Problems that require concepts from two or more chapters
The radius of the circle passing through the foci of the ellipse
x²/16 + y²/9 = 1 and having its centre at (0,3) is
a. 4
b. 3
c. √12
d. 7/2
JEE 1995
Saturday, September 27, 2008
Saturday, August 30, 2008
Binomial Theorem - Model Problems - Past JEE
1. The coefficient of x4 in [(x/2)- (3/x²)]10 is
a. 405/256
b. 504/259
c. 450/263
d. none of these
(JEE 1983)
2. Let n be a positive integer. If the coefficient of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)
3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals
a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)
(JEE 2001)
4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is
a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2
(JEE 2003)
Answers and Solutions
1. The coefficient of x4 in [(x/2)- (3/x²)]10 is
a. 405/256
b. 504/259
c. 450/263
d. none of these
(JEE 1983)
Answer (a)
Solution:
(r+1) th term of the expansion is given by
10Cr (x/2) 10-r(-3/x²) r
= 10Cr (x) 10-r(1/ x²)r(1/2) 10-r(-3) r
= 10Cr(-3) r(1/2) 10-r(x) 10-3r
In this term (x) 10-3r is equal to x4
10-3r = 4 or r = 2
So coefficient =
10Cr(-3) r(1/2) 10-r
10C2(-3) 2(1/2) 10-2
= [10*9/2]*9*(1/28)
= (45*9)/256
= 405/256
2. Let n be a positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)
Answer: 7
Solution:
The question requires concepts from binomial theorem and concept from arithmetic progression.
Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are
nC1, nC2, nC3
According to arithmetic progression properties, three terms can be taken as a-d, a, and a+d. Hence sum of first and third terms = 2*second term
=> 2(nC2) = nC1 + nC3
=> 2 n(n-1)/2 = n + [n(n-1)(n-2)/6]
=> n-1 = 1 +[(n² – 3n+2)/6]
=> n² – 3n+2 = 6n-12
=> n² – 9n+14 = 0
=> (n-2)(n-7) = 0
Since 4th term will be there only when n>2, n is equal to 7.
3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals
a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)
(JEE 2001)
Answer: (b)
Solution
5th term = 4+1th term = nC4 a n-4(-b) 4
6th term = 5+1th term = nC5 a n-5(-b) 5
As sum of 5th terms and 6th terms is zero
nC4 a n-4(-b) 4 + nC5 a n-5(-b) 5 = 0
nC4 a n-4(b) 4 = nC5 a n-5(b) 5
a n-4(b) 4/ a n-5(b) 5 = nC5/nC4
a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)
= (n-4)!/(5(n-5)!)
= (n-4)/5
4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is
a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2
(JEE 2003)
Answer: (d)
Solution:
Multiplying the last two expressions we have to find coefficient of t24 in (1+t²)12(1+t12+24)
t24 term will come as a sum of three terms in the expanded expression
1st term is t24 term in 1st expression multiplied by 1 in the second expression.
2nd term is t12 term in 1st expression multiplied by t12 term in the second expression.
3rd term is 1 in the first expression multiplied by t24 term in the second expression.
Coefficients of 1 and 3 terms are 1 and 1.
Coefficient of 2 term = coefficient t12 term in 1 st expression which is (1+t²)12
7th term or 6+1th term in the expansion will have t12 term
Hence coefficient of 7th term = 12C6
Therefore coefficient of t24 term will be
12C6 + 2
a. 405/256
b. 504/259
c. 450/263
d. none of these
(JEE 1983)
2. Let n be a positive integer. If the coefficient of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)
3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals
a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)
(JEE 2001)
4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is
a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2
(JEE 2003)
Answers and Solutions
1. The coefficient of x4 in [(x/2)- (3/x²)]10 is
a. 405/256
b. 504/259
c. 450/263
d. none of these
(JEE 1983)
Answer (a)
Solution:
(r+1) th term of the expansion is given by
10Cr (x/2) 10-r(-3/x²) r
= 10Cr (x) 10-r(1/ x²)r(1/2) 10-r(-3) r
= 10Cr(-3) r(1/2) 10-r(x) 10-3r
In this term (x) 10-3r is equal to x4
10-3r = 4 or r = 2
So coefficient =
10Cr(-3) r(1/2) 10-r
10C2(-3) 2(1/2) 10-2
= [10*9/2]*9*(1/28)
= (45*9)/256
= 405/256
2. Let n be a positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P. then the value of n is ________________ .
(JEE 1994)
Answer: 7
Solution:
The question requires concepts from binomial theorem and concept from arithmetic progression.
Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are
nC1, nC2, nC3
According to arithmetic progression properties, three terms can be taken as a-d, a, and a+d. Hence sum of first and third terms = 2*second term
=> 2(nC2) = nC1 + nC3
=> 2 n(n-1)/2 = n + [n(n-1)(n-2)/6]
=> n-1 = 1 +[(n² – 3n+2)/6]
=> n² – 3n+2 = 6n-12
=> n² – 9n+14 = 0
=> (n-2)(n-7) = 0
Since 4th term will be there only when n>2, n is equal to 7.
3. If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals
a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)
(JEE 2001)
Answer: (b)
Solution
5th term = 4+1th term = nC4 a n-4(-b) 4
6th term = 5+1th term = nC5 a n-5(-b) 5
As sum of 5th terms and 6th terms is zero
nC4 a n-4(-b) 4 + nC5 a n-5(-b) 5 = 0
nC4 a n-4(b) 4 = nC5 a n-5(b) 5
a n-4(b) 4/ a n-5(b) 5 = nC5/nC4
a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)
= (n-4)!/(5(n-5)!)
= (n-4)/5
4. The coefficient of t24 in (1+t²)12(1+t12)(1+24) is
a. 12C6 + 3
b. 12C6 + 1
c. 12C6
d. 12C6 + 2
(JEE 2003)
Answer: (d)
Solution:
Multiplying the last two expressions we have to find coefficient of t24 in (1+t²)12(1+t12+24)
t24 term will come as a sum of three terms in the expanded expression
1st term is t24 term in 1st expression multiplied by 1 in the second expression.
2nd term is t12 term in 1st expression multiplied by t12 term in the second expression.
3rd term is 1 in the first expression multiplied by t24 term in the second expression.
Coefficients of 1 and 3 terms are 1 and 1.
Coefficient of 2 term = coefficient t12 term in 1 st expression which is (1+t²)12
7th term or 6+1th term in the expansion will have t12 term
Hence coefficient of 7th term = 12C6
Therefore coefficient of t24 term will be
12C6 + 2
Wednesday, June 25, 2008
Logarithms - 6
If log0.3(x-1)0.09(x-1), then x lies in the interval:
a.(2,∞)
b. (1,2)
c. (-2,-1)
d. none of these
(JEE 1983)
Answer (a)
log0.3(x-1)0.09(x-1),
=> log (x-1)/log 0.3 < log (x-1)/log 0.09
=> log (x-1)/log 0.3 < log (x-1)/2log 0.3
=> 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,∞)
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 < log (x-1)/log 0.09
=> log (x-1)/log 0.3 < log (x-1)/2log 0.3
=> 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,∞)
Saturday, June 21, 2008
Straight Line - Model Problems - 6
Problems with complexity
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.
(Note: This problem belongs to “circle” topic. But as it is worded it seems to be a problem under straight line topic)
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.
(Note: This problem belongs to “circle” topic. But as it is worded it seems to be a problem under straight line topic)
Parabola - 1
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
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
Thursday, June 19, 2008
Hyperbola - 1
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
(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
Hyperbola - Past JEE
If x = 9 is the chord of contact of the hyperbola x²-y² = 9, then the equation of the corresponding pair of tangents is
a. 9x² -8y² +18x-9 = 0
b. 9x² -8y² -18x+9 = 0
c. 9x² -8y² -18x-9 = 0
d. 9x² -8y² +18x+9 = 0
(JEE 1999)
Answer: b.
