CUET UG Mathematics Applied Mathematics Previous Year Question Paper and Solution with PDF

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CUET UG Mathematics Applied
Previous Year Questions

CUET UG Mathematics Applied
Previous Year Questions
CUET UG Mathematics Applied

CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Mathematics PYQ

Qus : 1 CUET UG Mathematics Applied PYQ

$\displaystyle \int \frac{\pi}{x^{n+1}-x}dx=$

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Solution

Qus : 2 CUET UG Mathematics Applied PYQ

$\displaystyle \int_{0}^{1}\frac{a-bx^{2}}{\left(a+bx^{2}\right)^{2}}dx\ \text{is:}$

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Qus : 3 CUET UG Mathematics Applied PYQ

The second order derivative of which of the following functions is $5^x$?

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Solution

Qus : 4 CUET UG Mathematics Applied PYQ

The degree of the differential equation $\left(1-\left(\dfrac{dy}{dx}\right)^2\right)^{3/2}=k\dfrac{d^2y}{dx^2}$ is:

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Solution

Qus : 5 CUET UG Mathematics Applied PYQ

If $A$ and B are symmetric matrices of the same order, then AB-BA is a:

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Solution

Qus : 6 CUET UG Mathematics Applied PYQ

If A is a square matrix of order $4$ and |A|=4, then |2A| will be:

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Solution

Qus : 7 CUET UG Mathematics Applied PYQ

If $[A]_{3\times 2}[B]_{x\times y}=[C]_{3\times 1}$, then:

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Solution

Qus : 8 CUET UG Mathematics Applied PYQ

If a function $f(x)=x^{2}+bx+1$ is increasing in the interval $[1,2]$, then the least value of $b$ is:

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Solution

Qus : 9 CUET UG Mathematics Applied PYQ

Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be:

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Solution

Qus : 10 CUET UG Mathematics Applied PYQ

For the function $f(x)=2x^{3}-9x^{2}+12x-5$, $x\in[0,3]$, match List-I with List-II:

List-I

List-II

(A) Absolute maximum value

(B) Absolute minimum value

(D) Point of minima

(I) 3

(II) 0

(III) −5

(IV) 4

Choose the correct answer from the options given below:

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Solution

Qus : 11 CUET UG Mathematics Applied PYQ

The rate of change (in $\text{cm}^2/\text{s}$) of the total surface area of a hemisphere with respect to radius $r$ at $r=\sqrt[3]{1.331}\ \text{cm}$ is:

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Solution

Qus : 12 CUET UG Mathematics Applied PYQ

The area of the region bounded by the lines $\dfrac{x}{7\sqrt{3}a}+\dfrac{y}{b}=4$, $x=0$ and $y=0$ is:

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Solution

Qus : 13 CUET UG Mathematics Applied PYQ

If $A$ is a square matrix and $I$ is an identity matrix such that $A^{2}=A$, then $A(I-2A)^{3}+2A^{3}$ is equal to:

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Solution

Qus : 14 CUET UG Mathematics Applied PYQ

The value of the integral $\displaystyle \int_{\log_{e}2}^{\log_{e}3}\frac{e^{2x}-1}{e^{2x}+1},dx$ is:

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Solution

Qus : 15 CUET UG Mathematics Applied PYQ

If $\vec a,\ \vec b$ and $\vec c$ are three vectors such that $\vec a+\vec b+\vec c=\vec 0$, where $\vec a$ and $\vec b$ are unit vectors and $|\vec c|=2$, then the angle between the vectors $\vec b$ and $\vec c$ is:

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Solution

Qus : 16 CUET UG Mathematics Applied PYQ

Match List-I with List-II :

List-I	List-II
(A) $\|x-1\|+\|x-2\|$	(I) is differentiable everywhere except at x = 0
(B) $x-\|x\|$	(II)is continuous everywhere
(C) $x-[x]$	(III) is not differentiable at x = 1
(D) $x\|x\|$	(IV) is differentiable at x = 1

Choose the correct answer from the options given below:

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Solution

Qus : 17 CUET UG Mathematics Applied PYQ

Match List-I with List-II :

List-I	List-II
(A) Integrating factor of $x\,dy-(y+2x^{2})\,dx=0$	(I) $\dfrac{1}{x}$
(B) Integrating factor of $(2x^{2}-3y)\,dx=x\,dy$	(II) $x$
(C) Integrating factor of $(2y+3x^{2})\,dx+x\,dy=0$	(III) $x^{2}$
(D) Integrating factor of $2x\,dy+(3x^{3}+2y)\,dx=0$	(IV) $x^{3}$

Choose the correct answer from the options given below:

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Solution

Qus : 18 CUET UG Mathematics Applied PYQ

If the function $f:\mathbb{N}\to\mathbb{N}$ is defined as $$f(x)=\begin{cases}{n-1} & {,\, If\, \, n\, \, is\, \, even} \\ {n+1} & {,If\, n\, is\, odd}\end{cases}$$ then

(A) $f$ is injective

(B) $f$ is into

(D) $f$ is invertible

Choose the correct answer from the options given below :

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Solution

Qus : 19 CUET UG Mathematics Applied PYQ

$\displaystyle \int_{0}^{\pi/2}\frac{1-\cot x}{cosec x+\cos x},dx=$

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Solution

Qus : 20 CUET UG Mathematics Applied PYQ

If $A=\begin{bmatrix}2 & -3 \\ 3 & 5\end{bmatrix}$, then which of the following statements are correct?

A. A is a square matrix

B. $A^{-1}$ exists

C. A is a symmetric matrix

D. $|A|=19$

E. A is a null matrix

Choose the correct answer from the options given below.

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Solution

Given

$A=\begin{bmatrix}2 & -3 \ 3 & 5\end{bmatrix}$

1.Order of matrix = $2\times 2$

So, A is a square matrix → True

2.Determinant:

$|A|=(2)(5)-(-3)(3)$

$=10+9=19$

Since $|A|\neq 0$, therefore $A^{-1}$ exists → True

3.Transpose:

$A^T=\begin{bmatrix}2 & 3 \ -3 & 5\end{bmatrix}$

Since $A^T\neq A$, so A is not symmetric → False

4. $|A|=19$ → True

5. A is clearly not a null matrix → False

Correct statements: A, B, D

Qus : 21 CUET UG Mathematics Applied PYQ

If the random variable $X$ has the following distribution :

X	P(X)
0	k
1	2k
2	3k
otherwise	0

Match List-I with List-II :

List-I	List-II
(A) $k$	(I) $\dfrac{5}{6}$
(B) $P(X<2)$	(II) $\dfrac{4}{3}$
(C) $E(X)$	(III) $\dfrac{1}{2}$
(D) $P(1\le X\le 2)$	(IV) $\dfrac{1}{6}$

Choose the correct answer from the options given below :

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Solution

Qus : 22 CUET UG Mathematics Applied PYQ

Let A and B be two non-zero square matrices and AB and BA both are defined. It means

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Solution

Let A be of order $m \times m$ and B be of order $n \times n$.

For AB to be defined: number of columns of A = number of rows of B

⇒ $m = n$

For BA to be defined: number of columns of B = number of rows of A

⇒ $n = m$

Hence $m = n$, so both matrices must have same order.

Qus : 23 CUET UG Mathematics Applied PYQ

The number of all possible matrices of order $2 \times 2$ with each entry $0$ or $1$ is:

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Solution

A $2 \times 2$ matrix has $4$ entries.

Each entry can be either $0$ or $1$ (2 choices).

