Qus : 121
1
Let $f:\mathbb{R}\to(0,\infty)$ be a strictly increasing function such that
$\displaystyle \lim_{x\to\infty}\frac{f(7x)}{f(x)}=1$.
Then the value of $\displaystyle \lim_{x\to\infty}\Big[\frac{f(5x)}{f(x)}-1\Big]$ is:
✓ Solution
Qus : 122
4
$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $,
where [t] denotes greatest integer less than or equal to t, is equal to:
✓ Solution
Qus : 123
3
The value of the integral \(\displaystyle \int_{1}^{2} \left(\frac{t^{4}+1}{t^{6}+1}\right) dt\) is:
1
\(\tan^{-1}\!\left(\frac{1}{2}\right) - \frac{1}{3}\tan^{-1}(8) + \frac{\pi}{3}\)
2
\(\tan^{-1}(2) - \frac{1}{3}\tan^{-1}(8) + \frac{\pi}{3}\)
3
\(\tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3}\)
4
\(\tan^{-1}\!\left(\frac{1}{2}\right) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3}\)
✓ Solution
Qus : 124
3
Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y - 2x - 6 = 0$ in the second quadrant.
Then the integral $I = \int_a^b \dfrac{9x^2}{1 + 5x^4},dx$ is equal to:
✓ Solution
Qus : 125
1
The value of the integral \(\displaystyle \int_{1/2}^{2} \frac{\tan^{-1}x}{x}\,dx\) is equal to:
1
\(\displaystyle \frac{\pi}{2}\,\log_e 2\)
2
\(\displaystyle \frac{\pi}{4}\,\log_e 2\)
3
\(\displaystyle \frac{1}{2}\,\log_e 2\)
4
\(\displaystyle \pi\,\log_e 2\)
✓ Solution
Qus : 126
2
If [x] is the greatest integer $\le$ x, then ${\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx$ is equal to :
✓ Solution
Qus : 127
1
$\displaystyle 4 \int_0^1 \left(\dfrac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right) dx - 3 \log_e(\sqrt{3})$ is equal to:
1
$2 - \sqrt{2} - \log_e(1 + \sqrt{2})$
2
$2 + \sqrt{2} + \log_e(1 + \sqrt{2})$
3
$2 + \sqrt{2} - \log_e(1 + \sqrt{2})$
4
$2 - \sqrt{2} + \log_e(1 + \sqrt{2})$
✓ Solution
Qus : 128
3
If $\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $, then I equals
1
$ \int_{0}^{1} \left( 1 + \sqrt{1-y^2} \right) , dy $
2
$ \int_{0}^{1} \left( \dfrac{y^2}{2} - \sqrt{1-y^2} + 1 \right) , dy $
3
$ \int_{0}^{1} \left( 1 - \sqrt{1-y^2} \right) , dy $
4
$ \int_{0}^{1} \left( \dfrac{y^2}{2} + \sqrt{1-y^2} + 1 \right) , dy $
✓ Solution
Qus : 129
2
Let $f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $. Then f(3) – f(1) is eqaul to :
1
$ - {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}$
2
$ {\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}$
3
$ - {\pi \over {6}} + {1 \over 2} + {{\sqrt 3 } \over 4}$
4
$ {\pi \over {6}} + {1 \over 2} - {{\sqrt 3 } \over 4}$
✓ Solution
Qus : 130
1
$ \text{If [t] denotes the greatest integer } \le t, \text{ then the value of } \frac{3(e-1)}{e} \int_{1}^{2} x^2 e^{\lfloor x \rfloor + \lfloor x^3 \rfloor} dx \text{ is:} $
✓ Solution