Question

Please help me answer this calc A question

Answer 1

Answer:

B

Step-by-step explanation:

The limit in this problem is the definition of a derivative.

So basically we can say that the left side of the equation is the derivative of Arcsin(x).

From our tables or derivatives of inverse trig functions we know the derivative of arcsin(x), or arcsin(a) is:

$\frac{d}{dx}ArcSin(a)=\frac{1}{\sqrt{1-a^2} }$

So, we equate the Left Side to 2 (in the equation) and solve for a:

$\frac{1}{\sqrt{1-a^2} }=2\\1=2\sqrt{1-a^2} \\\frac{1}{2}=\sqrt{1-a^2}\\(\frac{1}{2})^2=(\sqrt{1-a^2})^2\\\frac{1}{4}=1-a^2\\a^2=1-\frac{1}{4}\\a^2=\frac{3}{4}\\a=\sqrt{\frac{3}{4}} \\a=\frac{\sqrt{3} }{\sqrt{4} }\\a=\frac{\sqrt{3} }{2}$

So, the value of a is  $\frac{\sqrt{3} }{2}$

The correct answer choice is B