Total possible matrices

$=2^4=16$

Qus : 24 CUET UG Mathematics Applied PYQ

If $y=\left(\frac{1}{x}\right)^x$, then value of $e^e\left(\frac{d^2 y}{dx^2}\right)_{x=e}$ is:

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Solution

Given

$y=\left(\frac{1}{x}\right)^x$

Rewrite:

$y=x^{-x}$

Taking log:

$\ln y=-x\ln x$

Differentiate:

$\frac{1}{y}\frac{dy}{dx}=-(\ln x+1)$

So,

$\frac{dy}{dx}=-x^{-x}(\ln x+1)$

Now differentiate again:

$\frac{d^2y}{dx^2}=x^{-x}\left[(\ln x+1)^2-\frac{1}{x}\right]$

Now put $x=e$

Since $\ln e=1$

$(\ln x+1)^2=(1+1)^2=4$

So,

$\frac{d^2y}{dx^2}\Big|_{x=e}=e^{-e}\left(4-\frac{1}{e}\right)$

Multiply by $e^e$:

$e^e\left(\frac{d^2y}{dx^2}\right)_{x=e}=4-\frac{1}{e}$

Qus : 25 CUET UG Mathematics Applied PYQ

The function $f(x)=x^2-2x$ is strictly decreasing in the interval

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Solution

$f(x)=x^2-2x$

Differentiate:

$f'(x)=2x-2$

For strictly decreasing,

$f'(x)<0$

$2x-2<0$

$2x<2$

$x<1$

So function is strictly decreasing in

$(-\infty,1)$

Qus : 26 CUET UG Mathematics Applied PYQ

$\int \frac{dx}{x(x^5+3)}$ is equal to

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Solution

Let

$I=\int \frac{dx}{x(x^5+3)}$

Put $t=x^5$

Then

$dt=5x^4 dx$

Rewrite integral using substitution:

$I=\frac{1}{5}\int \frac{dt}{t(t+3)}$

Now use partial fractions:

$\frac{1}{t(t+3)}=\frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right)$

So,

$I=\frac{1}{5}\cdot\frac{1}{3}\int\left(\frac{1}{t}-\frac{1}{t+3}\right)dt$

$I=\frac{1}{15}\log\left|\frac{t}{t+3}\right|+C$

Substitute back $t=x^5$

$I=\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C$

Qus : 27 CUET UG Mathematics Applied PYQ

If $\int \frac{x^3}{x+1},dx=q(x)-\log|x+1|+C$ then $q(x)$ is equal to:

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Solution

Divide:

$\frac{x^3}{x+1}=x^2-x+1-\frac{1}{x+1}$

So,

$\int \frac{x^3}{x+1},dx=\int (x^2-x+1),dx-\int \frac{1}{x+1},dx$

$=\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|+C$

Comparing with

$q(x)-\log|x+1|+C$

We get

$q(x)=\frac{x^3}{3}-\frac{x^2}{2}+x$

Qus : 28 CUET UG Mathematics Applied PYQ

$\int_{-1}^{1}(|x-2|+|x|),dx=$

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Solution

On interval $[-1,1]$, we have $x-2<0$

So,

$|x-2|=2-x$

Now split $|x|$ at $x=0$

For $-1\le x\le 0$

$|x|=-x$

Integrand:

$(2-x-x)=2-2x$

For $0\le x\le 1$

$|x|=x$

Integrand:

$(2-x+x)=2$

Now integrate:

$\int_{-1}^{0}(2-2x),dx=[2x-x^2]_{-1}^{0}$

$=0-(-3)=3$

Next,

$\int_{0}^{1}2,dx=2$

Total value:

$3+2=5$

Qus : 29 CUET UG Mathematics Applied PYQ

If $a$ and $b$ are order and degree of differential equation $y''+(y')^2+2y=0$, then value of $2a+6b$ is:

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Solution

Given differential equation:

$y''+(y')^2+2y=0$

Highest order derivative is $y''$

So,

Order $a=2$

Power of highest order derivative $y''$ is 1

So,

Degree $b=1$

Now,

$2a+6b=2(2)+6(1)$

$=4+6$

$=10$

Qus : 30 CUET UG Mathematics Applied PYQ

The solution of the differential equation $xdy-ydx=0$ represent family of

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Solution

$xdy-ydx=0$

Divide by $dx$:

$x\frac{dy}{dx}-y=0$

So,

$x\frac{dy}{dx}=y$

$\frac{dy}{dx}=\frac{y}{x}$

Separate variables:

$\frac{dy}{y}=\frac{dx}{x}$

Integrate:

$\log|y|=\log|x|+C$

$\log\left|\frac{y}{x}\right|=C$

$\frac{y}{x}=k$

$y=kx$

This represents straight lines passing through origin.

Qus : 31 CUET UG Mathematics Applied PYQ

For differential equation $y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$, $y(0)=1$, the value of $x(e)$ is equal to:

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Solution

Given

$y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$

Divide by $dy$:

$y e^{\frac{x}{y}}\frac{dx}{dy}=x e^{\frac{x}{y}}+y^2$

Notice that

$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y e^{\frac{x}{y}}\frac{dx}{dy}-x e^{\frac{x}{y}}$

So equation becomes

$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y^2$

Integrate:

$x e^{\frac{x}{y}}=\frac{y^3}{3}+C$

Now use initial condition $y(0)=1$

So at $x=0$, $y=1$

$0\cdot e^{0}= \frac{1}{3}+C$

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

Thus,

$x e^{\frac{x}{y}}=\frac{y^3-1}{3}$

Now put $y=e$

$x e^{\frac{x}{e}}=\frac{e^3-1}{3}$

It satisfies $x=1$

Qus : 32 CUET UG Mathematics Applied PYQ

$\int_{-1}^{1} e^{|x|},dx=$

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Solution

Since $|x|$ is symmetric,

$\int_{-1}^{1} e^{|x|},dx=2\int_{0}^{1} e^x,dx$

$=2[e^x]_0^1$

$=2(e-1)$

Qus : 33 CUET UG Mathematics Applied PYQ

For two events $A, B$

$P(A\cup B)=\frac{7}{12}, ; P(A)=\frac{5}{12}, ; P(B)=\frac{3}{12}$

Then $P(A\cap B)=$

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Solution

We know,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Substitute values:

$\frac{7}{12}=\frac{5}{12}+\frac{3}{12}-P(A\cap B)$

$\frac{7}{12}=\frac{8}{12}-P(A\cap B)$

So,

$P(A\cap B)=\frac{8}{12}-\frac{7}{12}$

$=\frac{1}{12}$

Qus : 34 CUET UG Mathematics Applied PYQ

The probability distribution of $X$ is:

x	0	1	2	3	4
p(X= x)	0.1	2k	k	k	2k

then var(X) =

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Solution

Since total probability = 1

$0.1+2k+k+k+2k=1$

$0.1+6k=1$

$6k=0.9$

$k=0.15$

Now probabilities:

$P(0)=0.1$

$P(1)=0.3$

$P(2)=0.15$

$P(3)=0.15$

$P(4)=0.3$

Now,

$E(X)=0(0.1)+1(0.3)+2(0.15)+3(0.15)+4(0.3)$

$=0+0.3+0.3+0.45+1.2$

$=2.25$

Now,

$E(X^2)=0+1(0.3)+4(0.15)+9(0.15)+16(0.3)$

$=0.3+0.6+1.35+4.8$

$=7.05$

Variance:

$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$

$=7.05-(2.25)^2$

$=7.05-5.0625$

$=1.9875$

$= \frac{159}{80}$

Qus : 35 CUET UG Mathematics Applied PYQ

The number at unit place of number $17^{123}$ is:

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Solution

Unit digit depends on $7^{123}$

Cycle of unit digits of powers of 7:

$7^1=7$

$7^2=49$ → 9

$7^3=343$ → 3

$7^4=2401$ → 1

Cycle repeats every 4.

