$, its derivative at $0$ is

${𝑓}^{\prime }\left(0\right)=0$

So it is differentiable at $𝑥=0$, but ${𝑓}^{\prime }\left(0\right)\ne 1$.

Hence the correct option is:

$\overline{)\text{Option B: }𝑓\left(𝑥\right)\text{ is continuous at }𝑥=0}$

Given,

$\sqrt{\surd 𝑦+𝑥}+\sqrt{\surd 𝑦-𝑥}=𝑐$

Square both sides:

$(\sqrt{\surd 𝑦+𝑥}+\sqrt{\surd 𝑦-𝑥}{)}^2={𝑐}^2$

$(𝑦+𝑥)+(𝑦-𝑥)+2\sqrt{\surd (𝑦+𝑥)(𝑦-𝑥)}={𝑐}^2$

$2𝑦+2\sqrt{\surd {𝑦}^2-{𝑥}^2}={𝑐}^2$

Divide by $2$:

$𝑦+\sqrt{\surd {𝑦}^2-{𝑥}^2}=\frac{{𝑐}^2}{2}$

Since $𝑐$ is constant, $\frac{{𝑐}^2}{2}$ is also constant. Differentiate both sides with respect to $𝑥$:

$\frac{𝑑𝑦}{𝑑𝑥}+\frac{1}{2\sqrt{\surd {𝑦}^2-{𝑥}^2}}(2𝑦\frac{𝑑𝑦}{𝑑𝑥}-2𝑥)=0$

$\frac{𝑑𝑦}{𝑑𝑥}+\frac{𝑦\frac{𝑑𝑦}{𝑑𝑥}-𝑥}{\sqrt{\surd {𝑦}^2-{𝑥}^2}}=0$

Let $\frac{𝑑𝑦}{𝑑𝑥}=𝑝$. Then,

$𝑝+\frac{𝑦𝑝-𝑥}{\sqrt{\surd {𝑦}^2-{𝑥}^2}}=0$

$𝑝\sqrt{\surd {𝑦}^2-{𝑥}^2}+𝑦𝑝=𝑥$

$𝑝(𝑦+\sqrt{\surd {𝑦}^2-{𝑥}^2})=𝑥$

$𝑝=\frac{𝑥}{𝑦+\sqrt{\surd {𝑦}^2-{𝑥}^2}}$

Rationalise the denominator:

$𝑝=\frac{𝑥}{𝑦+\sqrt{\surd {𝑦}^2-{𝑥}^2}}\cdot \frac{𝑦-\sqrt{\surd {𝑦}^2-{𝑥}^2}}{𝑦-\sqrt{\surd {𝑦}^2-{𝑥}^2}}$

$𝑝=\frac{𝑥(𝑦-\sqrt{\surd {𝑦}^2-{𝑥}^2})}{{𝑦}^2-({𝑦}^2-{𝑥}^2)}$

$𝑝=\frac{𝑥(𝑦-\sqrt{\surd {𝑦}^2-{𝑥}^2})}{{𝑥}^2}$

$𝑝=\frac{𝑦}{𝑥}-\frac{\sqrt{\surd {𝑦}^2-{𝑥}^2}}{𝑥}$

Now,

$\frac{\sqrt{\surd {𝑦}^2-{𝑥}^2}}{𝑥}=\sqrt{\surd \frac{{𝑦}^2}{{𝑥}^2}-1}$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑦}{𝑥}-\sqrt{\surd \frac{{𝑦}^2}{{𝑥}^2}-1}$

Comparing with

$\frac{𝑑𝑦}{𝑑𝑥}=𝐾-\sqrt{\surd \frac{{𝑦}^2}{{𝑥}^2}-1}$

we get

$𝐾=\frac{𝑦}{𝑥}$

So the correct option is Option D.

Given

$𝑦=({𝑥}^2+1{)}^{\sin 𝑥}$

Since the variable is present in both the base and the power, we use logarithmic differentiation.

Taking logarithm on both sides,

$\log 𝑦=\sin 𝑥\cdot \log ({𝑥}^2+1)$