KNOWLEDGE QUESTION AND ANSWER BOARD
http://knol.google.com/k/narayana-rao-kvss/-/2utb2lsm2k7a/654#
Thursday, January 1, 2009
Saturday, September 27, 2008
Analytical geometry - Combination problems
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
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, 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
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