x² = 4ay
In this case, the vertex is at the origin and the axis coincides with y-axis.
Focus is at F(0,a) and the equation of the directrix d: y = -a.
The parabola opens upward
x² = -4ay
Some other standard forms of parabola
x² = 4ay
In this case, the vertex is at the origin and the axis coincides with y-axis.
Focus is at F(0,a) and the equation of the directrix d: y = -a.
The parabola opens upward
x² = -4ay
Equation of parabola in its standard form
For this equation focus is at F(a,0) and the equation of the directrix is d: x=-a. It vertex is at (0,0).
If a is positive it open to the right.
Length of the latus rectum = |4p|
Equation of the chord joining any two points on the parabola
(y-y1) = m(x-x1)"
slope between (x1,y1) and (x2,y2) = (y2-y1)/(x2-x1)
Two points on parabola are A(at1²,2at1) and B(at2²,2at2)
So the equation joining these two points is
(y-2at1) = [(2at2-2at1)/(at2²-at1²)]*(x-at1²)
=> y - 2at1 = [2/(t2+ta)]*(x-at1²)
=> y(t1+t2) = 2x+2at1t2
Intersection of a straight line and a parabola
Straight line equation y = mx+c
At intersection point, both equations are satisfied
hence (mx+c)² = 4ax
=> m²x²+2x(mc-2a)+c² = 0
It is a quadratic equation. Solution gives intersection points
The intersection points are concident if
4(mc-2a)² - 4m²c²>0
=> a² - amc>0
=>a-mc>0
=>a>mc
=>a/m>c
=>c
The intersection points are real and distinct if
4(mc-2a)² - 4m²c²=0
=> 4(mc-2a)² = 4m²c²
The intersection points are imaginary if
4(mc-2a)² - 4m²c²<0
1. Equation of the chord of contacts of tangents to a parabola
From a point two tangents are drawn to a parabola. The chord between the contact points of these two tangents is chord of contact of tangents.
When Parabola equation y² = 4ax; point from which tangents are drawn is (x1,y1)
chord equation is yy1 = 2a(x+x1)
When Parabola equation y² = 4ax; point from which tangents are drawn is (x1,y1)
chord equation is yy1 = 2a(x+x1)
2. Equation of diameter of a parabola
The locus of bisectors of a system of parallel chords is termed diameter.
If Parabola equation is y² = 4ax, and system of parallel chords equation is y = mx+c,
The equation of the diameter is y = 2a/m
It is a line parallel to the X-axis.
If Parabola equation is y² = 4ax, and system of parallel chords equation is y = mx+c,
The equation of the diameter is y = 2a/m
It is a line parallel to the X-axis.
3.Lengths of tangent, subtangent, normal and subnormal
PT is termed the length of the tangent.
PN is termed the length of the normal.
Drop a perpendicular to the axis from the point P and call it PP'.
P'T = subtangent
P'N = subnormal
If the tangent makes an angle of ψ with the axis
length of the tangent = y1 cosec ψ
length of the normal = y1 sec ψ
length of the subtangent = y1 cot ψ
length of the subnormal = y1 tan ψ
tan ψ = 2a/y1 = m (slope of the tangent)
4. Equation of parabola in its standard form
y² = 4ax
For this equation focus is at F(a,0) and the equation of the directrix is d: x=-a. It vertex is at (0,0).
If a is positive it open to the right.
Length of the latus rectum = |4p|
5. Some other standard forms of parabola
y² = -4ax
x² = 4ay
In this case, the vertex is at the origin and the axis coincides with y-axis.
Focus is at F(0,a) and the equation of the directrix d: y = -a.
The parabola opens upward
x² = -4ay
6. Equation of a parabola in parametric form
x = at²
y = 2at
It satisfies y² = 4ax
y² = 4a²t²
4ax = 4a²t²
7. Equation of the chord joining any two points on the parabola
From the straight line chapter we know: "The equation of a line having slope m and passing through (x1,y1) is
(y-y1) = m(x-x1)"
slope between (x1,y1) and (x2,y2) = (y2-y1)/(x2-x1)
Two points on parabola are A(at1²,2at1) and B(at2²,2at2)
So the equation joining these two points is
(y-2at1) = [(2at2-2at1)/(at2²-at1²)]*(x-at1²)
=> y - 2at1 = [2/(t2+ta)]*(x-at1²)
=> y(t1+t2) = 2x+2at1t2
8. Intersection of a straight line and a parabola
Parabola equation y² = 4ax,
Straight line equation y = mx+c
At intersection point, both equations are satisfied
hence (mx+c)² = 4ax
=> m²x²+2x(mc-2a)+c² = 0
It is a quadratic equation. Solution gives intersection points
The intersection points are concident if
4(mc-2a)² - 4m²c²>0
=> a² - amc>0
=>a-mc>0
=>a>mc
=>a/m>c
=>c
The intersection points are real and distinct if
4(mc-2a)² - 4m²c²=0
=> 4(mc-2a)² = 4m²c²
The intersection points are imaginary if
4(mc-2a)² - 4m²c²<0
The parabola - definitions
Parabola is the locus of a point P which moves in a plane so that its distance from a fixed line of the plane and its distance from a fixed point of the plane, not on the line, are equal.
The fixed point F is called the focus and fixed line is called the directrix of the parabola.
The perpendicular to he directix from the focus is called the axis of the parabola.
The intersection of the parabola and the axis of the parabola is called vertex.
Vertex is the mid point of axis.
The line joining any two distinct points of the parabola is called a chord.
