VITEEE 2021 Question Paper with Solutions

The word 'bane' refers to something that causes great distress, ruin, or harm. Therefore, we're looking for an antonym that represents a positive or beneficial influence, the opposite of distress or harm.

• Option A: Boom - This can refer to a period of rapid growth, economic success, or increased popularity. Considering the definition of 'bane,' 'Boom' is a suitable antonym because it implies positive outcomes or benefits, which is the exact opposite of harm or ruin. Therefore, Boom is the correct antonym.


• Option B: Commotion - This refers to a state of confusion and disorder. It does not directly contrast with the notion of causing distress or harm as 'bane' does, thus not a suitable antonym.


• Option C: Penance - This is an act of self-punishment or religious discipline to atone for one's sins or wrongdoings. It does not embody the opposite concept of causing harm or distress; therefore, it is not the correct antonym.


• Option D: Curse - This actually synonymizes with 'bane,' as both imply something that causes harm or misfortune. As such, it cannot be the antonym.

Given these definitions, Option A: Boom is the correct antonym for 'bane' as it represents a positive impact or state of flourishing, which is the opposite of bane's implication of harm or distress.

The sentence "I can write with .............. hand." is a reference to the ability to use one of two hands, or possibly both hands, for writing. Thus, the most appropriate completion of the sentence should indicate the capability to use any one of the two hands (left or right) or both. The choices are analyzed as follows:

• Option A - every: This option is not suitable because "every" is used to refer to all members of a group of three or more. Since the sentence is about writing with a hand, and individuals typically have two hands, "every" is not the right choice.


• Option B - two: "Two" is also incorrect because it literally implies writing with two hands simultaneously, which, although possible, is not the common interpretation or intention behind the statement.


• Option C - either: This is the most appropriate choice. "Either" correctly implies a choice between two distinct alternatives, in this case, the left hand or the right hand, without stating a preference for one over the other. It conveys the ability to write with one hand, with the choice being open.


• Option D - both: "Both" indicates the use of two things together. While it's true some people might be able to write with both hands (ambidexterity), the statement aiming to be more general and inclusive of all individuals' abilities would best be served by indicating the potential to use one hand or the other.

Therefore, the correct completion for the sentence is Option C - either, which makes the complete sentence: "I can write with either hand."

The sentence "We bought ............. books." requires a word that can appropriately modify the noun "books" in a way that relates to quantity because "books" is a countable noun. Let's look at the options provided:

• Option A - much: "Much" is used with uncountable nouns to indicate a large quantity, so it is not suitable for countable nouns like "books."


• Option B - more: "More" is used to indicate a comparative quantity or degree. Without any context for comparison, "more" alone does not complete the sentence appropriately.


• Option C - any: "Any" is usually used in questions and negative statements, not typically for affirmative statements to indicate quantity.


• Option D - many: "Many" is used with countable nouns to indicate a large quantity. It fits perfectly in the context of the sentence, which talks about purchasing a quantity of books.

Therefore, the most appropriate choice to complete the sentence is Option D - many. The completed sentence should read: "We bought many books."

The correct synonym for the word 'Treason' is 'Disloyalty'. Therefore, the answer is Option A - Disloyalty.

'Treason' refers to the crime of betraying one's country, especially by attempting to kill the sovereign or overthrow the government. The essence of this term lies in the act of betrayal or disloyalty to one's nation or its principles. Thus, 'Disloyalty', which means a lack of loyalty or faithfulness, closely aligns with the conceptual meaning of 'Treason' more than the other options provided.

Option B, 'Passion', means a strong and barely controllable emotion, which doesn't relate to the betrayal or political context of treason. Option C, 'Miserable', means extremely unhappy or uncomfortable, which also doesn't fit the context of treason. Lastly, Option D, 'Disappear', means to cease to be visible or to no longer exist, which has no direct relevance to the concept of treason or betrayal.

The correct sentence completion requires understanding the subject-verb agreement in English grammar. In this case, the subject "Your friend" is a singular noun. The verb must accordingly be in its third person singular form in the simple present tense. Therefore, the appropriate choice is:

Option B: talks

The completed sentence is: "Your friend talks too much."

