WBJEE 2025 Paper 1 (Mathematics) Question Paper with Solutions

To determine the values of $𝑎$ and $𝑏$ for which the function

$𝑓\left(𝑥\right)=\{\begin{array}{cc}{𝑥}^2+3𝑥+𝑎,& 𝑥\le 1\\ 𝑏𝑥+2,& 𝑥>1\end{array}$

is differentiable everywhere, we need to ensure two things at the point $𝑥=1$:

Continuity at $𝑥=1$.

Equal left-hand and right-hand derivatives at $𝑥=1$.

Let's go through these steps:

Continuity at $𝑥=1$:

For $𝑓\left(𝑥\right)$ to be continuous at $𝑥=1$, the value of the function from the left-hand side must equal the value from the right-hand side. Thus, we equate:

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

From the left-hand side:

$𝑓\left(1\right)=1^2+3\left(1\right)+𝑎=1+3+𝑎=4+𝑎.$

From the right-hand side:

$𝑓\left(1\right)=𝑏\left(1\right)+2=𝑏+2.$

Setting these equal gives:

$4+𝑎=𝑏+2.$

Solving for $𝑎$:

$𝑎=𝑏-2.$

Differentiability at $𝑥=1$:

The derivative from the left must equal the derivative from the right.

For $𝑥\le 1$, differentiate $𝑓\left(𝑥\right)={𝑥}^2+3𝑥+𝑎$:

${𝑓}^{\prime }\left(𝑥\right)=2𝑥+3.$

Evaluating at $𝑥=1$:

${𝑓}^{\prime }\left(1^{-}\right)=2\left(1\right)+3=5.$

For $𝑥>1$, differentiate $𝑓\left(𝑥\right)=𝑏𝑥+2$:

${𝑓}^{\prime }\left(𝑥\right)=𝑏.$

So,

${𝑓}^{\prime }\left(1^{+}\right)=𝑏.$

Set the derivatives equal:

$5=𝑏.$

Now substitute $𝑏=5$ into the equation we found from continuity:

$𝑎=5-2=3.$

Thus, the function is differentiable everywhere when $𝑎=3$ and $𝑏=5$.

Final Answer: Option A

We're looking for a polynomial $𝑝\left(𝑥\right)$ with these properties:

It has a local max at $𝑥=1$ and a local min at $𝑥=3$.

$𝑝\left(1\right)=6$ and $𝑝\left(3\right)=2$.

We want to find the value of its derivative at $𝑥=0$, which is ${𝑝}^{\prime }\left(0\right)$.

Solution

Since $𝑝\left(𝑥\right)$ has a local maximum and a local minimum, it must have at least two turning points. This means the derivative ${𝑝}^{\prime }\left(𝑥\right)$ must have at least two roots. Therefore, the polynomial $𝑝\left(𝑥\right)$ must be at least a cubic (degree 3) polynomial.

Formulate the Derivative:

Since $𝑝\left(𝑥\right)$ has a local max at $𝑥=1$ and a local min at $𝑥=3$, we know that ${𝑝}^{\prime }\left(1\right)=0$ and ${𝑝}^{\prime }\left(3\right)=0$. Therefore, we can express ${𝑝}^{\prime }\left(𝑥\right)$ in the form:

${𝑝}^{\prime }\left(𝑥\right)=𝑘(𝑥-1)(𝑥-3)$

where $𝑘$ is some constant.

Find the Polynomial:

Integrate ${𝑝}^{\prime }\left(𝑥\right)$ to find $𝑝\left(𝑥\right)$:

$𝑝\left(𝑥\right)=\int {𝑝}^{\prime }\left(𝑥\right)𝑑𝑥=\int 𝑘(𝑥-1)(𝑥-3)𝑑𝑥=\int 𝑘({𝑥}^2-4𝑥+3)𝑑𝑥$

$𝑝\left(𝑥\right)=𝑘(\frac{{𝑥}^3}{3}-2{𝑥}^2+3𝑥)+𝐶$

where $𝐶$ is the constant of integration.