A chord which passes through the focus is called a focal chord.
The distance between the focus and any point of the focal chord is called focal radius.
The focal chord which is penpendicular to the axis is called the latus rectum
The fixed point F is called the focus and fixed line is called the directrix of the parabola.
The perpendicular to he directix from the focus is called the axis of the parabola.
The intersection of the parabola and the axis of the parabola is called vertex.
Vertex is the mid point of axis.
The line joining any two distinct points of the parabola is called a chord.
A chord which passes through the focus is called a focal chord.
The distance between the focus and any point of the focal chord is called focal radius.
The focal chord which is penpendicular to the axis is called the latus rectum
Logarthim
If log0.3(x-1)
a.(2,∞)
b. (1,2)
c. (-2,-1)
d. none of these
(JEE 1983)
Answer (a)
log0.3(x-1)
=> log (x-1)/log 0.3 < 09 =""> log (x-1)/log 0.3 < 3 =""> log (x-1)>1/2 log (x-1) (as log 0.3<0)> long (x-1)>0
=> x-1>1
=> x>2
thus x Є (2,∞)
Parabola
1. Find the focus, directrix, and vertex for the parabola x² = -16y
For x² = 4ay
Focus is (0,a), directrix is y = -a, and vertex is (0,0)
Write the given equation as x² = 4 (-4)y
Hence focus is (0,-4), directrix is y = 4, and vertex is (0,0).
The parabola opens downward.
2. Find the equation of the parabola that satisfied the following conditions.
Vertex (0,0), latus rectum = 16, opens to the right.
Latus rectum for standard parabola = |4a|
Hence |a| for the given parabola = 4
As vertex is at (0,0) and the parabola open to the right focus is (4,0)
The standard equation is y² = 4ax.
The equation of the given parabola = y² = 4*4x = 16x
Straight Line
1. The set of lines ax+by+c = 0 where a,b,c satisfy the relation 3a+2b+4c = 0 is concurrent at the point --------------. (JEE, 1982)
Answer (3/4, ½)
3a+2b+4c = 0
Dividing the equation by 4
3a/4+2b/4+4c/4 = 0
=> 3a/4 + b/2+c = 0
x = ¾ and y = ½ satisfied this relation
Hence the set of lines are concurrent at (3/4, ½)
2. The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is
a. isosceles
b. equilateral
c. right angled
d. none of these
(JEE 1984)
Answer: (a)
Solution:
x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3
So no two lines are perpendicular to each other as m1*m2 is not equal to -1 for any two lines.
From (1) and (2) intersection point is
A(2,-2)
From (2) and (3) intersection point is
B(1,1)
From (3) and (1) intersection point is
C(-2,2)
AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.
3. Three lines px+qy+r = 0, qx+ry+p =0, and rx+py+q +0 are concurrent if:
a. p+q+r + 0
b. p²+q²+r² = pq+rq+rp
c. p³+q³+r³ = 3pqr
d. none of these
Answer: (a), (b),©
Solution
The condition for concurrency of three lines
a1x +b1y+c1 = 0,
a2x+b2y+c2 = 0 and
a3x+b3y+c3=0 is
The determinant
|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3|
= 0
Hence the condition for given lines to be concurrent is
|p q r|
|q r p|
|r p q|
= 0
=> p(rq-p²) –q(q²-rp) +r(pq-r²) = 0
=> prq - p³ - q³+qrp +rpq - r³ = 0
= > -p³-q³-r³ = -3pqr
=> p³+q³+r³ = 3pqr
=> p³+q³+r³ - 3pqr = 0
=> (p+q+r)(p²+q²+r²-pq-qr-rp) = 0
=> p+q+r = 0 or
=> p²+q²+r² = pq+qr+rp
4. The points (0,8/3), (1,3) and (82,30) are vertices of:
a. an obtuse angled triangle
b. an acute angled triangle
c. a right angled triangle
d. an isosceles triangle
e. none of these
(JEE 1986)
Answer (e)
Solution
Concepts involed:
The slope of a line passing through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1), provided x1≠x2.
The equation of a line having slope m and passing through (x1,y1)and (x2,y2) is
(y-y1)/(x-x1) = (y1-y2)/(x1-x2)
The equation of the line passing through (0,8/3) and (1,3) is
(y - 8/3)/(x) = (8/3 – 3)/(0-1)
y- 8/3 = x/3
=> x/3 –y +8/3 = 0
=> x – 3y +8 = 0
Checking whether (82,30) is on this line
82 – 3(30)+8 = 90-90 turns out to be zero.
Hence given points are collinear.
5. State whether the following statement is true or false
The lines 2x +3y +19 = 0 and 9x+6y-17 = 0 cut the coordinate axes in concyclic points.
(JEE 1988)
Answer: True
Concyclic points are points on a circle.
If ac = bd = k, then the equation of the circle through A(a,0), B(0,b), C(c,0)) and D(0,d) is
x² + y² -(a+c)x – (b+d) y +k = 0
In this case the points are A(-19/2, 0), B(0,-19/3), C(17/9,0), D (0,17/6)
ac = -19/2*17/9 = -323/18
bd = -19/3*17/6 = -323/18
Hence the given lines cut the coordinate axes in concyclic points.
Answer (3/4, ½)
3a+2b+4c = 0
Dividing the equation by 4
3a/4+2b/4+4c/4 = 0
=> 3a/4 + b/2+c = 0
x = ¾ and y = ½ satisfied this relation
Hence the set of lines are concurrent at (3/4, ½)
2. The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is
a. isosceles
b. equilateral
c. right angled
d. none of these
(JEE 1984)
Answer: (a)
Solution:
x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3
So no two lines are perpendicular to each other as m1*m2 is not equal to -1 for any two lines.