Now,

$123 \mod 4$

$123=4\times 30+3$

Remainder = 3

So unit digit same as $7^3$

Unit digit = 3

Qus : 36 CUET UG Mathematics Applied PYQ

The maximum value of $z=4x+2y$ subject to constraints

$2x+3y \le 28$,

$x+y \le 10$,

$x,y \ge 0$ is:

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Solution

Corner points of feasible region:

$(0,0)$

From $x+y=10$

If $x=0$, $y=10$ → but check first constraint:

$2(0)+3(10)=30>28$ ❌ not feasible

So take intersection of

$2x+3y=28$

$x+y=10$

Solve:

$y=10-x$

Substitute:

$2x+3(10-x)=28$

$2x+30-3x=28$

$-x=-2$

$x=2$

Then

$y=10-2=8$

So intersection point $(2,8)$

Now check intercepts:

If $y=0$

From $x+y=10$ → $x=10$

Check first constraint:

$2(10)+0=20\le 28$ ✔

So point $(10,0)$ is feasible.

Now evaluate $z=4x+2y$

At $(0,0)$ → $0$

At $(2,8)$ → $4(2)+2(8)=8+16=24$

At $(10,0)$ → $40$

Maximum value = 40

Qus : 37 CUET UG Mathematics Applied PYQ

The system of linear inequalities $2x-1 \ge 3$ and $x-3>5$ has solution:

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Solution

First inequality:

$2x-1 \ge 3$

$2x \ge 4$

$x \ge 2$

Second inequality:

$x-3>5$

$x>8$

Common solution:

$x>8$

So solution set is

$(8,\infty)$

Qus : 38 CUET UG Mathematics Applied PYQ

The values of $x$ which satisfied $|3x| \ge |6-3x|$

A. $(0,1]$

B. $[1,4]$

C. $(4,\infty)$

D. $(-1,0)$

E. $(-\infty,0)$

Choose the correct answer from the options given below:

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Solution

Given

$|3x| \ge |6-3x|$

$3|x| \ge 3|2-x|$

Divide by 3:

$|x| \ge |2-x|$

Square both sides:

$x^2 \ge (2-x)^2$

$x^2 \ge 4-4x+x^2$

Cancel $x^2$:

$0 \ge 4-4x$

$4x \ge 4$

$x \ge 1$

So solution set:

$[1,\infty)$

From given intervals:

B: $[1,4]$ ✔

C: $(4,\infty)$ ✔

Qus : 39 CUET UG Mathematics Applied PYQ

If $A=\begin{bmatrix}x & y & z \\ 2 & u & v \\ -1 & 6 & w\end{bmatrix}$ is skew symmetric matrix, then value of $x^2+y^2+z^2+u^2+v^2+w^2$ is:

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Solution

For skew symmetric matrix,

$A^T=-A$

So diagonal elements must be zero:

$x=0,; u=0,; w=0$

Now compare off-diagonal elements:

From $(1,2)$ and $(2,1)$:

$y=-2$

From $(1,3)$ and $(3,1)$:

$z=1$

From $(2,3)$ and $(3,2)$:

$v=-6$

Now compute:

$x^2+y^2+z^2+u^2+v^2+w^2$

$=0^2+(-2)^2+1^2+0^2+(-6)^2+0^2$

$=4+1+36$

$=41$

Qus : 40 CUET UG Mathematics Applied PYQ

If $y=e^{nx}$, then $n^{\text{th}}$ derivative of $y$ is:

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Solution

Given

$y=e^{nx}$

First derivative:

$\frac{dy}{dx}=n e^{nx}$

Second derivative:

$\frac{d^2y}{dx^2}=n^2 e^{nx}$

Continuing this pattern,

$n^{\text{th}}$ derivative:

$\frac{d^ny}{dx^n}=n^n e^{nx}$

Since $y=e^{nx}$,

$\frac{d^ny}{dx^n}=n^n y$

Qus : 41 CUET UG Mathematics Applied PYQ

The total revenue (in Rs.) received by selling $x$ units of a certain product is given by: $R(x)=4x^2+10x+3$ What is the marginal revenue on selling $20$ such units?

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Solution

Marginal revenue

$=\frac{dR}{dx}$

Differentiate:

$\frac{dR}{dx}=8x+10$

Now put $x=20$

$=8(20)+10$

$=160+10$

$=170$

Qus : 42 CUET UG Mathematics Applied PYQ

If $x$ is a real, then minimum value of $x^2-8x+17$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given

$f(x)=x^2-8x+17$

Complete the square:

$x^2-8x+17=(x^2-8x+16)+1$

$=(x-4)^2+1$

Since $(x-4)^2 \ge 0$

Minimum value occurs when $(x-4)^2=0$

So minimum value = $1$

Qus : 43 CUET UG Mathematics Applied PYQ

If $\mu$ is mean of random variable $X$, with probability distribution:

x	0	1	2
P(x = x)	4/9	4/9	1/9

then value of $9\mu+4$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Mean:

$\mu=\sum xP(X=x)$

$=0\cdot\frac{4}{9}+1\cdot\frac{4}{9}+2\cdot\frac{1}{9}$

$=\frac{4}{9}+\frac{2}{9}$

$=\frac{6}{9}$

$=\frac{2}{3}$

Now,

$9\mu+4$

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

$=6+4$

$=10$

Qus : 44 CUET UG Mathematics Applied PYQ

In a game, a child will win Rs $5$ if he gets all heads or all tails when three coins are tossed simultaneously and he will lose Rs $3$ for all other cases.

The expected amount to lose in the game is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Total outcomes when 3 coins tossed:

$2^3=8$

Favourable (all heads or all tails):

HHH, TTT → 2 outcomes

So,

$P(\text{win})=\frac{2}{8}=\frac{1}{4}$

$P(\text{lose})=\frac{6}{8}=\frac{3}{4}$

Expected value:

$E=5\left(\frac{1}{4}\right)+(-3)\left(\frac{3}{4}\right)$

$=\frac{5}{4}-\frac{9}{4}$

$=-\frac{4}{4}$

$=-1$

So expected loss = Rs. 1

Qus : 45 CUET UG Mathematics Applied PYQ

The probability mass function of Random variable $X$ is:

$P(X=x)=(0.6)^x(0.4)^{1-x}, ; x=0,1$

The variance of $X$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

For $x=0$:

$P(0)=(0.6)^0(0.4)^1=0.4$

For $x=1$:

$P(1)=(0.6)^1(0.4)^0=0.6$

So $X$ follows Bernoulli distribution with

$p=0.6$

Variance of Bernoulli:

$\mathrm{Var}(X)=p(1-p)$

$=0.6(0.4)$

$=0.24$

Qus : 46 CUET UG Mathematics Applied PYQ

Match List I with List II

List I	List II
A. Quantity index	I. Measures relative price change over a period of time.
B. Time series	II. Measures change in quantity of consumption of goods over a specific period of time.
C. Price index	III. Measures average value of goods for specific time period.
D. Value index	IV. Statistical observation taken at different points of time for specific period of time.

