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For $a\in R,$ $\left| a \right|>1$ let $\displaystyle \lim_{n \to \infty}\dfrac{1+\sqrt[3]{2}+\sqrt[3]{3}+........+\sqrt[3]{n}}{{{n}^{\dfrac{7}{3}}}\left( \dfrac{1}{{{(na+1)}^{2}}}+\dfrac{1}{{{(na+2)}^{2}}}+.....+\dfrac{1}{{{(na+n)}^{2}}} \right)}=54$ , then possible value(s) of a is(are)
(a)8
(b)-9
(c)7
(d)-6

seo-qna
Last updated date: 14th Apr 2024
Total views: 33.6k
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MVSAT 2024
Answer
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Hint: To solve this question, firstly we will simplify the function in terms of which limit is to be applied and make it more simpler so that we can use the concept of limit as summation. After that we will get two integrals in numerator and denominator, we will solve them and then putting values back to fraction, we will fraction equal to 54 and solve the quadratic equation obtained to get values of a.

Complete step-by-step answer:
Let us see what the concept of limit as summation is.
Suppose we have to evaluate a limit question, say $\displaystyle \lim_{n \to \infty}f(n)$, if we can transform this limit into the form $\displaystyle \lim_{n \to \infty}\dfrac{1}{n}\sum\limits_{r=a}^{b}{f\left( \dfrac{r}{n} \right)}$, then $\displaystyle \lim_{n \to \infty}f(n)=\int\limits_{a}^{b}{f(x).dx}$.