From (1) and (2) intersection point is
A(2,-2)
From (2) and (3) intersection point is
B(1,1)
From (3) and (1) intersection point is
C(-2,2)
AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.
3. Three lines px+qy+r = 0, qx+ry+p =0, and rx+py+q +0 are concurrent if:
a. p+q+r + 0
b. p²+q²+r² = pq+rq+rp
c. p³+q³+r³ = 3pqr
d. none of these
Answer: (a), (b),©
Solution
The condition for concurrency of three lines
a1x +b1y+c1 = 0,
a2x+b2y+c2 = 0 and
a3x+b3y+c3=0 is
The determinant
|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3|
= 0
Hence the condition for given lines to be concurrent is
|p q r|
|q r p|
|r p q|
= 0
=> p(rq-p²) –q(q²-rp) +r(pq-r²) = 0
=> prq - p³ - q³+qrp +rpq - r³ = 0
= > -p³-q³-r³ = -3pqr
=> p³+q³+r³ = 3pqr
=> p³+q³+r³ - 3pqr = 0
=> (p+q+r)(p²+q²+r²-pq-qr-rp) = 0
=> p+q+r = 0 or
=> p²+q²+r² = pq+qr+rp
4. The points (0,8/3), (1,3) and (82,30) are vertices of:
a. an obtuse angled triangle
b. an acute angled triangle
c. a right angled triangle
d. an isosceles triangle
e. none of these
(JEE 1986)
Answer (e)
Solution
Concepts involed:
The slope of a line passing through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1), provided x1≠x2.
The equation of a line having slope m and passing through (x1,y1)and (x2,y2) is
(y-y1)/(x-x1) = (y1-y2)/(x1-x2)
The equation of the line passing through (0,8/3) and (1,3) is
(y - 8/3)/(x) = (8/3 – 3)/(0-1)
y- 8/3 = x/3
=> x/3 –y +8/3 = 0
=> x – 3y +8 = 0
Checking whether (82,30) is on this line
82 – 3(30)+8 = 90-90 turns out to be zero.
Hence given points are collinear.
5. State whether the following statement is true or false
The lines 2x +3y +19 = 0 and 9x+6y-17 = 0 cut the coordinate axes in concyclic points.
(JEE 1988)
Answer: True
Concyclic points are points on a circle.
If ac = bd = k, then the equation of the circle through A(a,0), B(0,b), C(c,0)) and D(0,d) is
x² + y² -(a+c)x – (b+d) y +k = 0
In this case the points are A(-19/2, 0), B(0,-19/3), C(17/9,0), D (0,17/6)
ac = -19/2*17/9 = -323/18
bd = -19/3*17/6 = -323/18
Hence the given lines cut the coordinate axes in concyclic points.
Hyperbola
1. Find the following properties of hyperbola for the given equation
(i) Points where hyperbola intersects the x-axis
(ii) eccentricity
(iii) foci
(iv) distance between the foci
(v) length of the latus rectum
Equation of the hyperbola is x2. Equation of hyperbola in its standard form
x²/16 - y²/9² = 1
Concepts to be used
For the hyperbola x²/a² - y²/b² = 1
The hyperbola intersects x axis at (a,0) and (-a,0).
b² = a²(e²-1)
Where e = eccentricity
Focus is (ae,0)
Solution:
Hyperbola given is x²/16 - y²/9 = 1
a =4, b= 3
Intersecting points on the x axis = (4,0) and (-4,0)
Eccentricity 9 = 16(e²-1)
=> 9/16 + 1 = e²
=> 25/16 = e²
=> e = 5/4
foci are (±ae,) = (5,0) and (-5,0)
distance between the foci = 10
Length of the latus rectum = 2b²/a = 18/4 = 9/2
INDEFINITE INTEGRATION
a function ф(x) is called a primitive or an anti-derivative of a function f(x) if ф'(x) = f(x).
For a function f(x), the collection of all its primitives is called the indefinite integral of f(x0 and is denoted by ∫ f(x) dx.
∫ f(x) dx = ф(x) + C (where C is a constant)
Here ∫ is the integral sign, f(x) is th integrand, x is the variable of integration and dx is the element of integration or differential of x.
The process of finding an indefinite integral of a given function is called integration of the function.
Problem based on direct use of rules of integration and formula for standard integrals.
WORKING RULE
Simplify the integrand till it becomes an algebraic sum of functions so that the rules of integration and formula for standard integral may be directly used.
Example I = ∫ (2 - 3 Sin x) / Cos2x dx
Solution : I = ∫ (2 / Cos2x) dx - 3 ∫(Sin x) / Cos2x dx
I = 2 ∫ Sec2x dx - 3 ∫ Tan x . Sec x dx
I = 2 Tan x - 3 Sec x + C
Problem based on direct use of rules of integration and formula for standard integrals.
WORKING RULE
Simplify the integrand till it becomes an algebraic sum of functions so that the rules of integration and formula for standard integral may be directly used.
Example I = ∫ (2 - 3 Sin x) / Cos2x dx
Solution : I = ∫ (2 / Cos2x) dx - 3 ∫(Sin x) / Cos2x dx
I = 2 ∫ Sec2x dx - 3 ∫ Tan x . Sec x dx
I = 2 Tan x - 3 Sec x + C
Integration By Substitution
If g(x) is a continuously differentiable function, then to solve
∫ f(g(x)) g'(x) dx; we substitute g(x) = t and g'(x) dx will be equal to dt.