The chord of contact x =9 meets the hyperbola at (9,6√2) and (9,-6√2).
the equations of tangent at these points are 3x-2√(2y)-3 = 0 and 3x+2√(2y)-3 = 0.
the combined equation of the two is:
9x² -8y² -18x+9 = 0
a. 9x² -8y² +18x-9 = 0
b. 9x² -8y² -18x+9 = 0
c. 9x² -8y² -18x-9 = 0
d. 9x² -8y² +18x+9 = 0
(JEE 1999)
Answer: b.
The chord of contact x =9 meets the hyperbola at (9,6√2) and (9,-6√2).
the equations of tangent at these points are 3x-2√(2y)-3 = 0 and 3x+2√(2y)-3 = 0.
the combined equation of the two is:
9x² -8y² -18x+9 = 0
Tuesday, June 17, 2008
Definite Integrals - 6
1. The value of the integral ∫0 π /2 [√(cot x)/[ √(cot x) +√(tan x)]]dx is
a. π/4
b. π/2
c. π
d. none of these
(JEE 1983)
Answer: (a)
Solution:
Concept from definite integration to be used:
∫0af(x) = ∫0 af(x-a)
I = ∫0 π /2 [√(cot x)/[ √(cot x) +√(tan x)]]dx ….(i)
= ∫0 π /2 [√(cot(π /2- x))/[ √(cot (π /2 -x)) +√(tan(π /2- x))]]dx
= ∫0 π /2 [√(tan x)/[ √(tan x) +√(cot x)]]dx … (ii)
Adding (i) and (ii)
2I = ∫0 π /2 [[√(cot x) +√(tan x)]/[ √(tan x) +√(cot x)]]dx
2I = ∫0 π /2dx = π /2
I = π /4
a. π/4
b. π/2
c. π
d. none of these
(JEE 1983)
Answer: (a)
Solution:
Concept from definite integration to be used:
∫0af(x) = ∫0 af(x-a)
I = ∫0 π /2 [√(cot x)/[ √(cot x) +√(tan x)]]dx ….(i)
= ∫0 π /2 [√(cot(π /2- x))/[ √(cot (π /2 -x)) +√(tan(π /2- x))]]dx
= ∫0 π /2 [√(tan x)/[ √(tan x) +√(cot x)]]dx … (ii)
Adding (i) and (ii)
2I = ∫0 π /2 [[√(cot x) +√(tan x)]/[ √(tan x) +√(cot x)]]dx
2I = ∫0 π /2dx = π /2
I = π /4
Differential Equations - 1
Prob: Find the solution of the differential equation
log (dy/dx) = 3x + 4y, y(0) = 0
Solution:
log (dy/dx) = 3x + 4y
=> dy/dx = e3x+4y = e3xe4y
=> e-4ydy = e3xdx
Integrating both sides
e-4y/-4 = e3x/3 +C
Using the condition y(0) = 0
-1/4 = 1/3 +C
=> C = -1/4 - 1/3 = -7/12
=>e-4y/-4 = e3x/3 -7/12
=> -(12/4)e-4y = (12/3)e3x - 7
=>3e-4y + 4e3x = 7
log (dy/dx) = 3x + 4y, y(0) = 0
Solution:
log (dy/dx) = 3x + 4y
=> dy/dx = e3x+4y = e3xe4y
=> e-4ydy = e3xdx
Integrating both sides
e-4y/-4 = e3x/3 +C
Using the condition y(0) = 0
-1/4 = 1/3 +C
=> C = -1/4 - 1/3 = -7/12
=>e-4y/-4 = e3x/3 -7/12
=> -(12/4)e-4y = (12/3)e3x - 7
=>3e-4y + 4e3x = 7
Monday, June 16, 2008
Limits-1
Limit - Definition
a given function f(x) is said to have a limit l as x approaches a, lim x→a f(x) =l uf f(x) is as near to l as we please for all values of x≠a but sufficiently near to a.
a given function f(x) is said to have a limit l as x approaches a, lim x→a f(x) =l uf f(x) is as near to l as we please for all values of x≠a but sufficiently near to a.
