Q.12) Parabola

Q : The point (2, 1) divides a chord of the parabola y² = 4x in two
       equal line segments. Find the equation of the line, 
       containing that chord.

SOLUTION :

The equation of the required chord (bisected at a given point ) is given by S' = T

=> 1² - 4 * 2 = y * 1 - 2 (x + 2)
=> 1 - 8 = y - 2x - 4
=> -7 = y - 2x - 4
=> 2x - y - 3 = 0  => (Ans)

[ Here, S' = y² - 4x at (2, 1)
and T = tangent at (2, 1) ; ----> yy1 = 2a (x + x1) ]

Alternatively :

Let the co-ordinates of the extremities of the chord be P (t1²,2t1) and Q (t2²,2t2)

slope of chord PQ = 2(t2 - t1)/(t2² - t1²) ==> 2/(t1 + t2)
but (2t1 + 2t2)/2 = 1 {mid point formula}
=> 2(t1 + t2) = 2
=> (t1 + t2) = 1
slope = 2/1 = 2

equation of the required chord {with slpoe 2 and passes through (2, 1) } = y - 1 = (2)(x - 2)
=> 2x - y = 3 ------> (Ans)

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Q.11) Inequality

Q : If a, b, c are positive real numbers, then prove that :----

(a + 1)^7 * (b + 1)^7 * (c + 1)^7  >  7^7 * a^4 * b^4 * c^4

SOLUTION :

L.H.S. = (a + 1)^7 * (b + 1)^7 * (c + 1)^7 = [(a + 1)(b + 1)(c + 1) ]^7


=> [1 + a + b + c + ab + bc + ca + abc]^7
now [1 + a + b + c + ab + bc + ca + abc]^7 > [a + b + c + ab + bc + ca + abc]^7
 {taking 1 out }

Using AM ≥ GM, for positive real numbers a, b, c, ab, bc, ca and abc we get,
(a + b + c + ab + bc + ca + abc)/7 ≥ [a^4 * b^4 * c^4]^(1/7)
=> [(a + b + c + ab + bc + ca + abc)/7 ]^7 ≥ [a^4 * b^4 * c^4]
=> (a + b + c + ab + bc + ca + abc)^7 ≥ 7^7 [a^4 * b^4 * c^4] - - - - (ii)

from (i) and (ii),
[1 + a + b + c + ab + bc + ca + abc]^7 > 7^7 [a^4 * b^4 * c^4]
=> (a + 1)^7 * (b + 1)^7 * (c + 1)^7 > 7^7 [a^4 * b^4 * c^4] - - - (Proved)

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Q.10) Definite integral

Q : Evaluate ∫[0 to 1] ln (x + 1)/(x^2 + 1) dx

SOLUTION :

Answer ► I = (π/8) * ln 2


Consider I = ∫[0 to 1] ln(x + 1)/(x^2 + 1) dx
Let x = tan θ, then dx = sec² θ dθ
For limits; when x = 0, θ = 0 and when x = 1,θ = π/4

therefore, I = ∫[0 to 1] ln (x + 1)/(x^2 + 1) dx
=> ∫[0 to π/4] ln (tan θ + 1) sec² θ dθ/(tan² θ + 1)
=> ∫[0 to π/4] ln (tan θ + 1) sec² θ dθ / sec² θ

=> ∫[0 to π/4] ln (1 + tan θ) dθ - - - - - (i)
I = ∫[0 to π/4] ln [1 + tan (π/4 - θ) ] dθ
=> ∫[0 to π/4] ln [1 + { tan (π/4) - tan θ}/ {1 + tan (π/4) * tan θ} ] dθ
=> ∫[0 to π/4] ln [1 + {1 - tan θ}/ {1 + tan θ} ] dθ
=> ∫[0 to π/4] ln [{1 + tan θ + 1 - tan θ}/ {1 + tan θ} ] dθ
=> ∫[0 to π/4] ln [ 2 / {1 + tan θ} ] dθ

=> ∫[0 to π/4] [ln 2 - ln (1 + tan θ) ] dθ - - - - - - (ii)

Adding (i) & (ii),
2I = ∫[0 to π/4] [ ln 2 ] dθ
=> [0 to π/4] [θ * ln 2 ]
=> π/4 * ln 2

therefore, I = π/8 * ln 2 ► Answer

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Q.9) Trigonometry

Q : Prove this trigonometric identity

sin 3x +sin 2x - sin x = 4 sin x cos (x/2) cos (3x/2)

SOLUTION:

L.H.S. = sin3x + (sin 2x - sin x)


=> 2 * sin (3x/2) * cos (3x/2) + 2 * cos (3x/2) *sin (x/2)
=> 2 * cos (3x/2) [sin (3x/2) + sin (x/2) ]
=> 2 * cos (3x/2) [2 * sin (x) * cos (x/2) ]

=> 4 * cos (3x/2) sin x cos (x/2) ==> R.H.S. (Proved)

