1. Hi friends! I recently gave BCA maths exam and it was bit tough for me as i didn't had maths at +2 level. It consisted of Integration,differentiation,matrices,advance stats. etc.

Differentiate $\sin^{-1} \left(2x \sqrt{1 - x^2}\right)$ w.r.t x.

2nd question:-

Find maximum and minimum values of f(x) in the interval [0,3], where:-

f(x)= 3x^4 - 8x^3 + 12x^2 - 48x + 25

I just learned about finding maxima and minima of 2nd degree function.

2. Originally Posted by vivek_master146
Hi friends! I recently gave BCA maths exam and it was bit tough for me as i didn't had maths at +2 level. It consisted of Integration,differentiation,matrices,advance stats. etc.

Differentiate $\sin^{-1} \left(2x \sqrt{1 - x^2}\right)$ w.r.t x.

2nd question:-

Find maximum and minimum values of f(x) in the interval [0,3], where:-

f(x)= 3x^4 - 8x^3 + 12x^2 - 48x + 25

I just learned about finding maxima and minima of 2nd degree function.
Q1. Let $w = 2x \sqrt{1 - x^2}$ and use the chain rule: $\frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dx}$.

To get $\frac{dw}{dx}$ use the product rule. Note that to get the derivative of $\sqrt{1 - x^2}$ you might need to use the chain rule (although you might be capable of doing it by inspection).

Q2. Solve $f'(x) = 0$ to get (as it happens) the single stationary point. Note: $x^3 - 2x^2 + 2x - 4 = (x - 2)(x^2 + 2)$. Test the nature of this stationary point.

Note that the maximum and minimum values of f(x) will occur either at the turning point or an endpoint of the interval.

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