Hence the problem is transformed to ∫ f(t) dt
Example I = ∫tan x dx = ∫(sin x/cos x)dx
If f(x) = t, f'(x)dx = dt
cos x = t;
-sin x dx = dt
sin x dx = -dt
I = ∫(sin x/cos x)dx = ∫-dt/t = -log |t|+c = - log|cos x|+C
= log |sec x|+C
Integration of ∫ Sin^m x . Cos^n x dx
if m power of Sin x is odd, put Cos x = t .
If n power of Cos x is odd, put Sin x = t.
If both m, and n are odd then put Sin x = t or Cos x = t.
If both m, and n are even then See the sum of power = (m + n).
( a ) if (m + n ) = +ive even Integer then convert trigonometric equation in linear form and integrate it.
( b ) if (m + n ) = -ive even Integer then put Tan x = t and integrate it.
Integrals of the form [1/(x²±a²)]dx
∫(1/(x²+a²)dx = (1/a)tanˉ¹(x/a) + C
∫(1/(x²-a²)dx = (1/2a)log|(x-a)/(x+a)|+C
Integrals of the form [ ∫1/√(ax²+bx+c)]dx or ∫√(ax²+bx+c)]dx
Make ax² + bx + c as perfect square (like A2 + X2, A2 - X2 etc )
Let X = t and integrate it.
Integrals of the form [(px+q)/(ax²+bx+c)]dx
Express numerator as
px + q = λ(derivative of denominator) + µ = λ(2ax+b)+µ
Integrals of the functional form [1/(a sin²x + b cos²x +c)]dx
Divide numerator and denominator by cos²x
Replace sec²x by (1 + tan² x)
Put tan x = t
dt = sec²xdx
The integral reduces to ∫[1/(At² +Bt +C)]dt
Integrals of [(a sin x + b cos x +c)/(p sin x + q cos x +r)] dx
Express the numerator as
λ (denominator) + µ (Differential of denominator) + υ
The solution will come as λx + µ log |denominator| + υ ∫ dx / (p sin x + q cos x + r)
Integrals of e^ax sin bx dx , e^ax cos bx dx
∫eax sinbx dx = [eax/(a²+b²)[[a sin bx - b cos bx) +C
∫eax cos bx dx = [eax/(a²+b²)[[a cos bx + b sin bx) +C
Integrals of √(a² ± x²) and √( x² - a²)
∫√(a²+x²)dx = (1/2) x√(a²+x²) + (1/2) a²log |x+ √(a²+x²)| +C
∫√(a²-x²) = (1/2) x√(a²-x²) + (1/2) a²sin-1(x/a) + C
∫√(x²-a²) = (1/2) x√(x²-a²)- (1/2) a²log | x+ √ ( x² - a² ) | +C
∫√(a²-x²) = (1/2) x√(a²-x²) + (1/2) a²sin-1(x/a) + C
∫√(x²-a²) = (1/2) x√(x²-a²)- (1/2) a²log | x+ √ ( x² - a² ) | +C
Integration of Function [G(x)/(P√Q)]dx
∫ [ G(x) /(P√Q)] dx
When both P and Q are linear functions of x put Q = t²
For more specific functional form
P√Q = (ax+b)√(cx+d)
Hence put cx+d = t²
When both P and Q are linear functions of x put Q = t²
For more specific functional form
P√Q = (ax+b)√(cx+d)
Hence put cx+d = t²
Integration of [( x² + 1 ) / ( x4+ λ x²+1 ) ] dx
Divide the numerator and denominator by x² and put x + 1/x at t or x - 1/x as t as required
Evaluation of Algebraic limits
Direct substitution method
Factorisation method
Rationalization method
Using standard formulas
Method for limits when x→∞
Evaluation of Trigonometric limits
Evaluation of Exponential and Logarithmic Limits
Evaluation of Exponential Limits of the Form 1ˉ
Factorisation method
Rationalization method
Using standard formulas
Method for limits when x→∞
Evaluation of Trigonometric limits
Evaluation of Exponential and Logarithmic Limits
Evaluation of Exponential Limits of the Form 1ˉ
Algebra of Limits
Algebra of Limits
1. lim x→a [f(x) ± g(x)] = lim x→a f(x) ± lim x→ag(x)
2. lim x→a [k.f(x)] = k lim x→af(x)
3. lim x→a[f(x).g(x)] = [lim x→a f(x)][ lim x→a g(x)]
4. lim x→a [f(x)/g(x)] = [lim x→a f(x)]/[ lim x→a g(x)] provided lim x→a g(x) ≠ 0.
5. If f(x) is l.t. g(x) then lim x→a f(x) ≤ lim x→a g(x)
1. lim x→a [f(x) ± g(x)] = lim x→a f(x) ± lim x→ag(x)
2. lim x→a [k.f(x)] = k lim x→af(x)
3. lim x→a[f(x).g(x)] = [lim x→a f(x)][ lim x→a g(x)]
4. lim x→a [f(x)/g(x)] = [lim x→a f(x)]/[ lim x→a g(x)] provided lim x→a g(x) ≠ 0.
5. If f(x) is l.t. g(x) then lim x→a f(x) ≤ lim x→a g(x)
Right Hand and Left Hand Limt
Right Hand and Left Hand Limit
The statement x→aˉ means that x is tending to a from the left hand side.
