Let a + b = 4, where a < 2, and let g(x) be a differentiable function. If dg/dx > 0 for all x ….

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

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 ….

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

Numerical

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

Find a polynomial f(x) of degree 5 which increases in the interval (-∞, 2] and [6, ∞) and decreases in the interval …

Q: Find a polynomial f(x) of degree 5 which increases in the interval (-∞, 2] and [6, ∞) and decreases in the interval [2, 6]. Given that f(0) = 3 and f ‘(4) = 0.

Sol: The wavy cure of derivative will be like

⇒ f ‘(x)= k(x– 2) (x – 4)2(x – 6) and k > 0

Numerical

$\displaystyle f(x) = k\int (x^2 -8x + 12)(x^2 -8x + 16)dx $

$\displaystyle = k \int x^4 – 16 x^3 + 64 x^2 + 28 (x^2 – 8 x) + 192 ) dx$

$\displaystyle = k ( \frac{x^5}{5} – 4 x^4 + 92 \frac{x^3}{3}-112 x^2 + 192 x ) + c ]$

Now , f(0) = 3

$\displaystyle f(x) = k ( \frac{x^5}{5} – 4 x^4 + 92 \frac{x^3}{3}-112 x^2 + 192 x ) + 3 $ , k > 0

Find the shortest distance between the curves 9x^2 + 9y^2 – 30y + 16 = 0 and y^2 = x^3 ….

Q: Find the shortest distance between the curves 9x2 + 9y2 – 30y + 16 = 0 and y2 = x3

Sol: 9x2 + 9y2 – 30y + 16 = 0 can be rewritten as $\displaystyle x^2 + (y-\frac{5}{3})^2 = 1$

Any point on the curve y2 = x3 can be taken as (t2 , t3 ).

Let l be the distance between the centre of the given circle and the point

( t2 , t3), then L = l2 = t4 + ( t3 –5/3)2.

Now, we calculate the minimum value of l,

required distance = l – radius of given circle.

Now, $\displaystyle \frac{dL}{dt} = 4 t^3 + 2(t^3 – \frac{5}{3})3 t^2 = 0$

for maximum or minimum, t = 0 or 1.

Now, $\displaystyle \frac{d^2L}{dt^2}= 12t^2 + 30t^4 – 20t $

$\displaystyle \frac{d^2L}{dt^2}|_{t=0}= 0$

But, $\displaystyle \frac{d^3L}{dt^3}|_{t=0} \ne 0$ ⇒ There is neither maxima nor minima at t = 0.

Also, $\displaystyle \frac{d^2L}{dt^2} > 0$ at t = 1 ⇒ L is minimum at t = 1.

So, shortest distance = (value of I at t = 1) – (radius of the circle) = $\displaystyle \frac{\sqrt{13}}{3} – 1$

Let A(p^2 , – p), B(q^2, q), C(r^2, – r) be the vertices of a triangle ABC. A parallelogram AFDE is drawn with D, E and F on the line segments …..

Q: Let A(p2 , – p), B(q2, q), C(r2, – r) be the vertices of a triangle ABC. A parallelogram AFDE is drawn with D, E and F on the line segments BC, CA and AB respectively. Show that the maximum area of the parallelogram is $\displaystyle \frac{1}{4}(p+q)(q+r)(p-r)$ , given p > r.

Sol: Let AF = λ AB, AE = μ AC

Area of parallelogram = AF. AE sinA

In similar Δ’s ABC and FBD ,

$\displaystyle \frac{FB}{AB} = \frac{BD}{BC} = \frac{FD}{AC}$

⇒ 1 – λ = μ

Area = λ μ AB. AC sinA

= λ(1-λ)AB.AC sinA

Area is max. when λ is 1/2 (vertex of parabola y = λ – λ2)

⇒ μ = λ = 1/2 i.e. F and E are mid points of AB and AC respectively.

$\displaystyle Area_{max} = \frac{1}{4}AB.AC sinA $

$\displaystyle = \frac{1}{2}Area \; of \; \Delta ABC $

$ \large = \frac{1}{2} \left| \begin{array}{ccc} p^2 & -p & 1 \\ q^2 & q & 1 \\ r^2 & -r & 1 \end{array} \right| $

$\displaystyle = \frac{1}{2} (p+q)(q+r)(p-r)$