Exams Joint Entrance Examination Mathematics Definite Integration Question 92 Question 92 of 246 Hard NUMERICAL
Let l i m n → ∞ ( n n 4 + 1 − 2 n ( n 2 + 1 ) n 4 + 1 + n n 4 + 16 − 8 n ( n 2 + 4 ) n 4 + 16 \underset{n\to \infty}{lim}(\frac{n}{\sqrt{{n}^{4}+1}}-\frac{2n}{({n}^{2}+1)\sqrt{{n}^{4}+1}}+\frac{n}{\sqrt{{n}^{4}+16}}-\frac{8n}{({n}^{2}+4)\sqrt{{n}^{4}+16}} n → ∞ l im ( n 4 + 1 n − ( n 2 + 1 ) n 4 + 1 2 n + n 4 + 16 n − ( n 2 + 4 ) n 4 + 16 8 n + … + n n 4 + n 4 − 2 n ⋅ n 2 ( n 2 + n 2 ) n 4 + n 4 ) +\ldots +\frac{n}{\sqrt{{n}^{4}+{n}^{4}}}-\frac{2n\cdot {n}^{2}}{({n}^{2}+{n}^{2})\sqrt{{n}^{4}+{n}^{4}}}) + … + n 4 + n 4 n − ( n 2 + n 2 ) n 4 + n 4 2 n ⋅ n 2 ) be π k \frac{\pi }{k} k π , using only the principal values of the inverse trigonometric functions. Then k 2 {k}^{2} k 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 32
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l i m n → ∞ ( n n 4 + 1 + n n 4 + 16 + … n n 4 + n 4 ) − l i m n → ∞ ( ( ∣ 2 n ( n 2 + 1 ) ( n 4 + 1 ) ) ) ∣ + 8 n ( n 2 + 4 ) n 4 + 1 + ⋯ 2 n ⋅ n 2 ( n 2 + n 2 ) n 4 + n 4 = l i m n → ∞ ∑ n = 1 n 1 n 1 + r 4 n 4 − l i m n → ∞ ∑ n = 1 n 1 n 2 ⋅ ( r / n ) 2 ( 1 + ( r n ) 2 ) 1 + r 4 n 4 = ∫ 0 1 d x 1 + x 4 − 2 ∫ 0 1 x 2 ( 1 + x 2 ) 1 + x 4 d x = ∫ 0 1 1 − x 2 ( 1 + x 2 ) 1 + x 4 d x \begin{gathered} \underset{n\to \infty}{lim}(\frac{n}{\sqrt{{n}^{4}+1}}+\frac{n}{\sqrt{{n}^{4}+16}}+\ldots \frac{n}{\sqrt{{n}^{4}+{n}^{4}}}) \\ -\underset{n\to \infty}{lim}((|\frac{2n}{({n}^{2}+1)(\sqrt{{n}^{4}+1})}))|+\frac{8n}{({n}^{2}+4)\sqrt{{n}^{4}+1}}+\cdots \frac{2n\cdot {n}^{2}}{({n}^{2}+{n}^{2})\sqrt{{n}^{4}+{n}^{4}}} \\ =\underset{n\to \infty}{lim}\sum _{n=1}^{n}\frac{1}{n\sqrt{1+\frac{{r}^{4}}{{n}^{4}}}}-\underset{n\to \infty}{lim}\sum _{n=1}^{n}\frac{1}{n}\frac{2\cdot (r/n{)}^{2}}{(1+{(\frac{r}{n})}^{2})\sqrt{1+\frac{{r}^{4}}{{n}^{4}}}} \\ =\int _{0}^{1}\frac{dx}{\sqrt{1+{x}^{4}}}-2\int _{0}^{1}\frac{{x}^{2}}{(1+{x}^{2})\sqrt{1+{x}^{4}}}dx \\ =\int _{0}^{1}\frac{1-{x}^{2}}{(1+{x}^{2})\sqrt{1+{x}^{4}}}dx \end{gathered} n → ∞ l im ( n 4 + 1 n + n 4 + 16 n + … n 4 + n 4 n ) − n → ∞ l im (( ∣ ( n 2 + 1 ) ( n 4 + 1 ) 2 n )) ∣ + ( n 2 + 4 ) n 4 + 1 8 n + ⋯ ( n 2 + n 2 ) n 4 + n 4 2 n ⋅ n 2 = n → ∞ l im n = 1 ∑ n n 1 + n 4 r 4 1 − n → ∞ l im n = 1 ∑ n n 1 ( 1 + ( n r ) 2 ) 1 + n 4 r 4 2 ⋅ ( r / n ) 2 = ∫ 0 1 1 + x 4 d x − 2 ∫ 0 1 ( 1 + x 2 ) 1 + x 4 x 2 d x = ∫ 0 1 ( 1 + x 2 ) 1 + x 4 1 − x 2 d x
∫ 0 1 ( 1 x 2 − 1 ) d x ( x + 1 x ) x 2 + 1 x 2 = ∫ 0 1 ( 1 x 2 − 1 ) d x ( x + 1 x ) ( x + 1 x ) 2 − 2 x + 1 x = t ⇒ ( 1 − 1 x 2 ) d x = d t − ∫ ∞ 2 d t t t 2 − 2 = ∫ 2 ∞ d t t t 2 − 2 = − 1 2 sin − 1 2 x ∣ ∞ 2 = − 1 2 ( 0 − π 4 ) = π 2 5 / 2 ∴ k = 2 5 / 2 ∴ k 2 = 32 \begin{gathered} \int _{0}^{1}\frac{(\frac{1}{{x}^{2}}-1)dx}{(x+\frac{1}{x})\sqrt{{x}^{2}+\frac{1}{{x}^{2}}}} \\ =\int _{0}^{1}\frac{(\frac{1}{{x}^{2}}-1)dx}{(x+\frac{1}{x})\sqrt{{(x+\frac{1}{x})}^{2}-2}} \\ x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{{x}^{2}})dx=dt \\ -\int _{\infty}^{2}\frac{dt}{t\sqrt{{t}^{2}-2}}=\int _{2}^{\infty}\frac{dt}{t\sqrt{{t}^{2}-2}}={\frac{-1}{\sqrt{2}}{\sin }^{-1}\frac{\sqrt{2}}{x}|}_{\infty}^{2} \\ =\frac{-1}{\sqrt{2}}(0-\frac{\pi }{4}) \\ =\frac{\pi }{{2}^{5/2}} \\ \therefore k={2}^{5/2} \\ \therefore {k}^{2}=32 \end{gathered} ∫ 0 1 ( x + x 1 ) x 2 + x 2 1 ( x 2 1 − 1 ) d x = ∫ 0 1 ( x + x 1 ) ( x + x 1 ) 2 − 2 ( x 2 1 − 1 ) d x x + x 1 = t ⇒ ( 1 − x 2 1 ) d x = d t − ∫ ∞ 2 t t 2 − 2 d t = ∫ 2 ∞ t t 2 − 2 d t = 2 − 1 sin − 1 x 2 ∣ ∞ 2 = 2 − 1 ( 0 − 4 π ) = 2 5/2 π ∴ k = 2 5/2 ∴ k 2 = 32 Answer the question to unlock