MHT CET 2026 17th April Morning Shift (Mathematics) Question Paper with Solutions

For the equation

${𝑥}^2-𝑥+1=0,$

the roots are $𝛼$ and $𝛽$.

For any root $𝑥$ of this equation,

${𝑥}^2-𝑥+1=0$

so,

${𝑥}^2=𝑥-1$

Now multiply the original equation by $(𝑥+1)$:

$({𝑥}^2-𝑥+1)(𝑥+1)=0$

Using the identity,

$({𝑥}^2-𝑥+1)(𝑥+1)={𝑥}^3+1$

So,

${𝑥}^3+1=0$

${𝑥}^3=-1$

Hence,

${𝑥}^6=1$

So powers of $𝛼$ and $𝛽$ repeat after every $6$ powers.

Now,

$200÷6\text{ leaves remainder }2$

so,

${𝛼}^{200}={𝛼}^2$

Also,

$206÷6\text{ leaves remainder }2$

so,

${𝛽}^{206}={𝛽}^2$

Thus,

${𝛼}^{200}+{𝛽}^{206}+2={𝛼}^2+{𝛽}^2+2$

From the equation ${𝑥}^2-𝑥+1=0$,

$𝛼+𝛽=1$

and

$𝛼𝛽=1$

Now,

${𝛼}^2+{𝛽}^2=(𝛼+𝛽{)}^2-2𝛼𝛽$

$=1^2-2\left(1\right)$

$=-1$

So,

${𝛼}^2+{𝛽}^2+2=-1+2=1$

The correct answer is

$\overline{)1}$

Option A.

In the word SOLAPUR, the letters are:

Vowels: $𝑂,𝐴,𝑈$

So, number of vowels $=3$

Consonants: $𝑆,𝐿,𝑃,𝑅$

So, number of consonants $=4$

Since consonants are one more than vowels, the only possible alternate arrangement is:

$𝐶𝑉𝐶𝑉𝐶𝑉𝐶$

Now arrange the $4$ consonants in the $4$ consonant places:

$4!=24$

Arrange the $3$ vowels in the $3$ vowel places:

$3!=6$

So, total number of arrangements is:

$4!\times 3!=24\times 6=144$

Hence, the correct option is:

$\overline{)144}$

Option C.

Given,

$\cos {43}^{\circ }+\sin {43}^{\circ }={𝑘}^3$

We use the identity:

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

Here $𝐴={43}^{\circ }$, so

$\cos {43}^{\circ }+\sin {43}^{\circ }=\sqrt{\surd 2}\cos ({45}^{\circ }-{43}^{\circ })$

$\cos {43}^{\circ }+\sin {43}^{\circ }=\sqrt{\surd 2}\cos 2^{\circ }$

But it is given that

$\cos {43}^{\circ }+\sin {43}^{\circ }={𝑘}^3$

So,

${𝑘}^3=\sqrt{\surd 2}\cos 2^{\circ }$

Dividing both sides by $\sqrt{\surd 2}$,

$\cos 2^{\circ }=\frac{{𝑘}^3}{\sqrt{\surd 2}}$

So the correct option is

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

Option B.

Put $𝑥=\sin 𝜃$.

Since $𝜃\in [0,𝜋]$, we have

$0\le \sin 𝜃\le 1$

So,

$𝑥\in [0,1]$

Now use

$\sin 3𝜃=3\sin 𝜃-4{\sin }^3𝜃$

and

${\cos }^2𝜃=1-{\sin }^2𝜃$

The given equation is

$\sin 3𝜃-{\cos }^2𝜃=\frac{1}{4}$

So,

$3𝑥-4{𝑥}^3-(1-{𝑥}^2)=\frac{1}{4}$

$3𝑥-4{𝑥}^3-1+{𝑥}^2=\frac{1}{4}$

$-4{𝑥}^3+{𝑥}^2+3𝑥-\frac{5}{4}=0$

Multiplying by $-4$,

$16{𝑥}^3-4{𝑥}^2-12𝑥+5=0$

Now factor it:

$16{𝑥}^3-4{𝑥}^2-12𝑥+5=(2𝑥-1)(8{𝑥}^2+2𝑥-5)$

So,

$(2𝑥-1)(8{𝑥}^2+2𝑥-5)=0$

Hence,

$𝑥=\frac{1}{2}$

or

$8{𝑥}^2+2𝑥-5=0$

Solving the quadratic,

$𝑥=\frac{-2\pm \sqrt{\surd 4+160}}{16}$

$𝑥=\frac{-2\pm 2\sqrt{\surd 41}}{16}$

$𝑥=\frac{-1\pm \sqrt{\surd 41}}{8}$

Now $𝑥=\sin 𝜃$ must lie in $[0,1]$.