Use the Given Conditions:

We know that $𝑝\left(1\right)=6$ and $𝑝\left(3\right)=2$. Plug these values into the equation for $𝑝\left(𝑥\right)$:

$𝑝\left(1\right)=𝑘(\frac{1}{3}-2+3)+𝐶=𝑘\left(\frac{4}{3}\right)+𝐶=6$

$𝑝\left(3\right)=𝑘(\frac{27}{3}-18+9)+𝐶=𝑘\left(0\right)+𝐶=𝐶=2$

Now we know $𝐶=2$. Substitute this back into the first equation:

$𝑘\left(\frac{4}{3}\right)+2=6$

$𝑘\left(\frac{4}{3}\right)=4$

$𝑘=3$

Complete the Polynomial:

Substitute the values of $𝑘$ and $𝐶$ back into the equation for $𝑝\left(𝑥\right)$:

$𝑝\left(𝑥\right)=3(\frac{{𝑥}^3}{3}-2{𝑥}^2+3𝑥)+2={𝑥}^3-6{𝑥}^2+9𝑥+2$

Find the Derivative:

Differentiate $𝑝\left(𝑥\right)$ to find ${𝑝}^{\prime }\left(𝑥\right)$:

${𝑝}^{\prime }\left(𝑥\right)=3{𝑥}^2-12𝑥+9$

Calculate ${𝑝}^{\prime }\left(0\right)$:

Substitute $𝑥=0$ into ${𝑝}^{\prime }\left(𝑥\right)$:

${𝑝}^{\prime }\left(0\right)=3(0{)}^2-12(0)+9=9$

Answer:

Therefore, ${𝑝}^{\prime }\left(0\right)=9$. So, the correct option is B.

The function is $𝑓\left(𝑥\right)=2{𝑥}^3-3{𝑥}^2-12𝑥+4$

To find local extrema, we calculate the first derivative and set it to zero.

$\begin{array}{cc}& {𝑓}^{\prime }\left(𝑥\right)=6{𝑥}^2-6𝑥-12\\ \Rightarrow & 6{𝑥}^2-6𝑥-12=0\\ \Rightarrow & {𝑥}^2-𝑥-2=0\\ \Rightarrow & (𝑥-2)(𝑥+1)=0\end{array}$

The critical points are $𝑥=2$ and $𝑥=-1$.

Now, use the second derivatives test.

${𝑓}^{\prime \prime }\left(𝑥\right)=12𝑥-6$

At $𝑥=-1$,

${𝑓}^{\prime \prime }(-1)=12(-1)-6=-18<0$

So, there's a local maximum at $𝑥=-1$.

At $𝑥=2$;

${𝑓}^{\prime \prime }\left(2\right)=12\left(2\right)-6=18>0$

So, there's a local minimum at $𝑥=2$

Thus, the function has one local maximum and one local minimum.

Given,

$𝜙\left(𝑥\right)=𝑓\left(𝑥\right)+𝑓(2𝑎-𝑥),𝑥\in [0,2𝑎]$

and ${𝑓}^{\prime }\left(𝑥\right)>0$ for all $𝑥\in [0,𝑎]$.

${𝜙}^{\prime }\left(𝑥\right)={𝑓}^{\prime }\left(𝑥\right)+{𝑓}^{\prime }(2𝑎-𝑥)\cdot (-1)={𝑓}^{\prime }\left(𝑥\right)-{𝑓}^{\prime }(2𝑎-𝑥)$

We are given ${𝑓}^{\prime }\left(𝑥\right)>0$ for all $𝑥\in [0,𝑎]$. This means $𝑓\left(𝑥\right)$ is increasing on $[0,𝑎]$.

Consider the interval $[0,𝑎]$. For $𝑥\in [0,𝑎]$, we have $𝑥\le 𝑎$.

$\begin{array}{c}𝐴𝑛𝑑2𝑎-𝑥\ge 𝑎.\\ 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑓\left(𝑥\right)={𝑥}^2\end{array}$

for $𝑥\in [0,𝑎]$

${𝑓}^{\prime }\left(𝑥\right)=2𝑥$. Since $𝑥\ge 0,{𝑓}^{\prime }\left(𝑥\right)\ge 0$. The condition ${𝑓}^{\prime }\left(𝑥\right)>0$ is satisfied for $𝑥\in (0,𝑎]$.

Now, let's find ${𝜙}^{\prime }\left(𝑥\right)$ for $𝑓\left(𝑥\right)={𝑥}^2$.

