Class 11th
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New answer posted
10 months agoLet the tangents drawn from the origin to the circle,
touch it at the points
and
. Then
is equal to:
Contributor-Level 10
OS = √S?
Radius = R = 2
Length of AB = (2RL)/√(L²+R²) = 16/√20
AB² = 64/5

New answer posted
10 months agoContributor-Level 10
120.00
(1 + x + x² + x?)? = (1 + x)? · (1 + x²)?
Coefficient of x? = ?C?·?C? + ?C?·?C? + ?C?·?C?
= 15 + 90 + 15
= 120
New answer posted
10 months agoContributor-Level 10
19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? * (3!/(2!1!)) * 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19
New answer posted
10 months agoNew answer posted
10 months agoContributor-Level 10
5+3+7+a+b = 25 ⇒ a+b=10
S.D. = √(((5²+3²+7²+a²+b²)/5) - 5²) = 2
√((a²+b²+83)/5) - 25 = 4 ⇒ a²+b² = 62
⇒ (a+b)² - 2ab = 62 ⇒ ab = 19
So equation whose roots are a and b is x² - 10x + 19 = 0
New answer posted
10 months agoContributor-Level 10
Let z = (3+isinθ)/(4-icosθ) x (4+icosθ)/(4+icosθ)
= (12 - sinθcosθ + i(4sinθ + 3cosθ))/(16 + cos²θ)
z is real
∴ 4sinθ + 3cosθ = 0
⇒ tanθ = -3/4 [? θ lies is 2nd quadrant]
arg(sinθ + icosθ) = π + tan?¹(cosθ/sinθ)
= π - tan?¹(4/3)
New answer posted
10 months agoContributor-Level 10
p? = α? + β?
= (α + 1)² . α + (β - 1)² . β
= 5α + 5β + 6
= 5(1) + 6 = 11
p? = α² + β² = α + β + 2 = 3
p? = α³ + β³ = (α + 1) . α + (β + 1) . β
= 1 + 3 = 4
Hence p? ≠ p? . p?
New answer posted
10 months agoContributor-Level 10
Given log?/? x + log?/? x + log?/? x + . 20 times = 460
⇒ (2+3+4+.+21)log?x = 460
⇒ (20/2)(2+21)log?x = 460
⇒ log?x = 2
⇒ x = 49
New answer posted
10 months agoContributor-Level 10
a? + a? = 4 ⇒ a? + a?r = 4
a? + a? = 16 ⇒ a?r² + a?r³ = 16
⇒ r = ±2
r = 2 ⇒ a? = 4/3
r = -2 ⇒ a? = -4
Σ(i=1 to 9) a? = (a(r?-1))/(r-1) = (-4)((-2)? - 1)/(-2-1)
= 4/3 (-513) = 4λ
⇒ λ = -171
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