Class 11th
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New answer posted
2 months agoContributor-Level 10
Na? SO? + H? SO? → SO? + Na? SO? + H? O
SO? + 2NaOH → Na? SO? + H? O
Na? SO? + SO? + H? O → 2NaHSO?
New answer posted
2 months agoContributor-Level 10
x? = Σf? x? / Σf? = (10 + 15x + 50) / (4+x)
= (60+15x)/ (4+x) = 15
σ² = 50 = Σf? x? ²/Σf? - (x? )²
50 = (50+225x+1250)/ (4+x) - (15)²
50 = (1300+225x)/ (4+x) - 225
⇒ 275 (4+x) = 1300 + 225x
⇒ 50x = 200 ⇒ x = 4
New answer posted
2 months agoContributor-Level 10
Let P (2cosθ, 2sinθ)
∴ Q (-2cosθ, -2sinθ)
Given line x+y-2=0
∴ α = |2cosθ + 2sinθ – 2| / √2
β = |-2cosθ - 2sinθ – 2| / √2
∴ αβ = √2 (cosθ + sinθ – 1) · √2 (cosθ + sinθ + 1)
= 2|cos²θ + sin²θ + 2sinθcosθ – 1| = 2|sin2θ|
Max |sin2θ| = 1
∴ maximum αβ = 2.
New answer posted
2 months agoContributor-Level 10
Ways of selecting correct questions =? C? = 15
Ways of doing them correct = 1
Ways of doing remaining 2 questions incorrect = 3² = 9
∴ No. Of ways = 15 * 1 * 9 = 135
New answer posted
2 months agoContributor-Level 10
Given,
300 = 1 + (N – 1)d
⇒ (N − 1)d = 299
∴ (N, d) = (24,13) is the only possible pair
∴ a? = 1 + 19 (13) = 248 and, S? = (1+248)/2 * 20
= 2490
New answer posted
2 months agoContributor-Level 10
Any point (x, y) on perpendicular bisector equidistant from p and q
∴ (x − 1)² + (y − 4)² = (x − k)² + (y − 3)²
At x = 0, y = -4
∴ 1 + 64 = k² + 49
k² = 16
New answer posted
2 months agoContributor-Level 10
Let number of elements in T is R.
∴ 20R = 500 ⇒ R = 25
and 6R = 5N ⇒ N = 30
New answer posted
2 months agoContributor-Level 10
Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1
New answer posted
2 months agoContributor-Level 10
tan 30° = x/y ⇒ y = √3x
and tan 60° = (x+400)/y ⇒ √3y = x+400
= x + 400
Solving (i) and (ii), we get
2x = 400, x = 200
sin 30° = x/PC = 200/PC ⇒ PC = 400
New answer posted
2 months agoContributor-Level 10
Applying L'Hôpital's Rule
Lim (t→x) [2tf² (x) – x² (2f (t)f' (t)] / 1
∴ 2xf² (x) – x² (2f (x)f' (x) = 0
⇒ f (x) – xf' (x) = 0
⇒ f' (x)/f (x) = 1/x ⇒ lnf (x) = lnx + C
At x=1, c=1
∴ lnf (x) = lnx + 1
when f (x) = 1
then lnx = -1
x = 1/e
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