Limits-5
Prob: Find limit of (sin-1x - tan-1x)/x³ when x→0
when x → 0 both numerator and denominator become zero
So L'Hospital rule becomes applicable
L'Hospital rule
(i). If f and g are differentiable functions on (0,δ) such that
(ii) g'(x)≠ 0 or any x Є (0,δ)
(iii) lim x→0+ f(x) = 0 = lim x→04 g(x)
(iv) lim x→0+ f'(x)/g'(x) = L, then lim x→0+ f(x)/g(x) = L
lim x→0(sin-1x - tan-1x)/x³ =
lim x→0 [1/√(1-x²) - 1/(1+x²)]/3x² =
1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]/[(1+x²)*√(1-x²)] =
1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]*[(1+x²) +√(1-x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]
1/3 lim x→0 (1/x²)[(1+x²)² -(1-x²)]/ [(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ 1 + 2x²+x4 -1 +x²]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ 3x²+x4]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ x²(3+x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (3+x²)/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]= (1/3)(3/2) = 1/2
when x → 0 both numerator and denominator become zero
So L'Hospital rule becomes applicable
L'Hospital rule
(i). If f and g are differentiable functions on (0,δ) such that
(ii) g'(x)≠ 0 or any x Є (0,δ)
(iii) lim x→0+ f(x) = 0 = lim x→04 g(x)
(iv) lim x→0+ f'(x)/g'(x) = L, then lim x→0+ f(x)/g(x) = L
lim x→0(sin-1x - tan-1x)/x³ =
lim x→0 [1/√(1-x²) - 1/(1+x²)]/3x² =
1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]/[(1+x²)*√(1-x²)] =
1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]*[(1+x²) +√(1-x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]
1/3 lim x→0 (1/x²)[(1+x²)² -(1-x²)]/ [(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ 1 + 2x²+x4 -1 +x²]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ 3x²+x4]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (1/x²)[ x²(3+x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=
1/3 lim x→0 (3+x²)/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]= (1/3)(3/2) = 1/2
Sunday, June 15, 2008
Limits-10 - Problems with Complexity
Lim x→∞ [√(x²+1) - ³√(x³+1)]/[4√(x4+1) - 5√(x3+1)
a. 0
b. -1
c. 1
d. limit does not exist
Answer: (a)
When limits x→∞ are to be calculated, polynomials have to divided by their highest power term throughout. In this case, dividing numerator and denominator by x will result in the expression with in the root symbols getting divided by highest power term.
Lim x→∞ [√(x²+1) - ³√(x³+1)]/[4√(x4+1) - 5√(x3+1)]
= Lim x→∞ [√(1+(1/x²)) - ³√(1+(1/x³))]/[4√(1+(1/x4)) - 5√((1/x2+(1/x5))]
= (1-1/(1-0) = 0
1/x and Other terms with highest powers of x in the denominator approach zero as x→∞
a. 0
b. -1
c. 1
d. limit does not exist
Answer: (a)
When limits x→∞ are to be calculated, polynomials have to divided by their highest power term throughout. In this case, dividing numerator and denominator by x will result in the expression with in the root symbols getting divided by highest power term.