$\begin{array}{cc}& 𝜙\left(𝑥\right)={𝑥}^2+(2𝑎-𝑥{)}^2={𝑥}^2+4{𝑎}^2\\ & -4𝑎𝑥+{𝑥}^2=2{𝑥}^2-4𝑎𝑥+4{𝑎}^2\\ & \Rightarrow {𝜙}^{\prime }\left(𝑥\right)=4𝑥-4𝑎=4(𝑥-𝑎)\end{array}$

For $𝑥\in [0,𝑎],𝑥-𝑎\le 0$, so, ${𝜙}^{\prime }\left(𝑥\right)\le 0$.

Therefore, $𝜙\left(𝑥\right)$ is decreasing on $[0,𝑎]$.

Given, $𝑔\left(𝑓\right(𝑥\left)\right)=|\sin 𝑥|$

and  $𝑓\left(𝑔\right(𝑥\left)\right)=(\sin \sqrt{\surd 𝑥}{)}^2$

Option (a), if $𝑓\left(𝑥\right)={\sin }^2𝑥$ and $𝑔\left(𝑥\right)=\sqrt{\surd 𝑥}$,

then  $𝑔\left(𝑓\right(𝑥\left)\right)=\sqrt{\surd {\sin }^2𝑥}=|\sin 𝑥|$

and  $𝑓\left(𝑔\right(𝑥\left)\right)=(\sin \sqrt{\surd 𝑥}{)}^2$

Both conditions are satisfied.

We need to determine which number $2^{4𝑛}-15𝑛-1$ is divisible by for $𝑛\in 𝑁$.

Let $𝑓\left(𝑛\right)=2^{4𝑛}-15𝑛-1={16}^{𝑛}-15𝑛-1$.

Using the binomial expansion of $(1+15{)}^{𝑛}$.

$\begin{array}{cc}& {16}^{𝑛}=(1+15{)}^{𝑛}=(\frac{𝑛}{0})+(\frac{𝑛}{1})15+(\frac{𝑛}{2}){15}^2\\ & +(\frac{𝑛}{3}){15}^3+\dots \\ & \Rightarrow {16}^{𝑛}=1+15𝑛+\frac{𝑛(𝑛-1)}{2}225+\frac{𝑛(𝑛-1)(𝑛-2)}{6}\\ & 3375+\dots \end{array}$

Substituting this into $𝑓\left(𝑛\right)$ :

$\begin{array}{cc}& 𝑓\left(𝑛\right)=(1+15𝑛+\frac{𝑛(𝑛-1)}{2}225+\dots .)-15𝑛-1\\ \Rightarrow & 𝑓\left(𝑛\right)=\frac{𝑛(𝑛-1)}{2}225+\frac{𝑛(𝑛-1)(𝑛-2)}{6}3375+\dots \\ \Rightarrow & 𝑓\left(𝑛\right)={15}^2(\frac{𝑛(𝑛-1)}{2}+\frac{𝑛(𝑛-1)(𝑛-2)}{6}\times 15+\dots .)\end{array}$

Since, all terms from $(\frac{𝑛}{2}){15}^2$ onwards are multiples of ${15}^2=225,𝑓\left(𝑛\right)$ is divisible by 225.

We are given that ${𝑧}_1$ and ${𝑧}_2$ are complex numbers such that $\frac{2{𝑧}_1}{3{𝑧}_2}$ is a purely imaginary number.

Let $\frac{2{𝑧}_1}{3{𝑧}_2}=𝑘𝑖$, where $𝑘$ as a non-zero real number.

Then, $2{𝑧}_1=3{𝑧}_2𝑘𝑖$.

$\frac{{𝑧}_1}{{𝑧}_2}=\frac{3}{2}𝑘𝑖.$

Now, consider the expansion $\frac{{𝑧}_1-{𝑧}_2}{{𝑧}_1+{𝑧}_2}$.

Divide the numerator and denominator by ${𝑧}_2$.