The statement x→a+ means that x is tending to a from the right hand side.
Steps to find left hand limit
Put x = a - h and replace x→aˉ by h→0. Find limit of f(a-h) as h→0
Steps to find right hand limit
Put x = a + h and replace x→a+ by h→0. Find limit of f(a+h) as h→0
The statement x→aˉ means that x is tending to a from the left hand side.
The statement x→a+ means that x is tending to a from the right hand side.
Steps to find left hand limit
Put x = a - h and replace x→aˉ by h→0. Find limit of f(a-h) as h→0
Steps to find right hand limit
Put x = a + h and replace x→a+ by h→0. Find limit of f(a+h) as h→0
Limits
Definition
Let f(x) be a function of x. If for every positive number ε ( tends to 0),
then there exist a positive number δ such that whenever
0 < | x - a |< δ => | f(x) - l | < ε ,
then we can say , f(x) tends to limit l as x tends to a
and we can write
lim (x→a) f(x) = l.
Let f(x) be a function of x. If for every positive number ε ( tends to 0),
then there exist a positive number δ such that whenever
0 < | x - a |< δ => | f(x) - l | < ε ,
then we can say , f(x) tends to limit l as x tends to a
and we can write
lim (x→a) f(x) = l.
Binomial theorem for any index (n is not natural number)
If n is a rational number and x is a real number such that |x|is less than 1, then
(1+x) n = 1 = nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3!+…+n(n-1)(n-2)…(n-r+1)xr/r!+...+ ∞
Remarks
The condition |x|<1 is unnecessary when n is a whole number.
When n is not a whole number, then the condition |x|<1 is necessary.
The terms are infinite when n is not an whole number. When is it an whole number the series become finite as one of the terms will become zero in the coefficient at some point in time.
When n is a positive integer, there will be n+1 terms
To expand (x+a)n proceed as follows:
(x+a)n = {a(1 + x/a)} n= an(1 + x/a) n; (substitute x’ = x/a and proceed)
(1+x) n = 1 = nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3!+…+n(n-1)(n-2)…(n-r+1)xr/r!+...+ ∞
Remarks
The condition |x|<1 is unnecessary when n is a whole number.
When n is not a whole number, then the condition |x|<1 is necessary.
The terms are infinite when n is not an whole number. When is it an whole number the series become finite as one of the terms will become zero in the coefficient at some point in time.
When n is a positive integer, there will be n+1 terms
To expand (x+a)n proceed as follows:
(x+a)n = {a(1 + x/a)} n= an(1 + x/a) n; (substitute x’ = x/a and proceed)
Greatest Term in the expansion of (x+a)n
(i) Write (r+1)th term = T(r+1) and rth term = T (r)from the given expression.
(ii) Find T(r+1)/T(r)
(iii) Put T(r+1)/T(r)>1
(iv) Solve the inequality for r to get an inequality of the form rm
If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.
If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term
(ii) Find T(r+1)/T(r)
(iii) Put T(r+1)/T(r)>1
(iv) Solve the inequality for r to get an inequality of the form rm
If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.
If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term
Middle Terms in Bi' expression
Binomial expression (1+x)n contains (n+1) terms when n is a natural number.
If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms
If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms
Important from Binomial Theorem
1. total number of terms in the expansion = n+1
2. The sum of indices of x and a in each term is n.
3. the coefficients of terms equidistant from the beginning and the end are equal.
4. (x-a)n = (r = 0 to n)Σ ((-1) r*nCrxn-rar
The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.
5. (1+x) n = (r = 0 to n)Σ nCrxr
you get it by putting x =1 and a = x in the expression for (x+a)n.
6. (x+1) n = (r = 0 to n)Σ nCrxn-r
7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr
8. (x+a) n +(x-a) n = 2[nC0xna0 + nC2xn-2a2 + nC4xn-4a4+ ...]
9. General term in a binomial expansion (r+1) term in (x+a) n= nCrxn-rar
2. The sum of indices of x and a in each term is n.
3. the coefficients of terms equidistant from the beginning and the end are equal.
4. (x-a)n = (r = 0 to n)Σ ((-1) r*nCrxn-rar
The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.
5. (1+x) n = (r = 0 to n)Σ nCrxr
you get it by putting x =1 and a = x in the expression for (x+a)n.
6. (x+1) n = (r = 0 to n)Σ nCrxn-r
7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr
8. (x+a) n +(x-a) n = 2[nC0xna0 + nC2xn-2a2 + nC4xn-4a4+ ...]
9. General term in a binomial expansion (r+1) term in (x+a) n= nCrxn-rar
Binomial theorem
If x and a are real numbers, then for all n Є N
(x+a)n
= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an
(x+a)n
= (r = 0 to n)Σ nCrxn-rar
(x+a)n
= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an
(x+a)n
= (r = 0 to n)Σ nCrxn-rar
Pascal’s Triangle of Binomial coefficient
Expansions of (x+a)n as n = 0 to n
Each row is bounded by 1 on both sides
--- (The first row has only one item 1)First row is for n = 0
--- (The second row has two items 1 1)Second row is for n = 1
Any entry except the first and last entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right.
So the third row (n = 2) is 1 2 1 (The second row elements are 1 1).
Each row is bounded by 1 on both sides
--- (The first row has only one item 1)First row is for n = 0
--- (The second row has two items 1 1)Second row is for n = 1
Any entry except the first and last entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right.
So the third row (n = 2) is 1 2 1 (The second row elements are 1 1).