Let position vector of P and Q are

$$\begin{aligned} & O P=6 \hat{i}+3 \hat{j}-2 \hat{k} \\ \text{and} \quad & O Q=3 \hat{i}+\hat{j}-4 \hat{k} \end{aligned}$$

Now, by section formula

$$\begin{aligned} \mathrm{OR} & =\left[\frac{2 \times O Q-1 \times O P}{2-1}\right] \\ & =\left(\frac{2(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}})-(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})}{1}\right) \\ & =\left\{\frac{6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}-6 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{1}\right\} \\ & =(-\hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \end{aligned}$$

The given condition describes a scenario where a point $P$ moves in such a way that its distance from a fixed point (let's call this point $F$) is always equal to its distance from a fixed straight line (let's call this line $l$). This condition directly corresponds to the definition of a parabola.

In mathematical terms, a parabola is the set of all points in a plane equidistant from a given point (called the focus) and a given line (called the directrix). The given criteria match this description perfectly: the fixed point $F$ acts as the focus of the parabola, and the fixed line $l$ acts as the directrix.

Thus, the correct answer is:

Option B: a parabola

$$\begin{aligned} & \text { } \sim(\sim s \vee(\sim r \wedge s)) \\ & \equiv s \wedge(\sim(\sim r \wedge s)) \equiv s \wedge(r \vee \sim s) \\ & \equiv(s \wedge r) \vee(s \wedge \sim s) \\ & \equiv(s \wedge r) \vee F \quad (\because s ~\wedge \sim s ~\text{is false})\\ & \equiv s \wedge r \end{aligned}$$

To determine which of the given options is not a function, we need to recall the definition of a function in the context of sets of ordered pairs. A set of ordered pairs is a function if, for every $x$ in the domain, there is exactly one $y$ in the codomain corresponding to $x$. This means that each input ($x$ value) must map to only one output ($y$ value). Let's analyze each option based on this criterion.

Option A: $\left\{(x, y): x, y \in \mathbb{R}, x^2=y\right\}$

This is a function because for each $x$ value, there is exactly one $y$ value such that $y = x^2$. This relationship passes the vertical line test, as no vertical line can intersect the graph at more than one point. Thus, Option A is indeed a function.

Option B: $\left\{(x, y): x, y \in \mathbb{R}, y^2=x\right\}$

This option is not a function. The reason is that for a given $x$ (except for $x=0$), there can be two different $y$ values such that $y^2=x$: one positive and one negative. For example, if $x=4$, both $y=2$ and $y=-2$ satisfy the equation $y^2=x$. This violates the definition of a function since one $x$ value corresponds to more than one $y$ value. Therefore, Option B is not a function.

Option C: $\left\{(x, y): x, y \in \mathbb{R}, x=y^3\right\}$

This is a function because for each $y$ value, there is exactly one $x$ value that can be calculated using $x = y^3$. No matter what $y$ value you choose, plugging it into $y^3$ will yield one unique $x$, making it a function.

Option D: $\left\{(x, y): x, y \in \mathbb{R}, y=x^3\right\}$

This also describes a function. For every $x$ value, $y = x^3$ produces a single $y$ value. This equation satisfies the function criterion as each input ($x$) has exactly one output ($y$).

Therefore, the correct answer to which of the following is not a function is Option B:

$\left\{(x, y): x, y \in \mathbb{R}, y^2=x\right\}$

We know that, the focal distance or radii of the point $(x, y)$ from the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$

$$\begin{aligned} & =a \pm e x \\ \Rightarrow \quad & a \pm \sqrt{\frac{a^2-b^2}{a^2}} \cdot x=a \pm \sqrt{1-\frac{b^2}{a^2}} \cdot x \end{aligned}$$

By first principle of mathematical induction, statement (d) is true.

$$\begin{aligned} & \text { Given, } \frac{d y}{d x}+\sec x \cdot y=\tan x\left(0 \leq x \leq \frac{\pi}{2}\right) \\ & \begin{aligned} I F=e^{\int \sec x d x} & =e^{\log \mid \sec x+\tan x+} \\ & =|\sec x+\tan x| \end{aligned} \end{aligned}$$

$\therefore$ General solution is

$$\begin{aligned} & y \cdot(\mathrm{IF})=\int Q \cdot(\mathrm{IF}) d x+C \\ \Rightarrow & y \cdot|\sec x+\tan x| \\ & =\int \tan x \cdot(\sec x+\tan x) d x+C \\ \Rightarrow & y(\sec x+\tan x) \\ & =\int \tan x \cdot \sec x d x+\int \tan ^2 x d x+C \\ \Rightarrow & y(\sec x+\tan x)=\sec x+\int\left(\sec ^2 x-1\right) d x \\ \Rightarrow & y(\sec x+\tan x)=(\sec x+\tan x)-x+C \\ \Rightarrow & y=1+\frac{(C-x)}{(\sec x+\tan x)}+C^{-1}+\frac{C}{(\sec x+\tan x)}+1 \end{aligned}$$

To solve this problem, let's understand the functions involved and then solve for the quotient of the given functions.

