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CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Indefinite Integration PYQ

CUET UG Mathematics Applied PYQ
$\displaystyle \int \frac{\pi}{x^{n+1}-x}dx=$





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

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 $t=x^5$

$I=\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C$
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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

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

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