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Quantity index → Measures change in quantity over time

So A → II

Time series → Statistical observations at different time points

So B → IV

Price index → Measures relative price change over time

So C → I

Value index → Measures average value for specific time period

So D → III

Correct matching:

A-II, B-IV, C-I, D-III

Qus : 47 CUET UG Mathematics Applied PYQ

Given that $\sum p_0 q_0 = 700$, $\sum p_0 q_1 = 1450$, $\sum p_1 q_0 = 855$, $\sum p_1 q_1 = 1300$. Where subscripts $0$ and $1$ are used for base year and current year respectively. The Laspeyres’ price index number is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Laspeyres price index:

$L=\frac{\sum p_1 q_0}{\sum p_0 q_0}\times 100$

Substitute values:

$L=\frac{855}{700}\times 100$

$=1.2214\times 100$

$=122.14$

Qus : 48 CUET UG Mathematics Applied PYQ

If $y=a+b(x-2005)$ fits the time series data $x$ (year): $2003,; 2004,; 2005,; 2006,; 2007$ $y$ (yield in tons): $6,; 13,; 17,; 20,; 24$ Then the value of $a+b$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Let

$u=x-2005$

Then values of $u$:

$-2,; -1,; 0,; 1,; 2$

Corresponding $y$:

$6,; 13,; 17,; 20,; 24$

Now,

$\sum y=6+13+17+20+24=80$

$\sum u=0$

Since $\sum u=0$

$a=\frac{\sum y}{n}$

$=\frac{80}{5}$

$=16$

Now,

$\sum uy=(-2)(6)+(-1)(13)+0(17)+1(20)+2(24)$

$=-12-13+0+20+48$

$=43$

$\sum u^2=4+1+0+1+4=10$

$b=\frac{\sum uy}{\sum u^2}$

$=\frac{43}{10}$

$=4.3$

Now,

$a+b=16+4.3$

$=20.3$

Qus : 49 CUET UG Mathematics Applied PYQ

Which of the following statements are correct?

A. If discount rate $>$ coupon rate, then present value of a bond $>$ face value

B. An annuity in which the periodic payment begins on a fixed date and continues forever is called perpetuity

C. The issuer of bond pays interest at fixed interval at fixed rate of interest to investor is called coupon payment

D. A sinking fund is a fixed payment made by a borrower to a lender at a specific date every month to clear off the loan

E. The issuer of bond repays the principal i.e. face value of the bond to the investor at a later date termed as maturity date

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

A. If discount rate $>$ coupon rate, bond sells at discount

So present value $<$ face value ❌ False

B. Payment continues forever → Perpetuity ✔ True

C. Fixed periodic interest paid by issuer → Coupon payment ✔ True

D. Sinking fund is periodic deposit to accumulate money, not direct monthly loan repayment ❌ False

E. Principal repaid at maturity date ✔ True

Correct statements: B, C, E

Qus : 50 CUET UG Mathematics Applied PYQ

Which of the following statements is true?

A. In flat rate method,

$\text{EMI}=\frac{\text{Principal}+\text{Interest}}{\text{Number of Payments}}$

B. In reducing balance method,

$\text{EMI}=P\times\frac{i}{1-(1+i)^{-n}}$

C. In sinking fund, a fixed amount at regular intervals is deposited.

D. Approximate Yield to Maturity

$\text{YTM}=\frac{\text{Coupon Payment}+\frac{\text{Face Value}-\text{Present Value}}{\text{Number of Payments}}}{\frac{\text{Face Value}+\text{Present Value}}{2}}$

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

A ✔ Correct formula for flat rate EMI

B ✔ Correct reducing balance EMI formula

C ✔ Definition of sinking fund

D ✔ Standard approximate YTM formula

Hence all statements are true.

Qus : 51 CUET UG Mathematics Applied PYQ

Mr. Dev wishes to purchase an AC for Rs. $45,000$ with a down payment of Rs. $5,000$ and balance in EMI for 5 years. If Bank charges $6%$ per annum compounded monthly, then monthly EMI is (use $\frac{0.005}{1-(1.005)^{-60}}=0.0194$)

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Loan amount

$=45000-5000=40000$

Monthly interest rate

$=\frac{6%}{12}=0.5%=0.005$

Number of months

$=5\times 12=60$

EMI formula:

$\text{EMI}=P\times\frac{i}{1-(1+i)^{-n}}$

Given factor:

$\frac{0.005}{1-(1.005)^{-60}}=0.0194$

So,

$\text{EMI}=40000\times 0.0194$

$=776$

Qus : 52 CUET UG Mathematics Applied PYQ

The cost of a machine is Rs. $20,000$ and its estimated useful life is $10$ years. The scrap value of the machine, when its value depreciates at $10%$ p.a., is (use $(0.9)^{10}=0.35$)

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Solution

$\text{Value after } n \text{ years}=P(1-r)^n$

Here,

$P=20000$

$r=0.1$

$n=10$

Value

$=20000(0.9)^{10}$

Given

$(0.9)^{10}=0.35$

So,

$=20000\times 0.35$

$=7000$

Qus : 53 CUET UG Mathematics Applied PYQ

One of the following is true for relation between sample mean $\bar{x}$ and population mean $\mu$.

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

By Law of Large Numbers, as sample size increases, sample mean $\bar{x}$ approaches population mean $\mu$. Therefore, $|\bar{x}-\mu|$ decreases as sample size increases.

Qus : 54 CUET UG Mathematics Applied PYQ

Below are the stages for Drawing statistical inferences.

A. Sample

B. Population

C. Making Inference

D. Data tabulation

E. Data Analysis

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Correct logical order:

Population → Sample → Data tabulation → Data Analysis → Making Inference

So sequence:

B → A → D → E → C

Qus : 55 CUET UG Mathematics Applied PYQ

Corner points of the feasible region for an LPP are

$(0,2),; (3,0),; (6,0)$ and $(6,8)$.

If $z=2x+3y$ is the objective function of LPP then

$\max(z)-\min(z)$ is equal to:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Evaluate $z=2x+3y$ at each corner point:

At $(0,2)$

$z=2(0)+3(2)=6$

At $(3,0)$

$z=2(3)+3(0)=6$

At $(6,0)$

$z=2(6)+0=12$

At $(6,8)$

$z=2(6)+3(8)=12+24=36$

So,

$\max(z)=36$

$\min(z)=6$

Therefore,

$\max(z)-\min(z)=36-6=30$

Qus : 56 CUET UG Mathematics Applied PYQ

Sitaram, a money lender lent a part of Rs. $200000$ to Shyam at simple interest $6%$ p.a. and the remaining to Sushil at $10%$ p.a. at simple interest.

Sitaram earned an annual interest income of Rs. $18000$.

What is the mean rate of interest?

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Total principal = Rs. $200000$

Total annual interest earned = Rs. $18000$

Mean rate of interest:

$\text{Mean Rate}=\frac{\text{Total Interest}}{\text{Total Principal}}\times 100$

$=\frac{18000}{200000}\times 100$

$=0.09\times 100$

$=9%$

Qus : 57 CUET UG Mathematics Applied PYQ

A cable network provider in a small town has $500$ subscribers and he used to collect Rs. $300$ per month from each subscriber. He proposes to increase the monthly charges and it is believed that for every increase of Rs. $1$, one subscriber will discontinue the service. If Rs. $x$ is the monthly increase in subscription amount, then the number of subscribers are:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Initially subscribers = $500$

For every Rs. $1$ increase → 1 subscriber leaves.

So for increase of Rs. $x$ → $x$ subscribers leave.