$\frac{\frac{{𝑧}_1}{{𝑧}_2}-1}{\frac{{𝑧}_1}{{𝑧}_2}+1}=\frac{\frac{3}{2}𝑘𝑖-1}{\frac{3}{2}𝑘𝑖+1}$

$\begin{array}{cc}\therefore \mid \frac{-1+\frac{3}{2}𝑘𝑖}{1+\frac{3}{2}𝑘𝑖}\mid & =\frac{\sqrt{\surd (-1{)}^2+{\left(\frac{3}{2}𝑘\right)}^2}}{\sqrt{\surd 1^2+{\left(\frac{3}{2}𝑘\right)}^2}}\\ & =\frac{\sqrt{\surd 1+\frac{9}{4}{𝑘}^2}}{\sqrt{\surd 1+\frac{9}{4}{𝑘}^2}}=1\end{array}$

Let the integral be $𝐼={\int }_6^3\frac{\sqrt{\surd 𝑥}}{\sqrt{\surd 9-𝑥}+\sqrt{\surd 𝑥}}𝑑𝑥$

We can use the property of definite inegrals

${\int }_{𝑏}^{𝑎}𝑓\left(𝑥\right)𝑑𝑥={\int }_{𝑏}^{𝑎}𝑓(𝑎+𝑏-𝑥)𝑑𝑥$

Applying this property to the integral.

$\begin{array}{cc}& 𝐼={\int }_6^3\frac{\sqrt{\surd 9-𝑥}}{\sqrt{\surd 9-(9-𝑥)}+\sqrt{\surd 9-𝑥}}𝑑𝑥\\ & 𝐼={\int }_6^3\frac{\sqrt{\surd 9-𝑥}}{\sqrt{\surd 𝑥}+\sqrt{\surd 9-𝑥}}𝑑𝑥\end{array}$

Now, add the original integral and this new integral.

$2𝐼={\int }_6^3\frac{\sqrt{\surd 𝑥}}{\sqrt{\surd 9-𝑥}+\sqrt{\surd 𝑥}}𝑑𝑥+{\int }_6^3\frac{\sqrt{\surd 9-𝑥}}{\sqrt{\surd 𝑥}+\sqrt{\surd 9-𝑥}}𝑑𝑥$

$\begin{array}{cc}\Rightarrow & 2𝐼={\int }_6^3\frac{\sqrt{\surd 𝑥}+\sqrt{\surd 9-𝑥}}{\sqrt{\surd 𝑥}+\sqrt{\surd 9-𝑥}}𝑑𝑥\\ \Rightarrow & 2𝐼={\int }_6^{3}1𝑑𝑥\\ \Rightarrow & 2𝐼=[𝑥{]}_6^3\\ \Rightarrow & 2𝐼=6-3\\ \Rightarrow & 2𝐼=3\\ \Rightarrow & 𝐼=\frac{3}{2}.\end{array}$

The line $𝑦-\sqrt{\surd 3}𝑥+3=0$ passes through $𝑋(\sqrt{\surd 3},0)$ because $0-\sqrt{\surd 3}\left(\sqrt{\surd 3}\right)+3=0$

The distance from $𝑋(\sqrt{\surd 3},0)$ to any point $(𝑥,𝑦)$ on the line is

             $𝑑=\sqrt{\surd (𝑥-\sqrt{\surd 3}{)}^2+{𝑦}^2}$

Since,  $𝑦=\sqrt{\surd 3}(𝑥-\sqrt{\surd 3})$

we have${𝑦}^2=3(𝑥-\sqrt{\surd 3}{)}^2$

$\begin{array}{cc}𝑆𝑜,𝑑& =\sqrt{\surd (𝑥-\sqrt{\surd 3}{)}^2+3(𝑥-\sqrt{\surd 3}{)}^2}\\ & =\sqrt{\surd 4(𝑥-\sqrt{\surd 3}{)}^2}\\ & =2|𝑥-\sqrt{\surd 3}|\end{array}$

The line intersects the parabola ${𝑦}^2=𝑥+2$

$\begin{array}{cc}& \text{ Substitute }𝑦=\sqrt{\surd 3}𝑥-3\\ & \begin{array}{cc}& (\sqrt{\surd 3}𝑥-3{)}^2=𝑥+2\\ & 3{𝑥}^2-6\sqrt{\surd 3}𝑥+9=𝑥+2\\ & \Rightarrow 3{𝑥}^2-(6\sqrt{\surd 3}+1)𝑥+7=0\end{array}\end{array}$

