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CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Definite Integration PYQ

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

Solution

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

Solution

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





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

Solution

CUET UG Mathematics Applied PYQ
$\int_{-1}^{1}(|x-2|+|x|),dx=$





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

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:

$3+2=5$
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 $=2$

CUET UG Mathematics Applied PYQ
$\int \tan x(\sec x - \tan x)\,dx =$





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

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 $\int \tan x(\sec x-\tan x)dx = \sec x - (\tan x - x)$ $= \sec x - \tan x + x + C$

CUET UG Mathematics Applied

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