1708\hspace{-16}$Ans:: $\bf{(1)\;\;\; I=\frac{\pi.\ln(2)}{4}}$
\hspace{-16}\bf{(1)\;\; \int_{0}^{\infty}\frac{\ln (x)}{x^2+4}dx}$\\\\\\ $\bf{(2)}\;\;$ Find Max. value of $\bf{\int_{0}^{1}f^3(x)dx}$\\\\\\ Given $\bf{\mid f(x)\mid \leq 1}$ and $\bf{\int_{0}^{1}f(x)dx=0}$\\\\\\ $\bf{(3)\;\;}$ A Diff. Function $\bf{g}$ satisfies $\bf{\int_{0}^{x}g(x-t+1)g(t)dt = x^4+x^2 \;\forall x\geq 0}$\\\\ Then Find $\bf{g(t)}$\\\\\\ $\bf{(4)\;\;}$ Find the ordered pair with non-Infinite $\bf{(\alpha,\beta)}$ with $\bf{\beta \neq 0}$ Such that \\\\\\ $\bf{\lim_{n\rightarrow \infty}\frac{\sqrt[n^2]{1!.2!.3!......n!}}{n^{\alpha}}=\beta}$ Holds\\\\\\ $\bf{(5)\;\;}$ Find Max. value of $\bf{\int_{\frac{-\pi}{2}}^{\frac{3\pi}{2}}\sin (x).f(x)}$\\\\ Given that $\bf{\mid f(x)\mid \leq 5}$
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11 Answers
2621) x = 1/t dx= -1/t2 dt
I = ∞∫0 - ln(1/t)(1/t2+4)dtt2
I = 0∫∞ - ln(t)(t2+4)dt = -I
I=0
11) let x = 2tanθ
I = \int_{0}^{\frac{\pi}{2}} \frac{ln(2)+ln(tan \theta)}{2} \cdot d\theta
second integral is zero as I = -I ( direct property)
=> I =12 ∫ ln(2) = \frac{\pi \cdot ln(2)}{4}
71It second one 0 ?
For third one : Can we directly use the Newton Leibnitz second lemma to get g(x) by differentiating both sides?
) Umm guess, 20Î ?
1since |f(x)|≤1 for all x..just considering replacement of x by f(f(x)) we get |f(f(f(x)))|≤1.
|∫ f(f(f(x)))dx|≤ ∫ | f(f(f(x))) | dx **[x goes from 0 to 1]
≤ ∫dx [ as |f(f(f(x)))|≤1 ]
= 1
point to note our claim holds true iff f is integrable in [0,1]
answer is 1
1|f(x).sinx|≤|f(x)|≤5
using the above property of reimann integrable functions.. we get the required maximum value of the integral to be
5. ∫ dx [bounds being -Π/2 to 3Π/2]
= 5.2Î
= 10Î
point to note..we could not have used the fact that |sinx|≤1 iff for NO value of x in the given range sinx achieves the value +1 or -1. fortunately here at x= Π/2 sinx =1and at x= -Π/2 sinx =-1
1yes if the integral is zero as per the given conditions the function is identically zero on [0,1]
