1
JEE Advanced 2023 Paper 2 Online
+3
-1
Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f(1)=\frac{1}{3}$ and $3 \int\limits_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is :
A
$\frac{e^2+4}{3}$
B
$\frac{\log _e 4+e}{3}$
C
$\frac{4 e^2}{3}$
D
$\frac{e^2-4}{3}$
2
JEE Advanced 2017 Paper 2 Offline
+3
-1
If y = y(x) satisfies the differential equation

$${8\sqrt x \left( {\sqrt {9 + \sqrt x } } \right)dy = {{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)}^{ - 1}}}$$

dx, x > 0 and y(0) = $$\sqrt 7$$, then y(256) =
A
16
B
3
C
9
D
80
3
JEE Advanced 2014 Paper 2 Offline
+4
-1
The function $$y=f(x)$$ is the solution of the differential equation
$${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}\,$$ in $$(-1,1)$$ satisfying $$f(0)=0$$.
Then $$\int\limits_{ - {{\sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {f\left( x \right)} \,d\left( x \right)$$ is
A
$${\pi \over 3} - {{\sqrt 3 } \over 2}$$
B
$${\pi \over 3} - {{\sqrt 3 } \over 4}$$
C
$${\pi \over 6} - {{\sqrt 3 } \over 4}$$
D
$${\pi \over 6} - {{\sqrt 3 } \over 2}$$
4
JEE Advanced 2013 Paper 1 Offline
+4
-1
A curve passes through the point $$\left( {1,{\pi \over 6}} \right)$$. Let the slope of
the curve at each point $$(x,y)$$ be $${y \over x} + \sec \left( {{y \over x}} \right),x > 0.$$
Then the equation of the curve is
A
$$sin\left( {{y \over x}} \right) = \log x + {1 \over 2}$$
B
$$cos\,ec\left( {{y \over x}} \right) = \log x + 2$$
C
$$\,s\,ec\left( {{{2y} \over x}} \right) = \log x + 2\,$$
D
$$\,cos\left( {{{2y} \over x}} \right) = \log x + {1 \over 2}$$
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