So the valid values are

$𝑥=\frac{1}{2}$

and

$𝑥=\frac{-1+\sqrt{\surd 41}}{8}$

Both lie between $0$ and $1$.

For each value of $\sin 𝜃$ in $(0,1)$, there are exactly two values of $𝜃$ in $[0,𝜋]$.

So total number of solutions is

$2+2=4$

Hence, the correct option is

$\overline{)4}$

Option C.

In a rhombus, the diagonals bisect each other at right angles.

Here, $𝐴(2,-3)$ and $𝐶(-6,7)$ are opposite vertices, so $𝐴𝐶$ is one diagonal. The other diagonal is $𝐵𝐷$.

First find the midpoint of $𝐴𝐶$:

$(\frac{2+(-6)}{2},\frac{-3+7}{2})=(\frac{-4}{2},\frac{4}{2})=(-2,2)$

So diagonal $𝐵𝐷$ passes through $(-2,2)$.

Now find the slope of $𝐴𝐶$:

${𝑚}_{𝐴𝐶}=\frac{7-(-3)}{-6-2}=\frac{10}{-8}=-\frac{5}{4}$

Since diagonals of a rhombus are perpendicular, the slope of $𝐵𝐷$ will be the negative reciprocal of $-\frac{5}{4}$:

${𝑚}_{𝐵𝐷}=\frac{4}{5}$

Now use the point-slope form of a line:

$𝑦-2=\frac{4}{5}(𝑥+2)$

Multiply by $5$:

$5𝑦-10=4𝑥+8$

Bring all terms to one side:

$4𝑥-5𝑦+18=0$

So the equation of diagonal $𝐵𝐷$ is

$\overline{)4𝑥-5𝑦+18=0}$

Correct option: Option C.

The given equation is

$2{𝑥}^2+𝑥𝑦-{𝑦}^2-𝑥+2𝑦-1=0$

This represents a pair of straight lines. To find the angle between the two lines, we can factorise it.

Notice the quadratic part:

$2{𝑥}^2+𝑥𝑦-{𝑦}^2$

can come from

$(2𝑥-𝑦)(𝑥+𝑦)$

Now try:

$(2𝑥-𝑦+1)(𝑥+𝑦-1)$

Multiplying,

$(2𝑥-𝑦+1)(𝑥+𝑦-1)$

$=2{𝑥}^2+2𝑥𝑦-2𝑥-𝑥𝑦-{𝑦}^2+𝑦+𝑥+𝑦-1$

$=2{𝑥}^2+𝑥𝑦-{𝑦}^2-𝑥+2𝑦-1$

So the two lines are

$2𝑥-𝑦+1=0$

and

$𝑥+𝑦-1=0$

From the first line,

$𝑦=2𝑥+1$

so its slope is

${𝑚}_1=2$

From the second line,

$𝑦=-𝑥+1$

so its slope is

${𝑚}_2=-1$

The angle $𝜃$ between two lines with slopes ${𝑚}_1$ and ${𝑚}_2$ is given by

$\tan 𝜃=\mid \frac{{𝑚}_1-{𝑚}_2}{1+{𝑚}_1{𝑚}_2}\mid$

So,

$\tan 𝜃=\mid \frac{2-(-1)}{1+2(-1)}\mid$

$=\mid \frac{3}{-1}\mid$

$=3$

Thus,

$\tan 𝜃=3$

Now,

$\cos 𝜃=\frac{1}{\sqrt{\surd 1+{\tan }^2𝜃}}$

$=\frac{1}{\sqrt{\surd 1+9}}$

$=\frac{1}{\sqrt{\surd 10}}$

So,

$𝜃={\cos }^{-1}\frac{1}{\sqrt{\surd 10}}$

Hence, the correct option is

$\overline{)\text{Option C }{\cos }^{-1}\frac{1}{\sqrt{\surd 10}}}$

The pair of lines passes through $(1,4)$.

