This set of Ordinary Differential Equations Question Paper focuses on “Special Functions -2 (Beta)”.
1. β(m, n) = β(n, m). Is the statement true?
a) True
b) False
View Answer
Explanation: L.H.S. = \(\beta(m, n) = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)}. \)
R.H.S. \( = \beta(n, m) = \frac{\Gamma(n).\Gamma(m)}{\Gamma(n+m)} = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)} = \) L.H.S.
Therefore, \(\beta(m, n) = \beta(n, m).\)
2. Which of the following function is not called the Euler’s integral of the first kind?
a) \(\beta(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) \)
b) \(\beta(m, n) = \int_0^{π/2} (sinθ)^{2m-1} (cosθ)^{2n-1} dθ \)
c) \(\beta(m, n) = \int_0^∞ \frac{y^{n+1}}{(1+y)^{m+n}} dy \)
d) \(\beta(m, n) = 2 \int_0^{π/2} (sinθ)^{2m-1} (cosθ)^{2n-1} dθ \)
View Answer
Explanation: Euler’s integral of the first kind is nothing but Beta function. So, here only \( \beta(m, n) = \int_0^{π/2}(sinθ)^{2m-1} (cosθ)^{2n-1} dθ\) is not the definition of Beta function.
3. Which of the following is not the definition of Beta function?
a) \(\beta(m, n) = 2\int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) \)
b) \(\beta(m, n) = 2\int_0^{π/2}(sinθ)^{2m-1} (cosθ)^{2n-1} dθ \)
c) \(\beta(m, n) = \int_0^∞ \frac{y^{n+1}}{(1+y)^{m+n}} dy \)
d) \(\beta(m, n) = \int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} dx \)
View Answer
Explanation: \(\beta(m, n)\) can be written as either \(\int_0^1 x^{m-1} (1-x)^{n-1} dx \, (m>0, n>0)\) (or)
\(= 2 \int_0^{π/2}(sinθ)^{2m-1} (cosθ)^{2n-1} dθ \)
(or)
\( = \int_0^∞ \frac{y^{n+1}}{\left(1+y\right)^{m+n}} dy \;or \int_0^1 \frac{x^{m-1}+x^{n-1}}{\left(1+x\right)^{m+n}} dx\).
So the correct answer is \(2\int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0) \) which is actually not the formula for Beta function.
4. What is the value of \(\beta(m, \frac{1}{2}) \)?
a) β(m, m)
b) 22m-1 β(m, m)
c) 22m+1 β(m, m)
d) 22m β(m, m)
View Answer
Explanation: \(\beta(m, \frac{1}{2}) = 2\int_0^{π/2}(sinθ)^{2m-1} dθ \)
\(\beta(m, m) = 2 \int_0^{π/2}(sinθ)^{2m-1} (cosθ)^{2m-1} dθ \)
\( = 2^{-2m+2} \int_0^{π/2}(2 sinθ cosθ)^{2m-1} dθ \)
Substituting 2θ=φ,
\( = 2^{-2m+1} \int_0^π sinφdφ \)
\( = 2^{-2m+1}.2.\int_0^{π/2}sinφdφ \)
\( = \frac{1}{2^{2m-1}} \beta(m, \frac{1}{2}). \)
5. What is the value of β(3,2)?
a) \(\frac{1}{14} \)
b) \(\frac{1}{16} \)
c) \(\frac{1}{12} \)
d) \(\frac{1}{10} \)
View Answer
Explanation: \(\beta(3, 2) = \frac{\Gamma(3).\Gamma(2)}{\Gamma(3+2)} \)
\( = \frac{2!1!}{4!} = \frac{1}{12}.\)
6. What is the value of \(\beta(\frac{1}{4},\frac{3}{4})\)?
a) \(\pi \)
b) \(\sqrt{2}\pi \)
c) \(\sqrt{2\pi} \)
d) \({2}\pi \)
View Answer
Explanation: \(\beta(\frac{1}{4}, \frac{3}{4}) = \frac{\Gamma(\frac{1}{4}).\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4}+\frac{3}{4})} \)
\( = \frac{\pi}{sin \frac{π}{4}} = \sqrt{2}\pi. \)
7. What is the value of \(\int_0^{π/2}\sqrt{sinθdθ} + \int_0^{π/2}\sqrt{cosθdθ} \)?
a) \(8\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \)
b) \(4\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \)
c) \(8\sqrt{\pi} \frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})} \)
d) \(4\sqrt{\pi} \frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})} \)
View Answer
Explanation: \(\int_0^{π/2}\sqrt{sinθdθ} + \int_0^{π/2}\sqrt{cosθdθ} \)
\( = \beta(\frac{3}{4}, \frac{1}{2}) + \beta(\frac{3}{4}, \frac{1}{2}) \)
\( = 2 \beta(\frac{3}{4}, \frac{1}{2}) \)
\( = 2\frac{\Gamma(\frac{1}{2}).\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2}+\frac{3}{4})} \)
\( = 2\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} . \frac{1}{(\frac{1}{4})} \)
\( = 8\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}. \)
8. What is the value of \(\int_0^∞ \frac{1}{(1+y)^5} dy \)?
a) 1⁄2
b) 1⁄3
c) 1⁄4
d) 1⁄5
View Answer
Explanation: \(\int_0^∞ \frac{y^{1-1}}{(1+y)^5} dy \)
n=1 and m + n = 5 which implies m=4.