Let the roots of this quadratic be ${𝑥}_{𝑃}$ and ${𝑥}_{𝑄}$

$\begin{array}{c}{𝑥}_{𝑃}\cdot {𝑥}_{𝑄}=\left(2\mid {𝑥}_{𝑃}-\sqrt{\surd 3}\mid \right)\left(2\mid {𝑥}_{𝑄}-\sqrt{\surd 3}\mid \right)=4\\ \mid ({𝑥}_{𝑃}-\sqrt{\surd 3})({𝑥}_{𝑄}-\sqrt{\surd 3})\mid =4\mid {𝑥}_{𝑃}{𝑥}_{𝑄}-\sqrt{\surd 3}({𝑥}_{𝑃}+{𝑥}_{𝑄})+3\mid \end{array}$

From vieta's formulas, ${𝑥}_{𝑃}+{𝑥}_{𝑄}=\frac{6\sqrt{\surd 3}+1}{3}$ and

$\begin{array}{cc}& {𝑥}_{𝑃}{𝑥}_{𝑄}=\frac{7}{3}.\\ & \text{ substitute these values }\\ & \begin{array}{cc}{𝑥}_{𝑃}\cdot {𝑥}_{𝑄}& =4\mid \frac{7}{3}-\sqrt{\surd 3}\left(\frac{6\sqrt{\surd 3}+1}{3}\right)+3\mid \\ & =4\mid \frac{7-(18+\sqrt{\surd 3})+9}{3}\mid \\ & =4\mid \frac{-2-\sqrt{\surd 3}}{3}\mid =4\left(\frac{2+\sqrt{\surd 3}}{3}\right)\end{array}\end{array}$

The function is $𝑓\left(𝑥\right)=|1-2𝑥|$

The point where the argument of the absolute value becomes zero is $1-2𝑥=0$

$𝑥=1/2$

Continuity at $𝑥=1/2$

A function of the form $\left|𝑢\right(𝑥\left)\right|$ is continuous if $𝑢\left(𝑥\right)$ is continuous. Since, $1-2𝑥$ is a polynomial, it is continuous for all $𝑥$. Therefore, $𝑓\left(𝑥\right)=|1-2𝑥|$ is continuous for all x including at

$𝑥=1/2$

Differentiability at $𝑥=1/2$

The absolute value function $\left|𝑢\right|$ is not differentiable where $𝑢=0$.

Here, $𝑢\left(𝑥\right)=1-2𝑥$. At $𝑥=1/2,𝑢\left(𝑥\right)=0$.

Also ${𝑢}^{\prime }\left(𝑥\right)=-2$, which is not zero at $𝑥=1/2$

Therefore, $𝑓\left(𝑥\right)=|1-2𝑥|$ is not differentiable at $𝑥=1/2$

Conclusion $𝑓\left(𝑥\right)$ is continuous but not differentiable at $𝑥=\frac{1}{2}$

We are given that

$𝑓$ is the inverse function of $𝑔$ and ${𝑔}^{\prime }\left(𝑥\right)=\frac{1}{1+{𝑥}^{𝑛}}$.

Using the formula for the derivative of an inverse function,

${𝑓}^{\prime }\left(𝑥\right)=\frac{1}{{𝑔}^{\prime }\left(𝑓\right(𝑥\left)\right)}$

Substitute the given expression for ${𝑔}^{\prime }\left(𝑥\right)$ into the formula.

${𝑔}^{\prime }\left(𝑓\right(𝑥\left)\right)=\frac{1}{1+\left(𝑓\right(𝑥){)}^{𝑛}}$

therefore,

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

For the matrix $𝐴=⎛⎝⎜\begin{array}{ccc}0& 𝑎& 𝑎\\ 2𝑏& 𝑏& -𝑏\\ 𝑐& -𝑐& 𝑐\end{array}⎞⎠⎟$ to be orthogonal, $𝐴{𝐴}^{𝑇}=𝐼$.

Calculating $𝐴{𝐴}^{𝑇}$.