One line is parallel to the $𝑋$-axis.

A line parallel to the $𝑋$-axis has equation $𝑦=\text{constant}$.

Since it passes through $(1,4)$, its equation is

$𝑦=4$

So,

$𝑦-4=0$

The other line makes an angle of ${45}^{\circ }$ with the positive direction of the $𝑋$-axis.

Its slope is

$𝑚=\tan {45}^{\circ }=1$

Using point-slope form through $(1,4)$,

$𝑦-4=1(𝑥-1)$

$𝑦-4=𝑥-1$

$𝑦=𝑥+3$

So,

$𝑦-𝑥-3=0$

The joint equation of two lines is obtained by multiplying their equations:

$(𝑦-4)(𝑦-𝑥-3)=0$

Expanding,

$𝑦(𝑦-𝑥-3)-4(𝑦-𝑥-3)=0$

${𝑦}^2-𝑥𝑦-3𝑦-4𝑦+4𝑥+12=0$

${𝑦}^2-𝑥𝑦-7𝑦+4𝑥+12=0$

Multiplying by $-1$,

$𝑥𝑦-{𝑦}^2-4𝑥+7𝑦-12=0$

So the correct option is

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

The centre of the circle lies on the line

$𝑦-4𝑥+3=0$

So if the centre is $(ℎ,𝑘)$, then

$𝑘-4ℎ+3=0$

or

$𝑘=4ℎ-3$

Since the circle passes through $(2,3)$ and $(4,5)$, the centre must be equidistant from both these points.

So,

$(ℎ-2{)}^2+(𝑘-3{)}^2=(ℎ-4{)}^2+(𝑘-5{)}^2$

Expanding,

${ℎ}^2-4ℎ+4+{𝑘}^2-6𝑘+9={ℎ}^2-8ℎ+16+{𝑘}^2-10𝑘+25$

Cancel ${ℎ}^2$ and ${𝑘}^2$:

$-4ℎ-6𝑘+13=-8ℎ-10𝑘+41$

Bring like terms together:

$4ℎ+4𝑘=28$

$ℎ+𝑘=7$

Now use $𝑘=4ℎ-3$:

$ℎ+4ℎ-3=7$

$5ℎ=10$

$ℎ=2$

Then,

$𝑘=4\left(2\right)-3=5$

So the centre is $(2,5)$.

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

${𝑟}^2=(2-2{)}^2+(5-3{)}^2=4$

Hence the equation of the circle is

$(𝑥-2{)}^2+(𝑦-5{)}^2=4$

Expanding,

${𝑥}^2-4𝑥+4+{𝑦}^2-10𝑦+25=4$

${𝑥}^2+{𝑦}^2-4𝑥-10𝑦+25=0$

So the correct option is

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

For the ellipse

$\frac{{𝑥}^2}{16}+\frac{{𝑦}^2}{9}=1$

we have

${𝑎}^2=16,{𝑏}^2=9$

The major axis is along the $𝑥$-axis because $16>9$.

The tangent to an ellipse

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

with slope $𝑚$ is given by

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

The tangent makes an inclination of ${30}^{\circ }$ with the major axis, so its slope is

$𝑚=\tan {30}^{\circ }=\frac{1}{\sqrt{\surd 3}}$

Now substitute ${𝑎}^2=16$, ${𝑏}^2=9$, and $𝑚=\frac{1}{\sqrt{\surd 3}}$:

$𝑦=\frac{𝑥}{\sqrt{\surd 3}}\pm \sqrt{\surd 16\left(\frac{1}{3}\right)+9}$

$𝑦=\frac{𝑥}{\sqrt{\surd 3}}\pm \sqrt{\surd \frac{16}{3}+\frac{27}{3}}$

$𝑦=\frac{𝑥}{\sqrt{\surd 3}}\pm \sqrt{\surd \frac{43}{3}}$

Multiply throughout by $\sqrt{\surd 3}$:

$\sqrt{\surd 3}𝑦=𝑥\pm \sqrt{\surd 43}$

Rearranging,

$𝑥-\sqrt{\surd 3}𝑦\pm \sqrt{\surd 43}=0$

So the correct option is

$\overline{)\text{Option B }𝑥-\sqrt{\surd 3}𝑦\pm \sqrt{\surd 43}=0}$

Put $𝑥=1$ in the numerator:

$\sin \left(3\right(1{)}^2-4\left(1\right)+1)-1^2+1=\sin 0-1+1=0$

For the limit to be a non-zero finite value like $-2$, the denominator must also become $0$ at $𝑥=1$.

