Problem : 20) Find the second derivative of (d²y/dx²) ,
if (x³ - 3xy + y³ ) = 1
Solution :
Differentiating, we get 3x² - 3x * dy/dx + 3y² * dy/dx ) = 0
=> dy/dx = ( y - x²) / (y² - x) - - - - - (1)
Now, note that:
(i) (y² - x) * (dy/dx) = ( y - x²) ;
(ii) ( y - x²) * (dy/dx) = (y - x²)² / (y² - x)
(iii) 3x²y - x⁴ - xy³ = -x (x³ - 3xy + y³ ) ==> -x
Differentiating (1) , we get,
[(y² - x) * (dy/dx - 2x) - (y - x²) * (2y * (dy/dx) - 1) ] / (y² - x)²
=> [(y² - x) * (dy/dx) - 2x * (y² - x) - 2y * (y - x²) (dy/dx) + (y - x²)] / (y² - x)²
Using (i) & (ii) ,
. . [(y - x²) - 2x *(y² - x) - 2y * {(y - x²)² / (y² - x)} + y - x² ] / (y² - x)²
. . [ 2y - 2x² - 2xy² + 2x² - 2y * {(y - x²)² / (y² - x)} ] / (y² - x)²
. . 2y [ (1 - xy) * (y² - x) - {y² + x⁴ - 2x²y} ] /(y² - x)³
. . 2y [y² - x - xy³ + x²y - y² - x⁴ + 2x²y ] /(y² - x)³
. .2y [ - x - xy³ + x²y - x⁴ + 2x²y ] /(y² - x)³
. . 2y [- x + (-x) * (x³ - 3xy + y³ ) ] /(y² - x)³
Using (iii) we have,
==> 2y * (-2x) /(y² - x)³
==> -4xy / (y² - x)³ ==>(Answer)
Monday, November 22, 2010
Algebra
Problem : 19 )
if x + y = 1, and (x² + y²) (x³ + y³ ) = 26, then
what is the value of (x² + y²) ?
Solution :
Problem : Given x + y = 1 and (x² + y²) (x³ + y³ ) = 26
. .(x² + y²) (x³ + y³ ) = 26
=> [(x + y)² - 2xy] [(x + y)³ - 3xy (x + y)] = 26
=> (1 - 2xy) (1 - 3xy) = 26
=> 1 - 5xy + 6x²y² = 26
=> 6x²y² - 5xy - 25 = 0
=> 6x²y² - 15xy + 10xy - 25 = 0
=> (3xy + 5)(2xy - 5) = 0
=> xy = -5/3 and xy = 5/2
Now, x² + y² = (x + y)² - 2xy
=> x² + y² = 1 - 2xy
=> x² + y² = 1 - 2 * (-5/3)
=> 13 / 3 ==> (Answer)
Note : If we take xy = 5/2, x² + y² = -ve , which is not possible as sum of two real squares will never be negative
if x + y = 1, and (x² + y²) (x³ + y³ ) = 26, then
what is the value of (x² + y²) ?
Solution :
Problem : Given x + y = 1 and (x² + y²) (x³ + y³ ) = 26
. .(x² + y²) (x³ + y³ ) = 26
=> [(x + y)² - 2xy] [(x + y)³ - 3xy (x + y)] = 26
=> (1 - 2xy) (1 - 3xy) = 26
=> 1 - 5xy + 6x²y² = 26
=> 6x²y² - 5xy - 25 = 0
=> 6x²y² - 15xy + 10xy - 25 = 0
=> (3xy + 5)(2xy - 5) = 0
=> xy = -5/3 and xy = 5/2
Now, x² + y² = (x + y)² - 2xy
=> x² + y² = 1 - 2xy
=> x² + y² = 1 - 2 * (-5/3)
=> 13 / 3 ==> (Answer)
Note : If we take xy = 5/2, x² + y² = -ve , which is not possible as sum of two real squares will never be negative
Sunday, November 21, 2010
Inequality
Problem : 18 )
Prove the inequality (1 + a) (1 + b) (1 + c) ≥ 8 (1 - a) (1 - b) (1 - c)
a, b, c are positive numbers and their sum is 1. Also indicate when equality occurs.