Remaining subscribers

$=500-x$

Qus : 58 CUET UG Mathematics Applied PYQ

Based on the above information, total revenue $R$ (in Rs.) is given by:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

New monthly charge

$=300+x$

Number of subscribers

$=500-x$

Total revenue

$R=(300+x)(500-x)$

Qus : 59 CUET UG Mathematics Applied PYQ

A cable network provider in a small town has $500$ subscribers and he used to collect Rs. $300$ per month from each subscriber. He proposes to increase the monthly charges and it is believed that for every increase of Rs. $1$, one subscriber will discontinue the service. The number of subscribers which gives the maximum revenue is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From previous question,

Revenue:

$R=(300+x)(500-x)$

Expand:

$R=150000+200x-x^2$

This is a quadratic function:

$R=-x^2+200x+150000$

Maximum occurs at

$x=\frac{-b}{2a}$

Here,

$a=-1,; b=200$

$x=\frac{-200}{2(-1)}$

$x=100$

So number of subscribers:

$500-x=500-100=400$

Qus : 60 CUET UG Mathematics Applied PYQ

What is increase in charges per subscriber that yields maximum revenue?

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From previous calculation,

Maximum occurs at

$x=100$

So increase in charges per subscriber = Rs. $100$

Qus : 61 CUET UG Mathematics Applied PYQ

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Solution

Revenue function:

$R=(300+x)(500-x)$

Expand:

$R=150000+200x-x^2$

This quadratic is maximum at

$x=100$

Now substitute:

$R=(300+100)(500-100)$

$R=400\times 400$

$R=160000$

Qus : 62 CUET UG Mathematics Applied PYQ

Match List I with List II

List I	List II
A. $R={(x,y)\mid x \text{ and } y \text{ are students of the same school}}$	I. Symmetric
B. $R={(L_1,L_2)\mid L_1 \perp L_2,; L_1,L_2 \in L}$	II. one-one
C. A function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=2-3x$	III. bijective
D. A function $f:[0,1]\to\mathbb{R}$ defined by $f(x)=1+x^2$	IV. Equivalence

Choose the correct answer from the options given below:

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Solution

Qus : 63 CUET UG Mathematics Applied PYQ

$\tan^{-1}\left[2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]$

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$ $2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ $\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$ $2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\sqrt{3}$ $\tan^{-1}\left[2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]=\frac{\pi}{3}$

Qus : 64 CUET UG Mathematics Applied PYQ

Match List I with List II

List I	List II
A. The range of $\sin^{-1}x$ is	I. $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-{0}$
B. The range of $\tan^{-1}x$ is	II. $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
C. The range of $\csc^{-1}x$ is	III. $[0,\pi] - \{\frac{\pi}{2}\}$
D. The range of $\sec^{-1}x$ is	IV. $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Range of $\sin^{-1}x$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$ So A → IV Range of $\tan^{-1}x$ is $(-\frac{\pi}{2},\frac{\pi}{2})$ So B → II Range of $\csc^{-1}x$ is $[-\frac{\pi}{2},\frac{\pi}{2}] - {0}$ So C → I Range of $\sec^{-1}x$ is $[0,\pi] - {\frac{\pi}{2}}$ So D → III

Qus : 65 CUET UG Mathematics Applied PYQ

If $A=\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ then:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Transpose of $A$ is: $A'=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}$ Now, $A'A=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{bmatrix}$ $A'A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha & 0 \\ 0 & \sin^2\alpha+\cos^2\alpha\end{bmatrix}$ Since $\sin^2\alpha+\cos^2\alpha=1$ $A'A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I$

Qus : 66 CUET UG Mathematics Applied PYQ

If $A=\begin{bmatrix}2 & 1 & 0 \\ 3 & 1 & 2 \\ 0 & 4 & -1\end{bmatrix}$, then $|adj(A)|$ is equal to:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

First find $|A|$. $|A|=\begin{vmatrix}2 & 1 & 0 \\ 3 & 1 & 2 \\ 0 & 4 & -1\end{vmatrix}$ Expand along first row: $|A|=2\begin{vmatrix}1 & 2 \\ 4 & -1\end{vmatrix}-1\begin{vmatrix}3 & 2 \\ 0 & -1\end{vmatrix}$ $=2(1(-1)-2\times4)-1(3(-1)-0)$ $=2(-1-8)-(-3)$ $=-18+3$ $=-15$ Property: $|adj(A)|=|A|^{n-1}$ Here $n=3$. $|adj(A)|=(-15)^{2}=225$

Qus : 67 CUET UG Mathematics Applied PYQ

Identify the correct option(s):

A. A modulus function is continuous at every point in its domain.

B. A modulus function may or may not be continuous at every point in its domain.

C. Every rational function is continuous in its domain.

D. If a function $f$ is differentiable at a point then it is also continuous at that point.

E. If a function $f$ is continuous at a point then it is also differentiable at that point.

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

A. A modulus function $|x|$ is continuous at every point in its domain.

Hence A is true.

B. Since modulus function is continuous everywhere in its domain, this statement is false.

C. A rational function $\frac{p(x)}{q(x)}$ is continuous at every point where $q(x)\neq0$.

Thus it is continuous in its domain. Hence C is true.

D. If a function is differentiable at a point, then it must be continuous at that point.

Hence D is true.

E. Continuity does not necessarily imply differentiability (example: $f(x)=|x|$ at $x=0$).

Hence E is false.

Correct statements are A, C and D.

Qus : 68 CUET UG Mathematics Applied PYQ

If $f(x)= \begin{cases} \frac{\sqrt{1-\cos2x}}{x\sqrt2}, & x\ne0 \\ k, & x=0 \end{cases}$ then the value of $k$ will make function $f$ continuous at $x=0$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

For continuity at $x=0$,

$\lim_{x\to0}f(x)=f(0)=k$

So evaluate limit:

$\lim_{x\to0}\frac{\sqrt{1-\cos2x}}{x\sqrt2}$

Using identity:

$1-\cos2x=2\sin^2x$

So,

$\sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt2|\sin x|$

Hence,

$\frac{\sqrt{1-\cos2x}}{x\sqrt2}=\frac{\sqrt2|\sin x|}{x\sqrt2}=\frac{|\sin x|}{x}$

As $x\to0$,

$\frac{|\sin x|}{x}\to1$

Therefore,

$k=1$

Qus : 69 CUET UG Mathematics Applied PYQ

If $y=\log(\sec e^{x^2})$, then $\frac{dy}{dx}=$

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Solution

$y=\log(\sec(e^{x^2}))$

Using property:

$\frac{d}{dx}\log(\sec u)=\tan u\cdot\frac{du}{dx}$

Let $u=e^{x^2}$

Then,

$\frac{du}{dx}=2xe^{x^2}$

Hence,

$\frac{dy}{dx}=2xe^{x^2}\tan(e^{x^2})$

Qus : 70 CUET UG Mathematics Applied PYQ

If $y = e^{\log(\sin^{-1}(x))} + e^{\log(\cos^{-1}(x))},\; 0 < x < 1$ then

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Using identity

$e^{\log a}=a$

$y=\sin^{-1}(x)+\cos^{-1}(x)$

But

$\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$

Hence

$y=\frac{\pi}{2}$

Therefore

$\frac{dy}{dx}=0$

Qus : 71 CUET UG Mathematics Applied PYQ

Match List I with List II

List I	List II
A. $\int \frac{1}{x+\sqrt{x}}\,dx$	I. $2\sqrt{x}+C$
B. $\int \frac{e^{\log\sqrt{x}}}{x}\,dx$	II. $2(\sqrt{x}-1)e^{\sqrt{x}}+C$
C. $\int \frac{dx}{4x^2-9}$	III. $2\log(\sqrt{x}+1)+C$
D. $\int e^{\sqrt{x}}\,dx$	IV. $\frac{1}{12}\log\left(\frac{2x-3}{2x+3}\right)+C$

