Application of derivative

Q: Find all the possible values of the parameter ‘a ‘ so that the function,
f (x) = x³ – 3 (7 – a) x² – 3 (9- a²) x + 2 , has a negative point of local minimum.

Sol. f(x) = x³ –3(7-a) x² –3(9-a²)x + 2

f ‘(x) = 3x² –6(7-a) x – 3(9-a²)

For distinct real roots D > 0

36(7-a)² + 4 ×3×3 (9 – a²) > 0

⇒ 49 + a² –14a + 9 –a² > 0

14a < 58 ⇒ a < 29/7 For local minima f ”(x) = 6x – 6(7 – a) > 0

⇒ x – 7 + a > 0

7 – a < x as x must be -ve

⇒ 7-a < 0 ⇒ a > 7

Thus, by contradiction i.e. for real roots a < and for negative point of local minimum a > 7. No possible value of a.

Q: $ \large Let \; f(x) = \left\{\begin{array}{ll} -x^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \; , 0 \leq x < 1 \\ 2x – 3 \; , 1 \leq x \leq 3 \end{array} \right. $

Find all possible real values of b such that f(x) has the smallest value at x = 1.

Sol. At x = 1, f(x) = -1

⇒ Smallest value of f(x) = -1

⇒ at all other points of the interval, f(x) > -1.

Now, for x ≥ 1, f(x) = 2x –3

⇒ f ‘(x) = 2 > 0 ⇒ f(x) is an increasing function

⇒ least value exists at x =1.

Again, for x < 1, f ‘(x) = -3x² < 0

⇒ f(x) is decreasing function in the interval 0 ≤ x < 1.

Therefore, f(x) is smallest at x =1 provided

$\displaystyle f(1-0) = \lim_{h\rightarrow 0} -(1-h)^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle -1 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq 0 $

$\displaystyle \frac{(b^2 + 1)(b – 1}{(b + 1)(b +2 )} \geq 0 $

$\displaystyle \frac{(b – 1}{(b + 1)(b +2 )} \geq 0 $ ; Since (b^2 + 1) is always +ve

⇒ b ∈ (-2 ,-1) ∪ [ 1 , ∞)

Q: Let a + b = 4, where a < 2, and let g(x) be a differentiable function. If dg/dx > 0 for all x , Prove that $\displaystyle \int_{0}^{a} g(x) dx + \int_{0}^{b} g(x) dx $ increases as (b – a) increases.

Sol. Let b – a = t

given a + b = 4

so a = 2 -t/2 , b = 2 + t/2

Let $\displaystyle f(t) = \int_{0}^{a} g(x) dx + \int_{0}^{b} g(x) dx $

$\displaystyle f(t) = \int_{0}^{2-\frac{t}{2}} g(x) dx + \int_{0}^{2+\frac{t}{2}} g(x) dx $

$\displaystyle f'(t) = g(2-\frac{t}{2})(-\frac{1}{2}) + g(2+\frac{t}{2})(\frac{1}{2}) $

$\displaystyle f'(t) = \frac{1}{2}[g(b)- g(a)]$

Since $\frac{dg}{dx} > 0$ for all x, so g(x) is increasing since b > a

g(b) > g(a)

Hence, f'(t) > 0

f(t) increasing as t increases

i.e. f(t) increases as (b – a) increases

Q: A conical vessel is to be prepared out of a circular sheet of gold of unit radius. How much sectorial area is to be removed from the sheet so that vessel has maximum volume.

Sol: Lateral height of cone = Radius of circle = 1

Lateral area of cone = Area of circle with sector removed

$\displaystyle \pi (r) (1) = \frac{\pi (1)^2}{2 \pi}(2 \pi – 2 \theta) $

$\displaystyle r = \frac{\pi – \theta}{\pi}$ , (here r is radius of cone)

Height of the cone $\displaystyle h = \sqrt{1^2 – r^2}$

Volume of the cone $\displaystyle = \frac{1}{3} \pi r^2 h $

$\displaystyle = \frac{1}{3} \pi (\frac{\pi – \theta}{\pi})^2 (\sqrt{1^2 – r^2}) $

$\displaystyle = \frac{1}{3} \pi (\frac{\pi – \theta}{\pi})^2 (\sqrt{1 – (\frac{\pi – \theta}{\pi})^2}) $

On maximizing V , we get

$\displaystyle \frac{\pi – \theta}{\pi} = \sqrt{\frac{2}{3}} $

$\displaystyle \theta = \pi (1 – \sqrt{\frac{2}{3}}) $

Area of sector removed $\displaystyle = \frac{1}{2}(1^2)(2 \theta) = \pi (1 – \sqrt{\frac{2}{3}})$