\(\beta(4, 1) = \frac{\Gamma(4).\Gamma(1)}{\Gamma(4+1)} \)
\(\frac{3!0!}{4!} = \frac{1}{4} \).
9. What is the value of \(\int_0^1 \frac{dx}{\sqrt{1+x^4}}\)?
a) \(\beta(\frac{1}{4}, \frac{1}{2}) \)
b) \(\frac{1}{4\sqrt{2}}\beta(\frac{1}{4}, \frac{1}{2}) \)
c) \(\frac{1}{3\sqrt{2}}\beta(\frac{1}{3}, \frac{1}{2}) \)
d) \(\frac{1}{4\sqrt{3}}\beta(\frac{1}{4}, \frac{1}{3}) \)
View Answer
Explanation: Substitute x2 = tanθ
Therefore θ varies from 0 to \(\frac{\pi}{4}.\)
\(\int_0^{π/4} \frac{(secθ)^2}{2 secθ \sqrt{tanθ}} dθ \)
\(= \int_0^{π/4}\frac{dθ}{2\sqrt{sinθ cosθ}} \)
\(= \frac{1}{4\sqrt{2}} \beta(\frac{1}{4}, \frac{1}{2}).\)
10. What is the value of \(\int_0^1\frac{(y^5+y^2)}{(1+y)^9} dy \)?
a) \(\frac{1}{158} \)
b) \(\frac{2}{167} \)
c) \(\frac{1}{146} \)
d) \(\frac{1}{168} \)
View Answer
Explanation: m-1 = 5 => m=6 and n-1 = 2 => n=3.
\(\beta(6, 3) = \frac{\Gamma(6).\Gamma(3)}{\Gamma(6+3)} \)
\(= \frac{5!2!}{8!} = \frac{1}{168}. \)
11. Solve using the Beta function. \(\int_0^1 x^{-2} (1-x)^{-3} dx. \)
a) Can be solved using a Beta function with m = -1 and n = -2
b) Can be solved using a Beta function with m = 1 and n = -2
c) Can be solved using a Beta function with m = -1 and n = 2
d) Can’t be solved using the Beta function
View Answer
Explanation: This function can’t be solved using the Beta function as m and n have negative values. Beta function can’t be solved if m and n are negative numbers.
12. What is the value of \(\int_0^∞ (sech x)^5 dx \)?
a) \(\frac{3\pi}{80} \)
b) \(\frac{3\pi}{240} \)
c) \(\frac{3\pi}{16} \)
d) \(\frac{\pi}{240} \)
View Answer
Explanation: \(\int_0^∞ (sech x)^5 dx = \frac{2^5}{4} \beta(\frac{5}{2}, \frac{5}{2}) \)
\(\displaystyle = \frac{8 \Gamma(\frac{5}{2})\Gamma(\frac{5}{2})}{\Gamma(5)} = \frac{(8*\frac{3}{2}*\frac{1}{2}*\sqrt{\pi}*\frac{3}{2}*\frac{1}{2}*\sqrt{\pi})}{24} = \frac{9π}{48}=\frac{3π}{16}. \)
13. What is the value of \(\beta(\frac{9}{2},3) \)?
a) \(\frac{16}{1287} \)
b) \(\frac{16}{1278} \)
c) \(\frac{14}{1287} \)
d) \(\frac{16}{127} \)
View Answer
Explanation: \(\beta(\frac{9}{2},3) = \frac{\Gamma(\frac{9}{2})\Gamma(3)}{\Gamma(\frac{9}{2}+3)} \)
\( = \frac{\Gamma(\frac{9}{2})2}{\frac{13}{2}*\frac{11}{2}*\frac{9}{2}*\Gamma(\frac{9}{2})} = \frac{16}{1287}.\)
14. What is the value of \(\int_0^1 x^5 (1-x)^6 dx\)?
a) \(\frac{1}{12*11*10*9*8*7} \)
b) \(\frac{1}{12*11*10*9*8} \)
c) \(\frac{1}{12*11*10*9*8*7*6} \)
d) \(\frac{1}{12*11*10*9*8*7*6*5} \)
View Answer
Explanation: Here, m-1 = 5 which implies m=6 and n-1 = 6 which implies n=7.
\(\beta(6,7)= \frac{\Gamma(6)\Gamma(7)}{\Gamma(13)} = \frac{1}{12*11*10*9*8*7*6}. \)
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