So,

$2(1{)}^3-7(1{)}^2+𝑎\left(1\right)+𝑏=0$

$2-7+𝑎+𝑏=0$

$𝑎+𝑏=5$

Now check the order of zero. The numerator is

$\sin (3{𝑥}^2-4𝑥+1)-{𝑥}^2+1$

Let

$3{𝑥}^2-4𝑥+1=(𝑥-1)(3𝑥-1)$

Near $𝑥=1$, this expression becomes very small, so using the standard limit idea $\sin 𝑦\sim 𝑦$ when $𝑦\to 0$,

$\sin (3{𝑥}^2-4𝑥+1)\sim 3{𝑥}^2-4𝑥+1$

So the numerator behaves like

$(3{𝑥}^2-4𝑥+1)-{𝑥}^2+1$

$=2{𝑥}^2-4𝑥+2$

$=2(𝑥-1{)}^2$

So the numerator has factor $(𝑥-1{)}^2$.

For the limit to be non-zero, the denominator must also have factor $(𝑥-1{)}^2$.

Denominator:

$2{𝑥}^3-7{𝑥}^2+𝑎𝑥+𝑏$

Since $𝑥=1$ is already a root,

$𝑎+𝑏=5$

For $(𝑥-1{)}^2$ to be a factor, derivative of denominator should also be zero at $𝑥=1$.

Differentiate:

$\frac{𝑑}{𝑑𝑥}(2{𝑥}^3-7{𝑥}^2+𝑎𝑥+𝑏)=6{𝑥}^2-14𝑥+𝑎$

At $𝑥=1$:

$6-14+𝑎=0$

$𝑎=8$

Using $𝑎+𝑏=5$,

$8+𝑏=5$

$𝑏=-3$

So the roots are $8$ and $-3$.

The quadratic equation with roots $8$ and $-3$ is

${𝑥}^2-(8+(-3\left)\right)𝑥+8(-3)=0$

${𝑥}^2-5𝑥-24=0$

Hence, the correct option is

$\overline{){𝑥}^2-5𝑥-24=0}$

Option C.

Take

$𝑉$: voltage increases

$𝐶$: current decreases

Now write each statement in symbolic form.

$𝑝$: If voltage increases then current decreases.

$𝑉\Rightarrow 𝐶$

$𝑞$: If voltage does not increase then current does not decrease.

$\sim 𝑉<text>\Rightarrow </text><text>\sim </text>𝐶$

$𝑟$: If current decreases then voltage increases.

$𝐶\Rightarrow 𝑉$

$𝑠$: If current does not decrease then voltage does not increase.

$\sim 𝐶<text>\Rightarrow </text><text>\sim </text>𝑉$

Now, in logic, a statement and its contrapositive have the same meaning.

The contrapositive of

$𝑉\Rightarrow 𝐶$

is

$\sim 𝐶<text>\Rightarrow </text><text>\sim </text>𝑉$

So, $𝑝$ and $𝑠$ have the same meaning.

Also, the contrapositive of

$\sim 𝑉<text>\Rightarrow </text><text>\sim </text>𝐶$

is

$𝐶\Rightarrow 𝑉$

So, $𝑞$ and $𝑟$ have the same meaning.

Hence the pairs having the same meaning are:

$𝑝,𝑠\text{and}𝑞,𝑟$

So the correct option is

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

For $𝑝\to (𝑞\vee \sim 𝑟)$ to be false, the antecedent must be true and the consequent must be false.

So,

$𝑝=𝑇$

and

$𝑞\vee \sim 𝑟=𝐹$

Now, $𝑞\vee \sim 𝑟$ is false only when both parts are false.

So,

$𝑞=𝐹$

and

$\sim 𝑟=𝐹$

This means

$𝑟=𝑇$

Now we have:

$𝑝=𝑇,𝑞=𝐹,𝑟=𝑇$

Evaluate the first statement:

$(𝑝\leftrightarrow 𝑞)\wedge 𝑟$

Since $𝑝=𝑇$ and $𝑞=𝐹$, we get

$𝑝\leftrightarrow 𝑞=𝐹$

So,

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

Now evaluate the second statement:

$\sim 𝑝\to \sim 𝑞$

Since $𝑝=𝑇$, $\sim 𝑝=𝐹$.

