JEE Advanced 2019

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: For $a \in R, ∣ a ∣ > 1$ , let $lim_{n \to \infty} \frac{1 + 3 2 + \dots + 3 n}{n ^{7/3} ( \frac{1}{( an + 1 ) ^{2}} + \frac{1}{( an + 2 ) ^{2}} + \dots + \frac{1}{( an + n ) ^{2}} )} = 54$ . Then the possible value(s) of $a$ is/are :

Q: How do I know when to factor out $n$?

Whenever you see a sum where the terms depend on $n$ and $r$, your goal is to create the ratio $r/n$. If you have $(an+r)$, dividing by $n$ gives $(a + r/n)$. This is the universal key to unlocking these problems. Always look for that $r/n$ pattern!

Q: What if the integral limits weren't 0 to 1?

The standard Riemann sum $\frac{1}{n} \sum_{r=1}^n f(r/n)$ always maps to $\int_0^1 f(x) dx$. If the sum went from $1$ to $2n$, the limits would change. Always check the range of $r/n$ as $n \to \infty$. In this problem, $r$ goes from $1$ to $n$, so $r/n$ goes from $1/n$ to $1$, which is $0$ to $1$ in the limit.

Select Answer:

* Multiple Correct

Visualized Solution

The Core Concept

The problem involves a limit of a complex series as $n \to \infty$ .
Key Concept: Definite Integral as a Limit of a Sum.
Formula: $lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$

Summation Form of Numerator

Numerator: $1 + 32 + \dots + 3 n$
Rewrite using sigma notation: $\sum_{r = 1}^{n} r^{\frac{1}{3}}$
To create the required $(\frac{r}{n})$ form, we can write $r = n \cdot (\frac{r}{n})$ .
$\sum_{r = 1}^{n} n^{\frac{1}{3}} (\frac{r}{n})^{\frac{1}{3}} = n^{\frac{1}{3}} \sum_{r = 1}^{n} (\frac{r}{n})^{\frac{1}{3}}$

Summation Form of Denominator

Denominator series: $\frac{1}{( an + 1 ) ^{2}} + \frac{1}{( an + 2 ) ^{2}} + \dots + \frac{1}{( an + n ) ^{2}}$
Rewrite using sigma notation: $\sum_{r = 1}^{n} \frac{1}{( an + r ) ^{2}}$
Factor out $n$ from the denominator inside the sum:
$\sum_{r = 1}^{n} \frac{1}{n ^{2} ( a + \frac{r}{n} ) ^{2}} = \frac{1}{n ^{2}} \sum_{r = 1}^{n} \frac{1}{( a + \frac{r}{n} ) ^{2}}$

Combining and Simplifying $n$

Original Expression: $\frac{Numerator}{n ^{\frac{7}{3}} \times Denominator Series}$
Substitute the derived forms:
$\frac{n ^{\frac{1}{3}} \sum _{r = 1}^{n} ( \frac{r}{n} ) ^{\frac{1}{3}}}{n ^{\frac{7}{3}} \cdot \frac{1}{n ^{2}} \sum _{r = 1}^{n} \frac{1}{( a + \frac{r}{n} ) ^{2}}}$
Simplify the powers of $n$ in the denominator: $n^{\frac{7}{3}} \cdot n^{- 2} = n^{\frac{7}{3} - \frac{6}{3}} = n^{\frac{1}{3}}$

Introducing the $\frac{1}{n}$ Factor

The $n^{\frac{1}{3}}$ terms cancel out completely:
$\frac{\sum _{r = 1}^{n} ( \frac{r}{n} ) ^{\frac{1}{3}}}{\sum _{r = 1}^{n} \frac{1}{( a + \frac{r}{n} ) ^{2}}}$
To apply the integral formula, we need a $\frac{1}{n}$ outside the sum.
Multiply numerator and denominator by $\frac{1}{n}$ :
$\frac{\frac{1}{n} \sum _{r = 1}^{n} ( \frac{r}{n} ) ^{\frac{1}{3}}}{\frac{1}{n} \sum _{r = 1}^{n} \frac{1}{( a + \frac{r}{n} ) ^{2}}} = 54$

Converting Sums to Integrals

Apply $lim_{n \to \infty} \frac{1}{n} \sum f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$
Numerator becomes: $I_{1} = \int_{0}^{1} x^{\frac{1}{3}} d x$
Denominator becomes: $I_{2} = \int_{0}^{1} \frac{1}{( a + x ) ^{2}} d x$
The equation transforms to: $\frac{I _{1}}{I _{2}} = 54$

Evaluating $I_{1}$

$I_{1} = \int_{0}^{1} x^{\frac{1}{3}} d x$
Use power rule: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$
$I_{1} = [\frac{x ^{\frac{4}{3}}}{\frac{4}{3}}]_{0}^{1}$
$I_{1} = \frac{3}{4} (1 - 0) = \frac{3}{4}$

