Maths-
The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is
Maths-General
Answer:The correct answer is:
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maths-
If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2t4s2, then the number of ordered pair (p, q) is –
If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2t4s2, then the number of ordered pair (p, q) is –
maths-General
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A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length
A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length
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nCr + 2nCr+1 + nCr+2 is equal to (2 
r n)
nCr + 2nCr+1 + nCr+2 is equal to (2 r n)
maths-General
maths-
The coefficient of 
in 
is
The coefficient of in is
maths-General
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If the tangents drawn from a point on the hyperbola 
to the ellipse 
make angles α and β with the transverse axis of the hyperbola, then
If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then
maths-General
maths-
If the line 
is a normal to the hyperbola then 
If the line is a normal to the hyperbola then
maths-General
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How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?
How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?
maths-General
maths-
The value of 50C4 + 
is -
The value of 50C4 + is -
maths-General
maths-
The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. The maximum number of triangles formed with vertices at these points are -
Total number of points = m +n + k. Therefore the total number of triangles formed by these points is m + n + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
m + n + kC3 – mC3 – nC3 – kC3.
m + n + kC3 – mC3 – nC3 – kC3.
The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. The maximum number of triangles formed with vertices at these points are -
maths-General
Total number of points = m +n + k. Therefore the total number of triangles formed by these points is m + n + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
m + n + kC3 – mC3 – nC3 – kC3.
m + n + kC3 – mC3 – nC3 – kC3.
maths-
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-
maths-General
maths-
The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 < K < 18) is -
The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K)
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 < K < 18) is -
maths-General
The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K)
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
physics-
A mass of 
strikes the wall with speed 
at an angle as shown in figure and it rebounds with the same speed. If the contact time is 
, what is the force applied on the mass by the wall
Force = Rate of change of momentum
Initial momentum
Final momentum
Substituting
Force on the ball
Negative sign indicates direction of the force
Initial momentum
Final momentum
Substituting
Force on the ball
Negative sign indicates direction of the force
A mass of strikes the wall with speed at an angle as shown in figure and it rebounds with the same speed. If the contact time is , what is the force applied on the mass by the wall
physics-General
Force = Rate of change of momentum
Initial momentum
Final momentum
Substituting
Force on the ball
Negative sign indicates direction of the force
Initial momentum
Final momentum
Substituting
Force on the ball
Negative sign indicates direction of the force
maths-
If the locus of the mid points of the chords of the ellipse 
, drawn parallel to 
is 
then 
If the locus of the mid points of the chords of the ellipse , drawn parallel to is then
maths-General
maths-
If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr+1 + nCr –1 + 2 × nCr equals-
If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr+1 + nCr –1 + 2 × nCr equals-
maths-General
maths-
The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -
The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -
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