Since $𝑞=𝐹$, $\sim 𝑞=𝑇$.

So,

$\sim 𝑝\to \sim 𝑞=𝐹\to 𝑇=𝑇$

Hence the truth values are

$𝐹,𝑇$

So the correct option is Option C.

In $△𝐴𝐵𝐶$, usual notation means side $𝑎$ is opposite angle $𝐴$, side $𝑏$ is opposite angle $𝐵$, and side $𝑐$ is opposite angle $𝐶$.

Given,

$\cos 𝐶=\frac{\sin 𝐴}{2\sin 𝐵}$

Using the sine rule,

$\frac{\sin 𝐴}{\sin 𝐵}=\frac{𝑎}{𝑏}$

So,

$\cos 𝐶=\frac{𝑎}{2𝑏}$

Now, using the cosine rule for angle $𝐶$,

$\cos 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}$

So,

$\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}=\frac{𝑎}{2𝑏}$

Cancel $2𝑏$ from both sides:

$\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{𝑎}=𝑎$

Multiply by $𝑎$:

${𝑎}^2+{𝑏}^2-{𝑐}^2={𝑎}^2$

Cancel ${𝑎}^2$ from both sides:

${𝑏}^2={𝑐}^2$

Since sides of a triangle are positive,

$𝑏=𝑐$

So the correct option is:

$\overline{)\text{Option C: }𝑏=𝑐}$

In $△𝐴𝐵𝐶$, the sides are given as

$𝑎=13,𝑏=14,𝑐=15$

Here, side $𝑎$ is opposite angle $𝐴$.

Using the cosine rule,

${𝑎}^2={𝑏}^2+{𝑐}^2-2𝑏𝑐\cos 𝐴$

So,

$\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}$

Substitute the values:

$\cos 𝐴=\frac{{14}^2+{15}^2-{13}^2}{2\times 14\times 15}$

$\cos 𝐴=\frac{196+225-169}{420}$

$\cos 𝐴=\frac{252}{420}=\frac{3}{5}$

Now,

${\sin }^2𝐴+{\cos }^2𝐴=1$

${\sin }^2𝐴=1-{\left(\frac{3}{5}\right)}^2$

${\sin }^2𝐴=1-\frac{9}{25}=\frac{16}{25}$

Since angle $𝐴$ is an angle of a triangle, $\sin 𝐴$ is positive.

$\sin 𝐴=\frac{4}{5}$

So,

$\sin 𝐴+\cos 𝐴=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$

Hence, the correct option is

$\overline{)\frac{7}{5}}$

Option B.

For any square matrix $𝐴$, it can be written as the sum of a symmetric matrix and a skew symmetric matrix as

$𝐴=\frac{𝐴+{𝐴}^{𝑇}}{2}+\frac{𝐴-{𝐴}^{𝑇}}{2}$

Here,

$𝐴=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right]$

So,

${𝐴}^{𝑇}=\left[\begin{array}{cc}4& 3\\ 1& 2\end{array}\right]$

The symmetric matrix is

$𝐵=\frac{𝐴+{𝐴}^{𝑇}}{2}$

$𝐵=\frac{1}{2}\left[\begin{array}{cc}8& 4\\ 4& 4\end{array}\right]=\left[\begin{array}{cc}4& 2\\ 2& 2\end{array}\right]$

The skew symmetric matrix is

$𝐶=\frac{𝐴-{𝐴}^{𝑇}}{2}$

$𝐶=\frac{1}{2}\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

Now find the determinants.

$\left|𝐴\right|=4\times 2-1\times 3=8-3=5$

$\left|𝐵\right|=4\times 2-2\times 2=8-4=4$

$\left|𝐶\right|=0\times 0-(-1)\left(1\right)=1$

So,

$\left|𝐵\right|+\left|𝐶\right|=4+1=5$

Hence,

$\left|𝐴\right|=\left|𝐵\right|+\left|𝐶\right|$

So the correct option is

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

Given,

${𝐴}^2-𝐴+𝐼=0$

Since $𝐴$ is non-singular, ${𝐴}^{-1}$ exists.

