A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:
A
$$x=2r$$
B
$$2x=r$$
C
$$2x = \left( {\pi + 4} \right)r$$
D
$$\left( {4 - \pi } \right)x = \pi \,\, r$$
Explanation
$$4x + 2\pi r = 2$$ $$\,\,\,$$ $$ \Rightarrow 2x + \pi r = 1$$
Let $$f(x)$$ be a polynomial of degree four having extreme values
at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :
If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$
A
$$f'\left( c \right) = g'\left( c \right)$$
B
$$f'\left( c \right) = 2g'\left( c \right)$$
C
$$2f'\left( c \right) = g'\left( c \right)$$
D
$$2f'\left( c \right) = 3g'\left( c \right)$$
Explanation
Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$
and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that
Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$
4
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
The intercepts on $$x$$-axis made by tangents to the curve,
$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :
A
$$ \pm 1$$
B
$$ \pm 2$$
C
$$ \pm 3$$
D
$$ \pm 4$$
Explanation
Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$
therefore $${{dy} \over {dx}} = \left| x \right|$$
But from $$y = 2x,{{dy} \over {dx}} = 2$$
$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$