The identity function is a function that always returns the same value that was used as its input. It is defined as $f(x) = x.$

The reciprocal function is a function that returns the reciprocal (or multiplicative inverse) of the input value. It is defined as $g(x) = \frac{1}{x},$ where $x \neq 0$ because the reciprocal of zero is undefined.

Now, the quotient of the identity function by the reciprocal function would be calculated by dividing $f(x)$ by $g(x)$. Formally, this quotient, let's call it $h(x)$, is calculated as:

$h(x) = \frac{f(x)}{g(x)}$

Substituting $f(x) = x$ and $g(x) = \frac{1}{x}$, we get:

$h(x) = \frac{x}{\frac{1}{x}}$

To simplify $h(x)$, remember that dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, we have:

$h(x) = x \cdot x = x^2$

This calculation is valid only for $x \in \mathbb{R}-\{0\}$, as $g(x)$ is undefined at $x = 0$. Therefore, the quotient of the identity function by the reciprocal function is $x^2$ for all real numbers except zero.

This matches Option C:

$x^2 \forall x \in \mathbb{R}-\{0\}$

Let the equation of straight line is

$\frac{x}{a}+\frac{y}{b}=1 \quad \text{.... (i)}$

where, $a=$ length of intercept on $x$-axis. and $b=$ length of intercept on $y$-axis. Now, by condition,

$a=2 b \Rightarrow b=\frac{a}{2} \quad \text{.... (ii)}$

Also, perpendicular distance from origin to line (i) $=1$

$$\begin{aligned} & \Rightarrow \quad \frac{|0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=1 \\ & \Rightarrow \quad \frac{1}{a^2}+\frac{1}{b^2}=1 \\ & \Rightarrow \quad \frac{1}{a^2}+\frac{4}{a^2}=1 \\ & \Rightarrow \quad a^2=5 \text {, } \\ & \text { [from Eq. (ii)]. } \\ & \Rightarrow \quad a= \pm \sqrt{5} \\ & \text { and } \\ & b= \pm \frac{\sqrt{5}}{2} \\ \end{aligned}$$

Now, from Eq. (i), we get

$\frac{x}{ \pm \sqrt{5}} \pm \frac{y}{\left(\frac{\sqrt{5}}{2}\right)}=1$

$$\begin{array}{lr} \Rightarrow & \frac{x}{( \pm \sqrt{5})}+\frac{2 y}{( \pm \sqrt{5})}=1 \\ \Rightarrow & x+2 y= \pm \sqrt{5} \\ \Rightarrow & x+2 y \pm \sqrt{5}=0 \end{array}$$

Given equation of circle is

$$\begin{aligned} & (x \cos \theta+y \sin \theta-a)^2+(x \sin \theta-y \cos \theta-b)^2 =k^2 \\ & \Rightarrow x^2 \cos ^2 \theta+y^2 \sin ^2 \theta+2 x y \sin \theta \cdot \cos \theta+a^2 \\ & -2 a(x \cos \theta+y \sin \theta)+x^2 \sin ^2 \theta+y^2 \cos ^2 \theta \\ & -2 x y \sin \theta \cdot \cos \theta+b^2-2 b(x \sin \theta-y \cos \theta)=k^2 \\ & \Rightarrow x^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+y^2\left(\sin ^2 \theta+\cos ^2 \theta\right) +(-2 a \cos \theta-2 b \sin \theta) x+(-2 a \sin \theta+2 b \cos \theta) y +\left(a^2+b^2-k^2\right)=0 \\ & \Rightarrow x^2+y^2-2(a \cos \theta+b \sin \theta) x +2(-a \sin \theta+b \cos \theta) y+\left(a^2+b^2-k^2\right)=0 \quad \text{.... (i)}\\ & \left(\because \sin ^2 \theta+\cos ^2 \theta=1\right) \\ \end{aligned}$$

