Math Solutions

Q: Let f (x) , x ≥ 0 , be a non negative continuous function and let $ F (x) = \int_{0}^{x}f(t)dt $ , x ≥ 0 . If for some c > 0 , f (x) ≤ c F (x) for all x ≥ 0 , then show that f (x) = 0 for all x ≥ 0.

Solution : We are given that for x ≥ 0

$ F (x) = \int_{0}^{x}f(t)dt $

⇒ F (0) = 0.

As f (x) ≤ c F (x) ∀ x ≥ 0 , we get

f (0) ≤ c F (0) ⇒ f (0) ≤ 0.

Since f (x) ≥ 0 ∀ x ≥ 0, we have

f (0) ≥ 0

⇒ f (0) = 0.

Since f is continuous on [0, ∞), F is differentiable on [0, ∞) and

F’ (x) = f (x) ∀ x ≥ 0.

Since f (x) ≤ c F (x) ∀ x ≥ 0, we get

F ‘ (x) – c F (x) ≤ 0 ∀ x ≥ 0.

Multiplying both the sides by e^–cx(the integrating factor) we get

e^–cx F’ (x) – c e^–cx F (x) ≤ 0 [as e^–cx > 0 ∀ x]

⇒ $\frac{d}{dx}[e^{-cx} F(x)] \le 0$

Thus g (x) = e^–cx F (x) is a decreasing function on [0, ∞).

That is g (x) ≤ g (0) for each x > 0.

But g (0) = e^{–c. 0} F (0) = 0

⇒ g (x) ≤ 0 ∀ x ≥ 0

⇒ e^–cx F (x) ≤ 0 ∀ x ≥ 0

⇒ F (x) ≤ 0 ∀ x ≥ 0

Thus f (x) ≤ c F (x) ≤ 0 ∀ x ≥ 0. (for c > 0)

But we are given that

f (x) ≥ 0 ∀ x ≥ 0.

Hence f (x) = 0 ∀ x ≥ 0.

Q: Function f (x) is defined as $\displaystyle f(x) = x^2 + \int_{-1}^{1} (x^2 + t^2 x^3 )(f (t) – a) dt + a $ , then find the value of ‘a ‘ for which f (x) = 0 has exactly one real root.

Solution : $\displaystyle f(x) = x^2 + \int_{-1}^{1} x^2 [f(t)-a]dt + x^3 \int_{-1}^{1} t^2 (f (t) – a) dt + a $

Let f (x) = x² A + Bx³ + a

Where $\displaystyle A = 1 + \int_{-1}^{1} [f(t)-a]dt $ and $ B = 1 + \int_{-1}^{1}t^2 [f(t)-a]dt $

$\displaystyle A = 1 + \int_{-1}^{1} ( t^2 A + B t^3 )dt $

$\displaystyle A = 1 + 2 \int_{0}^{1} t^2 A dt $

$\displaystyle A = 1 + 2 [ 2 A \frac{t^3}{3}]_{0}^{1} = 1 + \frac{2A}{3} $

$\displaystyle A = 1 + \frac{2A}{3} $

A = 3

And $\displaystyle B = \int_{-1}^{1}t^2 (t^2 A + B t^3) dt $

$\displaystyle B = 2A \int t^4 dt= \frac{2A}{5} $

$\displaystyle B = \frac{6}{5}$

Hence f (x) = 3x² + (6/5)x³ + a

⇒ f ‘ (x) = 6x + (18/5)x² = 6x(1 + 3x/5) = 0

⇒ x = 0 , -5/3

Also f ” (x) = 6 + x. Clearly, f (x) is minimum at x = 0 and maximum at -6/5

Now, from the graph it is clear that f (x) will have exactly one real root if either a > 0 or 17/25 + a < 0

a < -75/27

a < -25/9

Hence $a \in (-\infty , -\frac{25}{9}) \cup (0 , \infty)$

Q: Evaluate $\displaystyle \int_{0}^{t} [|sin^{-1}(sinx)|]dx $ Where t ∈ ( 2nπ , (8n+1)π/4 )