Areas of Bounded Regions - Revision Facilitator
Sections in the Chapter - R D Sharma Objective Mathematics
1 Curve sketching - Hints
2 Sketching of some common curves
3 Areas of Bounded regions - Procedure for find areas
Study Plan
Day 1
1 Curve sketching - Hints
Day 2
2 Sketching of some common curves
3 Areas of Bounded regions - Procedure for find areas
Day 3
Revision of concepts
Obective Type Exercises 1 to 10
Day 4
O.T.E.: 11 to 30
Day 5
O.T.E.: 31 to 39
Practice Exercises 1 to 10
Day 6
Practice Exercises 11 to 30
Day 7
Practice Exercises 31 to 52
Revision Period
Concept revision
Test paper problem solving
Try to recollect relevant points on the topic.
If required right click on the topic and click on open in a new window to read the relevant material.
Close the window and come back.
1 Curve sketching - Hints
2 Sketching of some common curves
3 Areas of Bounded regions - Procedure for find areas
Study Plan
Day 1
1 Curve sketching - Hints
Day 2
2 Sketching of some common curves
3 Areas of Bounded regions - Procedure for find areas
Day 3
Revision of concepts
Obective Type Exercises 1 to 10
Day 4
O.T.E.: 11 to 30
Day 5
O.T.E.: 31 to 39
Practice Exercises 1 to 10
Day 6
Practice Exercises 11 to 30
Day 7
Practice Exercises 31 to 52
Revision Period
Concept revision
Test paper problem solving
Try to recollect relevant points on the topic.
If required right click on the topic and click on open in a new window to read the relevant material.
Close the window and come back.
Area of Bounded Regions
if f(x) is a continuous function defined on [a,b]. then, the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b is given by
∫f(x)dx(between limits a to b) or ∫y dx (between limits a to b).
Steps to find the area of bounded regions
1. visualize a rough sketch.
2. Slice the area into horizontal or vertical strips appropriately.
3. Create a formula for area treating the strip as a rectangle.
If the strip is parallel to y-axis, the width of the rectangle will be Δx.
If the strip is parallel to x-axis, the width of the rectangle will be Δy.
4. Find the limits of x (for vertical strips) and limits of y for horizontal strips.
5. Do the integration ∫ydx or ∫xdy
y = f(x) is the given function or x = f(y) is the given function.
∫f(x)dx(between limits a to b) or ∫y dx (between limits a to b).
Steps to find the area of bounded regions
1. visualize a rough sketch.
2. Slice the area into horizontal or vertical strips appropriately.
3. Create a formula for area treating the strip as a rectangle.
If the strip is parallel to y-axis, the width of the rectangle will be Δx.
If the strip is parallel to x-axis, the width of the rectangle will be Δy.
4. Find the limits of x (for vertical strips) and limits of y for horizontal strips.
5. Do the integration ∫ydx or ∫xdy
y = f(x) is the given function or x = f(y) is the given function.
Sketching of some common curves
Straight line
General equation = ax+by+c = 0
Circle
General equation = x² + y² + 2gx+2fy+c = 0
Standard parabola
y² = 4ax
General parabola
y = ax²+bx+c
Standard ellipse
x²/a² + y²/b² = 1
General equation = ax+by+c = 0
Circle
General equation = x² + y² + 2gx+2fy+c = 0
Standard parabola
y² = 4ax
General parabola
y = ax²+bx+c
Standard ellipse
x²/a² + y²/b² = 1
Curve Sketchig - Some Useful Hints
To evaluate the areas of bounded regions, one has to know the rough sketch of the curve.
The principles/ideas that help in visualizing the curve or rough sketching the curve are:
1. Origin and tangents at the origin
Check whether the curve passes through the origin and finding out whether (0,0) satisfies the curve equation.
If the curve passes through (0,0) find the equation of tangent at (0,0)
2. Points of intersection of the curve with the coordinate axes
You can find the point at which the curve crosses the x-axis by putting y = 0 in the curve equation.
Similarly putting x = 0 gives the point at which the curve passes y-axis.
3. Symmetry
If all powers of y in the curve equation are even, then the curve is symmetric about x-axis.
The familiar curve y² = 4ax is symmetric about x-axis.
If all powers x in the equation of curve are even, it is symmetric about y-axis.
If by putting x = -x and y = -y, the equation of the curve remains the same, then it symmetric in opposite directions.
If the equation remains unaltered by interchanging x and y in the equation, then it is symmetric about the line y =x.
4. Regions where the curve does not exist
In the regions where the curve does not exist, x or y values will be imaginary.
In the case of y² = 4ax, when x is negative, y values will be imaginary. Hence curve does not exist on the left side of y-axis.
5. Find points where dy/dx = 0
At these points, the slope of the tangent to the curve is zero and the tangent will be parallel to the x axis.
6. Find maxima and minima points
Find points where dy/dx = 0 and check the sign of d²y/dx² at these points and find out maximum (d²y/dx² is negative) and minimum (d²y/dx² is positive) points of the curve.
Also find the interval in which the dy/dx is g.t. 0. In this interval, the function is monotonically increasing.
In the interval in which dy/dx is l.t. 0, the function is monotonically decreasing.
The principles/ideas that help in visualizing the curve or rough sketching the curve are:
1. Origin and tangents at the origin
Check whether the curve passes through the origin and finding out whether (0,0) satisfies the curve equation.
If the curve passes through (0,0) find the equation of tangent at (0,0)
2. Points of intersection of the curve with the coordinate axes
You can find the point at which the curve crosses the x-axis by putting y = 0 in the curve equation.