Solution :
Given : a, b, c > 0 and a + b + c = 1
=> (1 − a) > 0 ;(1 − b) > 0 ; (1 − c) > 0
Therefore, a = (1 − b − c ) ==> 1 + a = (1 − b) + (1 − c)
and. . . . . b = (1 − a − c ) ==> 1 + b = (1 − a) + (1 − c)
similarly, . c = (1 − a − b ) ==>1 + c = (1 − a) + (1 − b)
Using A.M.- G.M. Inequality ;
(1 + a) = (1 − b) + (1 − c) ≥ 2 √[(1 − b) × (1 − c)] - - -(1)
(1 + b) = (1 − a) + (1 − c) ≥ 2 √[(1 − a) × (1 − c)] - - - -(2)
(1 + c) = (1 − a) + (1 − b) ≥ 2 √[(1 − a) × (1 − b)] - - - - (3)
Multiplying (1), (2) and (3) we get,
(1 + a)(1 + b)(1 + c) ≥ 8 (1 − a)(1 − b)(1 − c)
Equality holds true if a = b = c = 1/3
=> (1 − a) > 0 ;(1 − b) > 0 ; (1 − c) > 0
Therefore, a = (1 − b − c ) ==> 1 + a = (1 − b) + (1 − c)
and. . . . . b = (1 − a − c ) ==> 1 + b = (1 − a) + (1 − c)
similarly, . c = (1 − a − b ) ==>1 + c = (1 − a) + (1 − b)
Using A.M.- G.M. Inequality ;
(1 + a) = (1 − b) + (1 − c) ≥ 2 √[(1 − b) × (1 − c)] - - -(1)
(1 + b) = (1 − a) + (1 − c) ≥ 2 √[(1 − a) × (1 − c)] - - - -(2)
(1 + c) = (1 − a) + (1 − b) ≥ 2 √[(1 − a) × (1 − b)] - - - - (3)
Multiplying (1), (2) and (3) we get,
(1 + a)(1 + b)(1 + c) ≥ 8 (1 − a)(1 − b)(1 − c)
Equality holds true if a = b = c = 1/3
Geometry : Circles touching externally
Problem : 17 ) Circles A, B, and C are externally tangent to each other and internally tangent to circle D. Circles B and C are congruent. Circle A has a radius of 1 unit and passes through the center of circle D. Find the radius of circle B.
Solution :
Radius (r) = 8/9 units.
here is the diagram in the link given below.
http://i854.photobucket.com/albums/ab104…
Note :
If two circles touch each other externally or internally then, the point of contact and the centers of the touching circles will be COLLINEAR.
Further, since the radius of circle A is 1 unit and it passes through the center of the bigger circle, therefore the radius of the bigger circle will be 2 units and hence DB = (2 - r).
Link to Y/A
here is the diagram in the link given below.
http://i854.photobucket.com/albums/ab104…
Note :
If two circles touch each other externally or internally then, the point of contact and the centers of the touching circles will be COLLINEAR.
Further, since the radius of circle A is 1 unit and it passes through the center of the bigger circle, therefore the radius of the bigger circle will be 2 units and hence DB = (2 - r).
Link to Y/A
Converse of Mid-Point Theorem
Problem 16 : ) Prove that :-
"The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side and half of the other side"
Solution :
Refer to the diagram given in the link below.
http://i854.photobucket.com/albums/ab104…
Let D is the mid-point of AB of Δ ABC and
DE drawn is parallel to BC.
To Prove That : E is the mid-point of AC.
Proof :
Extend DE and draw a line // BD ( or BA) which meets the extended DE at F .
=> BCFD is a //gm. {since opp. sides are parallel }
=> AD = DB = CF .
In Δs ADE and CFE,
1) ∠ADE = ∠CFE. . . . . . (Alt ∠s)
2) ∠AED = ∠ CEF . . . . . (Vert. opp. ∠s)
3) AD = CF
=> Δ ADE ≅ Δ CFE . . . . .(By A.A.S Test)
=> AE = CE and DE = EF . . . . . . .(C.S.C.T. )
Which shows that E is the mid-point of AC .
Further, BC = FD . . . {opp. sides are equal and parallel}
and DE = (1/2) FD
=> DE = (1/2) BC. ===> (Proved)
Link to Y/A
"The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side and half of the other side"
Solution :
Refer to the diagram given in the link below.
http://i854.photobucket.com/albums/ab104…
Let D is the mid-point of AB of Δ ABC and
DE drawn is parallel to BC.
To Prove That : E is the mid-point of AC.
Proof :
Extend DE and draw a line // BD ( or BA) which meets the extended DE at F .
=> BCFD is a //gm. {since opp. sides are parallel }
=> AD = DB = CF .
In Δs ADE and CFE,
1) ∠ADE = ∠CFE. . . . . . (Alt ∠s)
2) ∠AED = ∠ CEF . . . . . (Vert. opp. ∠s)
3) AD = CF
=> Δ ADE ≅ Δ CFE . . . . .(By A.A.S Test)
=> AE = CE and DE = EF . . . . . . .(C.S.C.T. )
Which shows that E is the mid-point of AC .