Lim x→∞ [√(x²+1) - ³√(x³+1)]/[4√(x4+1) - 5√(x3+1)]
= Lim x→∞ [√(1+(1/x²)) - ³√(1+(1/x³))]/[4√(1+(1/x4)) - 5√((1/x2+(1/x5))]
= (1-1/(1-0) = 0
1/x and Other terms with highest powers of x in the denominator approach zero as x→∞
Differentiation - 5 - Inverse Trigonometric Functions
d/dx of sin-1x = 1/√(1-x²)
d/dx of cos-1x = -1/√(1-x²)
d/dx of tan-1x = 1/(1+x²)
d/dx of cot-1x = -1/(1+x²)
d/dx of sec-1x = 1/|x|√(x²-1)
d/dx of cosec-1x = -1/|x|√(x²-1)
d/dx of cos-1x = -1/√(1-x²)
d/dx of tan-1x = 1/(1+x²)
d/dx of cot-1x = -1/(1+x²)
d/dx of sec-1x = 1/|x|√(x²-1)
d/dx of cosec-1x = -1/|x|√(x²-1)
Friday, June 13, 2008
Differentiation - Model Problems - 6
Problems having some complexity and requiring use multiple concepts from this chapter as well as other chapters of algebra, trigonometry and geometry
1. Prob:
If y = tan-1[(√(1+x²) -1)/x]
find y'(1)
To solve the problem, transform the given expression by putting x = tan θ or θ = tan-1x
[(√(1+x²) -1)/x] = [(√(1+tan²θ) - 1)/tan θ)
=> (√(sec²θ)-1)/tan θ = (sec θ - 1)/tan θ
writing sec θ = 1/cos θ and tan θ = sin θ/cos θ
(sec θ - 1)/tan θ = (1- cos θ)/sin θ
=> tan θ/2
Therefore y tan-1(tan θ/2) = θ/2
=> y = (tan-1x)/2
As d/dx of tan-1x = 1(1+x²)
dy/dx = (1/2)* 1(1+x²)
Therefore y'(1) = (1/2)* 1(1+1) = (1/2)*(1/2) = 1/4
2. Prob: if f(x) = logx(ln x), then find value of f'(x) at e. (JEE 1985)
Solution:
f(x) = logx(ln x),
=> xf(x) = ln(x)
Taking logaritm on both sides
f(x)ln x = ln (ln x)
Write y = f(x)
y (ln x) = ln (ln x)
Differentiating both sides with respect to x
y*(1/x)+(ln x)*(dy/dx) = (1/ln x)* (1/x)
substitute x = e
y*(1/e) + (ln e)*(dy/dx) = (1/ln e)*(1/e)
=> dy/dx = (1/e) - (y/e)
value of y at e = logx(ln e) = logx(1) = 0
Therefore dy/dx = 1/e
1. Prob:
If y = tan-1[(√(1+x²) -1)/x]
find y'(1)
To solve the problem, transform the given expression by putting x = tan θ or θ = tan-1x
[(√(1+x²) -1)/x] = [(√(1+tan²θ) - 1)/tan θ)
=> (√(sec²θ)-1)/tan θ = (sec θ - 1)/tan θ
writing sec θ = 1/cos θ and tan θ = sin θ/cos θ
(sec θ - 1)/tan θ = (1- cos θ)/sin θ
=> tan θ/2
Therefore y tan-1(tan θ/2) = θ/2
=> y = (tan-1x)/2
As d/dx of tan-1x = 1(1+x²)
dy/dx = (1/2)* 1(1+x²)
Therefore y'(1) = (1/2)* 1(1+1) = (1/2)*(1/2) = 1/4
2. Prob: if f(x) = logx(ln x), then find value of f'(x) at e. (JEE 1985)
Solution:
f(x) = logx(ln x),
=> xf(x) = ln(x)
Taking logaritm on both sides
f(x)ln x = ln (ln x)
Write y = f(x)
y (ln x) = ln (ln x)
Differentiating both sides with respect to x
y*(1/x)+(ln x)*(dy/dx) = (1/ln x)* (1/x)
substitute x = e
y*(1/e) + (ln e)*(dy/dx) = (1/ln e)*(1/e)
=> dy/dx = (1/e) - (y/e)
value of y at e = logx(ln e) = logx(1) = 0
Therefore dy/dx = 1/e
Differentiation - Model Problems - 7
1. There is exists a function f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x , and
a. f’’(x)>0 for all x
b. -1
d. f’’(x)<-2 for all x
(JEE, 1982)
Answer: (a)
Reason:
x² -x+1 = (x – ½)² + ¾ can be a solution to the f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x as f(0) = 1 and f’(0) = -1
f’’(x) = 2 and hence f’’(x)>0 for all x.
We may assume f(x) = e-x and get a similar conclusion. But e-x may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.
