Q: \displaystyle\lim_{x\to\frac{\pi}{4}}\dfrac{\int_{2}^{\sec^2x}f(t)\,dt}{x^2-\dfrac{\pi^2}{16}} equals

\dfrac{8}{\pi}f(2). This is a \tfrac00 form. By L'Hopital's rule and the Newton-Leibnitz theorem, the derivative of the numerator is f(\sec^2x)\cdot 2\sec^2x\tan x, and of the denominator is 2x. At x=\tfrac{\pi}{4}: \sec^2x=2,\ \tan x=1, so the limit =\dfrac{f(2)\cdot 2\cdot 2\cdot 1}{2\cdot \pi/4}=\dfrac{4f(2)}{\pi/2}=\dfrac{8f(2)}{\pi}.

Q: \displaystyle\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^{2n}\dfrac{r}{\sqrt{n^2+r^2}} equals

\sqrt5-1. Writing the sum as a Riemann sum with x=r/n, \dfrac1n\sum_{r=1}^{2n}\dfrac{r/n}{\sqrt{1+(r/n)^2}}\to\displaystyle\int_0^2\dfrac{x}{\sqrt{1+x^2}}\,dx=\Big[\sqrt{1+x^2}\Big]_0^2=\sqrt5-1.

Q: Let f(x)=\displaystyle\lim_{n\to\infty}\left[\dfrac{n^n(x+n)\left(x+\dfrac{n}{2}\right)\cdots\left(x+\dfrac{n}{n}\right)}{n!\left(x^2+n^2\right)\left(x^2+\dfrac{n^2}{4}\right)\cdots\left(x^2+\dfrac{n^2}{n^2}\right)}\right]^{x/n} for all x>0. Then

f\left(\dfrac13\right)\le f\left(\dfrac23\right). Writing x+\tfrac nk=\tfrac nk\left(1+\tfrac{kx}{n}\right) and x^2+\tfrac{n^2}{k^2}=\tfrac{n^2}{k^2}\left(1+\tfrac{k^2x^2}{n^2}\right) and simplifying using n!=\prod k, f(x)=\displaystyle\lim_{n\to\infty}\left[\prod_{k=1}^{n}\dfrac{1+kx/n}{1+(kx/n)^2}\right]^{x/n}. Taking logs and converting the Riemann sum to an integral gives \ln f(x)=x\displaystyle\int_0^1\ln\dfrac{1+xt}{1+x^2t^2}\,dt=(1+x)\ln(1+x)-x\ln(1+x^2)-2\tan^{-1}x+x. Differentiating, \dfrac{f'(x)}{f(x)}=\ln\dfrac{1+x}{1+x^2}. For 0 1+x^2, so f'(x)/f(x)>0 there, i.e. f is increasing on (0,1); hence f(1/2) 0, so f'(2)<0 (c) is true. At x=3, f'(3)/f(3)=\ln(2/5)<\ln(3/5)=f'(2)/f(2), so (d) is false. Hence (b) and (c) are correct.

Question 1

If \alpha and \beta are the roots of the equation 375x^2-25x-2=0, then \displaystyle\lim_{n	o\infty}\sum_{r=1}^{n}\alpha^r+\lim_{n	o\infty}\sum_{r=1}^{n}\beta^r is equal to

Accepted Answer

\dfrac{1}{12}. For 375x^2-25x-2=0: sum of roots \alpha+\beta=\dfrac{25}{375}=\dfrac{1}{15}, product \alpha\beta=\dfrac{-2}{375}. Since |\alpha|,|\beta|<1, the infinite geometric series give \displaystyle\sum_{r=1}^{\infty}\alpha^r+\sum_{r=1}^{\infty}\beta^r=\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}=\dfrac{(\alpha+\beta)-2\alpha\beta}{1-(\alpha+\beta)+\alpha\beta}=\dfrac{	frac{1}{15}+	frac{4}{375}}{	frac{14}{15}-	frac{2}{375}}=\dfrac{29/375}{348/375}=\dfrac{29}{348}=\dfrac{1}{12}.

Question 2

\displaystyle\lim_{x	o\frac{\pi}{4}}\dfrac{\int_{2}^{\sec^2x}f(t)\,dt}{x^2-\dfrac{\pi^2}{16}} equals

Accepted Answer

\dfrac{8}{\pi}f(2). This is a 	frac00 form. By L'Hopital's rule and the Newton-Leibnitz theorem, the derivative of the numerator is f(\sec^2x)\cdot 2\sec^2x	an x, and of the denominator is 2x. At x=	frac{\pi}{4}: \sec^2x=2,\ 	an x=1, so the limit =\dfrac{f(2)\cdot 2\cdot 2\cdot 1}{2\cdot \pi/4}=\dfrac{4f(2)}{\pi/2}=\dfrac{8f(2)}{\pi}.

Question 3

\displaystyle\lim_{n	o\infty}\dfrac{1}{n}\sum_{r=1}^{2n}\dfrac{r}{\sqrt{n^2+r^2}} equals

Accepted Answer

\sqrt5-1. Writing the sum as a Riemann sum with x=r/n, \dfrac1n\sum_{r=1}^{2n}\dfrac{r/n}{\sqrt{1+(r/n)^2}}	o\displaystyle\int_0^2\dfrac{x}{\sqrt{1+x^2}}\,dx=\Big[\sqrt{1+x^2}\Big]_0^2=\sqrt5-1.

Question 4

Let f(x)=\displaystyle\lim_{n	o\infty}\left[\dfrac{n^n(x+n)\left(x+\dfrac{n}{2}ight)\cdots\left(x+\dfrac{n}{n}ight)}{n!\left(x^2+n^2ight)\left(x^2+\dfrac{n^2}{4}ight)\cdots\left(x^2+\dfrac{n^2}{n^2}ight)}ight]^{x/n} for all x>0. Then

Accepted Answer

f\left(\dfrac13 ight)\le f\left(\dfrac23 ight). Writing x+ frac nk= frac nk\left(1+ frac{kx}{n} ight) and x^2+ frac{n^2}{k^2}= frac{n^2}{k^2}\left(1+ frac{k^2x^2}{n^2} ight) and simplifying using n!=\prod k, f(x)=\displaystyle\lim_{n o\infty}\left[\prod_{k=1}^{n}\dfrac{1+kx/n}{1+(kx/n)^2} ight]^{x/n}. Taking logs and converting the Riemann sum to an integral gives \ln f(x)=x\displaystyle\int_0^1\ln\dfrac{1+xt}{1+x^2t^2}\,dt=(1+x)\ln(1+x)-x\ln(1+x^2)-2 an^{-1}x+x. Differentiating, \dfrac{f'(x)}{f(x)}=\ln\dfrac{1+x}{1+x^2}. For 01+x^2, so f'(x)/f(x)>0 there, i.e. f is increasing on (0,1); hence f(1/2)0, so f'(2)<0 (c) is true. At x=3, f'(3)/f(3)=\ln(2/5)<\ln(3/5)=f'(2)/f(2), so (d) is false. Hence (b) and (c) are correct.

Squeeze, Newton-Leibnitz's Theorem and Limit Based on Converting Infinite Series into Definite Integrals — JEE Main Maths PYQ MCQs with Solutions

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