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 x

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)

(JEE 1994)

3. If the binomial expansion of (a-b)

a. (n-5)/6

b. (n-4)/5

c. 5/(n-4)

d. 6/(n-5)

(JEE 2001)

4. The coefficient of t

a.

b.

c.

d.

(JEE 2003)

Answers and Solutions

1. The coefficient of x

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

=

=

In this term (x)

10-3r = 4 or r = 2

So coefficient =

= [10*9/2]*9*(1/2

= (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)

(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)

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(

=> 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)

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 =

6th term = 5+1th term =

As sum of 5th terms and 6th terms is zero

a

a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)

= (n-4)!/(5(n-5)!)

= (n-4)/5

4. The coefficient of t

a.

b.

c.

d.

(JEE 2003)

Answer: (d)

Solution:

Multiplying the last two expressions we have to find coefficient of t

t

1st term is t

2nd term is t

3rd term is 1 in the first expression multiplied by t

Coefficients of 1 and 3 terms are 1 and 1.

Coefficient of 2 term = coefficient t

7th term or 6+1th term in the expansion will have t

Hence coefficient of 7th term =

Therefore coefficient of t

^{4}in [(x/2)- (3/x²)]^{10}isa. 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 equalsa. (n-5)/6

b. (n-4)/5

c. 5/(n-4)

d. 6/(n-5)

(JEE 2001)

4. The coefficient of t

^{24}in (1+t²)^{12}(1+t^{12})(1+^{24}) isa.

^{12}C_{6}+ 3b.

^{12}C_{6}+ 1c.

^{12}C_{6}d.

^{12}C_{6}+ 2(JEE 2003)

Answers and Solutions

1. The coefficient of x

^{4}in [(x/2)- (3/x²)]^{10}isa. 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

^{10}C_{r}(x/2)^{10-r}(-3/x²)^{r}=

^{10}C_{r}(x)^{10-r}(1/ x²)^{r}(1/2)^{10-r}(-3)^{r}=

^{10}C_{r}(-3)^{r}(1/2)^{10-r}(x)^{10-3r}In this term (x)

^{10-3r}is equal to x^{4} 10-3r = 4 or r = 2

So coefficient =

^{10}C_{r}(-3)^{r}(1/2)^{10-r}^{10}C_{2}(-3)^{2}(1/2)^{10-2}= [10*9/2]*9*(1/2

^{8})= (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^{n}C_{1},^{n}C_{2},^{n}C_{3}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(

^{n}C_{2}) =^{n}C_{1}+^{n}C_{3}=> 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 equalsa. (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 =

^{n}C_{4}a^{n-4}(-b)^{4}6th term = 5+1th term =

^{n}C_{5}a^{n-5}(-b)^{5}As sum of 5th terms and 6th terms is zero

^{n}C_{4}a^{n-4}(-b)^{4}+^{n}C_{5}a^{n-5}(-b)^{5}= 0^{n}C_{4}a^{n-4}(b)^{4}=^{n}C_{5}a^{n-5}(b)^{5}a

^{n-4}(b)^{4}/ a^{n-5}(b)^{5}=^{n}C_{5}/^{n}C_{4}a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)

= (n-4)!/(5(n-5)!)

= (n-4)/5

4. The coefficient of t

^{24}in (1+t²)^{12}(1+t^{12})(1+^{24}) isa.

^{12}C_{6}+ 3b.

^{12}C_{6}+ 1c.

^{12}C_{6}d.

^{12}C_{6}+ 2(JEE 2003)

Answer: (d)

Solution:

Multiplying the last two expressions we have to find coefficient of t

^{24}in (1+t²)^{12}(1+t^{12}+^{24})t

^{24}term will come as a sum of three terms in the expanded expression1st term is t

^{24}term in 1st expression multiplied by 1 in the second expression.2nd term is t

^{12}term in 1st expression multiplied by t^{12}term in the second expression.3rd term is 1 in the first expression multiplied by t

^{24}term in the second expression.Coefficients of 1 and 3 terms are 1 and 1.

Coefficient of 2 term = coefficient t

^{12}term in 1 st expression which is (1+t²)^{12}7th term or 6+1th term in the expansion will have t

^{12}termHence coefficient of 7th term =

^{12}C_{6}Therefore coefficient of t

^{24}term will be^{12}C_{6}+ 2## Wednesday, June 25, 2008

### Logarithms - 6

If log0.09(x-1), then x lies in the interval:

a.(2,∞)

b. (1,2)

c. (-2,-1)

d. none of these

(JEE 1983)

Answer (a)

log0.09(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,∞)

_{0.3}(x-1)a.(2,∞)

b. (1,2)

c. (-2,-1)

d. none of these

(JEE 1983)

Answer (a)

log

_{0.3}(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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