Multiply the equation by ${𝐴}^{-1}$:

${𝐴}^{-1}({𝐴}^2-𝐴+𝐼)={𝐴}^{-1}\cdot 0$

So,

${𝐴}^{-1}{𝐴}^2-{𝐴}^{-1}𝐴+{𝐴}^{-1}𝐼=0$

Now,

${𝐴}^{-1}{𝐴}^2=𝐴,{𝐴}^{-1}𝐴=𝐼,{𝐴}^{-1}𝐼={𝐴}^{-1}$

Hence,

$𝐴-𝐼+{𝐴}^{-1}=0$

So,

${𝐴}^{-1}=𝐼-𝐴$

Thus, the correct option is

$\overline{)𝐼-𝐴}$

Option C.

Let

$𝜃={\tan }^{-1}\left(\frac{1}{2}\right)$

So,

$\tan 𝜃=\frac{1}{2}$

We need the value of $3𝜃$.

Using the formula,

$\tan 3𝜃=\frac{3\tan 𝜃-{\tan }^3𝜃}{1-3{\tan }^2𝜃}$

Put $\tan 𝜃=\frac{1}{2}$:

$\tan 3𝜃=\frac{3\left(\frac{1}{2}\right)-{\left(\frac{1}{2}\right)}^3}{1-3{\left(\frac{1}{2}\right)}^2}$

$=\frac{\frac{3}{2}-\frac{1}{8}}{1-\frac{3}{4}}$

$=\frac{\frac{12}{8}-\frac{1}{8}}{\frac{1}{4}}=\frac{\frac{11}{8}}{\frac{1}{4}}$

$=\frac{11}{2}$

So,

$\tan 3𝜃=\frac{11}{2}$

Also, since $𝜃={\tan }^{-1}\left(\frac{1}{2}\right)$, we have $0<𝜃<\frac{𝜋}{6}$, so

$0<3𝜃<\frac{𝜋}{2}$

Hence $3𝜃$ lies in the principal range of ${\tan }^{-1}$.

Therefore,

$3{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(\frac{11}{2}\right)$

So the correct option is

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

For a rational function, $𝑓\left(𝑥\right)$ is not defined when its denominator is equal to zero.

Here,

$𝑓\left(𝑥\right)=\frac{𝑥+2}{{𝑥}^2-3𝑥+1}$

So we put the denominator equal to zero:

${𝑥}^2-3𝑥+1=0$

Using the quadratic formula,

$𝑥=\frac{-𝑏\pm \sqrt{\surd {𝑏}^2-4𝑎𝑐}}{2𝑎}$

For ${𝑥}^2-3𝑥+1=0$,

$𝑎=1,𝑏=-3,𝑐=1$

So,

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

$𝑥=\frac{3\pm \sqrt{\surd 9-4}}{2}$

$𝑥=\frac{3\pm \sqrt{\surd 5}}{2}$

So the values of $𝑥$ are

$𝑥=\frac{3+\sqrt{\surd 5}}{2},𝑥=\frac{3-\sqrt{\surd 5}}{2}$

Hence, the correct answer is Option A.

For the given functions,

$𝑓\left(𝑥\right)=2𝑥-\left|𝑥\right|$

and

$𝑔\left(𝑥\right)=2𝑥+\left|𝑥\right|$

First write them according to the sign of $𝑥$.

If $𝑥\ge 0$, then $\left|𝑥\right|=𝑥$.

So,

$𝑓\left(𝑥\right)=2𝑥-𝑥=𝑥$

and

$𝑔\left(𝑥\right)=2𝑥+𝑥=3𝑥$

If $𝑥<0$, then $\left|𝑥\right|=-𝑥$.

So,

$𝑓\left(𝑥\right)=2𝑥-(-𝑥)=3𝑥$

and

$𝑔\left(𝑥\right)=2𝑥+(-𝑥)=𝑥$

Now calculate the required values.

