જો a = lim_{n to infty} sum_{k=1}^{n} frac{2n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) પ્રથમ,$a$ ની કિંમત શોધો: $a = \lim_{n 	o \infty} \sum_{k=1}^{n} \frac{2}{n(1+(k/n)^2)} = \int_{0}^{1} \frac{2}{1+x^2} dx = 2[	an^{-1} x]_0^1 = 2

Vedclass · Accepted Answer

$\sqrt{2} f \left(\frac{a}{2}\right) = f'\left(\frac{a}{2}\right)$

Vedclass · Answer

(C) પ્રથમ,$a$ ની કિંમત શોધો: $a = \lim_{n 	o \infty} \sum_{k=1}^{n} \frac{2}{n(1+(k/n)^2)} = \int_{0}^{1} \frac{2}{1+x^2} dx = 2[	an^{-1} x]_0^1 = 2(\frac{\pi}{4}) = \frac{\pi}{2}$. હવે,$f(x)$ ને સરળ બનાવો: $f(x) = \sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an(x/2)$ જ્યાં $x \in (0, 1)$. તેથી $f'(x) = \frac{1}{2} \sec^2(x/2)$. હવે $x = a/2 = \pi/4$ માટે કિંમત શોધો: $f(\pi/4) = 	an(\pi/8) = \sqrt{2}-1$. $f'(\pi/4) = \frac{1}{2} \sec^2(\pi/8) = \frac{1}{2} (1 + 	an^2(\pi/8)) = \frac{1}{2} (1 + (\sqrt{2}-1)^2) = \frac{1}{2} (1 + 2 + 1 - 2\sqrt{2}) = \frac{4-2\sqrt{2}}{2} = 2-\sqrt{2} = \sqrt{2}(\sqrt{2}-1) = \sqrt{2} f(\pi/4)$. આમ,$f'(\frac{a}{2}) = \sqrt{2} f(\frac{a}{2})$.

જો $a = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2n}{n^2+k^2}$ અને $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}$,$x \in (0, 1)$,હોય તો:

Similar Questions

$\lim _{n \rightarrow \infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}}-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]=$

$\lim _{n \rightarrow \infty} n\left[\frac{1}{3 n^2+8 n+4}+\frac{1}{3 n^2+16 n+16}+\ldots+\frac{1}{15 n^2}\right]=$

નીચેના વિધાનો પૈકી:
$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$
$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$

જો $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$ હોય,તો $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ ની કિંમત શોધો:

જો $\lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n \cdot n!}\right)=\int_1^2 f(x) d x$ હોય, તો $f(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module