Solution : Since the function is periodic with period p, the given integral can be written as

$\displaystyle \int_{0}^{2n\pi} [|sin^{-1}(sinx)|]dx + \int_{2n\pi}^{t} [|sin^{-1}(sinx)|]dx $

Let I = I₁ + I₂ where

$\displaystyle I_1 = 2n \int_{0}^{\pi} [sin^{-1}(sinx)]dx $

$\displaystyle I_1 = 2n \int_{0}^{\pi/2} [x] dx + \int_{\pi/2}^{\pi} [\pi – x] dx $

$\displaystyle I_1 = 2n [ \int_{0}^{1} 0 dx + \int_{1}^{\pi/2} 1 dx + \int_{\pi/2}^{\pi -1} 1 dx + \int_{\pi -1}^{\pi} 0 dx] $

$\displaystyle I_1 = 2n [(\frac{\pi}{2}-1) + (\pi – 1 – \frac{\pi}{2}) ] $

$\displaystyle I_1 = 2n [\pi – 2 ] $

$\displaystyle I_2 = \int_{2n\pi}^{t} [sin^{-1}(sinx)]dx $

$\displaystyle I_2 = \int_{2n\pi}^{2n\pi + z} [sin^{-1}(sinx)]dx $ Where z ∈ (0 , π/2)

$\displaystyle = \int_{0}^{z}[sin^{-1} sinx]dx $

$\displaystyle = \int_{0}^{z}0 dx $ ; (0 < sin^-1x < 1 , ∀ x ∈ (0 , π/2) )

= 0

Thus , I =  2n [π – 2] +0

I =  2n [π – 2]

Q: $\displaystyle \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{cos^{-1}\frac{2x}{1+x^2} + tan^{-1}\frac{2x}{1-x^2}}{e^x +1}$

Solution : Let tan^-1x = θ

$\displaystyle – 1/\sqrt{3} \le x \le 1/\sqrt{3}$

$\displaystyle – \pi/6 \le \theta \le \pi/6 $

$\displaystyle cos^{-1}\frac{2x}{1+x^2} + tan^{-1}\frac{2x}{1-x^2} $

$\displaystyle = cos^{-1}\frac{2 tan\theta}{1+tan^2\theta} + tan^{-1}\frac{2 tan\theta}{1-tan^2\theta} $

$\displaystyle = cos^{-1}(sin2\theta) + tan^{-1}(tan2\theta)$

$\displaystyle = cos^{-1}cos(\frac{\pi}{2}-2\theta) + tan^{-1}(tan2\theta)$

$\displaystyle = \frac{\pi}{2}- 2 \theta + 2 \theta = \frac{\pi}{2} $

$\displaystyle I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{cos^{-1}\frac{2x}{1+x^2} + tan^{-1}\frac{2x}{1-x^2}}{e^x +1}$

$\displaystyle = \frac{\pi}{2} \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{dx}{e^x + 1} $

$\displaystyle = \frac{\pi}{2} \times \frac{1}{\sqrt{3}}$

Q: Prove that :  $\displaystyle \int_{0}^{\pi/2} \frac{a + b cosx}{(b + acosx)^2} dx = \frac{1}{b}$

Solution : We have to prove that $\displaystyle \int_{0}^{\pi/2} \frac{a + b cosx}{(b + acosx)^2} dx = \frac{1}{b}$

Multiplying numerator and denominator of LHS by cosec²x , we get

$\displaystyle L.H.S = \int_{0}^{\pi/2} \frac{a cosec^2 x + b cotx cosecx }{(b cosecx + a cotx)^2} dx $

$\displaystyle = – \int_{0}^{\pi/2} \frac{- a cosec^2 x – b cotx cosecx }{(b cosecx + a cotx)^2} dx $

$\displaystyle = – [\frac{1}{b cosecx + a cotx}]_{0}^{\pi/2} $

$\displaystyle = [\frac{sinx}{b + a cosx}]_{0}^{\pi/2} = \frac{1}{b}$