For $(𝑓\circ 𝑔)\left(2\right)$:

$𝑔\left(2\right)=3\left(2\right)=6$

Since $6>0$,

$𝑓\left(6\right)=6$

So,

$(𝑓\circ 𝑔)\left(2\right)=6$

For $(𝑔\circ 𝑓)\left(2\right)$:

$𝑓\left(2\right)=2$

Then,

$𝑔\left(2\right)=6$

So,

$(𝑔\circ 𝑓)\left(2\right)=6$

For $(𝑔\circ 𝑓)(-2)$:

Since $-2<0$,

$𝑓(-2)=3(-2)=-6$

Now,

$𝑔(-6)=-6$

So,

$(𝑔\circ 𝑓)(-2)=-6$

For $(𝑓\circ 𝑔)(-2)$:

Since $-2<0$,

$𝑔(-2)=-2$

Now,

$𝑓(-2)=3(-2)=-6$

So,

$(𝑓\circ 𝑔)(-2)=-6$

Now check the options.

Option D:

$(𝑔\circ 𝑓)\left(2\right)+(𝑔\circ 𝑓)(-2)=6+(-6)=0$

So the correct option is:

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

To check continuity at $𝑥=0$, we need

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

That means the left-hand limit and right-hand limit should both exist and should be equal to the value of the function at $0$.

For Option A,

$𝑓\left(𝑥\right)=(1+𝑥{)}^{\frac{2}{𝑥}}$

We know

$\underset{𝑥\to 0}{lim}(1+𝑥{)}^{\frac{1}{𝑥}}=𝑒$

So,

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

Also, $𝑓\left(0\right)={𝑒}^2$.

So Option A is continuous at $𝑥=0$.

For Option B,

$𝑓\left(𝑥\right)=\sin 𝑥-\cos 𝑥$

So,

$\underset{𝑥\to 0}{lim}(\sin 𝑥-\cos 𝑥)=\sin 0-\cos 0=0-1=-1$

Also, $𝑓\left(0\right)=-1$.

So Option B is continuous at $𝑥=0$.

For Option C,

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

First take the right-hand limit.

When $𝑥\to 0^{+}$, then $\frac{1}{𝑥}\to \infty$, so ${𝑒}^{\frac{1}{𝑥}}\to \infty$.

Thus,

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

Now take the left-hand limit.

When $𝑥\to 0^{-}$, then $\frac{1}{𝑥}\to -\infty$, so ${𝑒}^{\frac{1}{𝑥}}\to 0$.

Thus,

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

So,

$\underset{𝑥\to 0^{+}}{lim}𝑓\left(𝑥\right)\ne \underset{𝑥\to 0^{-}}{lim}𝑓\left(𝑥\right)$

The limit at $𝑥=0$ does not exist. Hence Option C is discontinuous at $𝑥=0$.

For Option D,

$𝑓\left(𝑥\right)=\frac{{𝑒}^{5𝑥}-{𝑒}^{2𝑥}}{\sin 3𝑥}$

Using standard limits,

${𝑒}^{5𝑥}-{𝑒}^{2𝑥}\approx 5𝑥-2𝑥=3𝑥$

and

$\sin 3𝑥\approx 3𝑥$

So,

$\underset{𝑥\to 0}{lim}\frac{{𝑒}^{5𝑥}-{𝑒}^{2𝑥}}{\sin 3𝑥}=1$

Also, $𝑓\left(0\right)=1$.

So Option D is continuous at $𝑥=0$.

Hence, the discontinuous function is

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

Given

$𝑦=𝑥\cdot 7^{𝑥}$

We need $\frac{𝑑𝑥}{𝑑𝑦}$ when $𝑥=1$.

First find $\frac{𝑑𝑦}{𝑑𝑥}$. Since $𝑦$ is a product of $𝑥$ and $7^{𝑥}$, use the product rule:

$\frac{𝑑𝑦}{𝑑𝑥}=𝑥\cdot \frac{𝑑}{𝑑𝑥}\left(7^{𝑥}\right)+7^{𝑥}\cdot \frac{𝑑}{𝑑𝑥}\left(𝑥\right)$

Now,

$\frac{𝑑}{𝑑𝑥}\left(7^{𝑥}\right)=7^{𝑥}\log 7$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=𝑥\cdot 7^{𝑥}\log 7+7^{𝑥}$

Taking $7^{𝑥}$ common,

$\frac{𝑑𝑦}{𝑑𝑥}=7^{𝑥}(𝑥\log 7+1)$

At $𝑥=1$,

$\frac{𝑑𝑦}{𝑑𝑥}=7^1(1\cdot \log 7+1)$