CUET UG Mathematics Applied 2022 Previous Year Questions (PYQs) – Page 1 of 9
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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
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.
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
$=2^4=16$
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 $e^e$:
$e^e\left(\frac{d^2y}{dx^2}\right)_{x=e}=4-\frac{1}{e}$
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
$(-\infty,1)$
$\int \frac{dx}{x(x^5+3)}$ 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 $t=x^5$
$I=\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C$
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
$q(x)=\frac{x^3}{3}-\frac{x^2}{2}+x$
$\int_{-1}^{1}(|x-2|+|x|),dx=$
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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:
$3+2=5$
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,
$2a+6b=2(2)+6(1)$
$=4+6$
$=10$
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.
CUET UG Mathematics Applied
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and More.
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
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