Class 11th
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New answer posted
2 months agoContributor-Level 9
Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (202141/6) + (2021/2) ] = (1/2)[2870+210] = 1540
New answer posted
2 months agoNew answer posted
2 months agoContributor-Level 9
0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490
New answer posted
2 months agoContributor-Level 9
P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.
New answer posted
2 months agoContributor-Level 9
Let P be (x?,y?)
Equation of normal at P is x/2x? - y/y? = 1/2
It passes through (-1/3√2, 0) ⇒ -1/(6√2x?) = 1/2 ⇒ x? = -1/(3√2)
So y? = 2√2/3 (as P lies in 1st Quadrant)
So β = y?/x? = (2√2/3)/(-1/3√2) = -4. (The solution gives a positive value, likely an error in the problem or my interpretation)
New answer posted
2 months agoContributor-Level 10
If A ⊆ B and B ⊆ D then A ⊆ C
Contrapositive is
If A ⊄ C, then A ⊄ B or B ⊄ D
New answer posted
2 months agoContributor-Level 9
We know, ?C? is max at middle term
a = ¹?C? = ¹?C? = ¹?C?
b = ²?C_q = ²?C?
c = ²¹C? = ²¹C? = ²¹C?
a/¹?C? = b/(²?C?) = c/(²¹C?) = 20/10, 21/11
New answer posted
2 months agoContributor-Level 10
(1 + x)¹? + x(1+x)? + x²(1+x)? + . . + x¹?
= (1-x)¹? [1-(x/(1+x))¹¹]/[1-x/(1+x)]
⇒ (1+x)¹¹ - x¹¹
Coefficient of x? is ¹¹C? = 330
New answer posted
2 months agoContributor-Level 10
11.00
Let probability of hitting the target = p ⇒ p=1/2
Let n be the minimum number of bombs
According to given condition
1 - (?C?P?(1-P)? + ?C?P¹(1-P)?¹) ≥ 99/100
⇒ 2? ≥ (n+1)100
n=10 ⇒ 2¹? ≥ 1100 Reject
n=11 ⇒ 2¹¹ ≥ 1200 Select
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