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New answer posted
2 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
2 months agoNew answer posted
2 months agoContributor-Level 10
Probability (at most two machines will be out of service) = (3/4)³ . k
⇒ ?C?(1/4)?(3/4)? + ?C?(1/4)¹(3/4)? + ?C?(1/4)²(3/4)³ = (3/4)³ . k
⇒ 17/8 (3/4)³ = (3/4)³ . k
⇒ k = 17/8
New answer posted
2 months agoContributor-Level 10
f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)
∴ maxima at x = -2/3
∴ minima at x = 1
New answer posted
2 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
2 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
2 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
2 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
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