1
JEE Advanced 2015 Paper 2 Offline
Numerical
+4
-0
Let $$f:R \to R$$ be a continuous odd function, which vanishes exactly at one point and $$f\left( 1 \right) = {1 \over {2.}}$$ Suppose that $$F\left( x \right) = \int\limits_{ - 1}^x {f\left( t \right)dt}$$ for all $$x \in \,\,\left[ { - 1,2} \right]$$ and $$G(x)=$$ $$\int\limits_{ - 1}^x {t\left| {f\left( {f\left( t \right)} \right)} \right|} dt$$ for all $$x \in \,\,\left[ { - 1,2} \right].$$ If $$\mathop {\lim }\limits_{x \to 1} {{F\left( x \right)} \over {G\left( x \right)}} = {1 \over {14}},$$ then the value of $$f\left( {{1 \over 2}} \right)$$ is
2
JEE Advanced 2015 Paper 1 Offline
Numerical
+4
-0
Let $$f:R \to R$$ be a function defined by
$$f\left( x \right) = \left\{ {\matrix{ {\left[ x \right],} & {x \le 2} \cr {0,} & {x > 2} \cr } } \right.$$ where $$\left[ x \right]$$ is the greatest integer less than or equal to $$x$$, if $$I = \int\limits_{ - 1}^2 {{{xf\left( {{x^2}} \right)} \over {2 + f\left( {x + 1} \right)}}dx,}$$ then the value of $$(4I-1)$$ is
3
JEE Advanced 2015 Paper 1 Offline
Numerical
+4
-0
Let $$F\left( x \right) = \int\limits_x^{{x^2} + {\pi \over 6}} {2{{\cos }^2}t\left( {dt} \right)}$$ for all $$x \in R$$ and $$f:\left[ {0,{1 \over 2}} \right] \to \left[ {0,\infty } \right]$$ be a continuous function. For $$a \in \left[ {0,{1 \over 2}} \right],\,$$ $$F'(a)+2$$ is the area of the region bounded by $$x=0, y=0, y=f(x)$$ and $$x=a,$$ then $$f(0)$$ is
4
JEE Advanced 2014 Paper 1 Offline
Numerical
+4
-0
The value of $$\int\limits_0^1 {4{x^3}\left\{ {{{{d^2}} \over {d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx}$$ is
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