$\begin{array}{cc}𝐴{𝐴}^{𝑇}& =⎛⎝⎜\begin{array}{ccc}0& 𝑎& 𝑎\\ 2𝑏& 𝑏& -𝑏\\ 𝑐& -𝑐& 𝑐\end{array}⎞⎠⎟⎛⎝⎜\begin{array}{ccc}0& 2𝑏& 𝑐\\ 𝑎& 𝑏& -𝑐\\ 𝑎& -𝑏& 𝑐\end{array}⎞⎠⎟\\ & =⎛⎝⎜\begin{array}{ccc}2{𝑎}^2& 0& 0\\ 0& 6{𝑏}^2& 0\\ 0& 0& 3{𝑐}^2\end{array}⎞⎠⎟\end{array}$

Equation this to the idendity matrix $𝐼$

$\begin{array}{cc}& 2{𝑎}^2=1\Rightarrow 𝑎=\pm \frac{1}{\sqrt{\surd 2}}\\ & 6{𝑏}^2=1\Rightarrow 𝑏=\pm \frac{1}{\sqrt{\surd 6}}\\ & 3{𝑐}^2=1\Rightarrow 𝑐\pm \frac{1}{\sqrt{\surd 3}}\end{array}$

The given matrix is $𝐴=⎡⎣⎢\begin{array}{ccc}5& 5𝑐𝑥& 𝑐𝑥\\ 0& 𝑎& 5𝑐𝑎\\ 0& 0& 5\end{array}⎤⎦⎥$

The determinant of an upper triangular matrix is the product of its diagonal elements.

So, $\left|𝐴\right|=5\times 𝑎\times 5=25𝑎$

We are given $|𝐴{|}^2=25$.

$\begin{array}{cc}(25𝑎{)}^2& =25\Rightarrow 625{𝑎}^2=25\\ {𝑎}^2& =\frac{25}{625}=\frac{1}{25}\\ \left|𝑎\right|& =\sqrt{\surd \frac{1}{25}}=\frac{1}{5}\end{array}$

Given function $𝑓:𝑅\to 𝑅$ satisfies $f\left(\frac{x+y}{3}\right)

The function ' $𝑓$ ' is differentiable at $𝑥=0$.

Differentiate the given equation with respect to $𝑥$, treating $𝑦$ as a constant

$\begin{array}{cc}& \frac{1}{3}{𝑓}^{\prime }\left(\frac{𝑥+𝑦}{3}\right)=\frac{{𝑓}^{\prime }\left(𝑥\right)}{3}\\ \Rightarrow & {𝑓}^{\prime }\left(\frac{𝑥+𝑦}{3}\right)={𝑓}^{\prime }\left(𝑥\right)\end{array}$

Now, differentiate the given equation with respect to $𝑦$, treating $𝑥$ as a constant.

$\begin{array}{c}\frac{1}{3}{𝑓}^{\prime }\left(\frac{𝑥+𝑦}{3}\right)=\frac{{𝑓}^{\prime }\left(𝑦\right)}{3}...\left(𝑖\right)\\ {𝑓}^{\prime }\left(\frac{𝑥+𝑦}{3}\right)={𝑓}^{\prime }\left(𝑦\right)...\left(𝑖𝑖\right)\end{array}$

From Eqs. (i) and (ii),

we have ${𝑓}^{\prime }\left(𝑥\right)={𝑓}^{\prime }\left(𝑦\right)$ for all $𝑥,𝑦\in 𝑅$.

$\Rightarrow {𝑓}^{\prime }\left(𝑥\right)$ is a constant . Let ${𝑓}^{\prime }\left(𝑥\right)=𝑘$.

We are given that $𝑓$ is a differentiable function for all real $𝑥,𝑓\left(2\right)=-4$

and ${𝑓}^{\prime }\left(𝑥\right)\ge 6$ for all $𝑥\in [2,4]$

By the mean value theorem, there exists $𝑎𝑐\in (2,4)$ such that

${𝑓}^{\prime }\left(𝑐\right)=\frac{𝑓\left(4\right)-𝑓\left(2\right)}{4-2}$

given

$\begin{array}{cc}& {𝑓}^{\prime }\left(𝑥\right)\ge 6\\ 𝑆𝑜& {𝑓}^{\prime }\left(𝑐\right)\ge 6.\end{array}$

$\begin{array}{c}𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,\frac{𝑓\left(4\right)-(-4)}{2}\ge 6\\ \frac{𝑓\left(4\right)+4}{2}\ge 6\\ 𝑓\left(4\right)+4\ge 12\\ 𝑓\left(4\right)\ge 8\end{array}$

Given $𝐸$ and $𝐹$ are two independent events,

$𝑃\left(𝐸\right)=0.3\text{ and }𝑃(𝐸\cup 𝐹)=0.5\text{. }$

Since, $𝐸$ and $𝐹$ are independent events.

