## Solve for the equation cos(π/4(n^2 + 2n)) =sin(π/4(n^2 + n +1)) , where n is an integer

Q: Solve for n the equation $cos(\frac{\pi}{4}(n^2 + 2 n)) = sin(\frac{\pi}{4}(n^2 + n + 1))$ , where n is an integer .

Solution : The given equation is

$cos(\frac{\pi}{4}(n^2 + 2 n)) = sin(\frac{\pi}{4}(n^2 + n + 1))$

$cos(\frac{\pi}{4}(n^2 + 2 n)) = cos(\frac{\pi}{2} – \frac{\pi}{4}(n^2 + n + 1))$

$sin\frac{\pi}{8}(2n^2 + 3 n -1) sin\frac{\pi}{8}(n+1) = 0$

Either $sin\frac{\pi}{8}(n+1) = 0$ or , $sin\frac{\pi}{8}(2n^2 + 3 n -1) = 0$

CaseI : $sin\frac{\pi}{8}(n+1) = 0 = sink\pi$ ; k ∈ I

Case II : $sin\frac{\pi}{8}(2n^2 + 3 n -1) = 0 = sink\pi$; k ∈ I

2n2 + 3n – 1= 8 k , k ∈ I

Let n1 and n2 be two integral values of n such that

$2n_1^2 + 3 n_1 – 1 = 8 k_1$ ….(i)

and $2n_2^2 + 3 n_2 – 1 = 8 k_2$ …(ii)

Subtracting (ii) from (i), we get

(n1 – n2) [2(n1 + n2) + 3] = 8(k1 – k2).

Since 2(n1 + n3) + 3 is odd, n1 – n2 must be a multiple of 8

i.e. n1 – n2 = 8p or n1 = 8p + n2.

For k2 = 1, (ii) gives n2 = -3. Hence n = 8p – 3.

Hence from case I and II, the solution is n = 8p – 1 or n = 8p – 3 for p ∈ I.

## If 2cosA = x + 1/x , 2cosB = y + 1/y . Show that 2cos(A-B) = x/y + y/x

Q: If $\displaystyle 2 cosA = x + \frac{1}{x}$ , $\displaystyle 2 cosB = y + \frac{1}{y}$ . Show that $\displaystyle 2 cos(A-B) = \frac{x}{y} + \frac{y}{x}$

Solution : $\displaystyle 2 cosA = x + \frac{1}{x}$

Since $\displaystyle 4 sin^2 A = 4 – 4 cos^2 A$

$\displaystyle4 sin^2 A = 4 – (x + \frac{1}{x})^2$

$\displaystyle 4 sin^2 A = – [(x + \frac{1}{x})^2 – 4]$

$\displaystyle 4 sin^2 A = i^2 (x-\frac{1}{x})^2$

$\displaystyle 2sinA = i (x-\frac{1}{x})$

$\displaystyle 2 cosB = y + \frac{1}{y}$

$\displaystyle 2sinB = i (y-\frac{1}{y})$

Now $\displaystyle 2 cos(A – B)= 2(cosA cosB + sinA sinB)$

$\displaystyle = 2 \times \frac{1}{4}[(x + \frac{1}{x})(y + \frac{1}{y}) + i^2 (x-\frac{1}{x})(y-\frac{1}{y})]$

$\displaystyle = \frac{1}{2}[xy + \frac{1}{xy} + \frac{y}{x} + \frac{x}{y}] – \frac{1}{2} [xy + \frac{1}{xy} – \frac{y}{x} – \frac{x}{y}]$

$\displaystyle = \frac{1}{2}. 2 (\frac{y}{x}+ \frac{x}{y})$

$\displaystyle = (\frac{y}{x}+ \frac{x}{y})$

## 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 roots of the equation x^2 – 10 c x – 11 d = 0 are a, b and those of x^2 – 10 a x – 11 b = 0 are c , d , then find the value of ….

Q: If roots of the equation x2 – 10 c x – 11 d = 0 are a, b and those of x2 – 10 a x – 11 b = 0 are c , d , then find the value of a + b + c + d. (a, b, c and d are distinct numbers)

Solution : As a + b = 10 c and c + d = 10 a

ab = -11 d , cd = -11 b

⇒ ac = 121 and (b + d) = 9(a + c)

a2 – 10 ac – 11d = 0

c2 – 10 ac – 11b = 0

⇒ a2 + c2 – 20ac – 11(b + d) = 0

⇒ (a + c)2 – 22(121) – 11 × 9(a + c) = 0

⇒ (a + c) = 121 or -22 (rejected)

Hence , a + b + c + d = 1210.

## If α be a root of the equation 4x^2 + 2 x – 1 = 0 , then prove that 4 α^3 – 3α is the other root .

Q: If α be a root of the equation 4 x2 + 2 x – 1 = 0 , then prove that 4 α 3 – 3α is the other root.

Solution : Let f (x) = 4 x2 + 2x – 1

then f (α) = 4α2 + 2α – 1 = 0 …(i)

Now 4 α3 – 3α

= 4 α3 .α – 3α

= (1 – 2α)α – 3α from (i)

= α – 2α2 – 3α

= –2(α2 + α)

$\displaystyle = -2 (\frac{1-2\alpha}{4} + \alpha)$

$\displaystyle = -2 (\frac{1-2\alpha + 4 \alpha}{4} )$

$\displaystyle = – \frac{2 \alpha + 1}{2} = – \alpha – \frac{1}{2}$

Sum of roots = (4 α3 – 3α ) + α = – 1/2

Hence other root is (4 α3 – 3α )

## If α is a root of equation 3ax^2 + 2bx + 6c = 0 and β is a root of equation –3ax^2 + 2bx + 6c = 0 then prove that ….

Q: If α is a root of equation 3ax2 + 2bx + 6c = 0 and β is a root of equation –3ax2 + 2bx + 6c = 0 then prove that the equation 2ax2 + 2bx + 6c = 0 has a root γ which lies between a and b .

Solution : α is root of 3ax2 + 2bx + 6c = 0

⇒ 3a α2 + 2b α + 6c = 0 … (i)

Also β is a root of –3ax2 + 2bx + 6c = 0

⇒ -3aβ2 + 2b β + 6c = 0 …(ii)

Let f(x) = 2ax2 + 2bx + 6c

f(α) = 2a α2 + 2b α + 6c

= 2a α2– 3 a α2 = – a α2 from (i)

f(β) = 2aβ2 + 2b β + 6c

= 2a β2 + 3 a β2 = 5a β2

f(α) f(β) = (- a α2) (5a β2)

= – 5 a2 α2 β2 < 0

2ax2 + 2bx + 6c = 0 has a root γ lying between α and β .