Ncert Solutions Maths class 11th
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New answer posted
a month agoContributor-Level 9
Equation of normal to the ellipse x²/a² + y²/b² = 1 at (x? , y? ) is a²x/x? - b²y/y? = a² - b².
At the point (ae, b²/a):
a²x/ (ae) - b²y/ (b²/a) = a² - b²
It passes through (0, -b).
a² (0)/ (ae) - b² (-b)/ (b²/a) = a² - b²
ab = a² - b²
Since b² = a² (1-e²), a²-b² = a²e².
ab = a²e²
a²b² = a? e?
a² (a² (1-e²) = a? e?
1 - e² = e?
e? + e² - 1 = 0
New answer posted
a month agoContributor-Level 9
Let the first A.P. be a? , a? + d, a? + 2d.
a? = a? + 39d = -159
a? = a? + 99d = -399
Subtracting the equations, 60d = -240 ⇒ d = -4.
Substituting d back, a? + 39 (-4) = -159 ⇒ a? - 156 = -159 ⇒ a? = -3.
Now, for the second A.P. with first term b? and common difference D = d+2 = -2.
b? = a?
⇒ b? + 99D = a? + 69d
⇒ b? + 99 (-2) = -3 + 69 (-4)
⇒ b? - 198 = -3 - 276
⇒ b? = -279 + 198 = -81
New answer posted
a month agoContributor-Level 9
Equation of line is x/3 + y/1 = 1
⇒ x + 3y - 3 = 0
The image (x? , y? ) of point (-1, -4) is given by:
(x? - (-1)/1 = (y? - (-4)/3 = -2 (1 (-1) + 3 (-4) - 3) / (1² + 3²)
(x? + 1)/1 = (y? + 4)/3 = -2 (-1 - 12 - 3)/10 = -2 (-16)/10 = 16/5
x? + 1 = 16/5 ⇒ x? = 11/5
(y? + 4)/3 = 16/5 ⇒ y? + 4 = 48/5 ⇒ y? = 28/5
New answer posted
a month agoContributor-Level 9
Applying Rolle's theorem in for function f (x), there exists c such that f' (c) = 0, c ∈ (0,1).
Again applying Rolle's theorem in [0, c] for function f' (x), there exists c? such that f' (c? ) = 0, c? ∈ (0, c).
Option A is correct.
New answer posted
a month agoContributor-Level 9
Given equation is 2x (2x + 1) = 1 ⇒ 4x² + 2x - 1 = 0. Roots of the equation are α and β.
∴ α + β = -2/4 = -1/2 ⇒ β = -1/2 - α
and
4α² + 2α - 1 = 0 ⇒ α² = (1-2α)/4 = 1/4 - α/2
Now
α = 1/2 - 2α²
Substituting into the expression for β:
β = -1/2 - (1/2 - 2α²) = -1 + 2α²
New answer posted
a month agoContributor-Level 9
In ΔCDF
sin 30° = z/1 [CD = 1 km (given)]
z = 1/2
cos 30° = y/1 ⇒ y = √3/2
Now in ΔABC
tan 45° = h/ (x+y)
⇒ h = x+y
⇒ x = h - √3/2
Now
In ΔBDE,
tan 60° = (h-z)/x
√3x = h - z
√3 (h - √3/2) = h - 1/2
√3h - 3/2 = h - 1/2
h (√3 - 1) = 1
h = 1/ (√3-1) km
New answer posted
a month 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
a month 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
a month 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
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