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Definite Integration question

2022 · 27 Jun · Shift 2 · Q27
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  5. /2022 · 27 Jun · Shift 2 · Q27

Definite Integration question

2022 · 27 Jun · Shift 2 · Q27

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The integral ∫0117[1x]dx\int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}0∫1​7[x1​]1​dx, where [ . ] denotes the greatest integer function, is equal to
  1. A
    1+6log⁡e(67)1 + 6{\log _e}\left( {{6 \over 7}} \right)1+6loge​(76​)
  2. B
    1−6log⁡e(67)1 - 6{\log _e}\left( {{6 \over 7}} \right)1−6loge​(76​)
  3. C
    log⁡e(76){\log _e}\left( {{7 \over 6}} \right)loge​(67​)
  4. D
    1−7log⁡e(67)1 - 7{\log _e}\left( {{6 \over 7}} \right)1−7loge​(76​)
View written solutionFree

Correct answer: A

  1. We need to evaluate I=∫0117[1x] dx.I=\int_0^1 \frac{1}{7^{\left[\frac1x\right]}}\,dx.I=∫01​7[x1​]1​dx.

  2. Since x∈(0,1]x\in(0,1]x∈(0,1], we have 1x∈[1,∞)\frac1x\in[1,\infty)x1​∈[1,∞). The value of [1x]\left[\frac1x\right][x1​] stays constant on intervals where n≤1x<n+1,n\le \frac1x < n+1,n≤x1​<n+1, for integers n≥1n\ge 1n≥1.

This is equivalent to 1n+1<x≤1n.\frac{1}{n+1}<x\le \frac{1}{n}.n+11​<x≤n1​. Hence, on the interval (1n+1,1n],\left(\frac{1}{n+1},\frac{1}{n}\right],(n+11​,n1​], we have [1x]=n.\left[\frac1x\right]=n.[x1​]=n.

  1. Therefore, split the integral as a sum: I=∑n=1∞∫1/(n+1)1/n17n dx.I=\sum_{n=1}^{\infty}\int_{1/(n+1)}^{1/n} \frac{1}{7^n}\,dx.I=∑n=1∞​∫1/(n+1)1/n​7n1​dx. Since 17n\frac{1}{7^n}7n1​ is constant on each interval, I=∑n=1∞17n(1n−1n+1).I=\sum_{n=1}^{\infty} \frac{1}{7^n}\left(\frac1n-\frac{1}{n+1}\right).I=∑n=1∞​7n1​(n1​−n+11​). Now, 1n−1n+1=1n(n+1).\frac1n-\frac{1}{n+1}=\frac{1}{n(n+1)}.n1​−n+11​=n(n+1)1​. So, I=∑n=1∞17nn(n+1).I=\sum_{n=1}^{\infty}\frac{1}{7^n n(n+1)}.I=∑n=1∞​7nn(n+1)1​.

  2. Use the identity 1n(n+1)=1n−1n+1.\frac{1}{n(n+1)}=\frac1n-\frac{1}{n+1}.n(n+1)1​=n1​−n+11​. Thus,

=∑n=1∞1n7n−∑n=1∞1(n+1)7n.=\sum_{n=1}^{\infty}\frac{1}{n7^n}-\sum_{n=1}^{\infty}\frac{1}{(n+1)7^n}.=n=1∑∞​n7n1​−n=1∑∞​(n+1)7n1​.

Let S1=∑n=1∞1n7n.S_1=\sum_{n=1}^{\infty}\frac{1}{n7^n}.S1​=∑n=1∞​n7n1​. Using the standard series ∑n=1∞rnn=−ln⁡(1−r),∣r∣<1,\sum_{n=1}^{\infty}\frac{r^n}{n}=-\ln(1-r), \quad |r|<1,∑n=1∞​nrn​=−ln(1−r),∣r∣<1, with r=17r=\frac17r=71​, we get S1=−ln⁡(1−17)=−ln⁡(67)=ln⁡(76).S_1=-\ln\left(1-\frac17\right)=-\ln\left(\frac67\right)=\ln\left(\frac76\right).S1​=−ln(1−71​)=−ln(76​)=ln(67​).

  1. Now compute S2=∑n=1∞1(n+1)7n.S_2=\sum_{n=1}^{\infty}\frac{1}{(n+1)7^n}.S2​=∑n=1∞​(n+1)7n1​. Let m=n+1m=n+1m=n+1. Then n=m−1n=m-1n=m−1, so
=7∑m=2∞1m7m.=7\sum_{m=2}^{\infty}\frac{1}{m7^m}.=7m=2∑∞​m7m1​.

Hence,

=7(S1−17)=7S1−1.=7\left(S_1-\frac17\right)=7S_1-1.=7(S1​−71​)=7S1​−1.
  1. Therefore, I=S1−S2=S1−(7S1−1)=1−6S1.I=S_1-S_2=S_1-(7S_1-1)=1-6S_1.I=S1​−S2​=S1​−(7S1​−1)=1−6S1​. Substitute S1=ln⁡(76)=−ln⁡(67)S_1=\ln\left(\frac76\right)=-\ln\left(\frac67\right)S1​=ln(67​)=−ln(76​):
=1+6\ln\left(\frac67\right).$$ 7. So the value of the integral is $$\boxed{1+6\ln\left(\frac67\right)}.$$ 8. Checking options: - **A:** $1+6\log_e\left(\frac67\right)$ ✅ - **B:** $1-6\log_e\left(\frac67\right)$ ❌ - **C:** $\log_e\left(\frac76\right)$ ❌ - **D:** $1-7\log_e\left(\frac67\right)$ ❌ Hence, the correct option is **A**.
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