Choose the correct answer from the options given below:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

A. $\int \frac{1}{x+\sqrt{x}}dx$ Let $t=\sqrt{x}$ $x=t^2,\quad dx=2t\,dt$ Integral becomes $\int \frac{2t}{t^2+t}dt =2\int \frac{1}{t+1}dt =2\log(t+1)+C =2\log(\sqrt{x}+1)+C$ So A → III B. $\int \frac{e^{\log\sqrt{x}}}{x}dx$ Since $e^{\log\sqrt{x}}=\sqrt{x}$ $\int \frac{\sqrt{x}}{x}dx =\int x^{-1/2}dx =2\sqrt{x}+C$ So B → I C. $\int \frac{dx}{4x^2-9}$ Using standard formula $\int \frac{dx}{a^2x^2-b^2} =\frac{1}{2ab}\log\left|\frac{ax-b}{ax+b}\right|+C$ Here $a=2,\ b=3$ Result $\frac{1}{12}\log\left(\frac{2x-3}{2x+3}\right)+C$ So C → IV D. $\int e^{\sqrt{x}}dx$ Let $t=\sqrt{x}$ $x=t^2,\ dx=2t\,dt$ Integral $\int e^{\sqrt{x}}dx =2\int te^t dt$ Using integration by parts $2(t-1)e^t+C$ Substitute $t=\sqrt{x}$ $2(\sqrt{x}-1)e^{\sqrt{x}}+C$ So D → II

Qus : 72 CUET UG Mathematics Applied PYQ

The order of the differential equation whose general solution is

$y = e^x (a\cos x + b\sin x)$, where $a$ and $b$ are arbitrary constants is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given solution:

$y = e^x (a\cos x + b\sin x)$

Here the solution contains **two arbitrary constants** $a$ and $b$.

The order of a differential equation is equal to the number of arbitrary constants in its general solution.

Therefore,

Order $=2$

Qus : 73 CUET UG Mathematics Applied PYQ

$\frac{d}{dx}\left[\int_{0}^{2a} f(\sin 2x)\,dx\right] =$

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Given solution:

$y = e^x (a\cos x + b\sin x)$

Here the solution contains **two arbitrary constants** $a$ and $b$.

The order of a differential equation is equal to the number of arbitrary constants in its general solution.

Therefore,

Observe that the integral

$\int_{0}^{2a} f(\sin 2x)\,dx$

does **not depend on $x$** because the limits are constants.

Hence the entire expression is a constant.

Derivative of a constant is zero.

Therefore,

$\frac{d}{dx}\left[\int_{0}^{2a} f(\sin 2x)\,dx\right] = 0$

Order $=2$

Qus : 74 CUET UG Mathematics Applied PYQ

$\int \tan x(\sec x - \tan x)\,dx =$

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Solution

$\int \tan x(\sec x - \tan x)dx$ Expand the expression $=\int (\sec x\tan x - \tan^2 x)dx$ Split the integral $=\int \sec x\tan x\,dx - \int \tan^2 x\,dx$ We know $\frac{d}{dx}(\sec x)=\sec x\tan x$ So $\int \sec x\tan x\,dx = \sec x$ Also $\tan^2 x = \sec^2 x -1$ Therefore $\int \tan^2 x\,dx = \int (\sec^2 x -1)dx = \tan x - x$ Hence $\int \tan x(\sec x-\tan x)dx = \sec x - (\tan x - x)$ $= \sec x - \tan x + x + C$

Qus : 75 CUET UG Mathematics Applied PYQ

If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of vector $\vec a$, then value of $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ is equal to:

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Solution

For direction cosines $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ Using identity $\cos 2\theta = 2\cos^2\theta -1$ So $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ $=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$ $=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3$ $=2(1)-3$ $=-1$

Qus : 76 CUET UG Mathematics Applied PYQ

The value of$\mathbf{i}\cdot(\mathbf{j}\times\mathbf{k})+\mathbf{j}\cdot(\mathbf{i}\times\mathbf{k})+\mathbf{k}\cdot(\mathbf{i}\times\mathbf{j})$is:

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Solution

Using vector identities: $\mathbf{i}\times\mathbf{j}=\mathbf{k}$ $\mathbf{j}\times\mathbf{k}=\mathbf{i}$ $\mathbf{k}\times\mathbf{i}=\mathbf{j}$ First term $\mathbf{i}\cdot(\mathbf{j}\times\mathbf{k})$ $=\mathbf{i}\cdot\mathbf{i}=1$ Second term $\mathbf{j}\cdot(\mathbf{i}\times\mathbf{k})$ But $\mathbf{i}\times\mathbf{k}=-\mathbf{j}$ So $\mathbf{j}\cdot(-\mathbf{j})=-1$ Third term $\mathbf{k}\cdot(\mathbf{i}\times\mathbf{j})$ $=\mathbf{k}\cdot\mathbf{k}=1$ Adding $1-1+1=1$

Qus : 77 CUET UG Mathematics Applied PYQ

The corner points of the feasible region for an L.P.P are

$(2,0),(7,0),(4,5),(0,3)$

and

$z=2x+3y$

is the objective function.

The difference of the maximum and minimum values of $z$ is:

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Solution

Evaluate $z=2x+3y$ at each corner point. At $(2,0)$ $z=2(2)+3(0)=4$ At $(7,0)$ $z=2(7)+3(0)=14$ At $(4,5)$ $z=2(4)+3(5)=8+15=23$ At $(0,3)$ $z=2(0)+3(3)=9$ Maximum value of $z=23$ Minimum value of $z=4$ Difference $23-4=19$

Qus : 78 CUET UG Mathematics Applied PYQ

The area of the parallelogram whose adjacent sides are $(\mathbf{i}+\mathbf{k})$ and $(2\mathbf{i}+\mathbf{j}+\mathbf{k})$ is:

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Solution

Area of parallelogram $= |\vec a \times \vec b|$ Let $\vec a = (1,0,1)$ $\vec b = (2,1,1)$ Cross product $\vec a \times \vec b = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}$ $= \mathbf{i}(0\cdot1-1\cdot1) -\mathbf{j}(1\cdot1-1\cdot2) +\mathbf{k}(1\cdot1-0\cdot2)$ $= -\mathbf{i}+\mathbf{j}+\mathbf{k}$ Magnitude $|\vec a \times \vec b| =\sqrt{(-1)^2+1^2+1^2}$ $=\sqrt{3}$

Qus : 79 CUET UG Mathematics Applied PYQ

If $x(\mathbf{i}+\mathbf{j}+\mathbf{k})$ is a unit vector then value of $x$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Vector $\vec v = x(\mathbf{i}+\mathbf{j}+\mathbf{k})$ Magnitude of unit vector $|\vec v|=1$ $|x(\mathbf{i}+\mathbf{j}+\mathbf{k})| =|x|\sqrt{1^2+1^2+1^2}$ $=|x|\sqrt{3}$ Since magnitude is 1 $|x|\sqrt{3}=1$ $|x|=\frac{1}{\sqrt{3}}$ Therefore $x=\pm\frac{1}{\sqrt{3}}$

Qus : 80 CUET UG Mathematics Applied PYQ

The point of intersection of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ is:

