Mathematics

# The value of $\int_{0}^{[x]} (x-[x])dx$, where $[x]$ is the greatest integer $|le x$ is equal to

$4[x]$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 One Word Medium
$\int { \frac { 2x+3 }{ 3x+2 } dx }$=$\dfrac{A}{3}x+\dfrac{B}{3}ln(3x+2)+c$
Find value of B-A

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int _ { - 4 } ^ { - 5 } e ^ { ( x + 5 ) ^ { 2 } } d x + 3 \int _ { 1 / 3 } ^ { 2 / 3 } e ^ { 9 ( x - 2 / 3 ) ^ { 2 } } d x$ is equal to-
• A. $e ^ { 5 }$
• B. $e ^ { 4 }$
• C.
• D. 3$e ^ { 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle \underset{0}{\overset{x}{\int}} \dfrac{(t - |t|)^2}{(1 + t^2)} dt$ is equal to
• A. $0 \, if \, x > 0$
• B. $\ln ( 1 + x^3) \, \ \text{if} \, x > 0$
• C. $4(x + \tan^{-1} x ) \, \text{if} \, x < 0$
• D. $4(x - \tan^{-1} x), \, \ \text{if} \, x < 0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate : $\displaystyle \int \sec x\tan x\sqrt{\sec ^{2}x+1}dx.$
• A. $\displaystyle \frac{1}{2}t\sqrt{t^{2}+1}-\frac{1}{2}\log \left [ t+\sqrt{t^{2}+1} \right ].$ where $t=\sec x$
• B. $\displaystyle t\sqrt{t^{2}+1}+\log \left [ t+\sqrt{t^{2}+1} \right ].$ where $t=\sec x$
• C. $\displaystyle \frac{1}{2}t\sqrt{t^{2}+1}+\frac{1}{2}\log \left [ t-\sqrt{t^{2}+1} \right ].$ where $t=\sec x$
• D. $\displaystyle \frac{1}{2}t\sqrt{t^{2}+1}+\frac{1}{2}\log \left [ t+\sqrt{t^{2}+1} \right ].$ where $t=\sec x$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Hard
$\displaystyle \lim_{n\rightarrow \infty} \left (\dfrac {(n + 1) (n + 2) .....3n}{n^{2n}} \right )^{\frac {1}{n}}$ is equal to:
• A. $\dfrac {18}{e^{4}}$
• B. $\dfrac {9}{e^{2}}$
• C. $3\log 3 - 2$
• D. $\dfrac {27}{e^{2}}$