JEE 2023 — Mathematics PYQ

To solve this integral, we use a algebraic manipulation technique to make the substitution easier.

Step 1: Simplify the Integral

Let the integral be $I$:

We can take $x^7$ out from the first bracket and move it into the second bracket (which is under a $1/7$ power). To enter the bracket with a $1/7$ exponent, $x^7$ becomes $(x^7{)}^7=x^{49}$ inside. However, a more elegant way is to factor $x^6$ from the first part and $x$ into the second:

Take $x^6$ inside the power $1/7$:

Actually, let's factor $x^{20},x^{13},x^6$ and take an $x$ inside:

Inside the $1/7$ power, $x$ becomes $x^7$:

Step 2: Substitution

Let $u=2x^{21}+3x^{14}+6x^7$.

Now, differentiate with respect to $x$:

Step 3: Change the Limits

• When $x=0$, $u=2(0)+3(0)+6(0)=0$.

• When $x=1$, $u=2(1{)}^{21}+3(1{)}^{14}+6(1{)}^7=2+3+6=11$.

Step 4: Integrate

Substitute the values back into the integral:

Step 5: Compare and Find $l+m+n$

The given form is $\frac{1}{l}(11{)}^{\frac{m}{n}}$.

Comparing this with our result $\frac{1}{48}(11{)}^{8/7}$:

• $l=48$

• $m=8$

• $n=7$

Check if $m$ and $n$ ($8$ and $7$) are coprime: Yes, $\gcd (8,7)=1$.

Finally, calculate the sum:

Correct Option: (B) 63