Evaluating $I_{2}$

$I_{2} = \int_{0}^{1} (a + x)^{- 2} d x$
$I_{2} = [\frac{- 1}{a + x}]_{0}^{1}$
Apply limits: $(\frac{- 1}{a + 1}) - (\frac{- 1}{a + 0})$
$I_{2} = \frac{1}{a} - \frac{1}{a + 1} = \frac{1}{a ( a + 1 )}$

Substituting Integrals Back

We had: $\frac{I _{1}}{I _{2}} = 54$
Substitute $I_{1} = \frac{3}{4}$ and $I_{2} = \frac{1}{a ( a + 1 )}$
$\frac{\frac{3}{4}}{\frac{1}{a ( a + 1 )}} = 54$
$\frac{3}{4} a (a + 1) = 54$

Solving the Quadratic Equation

$a (a + 1) = 54 \times \frac{4}{3}$
$a (a + 1) = 72$
$a^{2} + a - 72 = 0$
Factorize: $(a + 9) (a - 8) = 0$
Roots: $a = - 9 or a = 8$

Final Check and Conclusion

Possible values: $a = - 9, 8$
Check given condition: $∣ a ∣ > 1$
$∣ - 9∣ = 9 > 1$ (Valid)
$∣8∣ = 8 > 1$ (Valid)
Final Answer: Both $- 9$ and $8$ are correct.
Key Takeaway: Always look for $\frac{1}{n}$ and $\frac{r}{n}$ when dealing with limits of infinite series.

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The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

The given problem involves a limit as

n \to \infty

of a complex summation. In JEE Advanced mathematics, such expressions are frequently disguised Riemann Sums.

The fundamental theorem states that as

n \to \infty

, the discrete sum can be represented as:

n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x

This identity serves as our primary tool for simplifying the expression.

Transforming the Numerator

The numerator is given by

1 + 32 + \dots + 3 n

, which is the summation

\sum_{r = 1}^{n} r^{1/3}

. To align this with the Riemann form, we express

r

n \cdot (r / n)

r^{1/3} = n^{1/3} \cdot (\frac{r}{n})^{1/3}

Factoring out the

n^{1/3}

, the numerator becomes:

n^{1/3} r = 1 \sum n (\frac{r}{n})^{1/3}

Transforming the Denominator

The denominator involves terms of the form

\sum_{r = 1}^{n} \frac{1}{( an + r ) ^{2}}

. By factoring

n

out of the squared term, we obtain:

\frac{1}{n ^{2}} r = 1 \sum n (a + \frac{r}{n})^{- 2}

When combined with the

n^{7/3}

factor in the original expression, the powers of

n

cancel out perfectly. Multiplying the numerator and denominator by

\frac{1}{n}

allows us to transition into the integral form.

The Master Equation

The expression simplifies to the ratio of two definite integrals:

\frac{\int _{0}^{1} x ^{1/3} d x}{\int _{0}^{1} ( a + x ) ^{- 2} d x} = 54

Calculating the numerator integral:

\int_{0}^{1} x^{1/3} d x = [\frac{3}{4} x^{4/3}]_{0}^{1} = \frac{3}{4}

Calculating the denominator integral:

\int_{0}^{1} (a + x)^{- 2} d x = [- (a + x)^{- 1}]_{0}^{1} = \frac{1}{a} - \frac{1}{a + 1} = \frac{1}{a ( a + 1 )}

Final Calculation

Equating the ratio to 54, we have:

\frac{3/4}{1/ ( a ( a + 1 ))} = 54

\frac{3}{4} a (a + 1) = 54 \Rightarrow a (a + 1) = 72

Solving the quadratic equation

a^{2} + a - 72 = 0

, we find

(a + 9) (a - 8) = 0

. Given the constraint

∣ a ∣ > 1

, both

a = 8

and

a = - 9

are valid solutions.

The final values for

a

are $8$ and $- 9$ .

Common Traps & FAQs

Wait, why can't I just use L'Hopital's rule here?

L'Hopital's rule is designed for indeterminate forms of functions, like

0/0

\infty/\infty

. Here, you have a summation that grows with

n

. You need to convert the discrete sum into a continuous integral to evaluate the limit. Think of it as moving from a jagged staircase to a smooth ramp.

How do I know when to factor out

n

Whenever you see a sum where the terms depend on

n

and

r

, your goal is to create the ratio

r / n

. If you have

(an + r)

, dividing by

n

gives

(a + r / n)

. This is the universal key to unlocking these problems. Always look for that

r / n

pattern!

What if the integral limits weren't 0 to 1?

The standard Riemann sum

\frac{1}{n} \sum_{r = 1}^{n} f (r / n)

always maps to

\int_{0}^{1} f (x) d x

. If the sum went from

1

2 n

, the limits would change. Always check the range of

r / n

n \to \infty

. In this problem,

r

goes from

1

n

, so

r / n

goes from

1/ n

1

, which is

0

1

in the limit.