Now, first let us simplify the value on which we have to apply the limit.
So, we have $\dfrac{1+\sqrt[3]{2}+\sqrt[3]{3}+........+\sqrt[3]{n}}{{{n}^{\dfrac{7}{3}}}\left( \dfrac{1}{{{(na+1)}^{2}}}+\dfrac{1}{{{(na+2)}^{2}}}+.....+\dfrac{1}{{{(na+n)}^{2}}} \right)}$.
Now, let us take \[{{n}^{\dfrac{1}{3}}}\] from denominator to numerator, which will give us \[{{n}^{-\dfrac{1}{3}}}\] in numerator.
So, we have $\dfrac{{{n}^{-\dfrac{1}{3}}}\left( 1+\sqrt[3]{2}+\sqrt[3]{3}+........+\sqrt[3]{n} \right)}{{{n}^{2}}\left( \dfrac{1}{{{(na+1)}^{2}}}+\dfrac{1}{{{(na+2)}^{2}}}+.....+\dfrac{1}{{{(na+n)}^{2}}} \right)}$
Or, $\dfrac{\sqrt[3]{\dfrac{1}{n}}\left( 1+\sqrt[3]{2}+\sqrt[3]{3}+........+\sqrt[3]{n} \right)}{{{n}^{2}}\left( \dfrac{1}{{{(na+1)}^{2}}}+\dfrac{1}{{{(na+2)}^{2}}}+.....+\dfrac{1}{{{(na+n)}^{2}}} \right)}$
Multiplying, $\sqrt[3]{\dfrac{1}{n}}$ with terms in bracket, we get
$\dfrac{\left( \sqrt[3]{\dfrac{1}{n}}+\sqrt[3]{\dfrac{2}{n}}+\sqrt[3]{\dfrac{3}{n}}+........+\sqrt[3]{\dfrac{n}{n}} \right)}{{{n}^{2}}\left( \dfrac{1}{{{(na+1)}^{2}}}+\dfrac{1}{{{(na+2)}^{2}}}+.....+\dfrac{1}{{{(na+n)}^{2}}} \right)}$
Multiplying, $\sqrt[3]{\dfrac{1}{n}}$ with terms in bracket in denominator, we get
$\dfrac{\left( \sqrt[3]{\dfrac{1}{n}}+\sqrt[3]{\dfrac{2}{n}}+\sqrt[3]{\dfrac{3}{n}}+........+\sqrt[3]{\dfrac{n}{n}} \right)}{\left( \dfrac{{{n}^{2}}}{{{(na+1)}^{2}}}+\dfrac{{{n}^{2}}}{{{(na+2)}^{2}}}+.....+\dfrac{{{n}^{2}}}{{{(na+n)}^{2}}} \right)}$
Now, in denominator, multiplying denominator and numerator of each terms by $\dfrac{1}{{{n}^{2}}}$, we have
$\dfrac{\left( \sqrt[3]{\dfrac{1}{n}}+\sqrt[3]{\dfrac{2}{n}}+\sqrt[3]{\dfrac{3}{n}}+........+\sqrt[3]{\dfrac{n}{n}} \right)}{\left( \dfrac{1}{{{\left( a+\dfrac{1}{n} \right)}^{2}}}+\dfrac{1}{{{\left( a+\dfrac{2}{n} \right)}^{2}}}+.....+\dfrac{1}{{{\left( a+\dfrac{n}{n} \right)}^{2}}} \right)}$
On observing patterns in numerator and denominator, we can write the terms collectively in terms of summation.
So, we can write numerator $\sqrt[3]{\dfrac{1}{n}}+\sqrt[3]{\dfrac{2}{n}}+\sqrt[3]{\dfrac{3}{n}}+........+\sqrt[3]{\dfrac{n}{n}}$ as \[\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}\] and,
Denominator $\dfrac{1}{{{\left( a+\dfrac{1}{n} \right)}^{2}}}+\dfrac{1}{{{\left( a+\dfrac{2}{n} \right)}^{2}}}+.....+\dfrac{1}{{{\left( a+\dfrac{n}{n} \right)}^{2}}}$ as $\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}$.
So, we have whole fraction $\dfrac{\left( \sqrt[3]{\dfrac{1}{n}}+\sqrt[3]{\dfrac{2}{n}}+\sqrt[3]{\dfrac{3}{n}}+........+\sqrt[3]{\dfrac{n}{n}} \right)}{\left( \dfrac{1}{{{\left( a+\dfrac{1}{n} \right)}^{2}}}+\dfrac{1}{{{\left( a+\dfrac{2}{n} \right)}^{2}}}+.....+\dfrac{1}{{{\left( a+\dfrac{n}{n} \right)}^{2}}} \right)}$ as $\dfrac{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}}{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}}$.
So, $\dfrac{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}}{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}}$
Now, let $\dfrac{r}{n}=x$
Here r is variable and n is constant
So, differentiating $\dfrac{r}{n}=x$ both side with respect to x, we get
$\dfrac{dr}{n}=dx$
Putting limit in $\dfrac{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}}{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}}$, we have
$\displaystyle \lim_{n \to \infty}\dfrac{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}}{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}}$
Now, value of $\dfrac{r}{n}=x$, at r = 1 and $n\to \infty $, we have x = 0 as $\dfrac{1}{\infty }\to 0$ and value of $\dfrac{r}{n}=x$, at r = n and $n\to \infty $, we have x = 1.
So, we can rewrite $\displaystyle \lim_{n \to \infty}\dfrac{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt[3]{\dfrac{r}{n}}}}{\dfrac{1}{n}\sum\limits_{r=1}^{n}{\dfrac{1}{{{\left( a+\dfrac{r}{n} \right)}^{2}}}}}$ as \[\dfrac{\int\limits_{0}^{1}{\sqrt[3]{x}}dx}{\int\limits_{0}^{1}{\dfrac{1}{{{\left( a+x \right)}^{2}}}}dx}\].
So, let ${{I}_{1}}=\int\limits_{0}^{1}{\sqrt[3]{x}}$ and ${{I}_{2}}=\int\limits_{0}^{1}{\dfrac{1}{{{\left( a+x \right)}^{2}}}}$
We know that \[\int\limits_{a}^{b}{f(x)}=F(b)-F(a)\] and \[\int\limits_{{}}^{{}}{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}}+C\]
So, ${{I}_{1}}=\int\limits_{0}^{1}{\sqrt[3]{x}}$
${{I}_{1}}=\left( \dfrac{3}{4}{{x}^{\dfrac{4}{3}}} \right)_{0}^{1}$
${{I}_{1}}=\dfrac{3}{4}\left( 1-0 \right)$
So, ${{I}_{1}}=\dfrac{3}{4}$
Now, ${{I}_{2}}=\int\limits_{0}^{1}{\dfrac{1}{{{\left( a+x \right)}^{2}}}}$
Let, a + x = t
At, x = 0, t = a and at x = 1, t = a + 1
Differentiating both sides, we get
dx = dt
so, ${{I}_{2}}=\int\limits_{0}^{1}{\dfrac{1}{{{\left( a+x \right)}^{2}}}}$ is transformed to ${{I}_{2}}=\int\limits_{a}^{a+1}{\dfrac{1}{{{\left( t \right)}^{2}}}}dt$
${{I}_{2}}=\left( -\dfrac{1}{t} \right)_{a}^{a+1}$
${{I}_{2}}=\left( \dfrac{1}{a}-\dfrac{1}{a+1} \right)$
Putting back value of both integral in \[\dfrac{\int\limits_{0}^{1}{\sqrt[3]{x}}dx}{\int\limits_{0}^{1}{\dfrac{1}{{{\left( a+x \right)}^{2}}}}dx}=54\],
We get, \[\dfrac{\dfrac{3}{4}}{\left( \dfrac{1}{a}-\dfrac{1}{a+1} \right)}=54\]
On simplifying, we get
\[\dfrac{1}{72}=\left( \dfrac{1}{a}-\dfrac{1}{a+1} \right)\]
On solving, we get
${{a}^{2}}+a-72=0$
On factorising, we get
${{a}^{2}}+9a-8a-72=0$
On simplifying, we get
( a + 9 )( a – 8 ) = 0
Or, a = -9, 8

So, the correct answers are “Option a and b”.

Note: To solve such a question must know the concept of limit as summation and also, one should not put limit directly as it will give nothing so one must proceed ahead by simplifying the brackets and then use limit as summation to easily simplify the question. Always remember that \[\int\limits_{a}^{b}{f(x)}=F(b)-F(a)\] and \[\int\limits_{{}}^{{}}{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}}+C\] Try not to make any calculation error as this will make question more complex and hard to solve.