$\begin{array}{cc}\therefore & 𝑃(𝐸/𝐹)=𝑃\left(𝐸\right)\\ \therefore & 𝑃(𝐹/𝐸)=𝑃\left(𝐹\right)\end{array}$

Using the formula for the union of independent events

$\begin{array}{cc}𝑃(𝐸\cup 𝐹)& =𝑃\left(𝐸\right)+𝑃\left(𝐹\right)-𝑃\left(𝐸\right)𝑃\left(𝐹\right)\\ 0.5& =0.3+𝑃\left(𝐹\right)-\left(0.3\right)𝑃\left(𝐹\right)\\ 0.5& =0.3+0.7𝑃\left(𝐹\right)\\ 0.2& =0.7𝑃\left(𝐹\right)\\ 𝑃\left(𝐹\right)& =\frac{0.2}{0.7}=\frac{2}{7}\\ \therefore 𝑃(𝐸/𝐹)-𝑃(𝐹/𝐸)& =𝑃\left(𝐸\right)-𝑃\left(𝐹\right)\\ & =0.3-\frac{2}{7}\\ & =\frac{3}{10}-\frac{2}{7}\\ & =\frac{21-20}{70}\\ & =\frac{1}{70}\end{array}$

The function

$𝑓\left(𝑥\right)=𝑥-\left[𝑥\right]$

is just the fractional‐part function, usually denoted $\left\{𝑥\right\}=𝑥-\lfloor 𝑥\rfloor .$

On any open interval $(𝑛,𝑛+1)$ (where $𝑛\in \mathbb{Z}$), $\lfloor 𝑥\rfloor =𝑛$ is constant, so

$𝑓\left(𝑥\right)=𝑥-𝑛$

is continuous there.

At an integer $𝑥=𝑛$, the left‐ and right‐hand limits disagree:

As $𝑥\to {𝑛}^{-}$,

$𝑓\left(𝑥\right)=𝑥-(𝑛-1)\to 𝑛-(𝑛-1)=1.$

As $𝑥\to {𝑛}^{+}$,

$𝑓\left(𝑥\right)=𝑥-𝑛\to 0$

(and $𝑓\left(𝑛\right)=0$).

Hence each integer $𝑛$ is a jump discontinuity.

Answer: the points of discontinuity are exactly the integers, $\mathbb{Z}$.

Option D.

Let the roots of ${𝑥}^2-(𝑎-2)(𝑥-𝑎+1)=0$ be $𝛼$ and $𝛽$.

$\therefore 𝛼+𝛽=𝑎-2\text{ and }𝛼𝛽=-(𝑎-1)$

The sum of the squares of the roots is

${𝛼}^2+{𝛽}^2=(𝛼+𝛽{)}^2-2𝛼𝛽$

Substituting the values

$\begin{array}{cc}{𝛼}^2+{𝛽}^2& =(𝑎-2{)}^2-2(-(𝑎-1))\\ & =({𝑎}^2-4𝑎+4)+2𝑎-2\\ & ={𝑎}^2-2𝑎+2\end{array}$

To find the least value, we can rewrite this expression by completing the square

${𝑎}^2-2𝑎+2=(𝑎-1{)}^2+1$

The minimum value occurs when

$(𝑎-1{)}^2=0\Rightarrow 𝑎-1=0\text{ so }𝑎=1$

Let the integeral be

$𝐼={\int }_1^{-1}\frac{{𝑥}^3+\left|𝑥\right|+1}{{𝑥}^2+2\left|𝑥\right|+1}𝑑𝑥$

The integrand can be rewritten as,

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

The integral of an odd function over a symmetric interval $[-𝑎,𝑎]$ is 0 . The function $\frac{{𝑥}^3}{\left(\right|𝑥|+1{)}^2}$ is odd Since, ${𝑓}_1(-𝑥)=\frac{(-𝑥{)}^3}{(1-𝑥\mid +1{)}^2}=\frac{-{𝑥}^3}{\left(\right|𝑥|+1{)}^2}=-{𝑓}_1\left(𝑥\right)$.

