KNOWLEDGE QUESTION AND ANSWER BOARD
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Thursday, January 1, 2009
Monday, December 15, 2008
Binomial theorem - Past JEE Questions - Answers
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 express 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
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 express 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
Binomial theorem - Past JEE Questions - Problems
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, Scr)
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 Scr)
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 Scr)
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 Scr)
a. 405/256
b. 504/259
c. 450/263
d. none of these
(JEE 1983, Scr)
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 Scr)
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 Scr)
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 Scr)
Determinants - Past JEE Questions - Problems
1. The system of equations
λx+y+z = 0
-x+λy+z = 0
-x-y+λz = 0
will have a non-zero solution if real values of λ are given by(or if real values of λ are equal to) ________________________ .
(JEE 1984 Scr)
2. State whether the statement is true or false.
If
determinant
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|
is equal to determinant
|a1 b1 1|
|a2 b2 1|
|a3 b3 1|
then the two triangles with vertices 9x1,y1), (x2,y2), (x3,y3) and (a1,b1), (a2,b2) and (a3,b3) must be congruent.
λx+y+z = 0
-x+λy+z = 0
-x-y+λz = 0
will have a non-zero solution if real values of λ are given by(or if real values of λ are equal to) ________________________ .
(JEE 1984 Scr)
2. State whether the statement is true or false.
If
determinant
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|
is equal to determinant
|a1 b1 1|
|a2 b2 1|
|a3 b3 1|
then the two triangles with vertices 9x1,y1), (x2,y2), (x3,y3) and (a1,b1), (a2,b2) and (a3,b3) must be congruent.
Straight lines - Past JEE
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)
2. The straight lines x+y = 0, 3x+y-4 = 0, x+3y-4 = 0 for a triangle which is
a. isoscles
b. equilateral
c. right angled
d. none of these
(JEE 1983, Scr)
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
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)
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)
2. The straight lines x+y = 0, 3x+y-4 = 0, x+3y-4 = 0 for a triangle which is
a. isoscles
b. equilateral
c. right angled
d. none of these
(JEE 1983, Scr)
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
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)
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)
Sunday, December 14, 2008
Differentiation - Past JEE - Answers
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 is less than f’’’(x)<0 for all x
c. -2≤f’’(x) ≤-1 or all x
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 is less than f’’’(x)<0 for all x
c. -2≤f’’(x) ≤-1 or all x
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
Saturday, December 13, 2008
Differentiation - Past JEE
1. State true or false
The derivative of an even function is always an odd function (JEE 1983 Sc)
2. For the function f(x) = x/(1+e(1/x), x≠0; and f(x) = 0 if x = o
Find the derivative from the right,
f'(0+)
and the derivative from the left
f'(0-)
(JEE 1983 sc)
3. If f(x) = logx(ln x), then f'(x) at x = e is ______________
(JEE 1985, Sc)
4. The derivative of sec-1{1/(2x²-1)] with respect to √(1-x²) at x = 1/2 is _______________________. (JEE 1986, Sc)
5. If y² = P(x), is a polynomial of degree 3, then
2d/dx of [y³(d²y/dx²)] equals
a. P'''(x)+P'(x)
b. P'(x)P'''(x)
c. P(x)P'''(x)
d. a constant
6. If f(x0 = |x-2| and g(x) = f[f(x)], then g'(x) ________________ for x greater than 20. (JEE 1990, Sc)
7. If y = (sin x)tan x, then dy/dx is equal to
a. (sin x)tan x.( 1 + sec² x. log tan x)
b. tan x.(sin x)tan x - 1.cos x
c. (sin x)tan x.sec² x.log sin x
d. tan x.(sin x)tan x-1
(JEE 1994 Sc)
8. Let F(x) = f(x)g(x)h(x) for all real x, where f(x),g(x),h(x) are differentiable functions. At some points x 0
F'(x0)= 21F(x0)
f'(x0), = 4f(x0),
g'(x0), = -7g(x0),
h'(x0) = kh(x0)
then k = ?
(JEE 1997, SC)
9. the left hand derivative of f(x) [x]sin (πx) at x = k, k an integer is
a. (-1)k (k-1)π
b. (-1)k-1 (k-1)π
c. (-1)kkπ
d. (-1)k-1kπ
10. The domain of the derivative of the function
f(x) = tan -1x if -1x if |x|≤1 and 1/2(|x|-1) if |x| is greater than 1 is
a. R - {-1,1)
b. R - {1}
c. R - {-1}
d. R - {0}
(JEE, 2002, Sc)
The derivative of an even function is always an odd function (JEE 1983 Sc)
2. For the function f(x) = x/(1+e(1/x), x≠0; and f(x) = 0 if x = o
Find the derivative from the right,
f'(0+)
and the derivative from the left
f'(0-)
(JEE 1983 sc)
3. If f(x) = logx(ln x), then f'(x) at x = e is ______________
(JEE 1985, Sc)
4. The derivative of sec-1{1/(2x²-1)] with respect to √(1-x²) at x = 1/2 is _______________________. (JEE 1986, Sc)
5. If y² = P(x), is a polynomial of degree 3, then
2d/dx of [y³(d²y/dx²)] equals
a. P'''(x)+P'(x)
b. P'(x)P'''(x)
c. P(x)P'''(x)
d. a constant
6. If f(x0 = |x-2| and g(x) = f[f(x)], then g'(x) ________________ for x greater than 20. (JEE 1990, Sc)
7. If y = (sin x)tan x, then dy/dx is equal to
a. (sin x)tan x.( 1 + sec² x. log tan x)
b. tan x.(sin x)tan x - 1.cos x
c. (sin x)tan x.sec² x.log sin x
d. tan x.(sin x)tan x-1
(JEE 1994 Sc)
8. Let F(x) = f(x)g(x)h(x) for all real x, where f(x),g(x),h(x) are differentiable functions. At some points x 0
F'(x0)= 21F(x0)
f'(x0), = 4f(x0),
g'(x0), = -7g(x0),
h'(x0) = kh(x0)
then k = ?
(JEE 1997, SC)
9. the left hand derivative of f(x) [x]sin (πx) at x = k, k an integer is
a. (-1)k (k-1)π
b. (-1)k-1 (k-1)π
c. (-1)kkπ
d. (-1)k-1kπ
10. The domain of the derivative of the function
f(x) = tan -1x if -1x if |x|≤1 and 1/2(|x|-1) if |x| is greater than 1 is
a. R - {-1,1)
b. R - {1}
c. R - {-1}
d. R - {0}
(JEE, 2002, Sc)
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