Animated Solution for Mathematics - Definite Integration: For $a \in R, ∣ a ∣ > 1$ , let $lim_{n \to \infty} \frac{1 + 3 2 + \dots + 3 n}{n ^{7/3} ( \frac{1}{( an + 1 ) ^{2}} + \frac{1}{( an + 2 ) ^{2}} + \dots + \frac{1}{( an + n ) ^{2}} )} = 54$ . Then the possible value(s) of $a$ is/are :

Select Answer:

Visualized Solution

The Core Concept

Summation Form of Numerator

Summation Form of Denominator

Combining and Simplifying $n$

Introducing the $\frac{1}{n}$ Factor

Converting Sums to Integrals

Evaluating $I_{1}$

Evaluating $I_{2}$

Substituting Integrals Back

Solving the Quadratic Equation

Final Check and Conclusion

The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

Transforming the Numerator

Transforming the Denominator

The Master Equation

Final Calculation

Similar Questions

For $a \in R$ (the set of all real numbers), $a \neq = - 1, lim_{n \to \infty} \frac{( 1 ^{a} + 2 ^{a} + \dots + n ^{a} )}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + \dots + ( na + n )]} = \frac{1}{60}$ . Then $a = .........$

For positive integer $n$ , define $f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$ . Then, the value of $lim_{n \to \infty} f (n)$ is equal to

$lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots + \frac{1}{n} sec^{2} 1]$ equals

$lim_{n \to \infty} (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})$ is equal to :

If $f : R \to R$ is given by $f (x) = x + 1$ , then the value of $lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + \dots + f (\frac{5 ( n - 1 )}{n})]$ is:

$lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to

$lim_{n \to \infty} \frac{3}{n} \sum_{r = 0}^{n - 1} (2 + \frac{r}{n})^{2}$ is equal to

$lim_{n \to \infty} \frac{1 ^{4} + 2 ^{4} + 3 ^{4} + \dots + n ^{4}}{n ^{5}} - lim_{n \to \infty} \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + n ^{3}}{n ^{5}}$

$lim_{n \to \infty} \frac{1 ^{p} + 2 ^{p} + 3 ^{p} + \dots + n ^{p}}{n ^{p + 1}}$

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{j = 1}^{n} \frac{( 2 j - 1 ) + 8 n}{( 2 j - 1 ) + 4 n}$ is equal to :

Select Answer:

Visualized Solution

The Core Concept

Summation Form of Numerator

Summation Form of Denominator

Combining and Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Introducing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​ Factor

Converting Sums to Integrals

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Substituting Integrals Back

Solving the Quadratic Equation

Final Check and Conclusion

The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

Transforming the Numerator

Transforming the Denominator

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a∈R (the set of all real numbers), a=−1,limn→∞​(n+1)a−1[(na+1)+(na+2)+⋯+(na+n)](1a+2a+⋯+na)​=601​. Then a=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integer n, define f(n)=n+4n+3n216+5n−3n2​+8n+3n232+n−3n2​+12n+3n248−3n−3n2​+⋯+7n225n−7n2​. Then, the value of limn→∞​f(n) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​[n21​sec2n21​+n22​sec2n24​+⋯+n1​sec21] equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​np+11p+2p+3p+⋯+np​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limn→∞​n1​∑j=1n​(2j−1)+4n(2j−1)+8n​ is equal to :

Combining and Simplifying $n$

Introducing the $\frac{1}{n}$ Factor

Evaluating $I_{1}$

Evaluating $I_{2}$

For $a \in R$ (the set of all real numbers), $a \neq = - 1, lim_{n \to \infty} \frac{( 1 ^{a} + 2 ^{a} + \dots + n ^{a} )}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + \dots + ( na + n )]} = \frac{1}{60}$ . Then $a = .........$

For positive integer $n$ , define $f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$ . Then, the value of $lim_{n \to \infty} f (n)$ is equal to

$lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots + \frac{1}{n} sec^{2} 1]$ equals

$lim_{n \to \infty} (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})$ is equal to :

If $f : R \to R$ is given by $f (x) = x + 1$ , then the value of $lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + \dots + f (\frac{5 ( n - 1 )}{n})]$ is:

$lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to

$lim_{n \to \infty} \frac{3}{n} \sum_{r = 0}^{n - 1} (2 + \frac{r}{n})^{2}$ is equal to

$lim_{n \to \infty} \frac{1 ^{4} + 2 ^{4} + 3 ^{4} + \dots + n ^{4}}{n ^{5}} - lim_{n \to \infty} \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + n ^{3}}{n ^{5}}$

$lim_{n \to \infty} \frac{1 ^{p} + 2 ^{p} + 3 ^{p} + \dots + n ^{p}}{n ^{p + 1}}$

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{j = 1}^{n} \frac{( 2 j - 1 ) + 8 n}{( 2 j - 1 ) + 4 n}$ is equal to :