Let \(\alpha>0\) If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :
Let \(\alpha>0\) If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :
Show Hint
- 2
- $2 \sqrt{2}$
- 4
- $\sqrt{2}$
The Correct Option is A
Approach Solution - 1
After rationalising
\(\int\limits_{0}^{a}\frac{x}{\alpha}\left(\sqrt{x+\alpha}+\sqrt{x}\right)dx\)
\(=\int\limits_{0}^{a}\frac{1}{\alpha}[(x+\alpha)^{3/2}-\alpha(x+\alpha)^{1/2}+x^{3/2}]dx\)
\(=\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5/2}-\alpha\frac{2}{3}(x+\alpha)^{3/2}+\frac{2}{5}x^{5/2}\right]|^a_{0}\)
\(=\frac{1}\alpha{}[\frac{2}{5}(2\alpha)^{5/2}-\frac{2\alpha}{3}(2\alpha)^{3/2}+\frac{2}{5}\alpha^{5/2}-\frac{2}{5}\alpha^{5/2}+\frac{2}{3}\alpha^{5/2}]|^{a}_0\)
\(=\frac{1}{\alpha}[\frac{2^{7/2}\alpha^{5/2}}{5}-\frac{2^{5/2}\alpha^{5/2}}{3}+\frac{2}{3}\alpha^{5/2}]\)
\(=\alpha^{3/2}[\frac{2^{7/2}}{5}-\frac{2^{5/2}}{3}+\frac{2}{3}]\)
\(=\frac{\alpha^{3/2}}{15}(24\sqrt2-20\sqrt2+10)\)
\(=\frac{\alpha^{3/2}}{15}(4\sqrt2+10)\)
Now,
\(\frac{\alpha^{3/2}}{15}(4\sqrt2+10)=\frac{16+20\sqrt2}{15}\)
\(⇒\alpha=2\)
So, the correct option is (A) : 2
Approach Solution -2
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).