Similarly putting x = 0 gives the point at which the curve passes y-axis.
3. Symmetry
If all powers of y in the curve equation are even, then the curve is symmetric about x-axis.
The familiar curve y² = 4ax is symmetric about x-axis.
If all powers x in the equation of curve are even, it is symmetric about y-axis.
If by putting x = -x and y = -y, the equation of the curve remains the same, then it symmetric in opposite directions.
If the equation remains unaltered by interchanging x and y in the equation, then it is symmetric about the line y =x.
4. Regions where the curve does not exist
In the regions where the curve does not exist, x or y values will be imaginary.
In the case of y² = 4ax, when x is negative, y values will be imaginary. Hence curve does not exist on the left side of y-axis.
5. Find points where dy/dx = 0
At these points, the slope of the tangent to the curve is zero and the tangent will be parallel to the x axis.
6. Find maxima and minima points
Find points where dy/dx = 0 and check the sign of d²y/dx² at these points and find out maximum (d²y/dx² is negative) and minimum (d²y/dx² is positive) points of the curve.
Also find the interval in which the dy/dx is g.t. 0. In this interval, the function is monotonically increasing.
In the interval in which dy/dx is l.t. 0, the function is monotonically decreasing.
Applications of Derivatives - Part 1
List of the topics
Rates of Change The point of this section is to remind us of the application/interpretation of derivatives that we were dealing with in the previous chapter. Namely, rates of change.
Critical Points In this section we will define critical points. Critical points will show up in many of the sections in this chapter so it will be important to understand them.
Minimum and Maximum Values In this section we will take a look at some of the basic definitions and facts involving minimum and maximum values of functions.
Finding Absolute Extrema Here is the first application of derivatives that we’ll look at in this chapter. We will be determining the largest and smallest value of a function on an interval.
The Shape of a Graph, Part I We will start looking at the information that the first derivatives can tell us about the graph of a function. We will be looking at increasing/decreasing functions as well as the First Derivative Test.
The Shape of a Graph, Part II In this section we will look at the information about the graph of a function that the second derivatives can tell us. We will look at inflection points, concavity, and the Second Derivative Test.
The Mean Value Theorem Here we will take a look that the Mean Value Theorem.
Optimization Problems This is the second major application of derivatives in this chapter. In this section we will look at optimizing a function, possible subject to some constraint.
More Optimization Problems Here are even more optimization problems.
L’Hospital’s Rule and Indeterminate Forms This isn’t the first time that we’ve looked at indeterminate forms. In this section we will take a look at L’Hospital’s Rule. This rule will allow us to compute some limits that we couldn’t do until this section.
Linear Approximations Here we will use derivatives to compute a linear approximation to a function. As we will see however, we’ve actually already done this.
Differentials We will look at differentials in this section as well as an application for them.
Newton’s Method With this application of derivatives we’ll see how to approximate solutions to an equation.
Business Applications Here we will take a quick look at some applications of derivatives to the business field.
Rates of Change The point of this section is to remind us of the application/interpretation of derivatives that we were dealing with in the previous chapter. Namely, rates of change.
Critical Points In this section we will define critical points. Critical points will show up in many of the sections in this chapter so it will be important to understand them.
Minimum and Maximum Values In this section we will take a look at some of the basic definitions and facts involving minimum and maximum values of functions.
Finding Absolute Extrema Here is the first application of derivatives that we’ll look at in this chapter. We will be determining the largest and smallest value of a function on an interval.
The Shape of a Graph, Part I We will start looking at the information that the first derivatives can tell us about the graph of a function. We will be looking at increasing/decreasing functions as well as the First Derivative Test.
The Shape of a Graph, Part II In this section we will look at the information about the graph of a function that the second derivatives can tell us. We will look at inflection points, concavity, and the Second Derivative Test.
The Mean Value Theorem Here we will take a look that the Mean Value Theorem.
Optimization Problems This is the second major application of derivatives in this chapter. In this section we will look at optimizing a function, possible subject to some constraint.
More Optimization Problems Here are even more optimization problems.
L’Hospital’s Rule and Indeterminate Forms This isn’t the first time that we’ve looked at indeterminate forms. In this section we will take a look at L’Hospital’s Rule. This rule will allow us to compute some limits that we couldn’t do until this section.
Linear Approximations Here we will use derivatives to compute a linear approximation to a function. As we will see however, we’ve actually already done this.
Differentials We will look at differentials in this section as well as an application for them.
Newton’s Method With this application of derivatives we’ll see how to approximate solutions to an equation.
Business Applications Here we will take a quick look at some applications of derivatives to the business field.