2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
(JEE 1982)
Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
Solution: f’(x) = sin x²
=> f’(t) = sin t²
Given problem is visualized as
y = f(t)
t = [(2x-1)/( x²+1)]
dy/dx = dy/dt*dt/dx
=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
a. f’’(x)>0 for all x
b. -1
d. f’’(x)<-2 for all x
(JEE, 1982)
Answer: (a)
Reason:
x² -x+1 = (x – ½)² + ¾ can be a solution to the f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x as f(0) = 1 and f’(0) = -1
f’’(x) = 2 and hence f’’(x)>0 for all x.
We may assume f(x) = e-x and get a similar conclusion. But e-x may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.
2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
(JEE 1982)
Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
Solution: f’(x) = sin x²
=> f’(t) = sin t²
Given problem is visualized as
y = f(t)
t = [(2x-1)/( x²+1)]
dy/dx = dy/dt*dt/dx
=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
Tuesday, June 10, 2008
Progressions - Model Problems - 1
Chapter: Sequences, Series and Progressions
Arithmetic Progressions
Finding nth term of arithmetic progessions
nth term = a +(n-1)d
Prob: Find the 20th term of the A.P. starting with 4 and common difference 5.
a = 4
d = 5
n = 20
20th term = 4 + (20-1)*5 = 4+19*5 = 4 + 95 = 99.
Finding sum to n terms of AP
Sn = ½ n(a + l)
Sn = ½ n{2a+(n-1)d}
Prob: Find the sum to 20th term of the A.P. starting with 4 and common difference 5.
We found in the earlier problem the 20th term.
a = 4
n = 20
l = 99
Sn = ½ *20(4+99) = 10*103 = 1030.
Insertion of arithmetic means
Prob: Insert 4 arithmetic means between 4 and 19.
We have a = 4, l = 19 and 4 arithmetic means are to be inserted. It means the last term l is the 6th term.
19 = 4+5d
d = 3
the firm A.M. = second term = 4+3 = 7
second A.M. = 7+3 = 10
third A.M. = 10+3 = 13
fourth A.M. = 13+3 = 16
Arithmetic Progressions
Finding nth term of arithmetic progessions
nth term = a +(n-1)d
Prob: Find the 20th term of the A.P. starting with 4 and common difference 5.
a = 4
d = 5
n = 20
20th term = 4 + (20-1)*5 = 4+19*5 = 4 + 95 = 99.
Finding sum to n terms of AP
Sn = ½ n(a + l)
Sn = ½ n{2a+(n-1)d}
Prob: Find the sum to 20th term of the A.P. starting with 4 and common difference 5.
We found in the earlier problem the 20th term.
a = 4
n = 20
l = 99
Sn = ½ *20(4+99) = 10*103 = 1030.
Insertion of arithmetic means
Prob: Insert 4 arithmetic means between 4 and 19.
We have a = 4, l = 19 and 4 arithmetic means are to be inserted. It means the last term l is the 6th term.
19 = 4+5d
d = 3
the firm A.M. = second term = 4+3 = 7
second A.M. = 7+3 = 10
third A.M. = 10+3 = 13
fourth A.M. = 13+3 = 16
Progressions - Model Problems - 6
Complex problems in progressions
Prob 1: Find the sum of the series
S = 1²-2²+3²-4²+…-2002²+2003²
We need transform the given series into
S = (1-2)(1+2) + (3-4)(3+4)+…+(2001-2002)(2001+2002)+2003²
=> (-1)(3) + (-1)(7) + (-1)(11) +(-1)(15)+…+(-1)(4003) + 2003²
=> (-1)[3+7+11+…+4003]+2003²
So we have an arithmetic progression in the brackets with a =3, d = 4 and l = 4003 and n = 1001 terms.
Sum to n terms in AP is
Sn = ½ n(a + l)
Hence the sum of 1001 terms in the brackets = 1001*[3+4003]/2
= 1001*4006/2 = 1001*2003
So S = -1001*2003 + 2003²
= 2003(2003-1001) = 2003*1002 = 2007006
Prob 2: Find the value of n for which
704 + ½ *(704)+1/4 * (704) +… up to n terms =
1984 – ½ *(1984) + ¼* (1984)- … up to n terms
Actually there is no complexity in the problem. But for a first look it looks to be a complex problem. You have to identify both LHS and RHS as geometric progressions.