On comparing with

$x^2+y^2+2 g x+2 f y+c=0 \text {, we get }$

$$\begin{aligned} g & =-(a \cos \theta+b \sin \theta), \\ f & =+(-a \sin \theta+b \cos \theta) \\ \text { and } \quad c & =a^2+b^2-k^2 \end{aligned}$$

$$\begin{aligned} & \therefore \text { Radius }=\sqrt{g^2+f^2-c} \\ & =\sqrt{\begin{array}{l} (a \cos \theta+b \sin \theta)^2+(-a \sin \theta+b \cos \theta)^2 \\ -\left(a^2+b^2-k^2\right) \end{array}} \\ & =\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta+2 a b \sin \theta \cdot \cos \theta} \\ & +a^2 \sin ^2 \theta+b^2 \cos ^2 \theta-2 a b \sin \theta \cdot \cos \theta \\ & -\left(a^2+b^2-k^2\right) \\ & \Rightarrow \sqrt{a^2+b^2-a^2-b^2+k^2}=\sqrt{k^2}=k \\ \end{aligned}$$

Given, $$f(x)=\left\{\begin{array}{cc}\frac{(1-\cos 4 x)}{x^2}, & \text { if } x < 0 \\ a, & \text { if } x=0, \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}, & \text { if } x > 0\end{array}\right.$$

$\mathrm{LHL}=f(0-0)=\lim _{h \rightarrow 0} f(0-h)$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\cos 4(0-h)}{(0-h)^2} \\ & =\lim _{h \rightarrow 0} \frac{\left\{\frac{(-4 h)^2}{2 !}-\frac{(-4 h)^4}{4 !}+\ldots\right\}}{(-h)^2}\end{aligned}$$

=\lim _\limits{h \rightarrow 0}\left\{\frac{\frac{16 h^2}{2 !}-\frac{(4 h)^4}{4 !}+\ldots}{h^2}\right\}

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{16}{2 !}-\frac{4^4 h^2}{4 !}+\ldots \\ & =\frac{16}{2}=8\end{aligned}$$

$$\begin{aligned} \text { RHL }=f(0+0) & =\lim _{h \rightarrow 0} f(0+h) \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{h}}{\sqrt{16+\sqrt{h}}-4} \quad\left(\frac{0}{0} \text { form }\right) \end{aligned}$$

Using 'L' Hospital rule,

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{2 \sqrt{h}} \times \frac{1}{2 \sqrt{16+\sqrt{h}} \frac{1}{2 \sqrt{h}}} \\ & =\lim _{h \rightarrow 0} 2 \sqrt{16+\sqrt{h}}=2 \sqrt{16+0} \\ & =2 \times 4=8 \end{aligned}$$

Since, $f(x)$ is continuous at $x=0$.

$$\begin{array}{lc} \therefore & \mathrm{LHL}=\mathrm{RHL}=f(0) \\ \Rightarrow & \mathrm{a}=8 \end{array}$$

Given, $y=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots \infty=e^x$

On differentiation w.r.t. to $x$.

we get $\frac{d y}{d x}=e^x$

Given curve is

$$\begin{aligned} & 25 x^2+9 y^2-150 x-90 y+225=0 \\ \Rightarrow \quad & \left(25 x^2-150 x\right)+\left(9 y^2-90 y\right)+225=0 \\ \Rightarrow \quad & 25\left(x^2-6 x\right)+9\left(y^2-10 y\right)+225=0 \\ \Rightarrow \quad & 25\left\{x^2-6 x+9-9\right\}+9\left\{y^2-10 y+25-25\right\}+225=0 \\ \Rightarrow \quad & 25(x-3)^2-225+9(y-5)^2-225+225=0 \\ \Rightarrow \quad & 25(x-3)^2+9(y-5)^2=225 \\ \Rightarrow \quad & \frac{(x-3)^2}{9}+\frac{(y-5)^2}{25}=1 \quad[\because(b>a)] \end{aligned}$$

Which represent an ellipse.

