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

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

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,∞)

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)

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

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