1
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
The common tangents to the circle $${x^2} + {y^2} = 2$$ and the parabola $${y^2} = 8x$$ touch the circle at the points $$P, Q$$ and the parabola at the points $$R$$, $$S$$. Then the area of the quadrilateral $$PQRS$$ is
A
$$3$$
B
$$6$$
C
$$9$$
D
$$15$$
2
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let $$a, r, s, t$$ be nonzero real numbers. Let $$P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)$$ and $$S\,\,\left( {a{s^2},2as} \right)$$ be distinct points on the parabola $${y^2} = 4ax$$. Suppose that $$PQ$$ is the focal chord and lines $$QR$$ and $$PK$$ are parallel, where $$K$$ is the point $$(2a,0)$$

The value of $$r$$ is

A
$$ - {1 \over t}$$
B
$${{{t^2} + 1} \over t}$$
C
$$ {1 \over t}$$
D
$${{{t^2} - 1} \over t}$$
3
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-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 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let $$f:\left[ {0,2} \right] \to R$$ be a function which is continuous on $$\left[ {0,2} \right]$$ and is differentiable on $$(0,2)$$ with $$f(0)=1$$. Let
$$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} $$ for $$x \in \left[ {0,2} \right]$$. If $$F'\left( x \right) = f'\left( x \right)$$ for all $$x \in \left[ {0,2} \right]$$, then $$F(2)$$ equals
A
$${e^2} - 1$$
B
$${e^4} - 1$$
C
$$e - 1$$
D
$${e^4}$$
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