Whose eccentricity

$e=\sqrt{\frac{b^2-a^2}{b^2}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Given equation of lines are $\mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \quad($vector form$)$

$\Rightarrow \quad \frac{x-1}{1}=\frac{y-1}{2}=\frac{z-0}{-1} \quad$ (cartesian form) .... (i)

and $\quad \mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\mu(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ (vector form)

$\Rightarrow \frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-0}{-2}$ (cartesian form) .....(ii)

If two lines are coplanar, then

$$\left|\begin{array}{ccc} 1-1 & 1-1 & 0-0 \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{array}\right|=\left|\begin{array}{ccc} 0 & 0 & 0 \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{array}\right|=0$$

Hence, given lines are coplanar lines

$$\left(\because\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ I_1 & m_1 & n_1 \\ I_2 & m_2 & n_2 \end{array}\right|=0\right)$$

Given function is, $f(x)=\sin x-\cos x$, where $x \in[0,2 \pi]$

Now, $f^{\prime}(x)=\cos x+\sin x$.

For strictly decreasing of $f(x)$; put $f^{\prime}(x)<0$

$\Rightarrow \quad \sin x+\cos x<0$

$$\begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x< 0 \\ & \Rightarrow \quad\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)< 0 \\ & \Rightarrow \quad \sin \left(x+\frac{\pi}{4}\right)< 0 \\ & \because \quad \sin x< 0 \text { when } \pi < x < 2 \pi \\ & \therefore \quad \sin \left(x+\frac{\pi}{4}\right)<0 \text { when } \pi< x+\frac{\pi}{4}<2 \pi \\ & \Rightarrow \quad \pi-\frac{\pi}{4}< x <2 \pi-\frac{\pi}{4} \\ & \Rightarrow \quad \frac{3 \pi}{4}< x <\frac{7 \pi}{4} \end{aligned}$$

Let $E=$ Event of getting 5 tickets and arranged in a ascending order.

Since, it is given that number 30 is on 3rd position. So we have to choose two numbers from 1 to 29 numbers and choose two numbers from 31 to 50

$\therefore \quad n(E)={ }^{29} C_2 \times{ }^{20} C_2$

Total number of selection of 5 cards From 1 to 50 is, $n(S)={ }^{50} \mathrm{C}_5$

$\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{{ }^{29} C_2 \times{ }^{20} C_2}{{ }^{50} C_5}$

$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left\{\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right\}$

$$\begin{aligned} & =\sin x \cdot \cos x\{\tan x+\cot x\} \\ & =\sin x \cdot \cos x\left\{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right\} \\ & =\sin x \cdot \cos x\left(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cdot \cos x}\right) \\ & =1 \quad\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right) \end{aligned}$$

$$\begin{aligned} & \text { } \frac{d}{d x}\left[x^n \cdot \log _a x \cdot e^x\right] \\ & =x^n \cdot \frac{d}{d x}\left\{\log _a x \cdot e^x\right\}+\log _a x \cdot e^x \frac{d}{d x}\left(x^n\right) \\ & =x^n\left\{\log _a x \cdot \frac{d}{d x} e^x+e^x \frac{d}{d x} \log _a x\right\} \\ & +e^x \log _a x \cdot n x^{n-1} \\ & =x^n\left\{e^x \log _a x+\frac{e^x}{x} \cdot \frac{1}{\log _e a}\right\} \\ & +n x^{n-1} \cdot e^x \log _a x \\ & =x^{n-1} e^x\left\{x \log _a x+\frac{1}{\log _e a}\right\}+n x^{n-1} \cdot e^x \log _a x \\ & =x^{n-1} \cdot e^x\left\{x \log _a x+\frac{1}{\log _e a}+n \log _a x\right\} \\ \end{aligned}$$

Given, $\quad n(A \times A)=9 \Rightarrow n(A) \times n(A)=9$ which is possible when, $n(A)=3$.

i.e., set $A$ contain only three elements.

i.e., $A=\{1,-1,0\}$

So, $A \times A$ contain $(-1,0)$ and $(0,1)$.

$$\begin{aligned} & \text { Given, } x^2 y d x-\left(x^3+y^3\right) d y=0 \\ \Rightarrow \quad & \frac{d x}{d y}=\frac{x^3+y^3}{x^2 y} \quad \text{.... (i)}\\ \end{aligned}$$

Which is a homogeneous differential equation

Now, put $x=v y\quad \text{.... (ii)}$