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New answer posted
a year agoContributor-Level 10
m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies
New answer posted
a year agoContributor-Level 9
(x+1)/ (x²/³-x¹/³+1) - (x-1)/ (x-x¹/²) = (x¹/³+1) - (1+x? ¹/²)
= (x¹/³ - x? ¹/²)¹?
general term = ¹? C? (x¹/³)¹? (-x? ¹/²)? = ¹? C? x^ (20-5r)/6)
For independent of x, 20-5r=0 ⇒ r=4
∴ Coefficient = ¹? C? = 210
New answer posted
a year agoContributor-Level 9
p = 2i + 3j + k, q = I + 2j + k
r = αi + βj + γk is ⊥ to p+q and p-q
∴ r is collinear with (p+q) * (p-q) = -2 (p*q)
p*q = |i, j, k; 2,3,1; 1,2,1| = I - j + k
∴ r = λ (i - j + k)
|r| = √3 ⇒ λ = 1
∴ r = I - j + k = αi + βj + γk
⇒ α=1, β=-1, γ=1
|α|+|β|+|γ| = 3
New answer posted
a year agoContributor-Level 9
| Class | No. of students | Number of possible cases |
| :-: | :-: | :- |
| 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
| 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
| 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
Total cases =? C? *? C? *? C? +? C? *? C? *? C? +? C? *? C? *? C?
= 23,800 = 100K
∴ K = 238
New answer posted
a year agoContributor-Level 10
f (x+y)=f (x)f (y). f (n)=f (1)?
Σf (x)=f (1)/ (1-f (1)=2 ⇒ f (1)=2/3.
f (4)/f (2) = (f (1)? )/ (f (1)²) = f (1)² = 4/9.
New answer posted
a year agoContributor-Level 10
Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.
New answer posted
a year agoContributor-Level 10
√ (1+x²) (1+y²) + xy (dy/dx)=0.
√ (1+x²)/x dx + √ (1+y²)/y dy = 0.
√ (1+x²)+½ln| (√ (1+x²)-1)/ (√ (1+x²)+1)|+√ (1+y²)=C.
New answer posted
a year agoContributor-Level 10
α+β=64; αβ=256=2?
(α³/β? )¹/? + (β³/α? )¹/? = (α+β)/ (αβ)? /? = 64/32=2.
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