A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is
A
$${3 \over 2}{x^2}$$
B
$$\sqrt {{{{x^3}} \over 8}} $$
C
$${1 \over 2}{x^2}$$
D
$$\pi {x^2}$$
Explanation
Area $$ = {1 \over 2}{x^2}\,\sin \,\theta $$
Maximum value of $$\sin \theta $$ is $$1$$ at $$\theta = {\pi \over 2}$$
$${A_{\max }} = {1 \over 2}{x^2}$$
2
AIEEE 2006
MCQ (Single Correct Answer)
The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at
$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-
tween $$\left( {0,\alpha } \right)$$
Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$
4
AIEEE 2005
MCQ (Single Correct Answer)
The normal to the curve
$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point
$$\theta\, '$$ is such that
A
it passes through the origin
B
it makes an angle $${\pi \over 2} + \theta $$ with the $$x$$-axis
C
it passes through $$\left( {a{\pi \over 2}, - a} \right)$$