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Q.8) Integration

Q : Evaluate ∫ (cos 7x - cos 8x) / (1 + 2 cos 5x) dx


SOLUTION :

Let I = ∫ (cos 7x - cos 8x) / (1 + 2 cos 5x) dx


Numerator of the integrand = (cos 7x - cos 8x) = 2 sin (15x/2) * sin (x/2)
Denominator of the integrand = (1 + 2 cos 5x) = cosec (5x/2) [ sin (5x/2) + 2 sin (5x/2) * cos (5x) ]
Therefore, I = ∫ (cos 7x - cos 8x) / (1 + 2 cos 5x) dx

=> I = ∫ (2 sin (15x/2) * sin (x/2) / cosec (5x/2) [ sin (5x/2) + 2 sin (5x/2) * cos (5x) ] dx
=> I = ∫ (2 sin (15x/2) * sin (5x/2) * sin (x/2) / [ sin (5x/2) + 2 sin (5x/2) * cos (5x) ] dx
=> I = ∫ (2 sin (15x/2) * sin (5x/2) * sin (x/2) / [ sin (5x/2) + sin (15x/2) - sin (5x/2) ] dx

=> I = ∫ (2 sin (15x/2) * sin (5x/2) * sin (x/2) / [sin (15x/2) ] dx
=> I = ∫ (2 sin (5x/2) * sin (x/2) dx
=> I = ∫ (cos 2x - cos 3x ) dx

=> (1/2) sin 2x - (1/3) sin 3x + c => (Ans)


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Q.7) Trigonometry

Q : Prove That :--
      (cosα cosβ - sinα sinβ)(cosα cosβ + sinα sinβ) = cos²α - sin²β 

SOLUTION :

L.H.S ► (cosα cosβ - sinα sinβ) (cosα cosβ + sinα sinβ)

⇒ cos²α cos²β - sin²α sin²β - - - { Using, (a - b)(a + b) = a² - b² }
⇒ cos²α cos²β - (1 - cos²α) (1 - cos²β)
⇒ cos²α cos²β - [1 - cos²α - cos²β + cos²α cos²β]
⇒ cos²α cos²β - cos²α cos²β - 1 + cos²α + cos²β

⇒ - 1 + cos²α + cos²β
⇒ (- 1 + cos²β ) + cos²α
⇒ cos²α - (1 - cos²β)
⇒ cos²α - sin²β ► R.H.S. (Proved)


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Q.6) Quadratic Equation

Q : Solve for x ; (x² - 3x + 2)² - 3(x² - 3x + 2) + 2 = x

SOLUTION :
Answer ► x = { 0, 2, 2 ± √2 }


Given : (x² - 3x + 2)² - 3(x² - 3x + 2) + 2 = x

now, (x² - 3x + 2) = (x - 2)(x - 1)
let (x - 2) = y, therefore, (x - 1) = (y + 1)
so, the equation reduces to;
[y(y + 1)]² - 3y(y + 1) - y = 0

=> [y(y + 1)]² - 3y² - 3y - y = 0
=> y² (y² + 2y + 1) - 3y² - 4y = 0
=> y [y^3 + 2y² + y - 3y - 4] = 0
=> y [ y^3 + 2y^2 - 2y - 4] = 0
=> y (y² - 2)(y + 2) = 0
=> y = 0; y = -2; y = ±√2

therefore, x - 2 = 0 ; x - 2 = -2; x - 2 = ±√2
=> x = 2 , x = 0, x = 2 ± √2 ► (Answer)

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Q.5) Binomial Theorem

Q : Two consecutive terms in the expansion of (3 + 2x)^(74) whose coeff. are equal, are ________?



SOLUTION :  Let (r+1)th and (r+2)th terms are two consecutive terms in the expansion of (3+2x)^74.

By condition, T(r + 1) = T(r + 2)
and (r + 1)th term in the expansion of (a + x)^n is given by (nCr) * a^(n - r) * (x)^r

=> (74Cr) * 3^(74 - r) * (2x)^r = [74C(r+1)] * 3^(74 - r - 1) * (2x)^(r+1)
=> (74Cr) * 3^(74 - r) * (2)^r = [74C(r+1)] * 3^(74 - r - 1) * (2)^(r+1) - - {only coefficients}

=> (2^r) / (74 - r)! = 2^(r + 1) / 3 (r + 1) (73 - r)!
=> 1 / (74 - r) = 2 / [3 (r + 1)]
=> 3r + 3 = 148 - 2r
=> 5r = 145
=> r = 29

so, (r + 1)th and (r + 2)th terms are 30th and 31st terms. (Ans)

