Question

If y = log_ex³ + 3 sin^–1x + kx² and y ¢$\left( \frac{1}{2} \right) = 2\sqrt{3}$ then k =

• A. 6
• B. – 6
• C. $2\sqrt{3}$
• D. None

Answer 1

Answer: (B)

– 6

Answer 2

$\frac{dy}{dx} = \frac{3x^{2}}{x^{3}} + \frac{3}{\sqrt{1 - x^{2}}}$+ k(2x)

Ž$\frac{dy}{dx}_{\left( x = \frac{1}{2} \right)} = 6 + \frac{6}{\sqrt{3}}$ + k

Now $\frac{dy}{dx}_{\left( x = \frac{1}{2} \right)} = 2\sqrt{3}$ given

\ 6 + ($6/\sqrt{3}$) + k = $2\sqrt{3}$ Ž k = – 6