For LHS a = 704 and r = ½.
For RHS a = 1984 and r = -1/2
sum of n terms of GP
Sn = a(1-rn)/(1-r)
704[1 – (1/2)n]/(1 – ½ ) = 1984 [1 – (-1/2)n]/(1 – (-½ ))
704*2(1 – 1/2n] = 1984*2/3*[1 – (-1) n/2n]
704*6(1 – 1/2n] = 1984*2[1 – (-1) n/2n]
4224 – 4224/2n = 3968 - 3968(-1) n/2n
4224-3968 = 4224/2n - 3968(-1) n/2n
256 = 4224/2n - 3968(-1) n/2n
128 = 2112/2n - 1984(-1) n/2n
If n is assumed as odd
128 = (2112+1984)/ 2n
=> 2n = 4096/128 = 1024/32 = 128/4 = 32
=> 2n = 32
n = 5
If n is assumed as even
128 = (2112 – 1984)/ 2n = 128/2n
=> 2n = 128/128 = 1
Implies n = 0
Hence the answer is n =5 or n = 0. The logical answer is 5 as n = 0 is a trivial solution..
Prob 1: Find the sum of the series
S = 1²-2²+3²-4²+…-2002²+2003²
We need transform the given series into
S = (1-2)(1+2) + (3-4)(3+4)+…+(2001-2002)(2001+2002)+2003²
=> (-1)(3) + (-1)(7) + (-1)(11) +(-1)(15)+…+(-1)(4003) + 2003²
=> (-1)[3+7+11+…+4003]+2003²
So we have an arithmetic progression in the brackets with a =3, d = 4 and l = 4003 and n = 1001 terms.
Sum to n terms in AP is
Sn = ½ n(a + l)
Hence the sum of 1001 terms in the brackets = 1001*[3+4003]/2
= 1001*4006/2 = 1001*2003
So S = -1001*2003 + 2003²
= 2003(2003-1001) = 2003*1002 = 2007006
Prob 2: Find the value of n for which
704 + ½ *(704)+1/4 * (704) +… up to n terms =
1984 – ½ *(1984) + ¼* (1984)- … up to n terms
Actually there is no complexity in the problem. But for a first look it looks to be a complex problem. You have to identify both LHS and RHS as geometric progressions.
For LHS a = 704 and r = ½.
For RHS a = 1984 and r = -1/2
sum of n terms of GP
Sn = a(1-rn)/(1-r)
704[1 – (1/2)n]/(1 – ½ ) = 1984 [1 – (-1/2)n]/(1 – (-½ ))
704*2(1 – 1/2n] = 1984*2/3*[1 – (-1) n/2n]
704*6(1 – 1/2n] = 1984*2[1 – (-1) n/2n]
4224 – 4224/2n = 3968 - 3968(-1) n/2n
4224-3968 = 4224/2n - 3968(-1) n/2n
256 = 4224/2n - 3968(-1) n/2n
128 = 2112/2n - 1984(-1) n/2n
If n is assumed as odd
128 = (2112+1984)/ 2n
=> 2n = 4096/128 = 1024/32 = 128/4 = 32
=> 2n = 32
n = 5
If n is assumed as even
128 = (2112 – 1984)/ 2n = 128/2n
=> 2n = 128/128 = 1
Implies n = 0
Hence the answer is n =5 or n = 0. The logical answer is 5 as n = 0 is a trivial solution..
Progressions - Model Problems - 7
1. The third term of a geometric progression is 4. The product of the first five terms is:
a. 4³
b. 44
c. 45
d. none of these
(JEE 1982)
Answer ©
Select the five terms as a/r², a/r,a, ar,ar².
Product of the five terms is a/r² * a/r * a *ar*ar² = a5
As the third term is 4, a = 4
Hence product is 45
2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.