Q.4) Binomial Theorem

Q : Fractional part of 2^(78 / 31) is ______?
SOLUTION :
Since 2^78 = 2^75 * 2^3


=> 2^78 = (2^5)^15 * 2^3

=> 2^78 = (32)^15 * 2^3

=> 2^78 = (31 + 1)^15 * 2^3


Now, expanding (31 + 1)^15 using Binomial Theorem,

(31 + 1)^15 = C0 * (31^15) + C1 * (31^14) *1 + terms of descending powers of 31. . .+ Cn * 31^0 *1^15 . . {where, C0 , C1, . . Cn are binomial coefficients. }

=> (31 + 1)^15 = (sum of the terms divisible by 31) + 1


(2^78) / 31 = 2^3 * [ (sum of the terms divisible by 31) + 1] / 31

=> 2^3 * (sum of the terms divisible by 31) / 31 + (2^3) / 31

=> 8k + 8 / 31- - - - { where k is any integer}


so, the fractional part = 8 / 31 => (Ans)

Q. 3) INTEGRATION.

Integration Question ?


How do you do this ?



∫ e^(x) dx * [(1 - x)/(1 + x)]²

SOLUTION :  Let I = ∫ e^(x) * [(1 - x)/(1 + x)]² dx


=> ∫ e^(x) * (1 - 2x + x²) / (1 + x)² dx

=> ∫ e^(x) * (4 + x² - 2x - 3) / (1 + x)² dx

=> ∫ e^(x) * [ 4 / (1 + x)² + (x² - 2x - 3) / (1 + x)² ] dx

=> ∫ e^(x) * [ 4 / (1 + x)² + (x -3)(x + 1) / (1 + x)² ] dx

=> ∫ e^(x) * [ 4 / (1 + x)² + (x - 3) / (1 + x) ] dx



Now, using the fact, ∫ e^(x) * [ f(x) + f'(x)] dx = e^(x) * f(x) + C

here, f(x) = (x - 3) / (1 + x) and f'(x) = 4 / (1 + x)²



so, I = e^(x) * (x - 3) / (1 + x) + C ==> (Ans)

Q.2) Parabola.

Is the locus of point of intersection of two normals to a given parabola y^2 = 4ax which meet at right angles



given by y^2 = a (x - 3a) ?

SOLUTION : ►Yes, it is correct! Locus = y² = a(x - 3a)




Let the equations of normals to the parabola y² = 4ax at point P (at²₁, 2at₁)

and Q (at²₂, 2at₂) are

y + t₁x = 2at₁+ at³₁ - - - - (i) and y + t₂x = 2at₂+ at³₂ - - - - (ii)

Let (h, k) be the point of intersection of two perpendicular normals.



so, k + t₁h = 2at₁+ at³₁ - - - - (iii) and k + t₂h = 2at₂+ at³₂ - - - - (iv)

solving simultaneously (iii) & (iv) for intersection point (h,k) we get

h = 2a + a(t²₁+ t₁t₂+ t²₂) and k = t₁(1 - t²₂)

using t₁t₂= -1, {as the normals meet at right angles}



=> h/a = 1 + t²₁+ t²₂and k/a = ( t₁+ t₂)

now, using, ( t₁+ t₂)² = t²₁+ t²₂+ 2 t₁t₂

=> ( t₁+ t₂)² = t²₁+ t²₂- 2

=> ( t₁+ t₂)² = (t²₁+ t²₂+ 1) - 3

=> (k/a)² = (h/a) - 3

=> k² = ah - 3a² - - - - {multiply throughout by a²}

=> y² = ax - 3a²

=> y² = a(x - 3a) ► (Proved)

Q. 1) Binomial Theorem.

Fill in the blank. (Binomial Theorem)
If (1 + x)^n = C(0) + C(1) * x + C(2) * x^2 + . . . . . . + C(n) * x^n, then

(1/2) * C(1) + (1/4) * C(3) + (1/6) * C(5) + . . . is equal to - - - - - - - - - -

[Where, C(1) , C(2) , C(3) . . . C(n) are binomial coefficients ]

SOLUTION ► (2^n - 1) / (n + 1) Answer.

Integrating it from 0 to 1, and from (-1) to 0 and
subtracting second from first result we get,

∫(0 to 1) (1 + x)^n dx = ∫(0 to 1) [ C(0) + C(1) * x + C(2) * x^2 + . . + C(n) * x^n ] dx
=> [2^(n + 1) - 1]/(n + 1) = C(0)/1 + C(1)/2 + C(2)/3 + C(3)/4 + C(4)/5 .+ .+ . - - - - (i)

∫(-1 to 0) (1 + x)^n dx = ∫(-1 to 0) [ C(0) + C(1) * x + C(2) * x^2 + . . + C(n) * x^n ] dx
=>1/(n + 1) = C(0)/1 - C(1)/2 + C(2)/3 - C(3)/4 + C(4)/5 - C(5)/6.+ .+ . - - - - (ii)
=> use, (ii) - (i) and divide by 2,

=> C(1)/2 + C(3)/4 + C(5)/6 + C(7)/8 + .. + ..+ = [2^n - 1]/(n + 1) ► (Ans)