Further, BC = FD . . . {opp. sides are equal and parallel}
and DE = (1/2) FD
=> DE = (1/2) BC. ===> (Proved)
Link to Y/A
Saturday, November 20, 2010
Trigonometry
Problem : 15) Prove that :
sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)
Solution :
Problem : sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)
L.H.S. = sinA + sin5A + sin3A + sin7A
=> (sinA + sin7A) + (sin5A + sin3A) - { Rearranging the terms }
=> 2 sin 4A cos 3A + 2 sin 4A cos A
=> 2 sin 4A [ cos 3A + cos A ]
=> 2 sin 4A * 2 cos 2A cos A
=> 4 sin 2A cos 2A * 2 cos 2A cos A
=> 8 sin A * cos A *cos 2A * 2 cos 2A cos A
=> 16 sin A cos² A * cos² (2A) ==> R.H.S.
Note : 1) sin C + sin D = 2 * sin [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 2)cos C + cos D = 2 * cos [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 3) sin 2A = 2sin A * cos A
Link to Y/A
sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)
Solution :
Problem : sinA + sin5A + sin3A + sin7A = 16 sin A cos²A cos² (2A)
L.H.S. = sinA + sin5A + sin3A + sin7A
=> (sinA + sin7A) + (sin5A + sin3A) - { Rearranging the terms }
=> 2 sin 4A cos 3A + 2 sin 4A cos A
=> 2 sin 4A [ cos 3A + cos A ]
=> 2 sin 4A * 2 cos 2A cos A
=> 4 sin 2A cos 2A * 2 cos 2A cos A
=> 8 sin A * cos A *cos 2A * 2 cos 2A cos A
=> 16 sin A cos² A * cos² (2A) ==> R.H.S.
Note : 1) sin C + sin D = 2 * sin [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 2)cos C + cos D = 2 * cos [(C + D)/2 ] * cos [(C - D)/2]
. . . . . 3) sin 2A = 2sin A * cos A
Link to Y/A
A farmer's Land compensation
Problem : 14 ) A villager has his land in the shape of quadrilateral ABCD. The government wants to procure some land from ONE corner of his plot. He obliges but with a condition that an EQUAL land should be given to him CONTIGUOUS to his plot and he must finally have his plot in the shape of a triangle. Analyze and show at to how he was compensated in the above agreed manner.
Solution : Please see the link given below
http://i854.photobucket.com/albums/ab104…
Here is the more elegant solution given by one of my most valuable contact Mr. Anil Bakshi .
area lost = area gained (as these lie between same parallels)
http://www.flickr.com/photos/53770577@N0…
Link to Y/A
Here is the more elegant solution given by one of my most valuable contact Mr. Anil Bakshi .
area lost = area gained (as these lie between same parallels)
http://www.flickr.com/photos/53770577@N0…
Link to Y/A
Wednesday, March 3, 2010
INTEGRATION
Problem : 13
Integrate ∫ [24 dy / y √(y² - 16)]
Answer :
Let I = ∫ [ 24 dy / y√(y² - 16) ]
=> 24 ∫ [ y dy / y²√(y² - 16) ]
put y² - 16 = t² => y dy = t dt
=> 24 ∫ [ t dt / (t² + 16)√(t²) ]
=> 24 ∫ [ t dt / (t² + 16) * t ]
=> 24 ∫ [ dt / (t² + 16) ]
=> 24 ∫ [ dt / (t² + 4²) ]
=> 24 * (1/4) tan^(-1) (t/4) + C
=> 6 * tan^(-1) [√(y² - 16) / 4] + C (Answer)
OR
6 * cos^(-1) ( 4/y ) + C (Answer)
OR
6 * sec^(-1) ( y/4) + C (Answer)
Integrate ∫ [24 dy / y √(y² - 16)]
Answer :
Let I = ∫ [ 24 dy / y√(y² - 16) ]
=> 24 ∫ [ y dy / y²√(y² - 16) ]
put y² - 16 = t² => y dy = t dt
=> 24 ∫ [ t dt / (t² + 16)√(t²) ]
=> 24 ∫ [ t dt / (t² + 16) * t ]
=> 24 ∫ [ dt / (t² + 16) ]
=> 24 ∫ [ dt / (t² + 4²) ]
=> 24 * (1/4) tan^(-1) (t/4) + C
=> 6 * tan^(-1) [√(y² - 16) / 4] + C (Answer)
OR
6 * cos^(-1) ( 4/y ) + C (Answer)
OR
6 * sec^(-1) ( y/4) + C (Answer)
Wednesday, January 27, 2010
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)
Link to Y/A
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)
Link to Y/A
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)
Link to Y/A
(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)
Link to Y/A
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
Link to Y/A
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
Link to Y/A
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)
Link to Y/A
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)
Link to Y/A
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)
Link to Y/A
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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(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)
Link to Y/A
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)
Link to Y/A
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)
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)
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)
Tuesday, January 26, 2010
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)
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)
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)
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)
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