(JEE 1984)
Answer: 3050
Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)
Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)
Integers divisible by 10 will be in integers divisible by 2. We can remove them from integers divisible by 5 to find out the sum required
So integers for which sum is to be taken 2,4,6…,100 and
5,15,25,…,95
First progression sum 2(sum of 1,2,3,…,50)
= 2*50*51/2 = 2550
Second progression is an A.P. with d = 10 and a = 5
Hence sum = n/2(a+l) = 10/2(100) = 500
Total sum = 2550 +500 = 3050
a. 4³
b. 44
c. 45
d. none of these
(JEE 1982)
Answer ©
Select the five terms as a/r², a/r,a, ar,ar².
Product of the five terms is a/r² * a/r * a *ar*ar² = a5
As the third term is 4, a = 4
Hence product is 45
2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.
(JEE 1984)
Answer: 3050
Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)
Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)
Integers divisible by 10 will be in integers divisible by 2. We can remove them from integers divisible by 5 to find out the sum required
So integers for which sum is to be taken 2,4,6…,100 and
5,15,25,…,95
First progression sum 2(sum of 1,2,3,…,50)
= 2*50*51/2 = 2550
Second progression is an A.P. with d = 10 and a = 5
Hence sum = n/2(a+l) = 10/2(100) = 500
Total sum = 2550 +500 = 3050
Monday, June 9, 2008
Complex Numbers -Model Problems - 1
To find in divide n by 4 to get 4m+r where m is the quotient and r is the remainder.
in will be equal to ir
Prob: The value of i53/i121 is
a. 2i
b. i
c. -2i
d. 2
53 = 13*4 + 1
So i53 = i
121 = 30*4 + 1
So i121 = i
i53/i121 is equal to i/i = 1
Answer (b)
Multiplication of complex numbers
(a1+ib1) (a2+ib2) by multiplying and simplifying we get
(a1a2 – b1b2) + i(a1b2+a2b1)
Prob: Multiply (2+i) by (2+i)
(2+i)(2+i) = 2*2 - 1*1 + i(2*1+2*1) = 3+4i
Division of complex numbers
z1/z2 = z1* Multiplicative inverse of z2
Multiplicative inverse of a+ib = a/(a² + b²) - ib/(a² + b²)
Prob: Find the result of (7+i)/(1+3i)
First step; Find multiplicative inverse of 1+3i = 1/(1²+3²) - i 3/(1²+3²)
= 1/10 - i 3/10
Therefore (7+i)/(1+3i) = (7+i)(1/10 - 3i/10)
= 7*1/10 - (1)(-3/10) + i(7*(-3/10)+1(1/10))
= 7/10 + 3/10 + i(-21/10+1/10)
= 10/10 + i(-20/10)
= 1 - 2i
in will be equal to ir
Prob: The value of i53/i121 is
a. 2i
b. i
c. -2i
d. 2
53 = 13*4 + 1
So i53 = i
121 = 30*4 + 1
So i121 = i
i53/i121 is equal to i/i = 1
Answer (b)
Multiplication of complex numbers
(a1+ib1) (a2+ib2) by multiplying and simplifying we get
(a1a2 – b1b2) + i(a1b2+a2b1)
Prob: Multiply (2+i) by (2+i)
(2+i)(2+i) = 2*2 - 1*1 + i(2*1+2*1) = 3+4i
Division of complex numbers
z1/z2 = z1* Multiplicative inverse of z2
Multiplicative inverse of a+ib = a/(a² + b²) - ib/(a² + b²)
Prob: Find the result of (7+i)/(1+3i)
First step; Find multiplicative inverse of 1+3i = 1/(1²+3²) - i 3/(1²+3²)
= 1/10 - i 3/10
Therefore (7+i)/(1+3i) = (7+i)(1/10 - 3i/10)
= 7*1/10 - (1)(-3/10) + i(7*(-3/10)+1(1/10))
= 7/10 + 3/10 + i(-21/10+1/10)
= 10/10 + i(-20/10)
= 1 - 2i
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