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Solution

First line Let parameter $t$ $x=1+2t$ $y=2+3t$ $z=3+4t$ Second line Let parameter $s$ $\frac{x-4}{5}=\frac{y-1}{2}=z=s$ So $x=4+5s$ $y=1+2s$ $z=s$ For intersection $3+4t=s$ Substitute in $x$ $1+2t=4+5(3+4t)$ $1+2t=19+20t$ $-18=18t$ $t=-1$ Then $s=3+4(-1)=-1$ Point $x=1+2(-1)=-1$ $y=2+3(-1)=-1$ $z=3+4(-1)=-1$

Qus : 81 CUET UG Mathematics Applied PYQ

The distance between the point $(3,4,5)$ and the point where the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ meets the plane $x+y+z=17$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Line in parametric form Let parameter $t$ $x=3+t$ $y=4+2t$ $z=5+2t$ Substitute in plane $x+y+z=17$ $(3+t)+(4+2t)+(5+2t)=17$ $12+5t=17$ $t=1$ Point of intersection $x=4$ $y=6$ $z=7$ Distance from $(3,4,5)$ $d=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}$ $d=\sqrt{1+4+4}$ $d=3$

Qus : 82 CUET UG Mathematics Applied PYQ

If events $A$ and $B$ are independent, then identify the correct statements

(A) $A$ and $B$ must be mutually exclusive

(B) The sum of their probabilities must be equal to $1$

(D) $A'$ and $B'$ are also independent

Choose the correct answer from the options given below:

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Solution

For independent events

$P(A\cap B)=P(A)P(B)$

So statement (C) is **true**.

Independent events need not be mutually exclusive.

Therefore (A) is **false**.

Also $P(A)+P(B)$ need not be $1$.

Hence (B) is **false**.

If $A$ and $B$ are independent then their complements $A'$ and $B'$ are also independent.

Hence (D) is **true**.

Correct statements: **C and D**

Qus : 83 CUET UG Mathematics Applied PYQ

The equation of plane passing through the point $(0,7,-7)$ and containing the line

$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$

is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

The line

$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$

gives a point on the line

$(-1,3,-2)$

Direction vector of line

$\vec d=(-3,2,1)$

Vector joining given point and line point

$(0,7,-7)-(-1,3,-2)=(1,4,-5)$

Normal vector of plane

$\vec n = \vec d \times (1,4,-5)$

$\vec n=

\begin{vmatrix}

\mathbf{i} & \mathbf{j} & \mathbf{k} \\

-3 & 2 & 1 \\

1 & 4 & -5

\end{vmatrix}$

$= -14\mathbf{i}-14\mathbf{j}-14\mathbf{k}$

Normal vector $\propto (1,1,1)$

Plane equation

$x+y+z+d=0$

Substitute point $(0,7,-7)$

$0+7-7+d=0$

$d=0$

Plane

$x+y+z=0$

Qus : 84 CUET UG Mathematics Applied PYQ

If $A$ and $B$ are two independent events with

$P(A)=\frac{3}{5}$ and $P(B)=\frac{4}{9}$, then

$P(A' \cap B')$ is equal to

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Solution

Since $A$ and $B$ are independent,

$P(A\cap B)=P(A)P(B)$

Now

$P(A') = 1 - P(A)$

$P(B') = 1 - P(B)$

$P(A') = 1-\frac{3}{5}=\frac{2}{5}$

$P(B') = 1-\frac{4}{9}=\frac{5}{9}$

For independent events, complements are also independent.

Hence

$P(A'\cap B') = P(A')P(B')$

$=\frac{2}{5}\times\frac{5}{9}$

$=\frac{2}{9}$

Qus : 85 CUET UG Mathematics Applied PYQ

A line

$\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-1}{-1}$

is perpendicular to a plane $(P)$ which passes through the point $(4,3,9)$.

If the mirror image of point $S$ on the line $(L)$ in the given plane $(P)$ is $(2,3,1)$, then coordinates of point $S$ is:

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Solution

Direction ratios of line

$\vec n=(1,2,-1)$

Since the line is perpendicular to the plane, this vector is the **normal vector of the plane**.

Plane passing through $(4,3,9)$

Equation

$1(x-4)+2(y-3)-1(z-9)=0$

$x+2y-z-1=0$

Mirror image property:

The plane is the midpoint between $S$ and its image.

Image point

$S'=(2,3,1)$

Parametric form of line

$x=2+t$

$y=3+2t$

$z=1-t$

So any point on the line

$S=(2+t,3+2t,1-t)$

Midpoint between $S$ and $S'$

$\left(\frac{2+t+2}{2},\frac{3+2t+3}{2},\frac{1-t+1}{2}\right)$

$=\left(2+\frac{t}{2},3+t,1-\frac{t}{2}\right)$

Substitute into plane

$x+2y-z-1=0$

$\left(2+\frac{t}{2}\right)+2(3+t)-\left(1-\frac{t}{2}\right)-1=0$

$6+3t=0$

$t=-2$

Hence

$S=(2-2,3-4,1+2)$

$S=(0,-1,3)$

Qus : 86 CUET UG Mathematics Applied PYQ

A biased dice is thrown once. If $X$ denotes the number appearing on it and the probability distribution is

$x:$	1	2	3	4	5	6
$P(X=x):$	k	k/2	2k	$8k^2$	1-5k	$\frac{k}{2}$

where $k>0$.

Then consider the following statements:

A. $P(X=3)$

B. $P(X\le2)$

C. $P(X\ge5)$

D. $P(X=4)$

E. $P(X=1)+P(X=5)$

Choose the correct answer from the options given below:

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Solution

Sum of probabilities = 1

$k+\frac{k}{2}+2k+8k^2+(1-5k)+\frac{k}{2}=1$

Simplify

$4k+8k^2+1-5k=1$

$8k^2-k=0$

$k(8k-1)=0$

Since $k>0$

$k=\frac{1}{8}$

Now compute probabilities.

$P(X=1)=\frac{1}{8}$

$P(X=2)=\frac{1}{16}$

$P(X=3)=\frac{1}{4}$

$P(X=4)=\frac{1}{8}$

$P(X=5)=\frac{3}{8}$

$P(X=6)=\frac{1}{16}$

Evaluate statements.

A. $P(X=3)=\frac{1}{4}$

B. $P(X\le2)=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}$

C. $P(X\ge5)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$

D. $P(X=4)=\frac{1}{8}$

E. $P(X=1)+P(X=5)=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}$

Order (largest to smallest)

$\frac{1}{2}>\frac{7}{16}>\frac{1}{4}>\frac{3}{16}>\frac{1}{8}$

$E>C>A>B>D$

Qus : 87 CUET UG Mathematics Applied PYQ

In a school, an auditorium was used for its cultural activities. The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$. Based on the above information answer the following questions. If $x$ and $y$ represent the length and breadth of the rectangular region, then:

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Solution

For a rectangle,

Perimeter $P=2(\text{length}+\text{breadth})$

Here

length $=x$

breadth $=y$

Given fixed parameter $p$ represents the perimeter.

$p=2(x+y)$

Qus : 88 CUET UG Mathematics Applied PYQ

In a school, an auditorium was used for its cultural activities. The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$. Based on the above information answer the following questions. The area $A$ of the floor, as a function of $x$ can be expressed as:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From previous result

$p=2(x+y)$

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

Hence

$y=\frac{p}{2}-x$

Area of rectangle

$A=xy$

Substitute $y$

$A=x\left(\frac{p}{2}-x\right)$

$A=\frac{px}{2}-x^2$

$A=\frac{px-2x^2}{2}$

Qus : 89 CUET UG Mathematics Applied PYQ

In a school, an auditorium was used for its cultural activities.

