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Limits Continuity and Differentiability question

2021 · 20 Jul · Shift 2 · Q30
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  5. /2021 · 20 Jul · Shift 2 · Q30

Limits Continuity and Differentiability question

2021 · 20 Jul · Shift 2 · Q30

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If f:R→Rf:R \to Rf:R→R is given by f(x)=x+1f(x) = x + 1f(x)=x+1, then the value of lim⁡n→∞1n[f(0)+f(5n)+f(10n)+......+f(5(n−1)n)]\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]n→∞lim​n1​[f(0)+f(n5​)+f(n10​)+......+f(n5(n−1)​)] is :
  1. A
    32{3 \over 2}23​
  2. B
    52{5 \over 2}25​
  3. C
    12{1 \over 2}21​
  4. D
    72{7 \over 2}27​
View written solutionFree

Correct answer: D

  1. We are given f(x)=x+1.f(x)=x+1.f(x)=x+1.

The expression is lim⁡n→∞1n[f(0)+f(5n)+f(10n)+⋯+f(5(n−1)n)].\lim_{n\to\infty}\frac1n\left[f(0)+f\left(\frac5n\right)+f\left(\frac{10}n\right)+\cdots+f\left(\frac{5(n-1)}n\right)\right].limn→∞​n1​[f(0)+f(n5​)+f(n10​)+⋯+f(n5(n−1)​)].

  1. Write the general term.

The terms inside the bracket are f(5kn),k=0,1,2,…,n−1.f\left(\frac{5k}{n}\right),\quad k=0,1,2,\dots,n-1.f(n5k​),k=0,1,2,…,n−1.

So the sum is ∑k=0n−1f(5kn).\sum_{k=0}^{n-1} f\left(\frac{5k}{n}\right).∑k=0n−1​f(n5k​).

Hence the limit becomes lim⁡n→∞1n∑k=0n−1f(5kn).\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1} f\left(\frac{5k}{n}\right).limn→∞​n1​∑k=0n−1​f(n5k​).

  1. Substitute f(x)=x+1f(x)=x+1f(x)=x+1.

f(5kn)=5kn+1.f\left(\frac{5k}{n}\right)=\frac{5k}{n}+1.f(n5k​)=n5k​+1.

Therefore,

=\frac1n\left(\frac5n\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}1\right).$$ 4. Use standard sums. We know $$\sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}, \qquad \sum_{k=0}^{n-1}1=n.$$ So, $$\frac1n\left(\frac5n\cdot \frac{(n-1)n}{2}+n\right) =\frac1n\left(\frac{5(n-1)}{2}+n\right).$$ Now simplify: $$=\frac{5(n-1)}{2n}+1 =\frac{5n-5}{2n}+1.
  1. Take the limit as n→∞n\to\inftyn→∞.
=\frac52+1 =\frac72.$$ 6. Compare with options. $$\frac72$$ corresponds to **Option D**. 7. Verification with stored answer. Stored correct answer: **D**. Our derived answer is also **D**. Hence they agree.
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