## 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 ….

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.

## Function f (x) is defined as $f(x) = x^2 + \int_{-1}^{1} (x^2 + t^2 x^3 )(f (t) – a) dt + a$ , then find the value of a ….

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

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

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]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## If value of $\int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0)$ the values of A and B are

Q: If value of $\displaystyle \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0)$ the values of A and B are

(a) $\displaystyle A =\frac{\pi}{2} , B = 0$

(b) $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha}$

(c) $\displaystyle A =\frac{\pi}{6} , B = \frac{\pi}{ sin\alpha}$

(d) $\displaystyle A =\pi , B = \frac{\pi}{ sin\alpha}$

Ans: (a) , (b)

Sol: $\displaystyle I = \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx}$

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{cos^2 \frac{x}{2} + sin^2 \frac{x}{2}-cos\alpha (cos^2 \frac{x}{2} – sin^2 \frac{x}{2})}$

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{(1-cos\alpha) cos^2 \frac{x}{2} + (1 + cos\alpha) sin^2 \frac{x}{2}}$

$\displaystyle I = \int_{0}^{\alpha} \frac{sec^2 \frac{x}{2} dx}{2 sin^2 \frac{\alpha}{2} + 2 cos^2 \frac{\alpha}{2} tan^2 \frac{x}{2}}$

$\displaystyle I = \frac{1}{2} \int_{0}^{\alpha} \frac{sec^2 \frac{\alpha}{2} sec^2 \frac{x}{2} dx}{tan^2 \frac{\alpha}{2} + tan^2 \frac{x}{2}}$

Let $tan\frac{x}{2} = t$

$\displaystyle I = \int_{0}^{tan(\alpha/2)} \frac{sec^2 \frac{\alpha}{2} dt }{t^2 + tan^2 \frac{\alpha}{2}}$

$\displaystyle I = sec^2 \frac{\alpha}{2} cot\frac{\alpha}{2} [tan^{-1}\frac{t}{tan(\alpha/2)}]_{0}^{tan(\alpha/2)}$

$\displaystyle = \frac{2}{sin\alpha}.\frac{\pi}{4}= \frac{\pi}{2 sin\alpha}$

$\displaystyle A =\frac{\pi}{2} , B = 0$

Also , $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha}$

Hence (a), (b) are correct answers.

## The value of $\int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x}$ is

Q: The value of $\displaystyle \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x}$ is

(A) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a > 0 , b > 0$

(B) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a < 0 , b < 0$

(C) $\displaystyle \frac{\pi}{4} ; a = 1 , b = 1$

(D) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{a}{b}) + \frac{1}{ab}$

Ans: (A) , (B) , (C)

Sol: $\displaystyle I = \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x}$

Dividing Nr and Dr by cos2x ,

$\displaystyle I = \int_{0}^{\pi/4} \frac{sec^2 x dx}{a^2 + b^2 tan^2 x}$

Put tanx = t ; sec2x dx = dt

$\displaystyle I = \int_{0}^{1} \frac{dt}{a^2 + b^2 t^2}$

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + \frac{a^2}{b^2}}$

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + (\frac{a}{b})^2}$

$\displaystyle = \frac{b}{a b^2} [tan^{-1}\frac{b t}{a}]_{0}^{1}$

$\displaystyle = \frac{1}{a b} [tan^{-1}\frac{b}{a}]$ ; (a ≠0 , b≠ 0)

Also for a = 1, b = 1

$\displaystyle I = \int_{0}^{\pi/4}dx = \frac{\pi}{4}$

Hence (A), (B), (C) are correct answers.

## If $f(x) = \int_{x}^{x^2} \frac{dt}{(logt)^2} , x \ne 0$ then f(x) is

Q: If $\displaystyle f(x) = \int_{x}^{x^2} \frac{dt}{(logt)^2} , x \ne 0$ then f(x) is

(A) monotonically increasing in (2 , ∞)

(B) monotonically decreasing in (1 , 2)

(C) monotonically decreasing in (2 , ∞)

(D) monotonically decreasing in (0 , 1)

Ans: (A) , (B)

Sol: $\displaystyle f'(x) = \frac{2x}{(logx^2)^2} – \frac{1}{(logx)^2}$

$\displaystyle f'(x) = \frac{x-2}{2(logx)^2}$

f'(x) > 0 for x > 2

and f'(x) < 0 for x < 2

Hence (A), (B) are correct answers.

## $\int_{0}^{\pi} x f(sin x) dx$ is equal to

Q: $\displaystyle \int_{0}^{\pi} x f(sin x) dx$ is equal to

(A) $\frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx$

(B) $\pi \int_{0}^{\pi/2} f(sin x) dx$

(C) $2 \pi \int_{0}^{\pi/2} f(sin x) dx$

(D) none of these

Ans: (A) , (B)

Sol: Let $\displaystyle I = \int_{0}^{\pi} x f(sin x) dx$

$\displaystyle = \int_{0}^{\pi} (\pi – x ) f sin (\pi-x) dx$

$\displaystyle I = \pi \int_{0}^{\pi} f(sin x) dx – I$

$\displaystyle 2 I = \pi \int_{0}^{\pi} f(sin x) dx$

$\displaystyle I = \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx$

$\displaystyle I = 2 \times \frac{\pi}{2} \int_{0}^{\pi/2} f(sin x) dx$

$\displaystyle I = \pi \int_{0}^{\pi/2} f(sin x) dx$

Hence (A) and (B) are the correct answers.

## If $I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}}$ , then

Q: If $\displaystyle I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}}$ , then

(A) 0 < I < 1

(B) $I > \frac{\pi}{2\sqrt{2}}$

(C) I < √2 π

(D) I > 2 π

Ans: (B) , (C)

Sol: Since x ∈ [0 , π/2 ]

⇒ 1 ≤ 1 + sin3 x  ≤  2

$\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + sin^3 x}} \le 1$

$\int_{0}^{\pi/2} \frac{1}{\sqrt{2}} dx \le \int_{0}^{\pi/2} \frac{1}{\sqrt{1 + sin^3 x}} dx \le \int_{0}^{\pi/2} 1 dx$

$\frac{\pi}{2 \sqrt{2}} \le I \le \frac{\pi}{2}$

Hence (B) and (C) are the correct answers.