The number of tangent lines from the point $(1,-1)$ to the parabola $y=x^2$ is:
0
1
2
3
Explanation: Tangent at $(t,t^2)$: $y-t^2=2t(x-t)$. Passing through $(1,-1)$: $-1-t^2=2t(1-t)=2t-2t^2$. $t^2-2t-1=0$. Discriminant $=4+4=8>0$. Two real solutions, so 2 tangent lines.
Question 2 of 6easy
The minimum value of $f(x) = x + \frac{1}{x}$ for $x>0$ is:
$0$
$1$
$2$
$-2$
Explanation: $f'=1-1/x^2=0\Rightarrow x=1$. $f''=2/x^3>0$, so minimum. $f(1)=2$.
Question 3 of 6hard
The largest area of a rectangle inscribed in a triangle with base $b$ and height $h$ is:
$bh/4$
$bh/2$
$bh$
$bh/8$
Explanation: For a triangle, the inscribed rectangle of max area has dimensions $b/2$ (width) and $h/2$ (height), giving area $bh/4$.
Question 4 of 6hard
The point on the curve $y^2=4x$ closest to the point $(2,1)$ is:
$(1,2)$
$(4,4)$
$(1,-2)$
$(0,0)$
Explanation: Parametrize as $(t^2,2t)$. $D^2=(t^2-2)^2+(2t-1)^2$. Differentiate: $4t(t^2-2)+2(2t-1)\cdot2=0$. $4t^3-8t+8t-4=0$. $4t^3=4$, $t=1$. Point $(1,2)$.
Question 5 of 6hard
Number of critical points of $f(x)=|x(x-1)|$ is:
1
2
3
0
Explanation: $f'=0$ at $x=0$ and $x=1$ (where $|x(x-1)|=0$, corners exist). And at $x=1/2$ where $x(x-1)$ is minimum. So 3 critical points.
Question 6 of 6hard
The semi-vertical angle of a cone of maximum volume inscribed in a sphere of radius $R$ is:
$\sin^{-1}(1/3)$
$\tan^{-1}\sqrt{2}$
$\cos^{-1}(1/3)$
$\pi/4$
Explanation: Maximising $V=\frac{\pi}{3}(R^2-(R-h)^2)h$ (where $h$ is height from base to apex through sphere). Setting $dV/dh=0$ gives $h=4R/3$, and the semi-vertical angle satisfies $\cos\alpha=1/3$.
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