Thus, ${\int }_1^{-1}\frac{{𝑥}^3}{\left(\right|𝑥|+1{)}^2}𝑑𝑥=0$

The integral of an even function over a symmetric interval $[-𝑎,𝑎]$ is $2{\int }_{𝑎}^0𝑓\left(𝑥\right)𝑑𝑥$.

The function $\frac{1}{\left|𝑥\right|+1}$ is even since,

${𝑓}_2(-𝑥)=\frac{1}{|-𝑥|+1}=\frac{1}{\left|𝑥\right|+1}={𝑓}_2\left(𝑥\right)$

So, ${\int }_1^{-1}\frac{1}{\left|𝑥\right|+1}𝑑𝑥=2{\int }_1^0\frac{1}{𝑥+1}𝑑𝑥$

(because for $𝑥\in [0,1],\left|𝑥\right|=𝑥$ ).

Evaluating the integral;

$\begin{array}{cc}2{\int }_1^0\frac{1}{𝑥+1}𝑑𝑥& =2[\log |𝑥+1|{]}_1^0\\ & =2\log \left(2\right)-\log \left(1\right))=2\log 2\end{array}$

therefore, $𝐼=0+2\log 2=2\log 2$

Given $𝐚,𝐛,𝐜$ are non-coplanar vectors.

The three vectors are $𝐮=𝐚+2𝐛+3𝐜,𝐯=𝜆𝐛+4𝐜$ and $𝐰=(2𝜆-1)𝐜$

For these vectors to be non-coplanar, their scalar triple product must be non-zero.

$\left[𝐮𝐯𝐰\right]\ne \mathrm{𝟎}$

This can be expressed as the determinant of the coefficients of $𝐚,𝐛,𝐜$

$\mid \begin{array}{ccc}1& 2& 3\\ 0& 𝜆& 4\\ 0& 0& 2𝜆-1\end{array}\mid \left[𝐚𝐛𝐜\right]\ne 0$

Since, $𝐚,𝐛,𝐜$ are non-coplanar, $\left[𝐚𝐛𝐜\right]\ne 0$

Therefore, we require the determinant to be non-zero.

$\begin{array}{c}1\cdot \left(𝜆\right(2𝜆-1)-4\cdot 0)\ne 0\\ 𝜆(2𝜆-1)\ne 0\end{array}$

This implies $𝜆\ne 0$ and $2𝜆-1\ne 0$

$\Rightarrow 𝜆\ne \frac{1}{2}$

Thus, the vectors are non-coplanar for all real values of $𝜆$ except 0 and $\frac{1}{2}$

This corresponds to "all except two values of $𝜆$ ".

Let $𝜔$ be a cubic root of unity $(𝜔\ne 1)$.

We need to find the minimum value of ${\mid 𝑎+𝑏𝜔+𝑐{𝜔}^2\mid }^2$, where $𝑎,𝑏,𝑐$ are distinct non-zero integers.

We know that

${\mid 𝑎+𝑏𝜔+𝑐{𝜔}^2\mid }^2={𝑎}^2+{𝑏}^2+{𝑐}^2-𝑎𝑏-𝑏𝑐-𝑐𝑎$

This expression can be rewritten as

$\frac{1}{2}\left(\right(𝑎-𝑏{)}^2+(𝑏-𝑐{)}^2+(𝑐-𝑎{)}^2)$

The smallest possible absolute difference between two distinct integers is 1 .

Consider consecutive integers, for example,

$𝑎=1,𝑏=2,𝑐=3.$

$\begin{array}{c}𝑇ℎ𝑒𝑛(𝑎-𝑏{)}^2-(1-2{)}^2=(-1{)}^2=1\\ (𝑏-𝑐{)}^2=(2-3{)}^2=(-1{)}^2=1\end{array}$

$\Rightarrow (𝑐-𝑎{)}^2=(3-1{)}^2=(2{)}^2=4$

The sum of squares of differences is $1+1+4=6$ The minimum value of $\frac{1}{2}\left(6\right)=3$.

The integral is ${\int }_{1.5}^0\left[{𝑥}^2\right]𝑑𝑥$.