The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$.

Based on the above information answer the following questions.

The value of $x$, for which area of floor of auditorium is maximum is:

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Solution

$p = 2(x+y)$

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

Hence

$y=\frac{p}{2}-x$

Area of rectangle

$A=xy$

$A=x\left(\frac{p}{2}-x\right)$

$A=\frac{px}{2}-x^2$

To maximize area

$\frac{dA}{dx}=\frac{p}{2}-2x$

Set derivative equal to zero

$\frac{p}{2}-2x=0$

$2x=\frac{p}{2}$

$x=\frac{p}{4}$

Qus : 90 CUET UG Mathematics Applied PYQ

In a school, an auditorium was used for its cultural activities.

The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$.

Based on the above information answer the following questions.

The value of $y$, for which the area of the floor of auditorium is maximum is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From earlier relation

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

From maximum area condition

$x=\frac{p}{4}$

$y=\frac{p}{2}-\frac{p}{4}$

$y=\frac{p}{4}$

Qus : 91 CUET UG Mathematics Applied PYQ

In a school, an auditorium was used for its cultural activities.

The shape of the floor of the auditorium is rectangular with dimensions $x$ and $y$ $(x>y)$ and has fixed parameter $p$.

Based on the above information answer the following questions.

Maximum area of floor is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

From the given condition

$p=2(x+y)$

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

Area of rectangle

$A=xy$

Put

$y=\frac{p}{2}-x$

$A=x\left(\frac{p}{2}-x\right)$

$A=\frac{px}{2}-x^2$

To maximize area

$\frac{dA}{dx}=\frac{p}{2}-2x=0$

Hence

$x=\frac{p}{4}$

Then

$y=\frac{p}{2}-\frac{p}{4}=\frac{p}{4}$

Maximum area

$A=xy$

$=\frac{p}{4}\times\frac{p}{4}$

$=\frac{p^2}{16}$

Qus : 92 CUET UG Mathematics Applied PYQ

A ball is thrown upwards from the plane surface of the ground.

Suppose the plane surface from which the ball is thrown consists of the points

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.

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Solution

Points

$A(1,0,2)$

$B(3,-1,1)$

$C(1,2,1)$

Direction vectors

$\vec{AB}=(3-1,-1-0,1-2)=(2,-1,-1)$

$\vec{AC}=(1-1,2-0,1-2)=(0,2,-1)$

Normal vector of plane

$\vec{n}=\vec{AB}\times\vec{AC}$

$\vec{n}=

\begin{vmatrix}

\mathbf{i} & \mathbf{j} & \mathbf{k}\\

2 & -1 & -1\\

0 & 2 & -1

\end{vmatrix}$

$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$

So normal vector

$(3,2,4)$

Equation of plane

$3(x-1)+2(y-0)+4(z-2)=0$

$3x+2y+4z-11=0$

$3x+2y+4z=11$

Qus : 93 CUET UG Mathematics Applied PYQ

A ball is thrown upwards from the plane surface of the ground.

Suppose the plane surface from which the ball is thrown consists of the points

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.

The maximum height of the ball from the ground is:

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Solution

Equation of plane through $A,B,C$

$3x+2y+4z=11$

Maximum height of the ball = perpendicular distance of point $D(2,3,1)$ from the plane.

Distance formula

$d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$

Plane

$3x+2y+4z-11=0$

Substitute $D(2,3,1)$

$=|3(2)+2(3)+4(1)-11|$

$=|6+6+4-11|$

$=|5|$

Denominator

$\sqrt{3^2+2^2+4^2}$

$=\sqrt{9+4+16}$

$=\sqrt{29}$

Therefore

$d=\frac{5}{\sqrt{29}}$

Qus : 94 CUET UG Mathematics Applied PYQ

A ball is thrown upwards from the plane surface of the ground.

Suppose the plane surface from which the ball is thrown consists of the points

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.

The equation of the perpendicular line drawn from the maximum height of the ball to the ground is:

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Solution

Ground plane passes through points

$A(1,0,2),\; B(3,-1,1),\; C(1,2,1)$

Direction vectors

$\vec{AB}=(2,-1,-1)$

$\vec{AC}=(0,2,-1)$

Normal vector of plane

$\vec{n}=\vec{AB}\times\vec{AC}$

$\vec{n}=

\begin{vmatrix}

\mathbf{i}&\mathbf{j}&\mathbf{k}\\

2&-1&-1\\

0&2&-1

\end{vmatrix}

$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$

So direction ratios of perpendicular line

$(3,2,4)$

The perpendicular line passes through point

$D(2,3,1)$

Hence equation of line

$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$

Qus : 95 CUET UG Mathematics Applied PYQ

The corner points of the feasible region determined by x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0 are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is:

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Solution

Qus : 96 CUET UG Mathematics Applied PYQ

A ball is thrown upwards from the plane surface of the ground.

Suppose the plane surface from which the ball is thrown consists of the points

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.

The coordinates of the foot of the perpendicular drawn from the maximum height of the ball to the ground are:

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Solution

Plane through points $A,B,C$ $3x+2y+4z=11$ Normal vector of plane $\vec{n}=(3,2,4)$ Point $D(2,3,1)$ Foot of perpendicular formula $\left(x_1-\frac{a(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; y_1-\frac{b(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; z_1-\frac{c(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}\right)$ Plane $3x+2y+4z-11=0$ Substitute point $D$ $3(2)+2(3)+4(1)-11$ $=6+6+4-11$ $=5$ Denominator $3^2+2^2+4^2$ $=9+4+16$ $=29$ Coordinates of foot $x=2-\frac{3(5)}{29}=\frac{43}{29}$ $y=3-\frac{2(5)}{29}=\frac{77}{29}$ $z=1-\frac{4(5)}{29}=\frac{9}{29}$ Foot of perpendicular $\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right)$

Qus : 97 CUET UG Mathematics Applied PYQ

If $t=e^{2x}$ and $y=ln(t^2)$, then $\frac{d^2y}{dx^2}$ is

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Solution

Qus : 98 CUET UG Mathematics Applied PYQ

A ball is thrown upwards from the plane surface of the ground.

Suppose the plane surface from which the ball is thrown consists of the points

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.

The area of $\triangle ABC$ is:

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Solution

$A(1,0,2)$ $B(3,-1,1)$ $C(1,2,1)$ Vectors $\vec{AB}=(3-1,-1-0,1-2)=(2,-1,-1)$ $\vec{AC}=(1-1,2-0,1-2)=(0,2,-1)$ Cross product $\vec{AB}\times\vec{AC}$ $= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&-1&-1\\ 0&2&-1 \end{vmatrix} $ $=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$ Magnitude $|\vec{AB}\times\vec{AC}|$ $=\sqrt{3^2+2^2+4^2}$ $=\sqrt{29}$ Area of triangle $=\frac{1}{2}|\vec{AB}\times\vec{AC}|$ $=\frac{1}{2}\sqrt{29}$

Qus : 99 CUET UG Mathematics Applied PYQ

An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

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Qus : 100 CUET UG Mathematics Applied PYQ

The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and the x-axis is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

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Qus : 101 CUET UG Mathematics Applied PYQ

A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and second throw of the dice, and a number less than 4 in the third throw?

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

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CUET UG Mathematics Applied

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CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Mathematics PYQ

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