Ch3 PERMUTATION AND COMBINATION
JEE Question 07
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is
(A) 360 (B) 192 (C) 96 (D) 48
-------------------------------
JEE question
The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0), is
(A) 133 (B) 190 (C) 233 (D) 105
answer B
----------------
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is
(A) 360 (B) 192 (C) 96 (D) 48
-------------------------------
JEE question
The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0), is
(A) 133 (B) 190 (C) 233 (D) 105
answer B
----------------
Ch1 Theory Of Equations
| The trinomial expression of the form ax2 + bx + c = 0, a ≠ 0 a, b, c are R is called quadratic equation, because the highest order term in it is of second degree. We can say… (a) ax2 + b x + c = 0, a ≠ 0 has exactly two roots which may be real or unequal or equal or complex. (b) If ax2 + b x + c = 0, a ≠ 0 can not have three or more roots & if it has, it becomes an identity. If ax2 + b x + c = 0 is an identity then a = b = c = 0. SOLUTION OF Q.E. Let α and β be the roots of the equation ax2 + b x + c = 0, a ≠ 0. then (a) x1 & x2 = [-b +-(b2 - 4ac)1/2] / 2a(b) b2 - 4ac= D is called Discriminant of the Q.E. (d) x2 – (sum of roots) x + (product of roots) = 0 NATURE OR TYPE OF ROOTS D and α , β are the roots of the ax2 + b x + c = 0, a ≠ 0 If D greater than; 0 then roots are real, distinct.If D = 0 then roots are real, equal. If D less than 0 then roots are imaginary.If p + iq is one root of equation, then the other root is p – iq(conjugate of each other). If D is perfect square then the roots are rational.If p + √q is one root of equation, then the other root is p – √q. SIGN OF ROOTS OF Q.E : conditions for sign of roots (a) For Both roots are +ive then (i) D greater than 0(ii) sum of roots > 0 (iii) product of roots greater than 0 , these three conditions simultaneous true. (b) For Both roots are –ive (i) D greater than 0 (ii) sum of roots less than 0 (iii) product of roots greater than 0 , these three conditions simultaneous true. (c) For one root is +ive & one is –ive (i) D greater than 0 (ii) product of roots less than 0, these two conditions simultaneous true. |
Ch2 Progression
Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers.
The conceptual portion is very limited book. But the first problem in the examples itself is complicated. Every problem thereon is a complicated problem.
In Mathematics and Physics, the conceptual portion is going to be limited but the problems are going to be complicated. One has to sit down and do all the problems in examples and exercises to develop the sharp brain that can discover the structure in the problem given in the examination.
The conceptual portion is very limited book. But the first problem in the examples itself is complicated. Every problem thereon is a complicated problem.
In Mathematics and Physics, the conceptual portion is going to be limited but the problems are going to be complicated. One has to sit down and do all the problems in examples and exercises to develop the sharp brain that can discover the structure in the problem given in the examination.
IIT JEE Learning Mathematics
MINDLESS rote learning of things that can be learned effectively, possibly MUCH more effectively, in other ways needs to be stopped.
Mneomonic methods are, however, extremely easy to understand and put into practical use for a wide range of applications. These methods pay off in inverse proportion to the arbitrariness of the material being memorized. It means is that when faced with , say, a random or arbitrary list of items, dates, facts, etc., the more random the list and therefore the less conceptual links or "common knowledge" might be involved, the more a person using mnemonics would gain from using these techniques. Because otherwise the main option would be some variation on pure rote.
However, less time was needed for memorizing information with more structure, because the "inherent logic" or interconnectedness of the information helped one memorize.
Mathematics already is based on logical and conceptual links. Hence, it is often the case that what needs to be "memorized" in the sense mentioned above is minimal.
What sorts of things would need to be memorized in mathematics? Well, things like Order of Operations, which consists of conventions, not something that simply HAS to be. Terminology. Notation. Axioms. Things that do not follow from first precepts.
Even going beyond that, it is undoubtedly true that we need to "memorize" certain fundamental relationships and identities in specific areas of mathematics in order to not have to tediously look them up for every single instance in which they arise. In trigonometry, for example, understanding the definition of sine, cosine, and tangent in right triangle trigonometry is a key "fact" that one does much better to have at one's mental fingertips than not.
The amount that "must" be memorized is often far smaller than one originally believes, because of underlying relationships and concepts that create natural connections among a smaller set of facts. Anyone who is led to believe that entire chapters of a mathematics book and solutions of all problems need to be memorized by drill or rote is being mistaught.
That which is understood conceptually has a better chance of lasting, and can be more readily recreated through the concepts even if the "at one's fingertips" recall has been weakened or extinguished. Most people are well aware of this through personal experience.
Mneomonic methods are, however, extremely easy to understand and put into practical use for a wide range of applications. These methods pay off in inverse proportion to the arbitrariness of the material being memorized. It means is that when faced with , say, a random or arbitrary list of items, dates, facts, etc., the more random the list and therefore the less conceptual links or "common knowledge" might be involved, the more a person using mnemonics would gain from using these techniques. Because otherwise the main option would be some variation on pure rote.
However, less time was needed for memorizing information with more structure, because the "inherent logic" or interconnectedness of the information helped one memorize.
Mathematics already is based on logical and conceptual links. Hence, it is often the case that what needs to be "memorized" in the sense mentioned above is minimal.
What sorts of things would need to be memorized in mathematics? Well, things like Order of Operations, which consists of conventions, not something that simply HAS to be. Terminology. Notation. Axioms. Things that do not follow from first precepts.
Even going beyond that, it is undoubtedly true that we need to "memorize" certain fundamental relationships and identities in specific areas of mathematics in order to not have to tediously look them up for every single instance in which they arise. In trigonometry, for example, understanding the definition of sine, cosine, and tangent in right triangle trigonometry is a key "fact" that one does much better to have at one's mental fingertips than not.
The amount that "must" be memorized is often far smaller than one originally believes, because of underlying relationships and concepts that create natural connections among a smaller set of facts. Anyone who is led to believe that entire chapters of a mathematics book and solutions of all problems need to be memorized by drill or rote is being mistaught.
That which is understood conceptually has a better chance of lasting, and can be more readily recreated through the concepts even if the "at one's fingertips" recall has been weakened or extinguished. Most people are well aware of this through personal experience.
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