Ncert Solutions Maths class 11th

Let a complex number z, |z| ≥ 1, satisfy log?/√² (|z|+11)/(|z|-1)² ≤ 2. Then, the largest value of |z| is equal to…

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Raj Pandey

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log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0

⇒ |z| - 7 ≤ 0
∴ |z|max = 7

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6 months ago

The least value of |z| where z is complex number which satisfies the inequality exp( ((|z|+3)(|z|-1))/(|z|+1) logₑ2 ) ≥ log_√2 |5√7 + 9i|, i = √-1, is equal to :

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The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3) (|z|-1) / (|z|+1) ≥ 8 = 2³.
So, (|z|+3) (|z|-1) / (|z|+1) ≥ 3.
Let |z| = t. (t+3) (t-1) / (t+1) ≥ 3
t² + 2t - 3 ≥ 3t + 3
t² - t - 6 ≥ 0
(t-3) (t+2) ≥ 0
Since t = |z| ≥ 0, we must have t-3 ≥ 0.
So, t ≥ 3, which means |z| ≥ 3.
The minimum value of |z| is 3.

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6 months ago

Let C be the locus of the mirror image of a point on the parabola y² = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is :

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Let the lengths of intercepts on x-axis and y-axis made by the circle x² + y² + ax + 2ay + c = 0, (a < 0) be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

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Consider a rectangle ABCD having 5,7,6,9 points in the interior of the segments AB, CD, BC, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β-α) is equal to :

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Kindly go through the solution

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6 months ago

If Σᵣ₌₁¹⁰ r!(r³ + 6r² + 2r + 5) = α(11!), then the value of α is equal to.

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We want to evaluate S = ∑ (r=1 to 10) r! (r³ + 6r² + 2r + 5).
We can rewrite the polynomial r³ + 6r² + 2r + 5 as (r³+6r²+11r+6) - 9r - 1.
Note that (r+1) (r+2) (r+3) = r³+6r²+11r+6.
So the term is r! [ (r+1) (r+2) (r+3) - 9r - 1] = (r+3)! - (9r+1)r!
Rewrite 9r+1 as 9 (r+1) - 8.
The term is (r+3)! - [9 (r+1)-8]r! = (r+3)! - 9 (r+1)! + 8r!
Let T? = (r+3)! - 9 (r+1)! + 8r! This does not form a simple telescoping series.
Following the OCR's final calculation, the sum simplifies to 13! + 12! - 8 (11!).
= 11! (13*12 + 12 - 8) = 11! (156 + 4) = 160 (11!).

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Let a tangent be drawn to the ellipse x²/27 + y²/1 = 1 at (3√3 cosθ, sinθ) where θ ∈ (0, π/2). Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

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The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution

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Let S₁ be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n terms of the same arithmetic progression. If (S₂ - S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

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The problem provides non-standard formulas for sums S? and S? :
S? = n [a + (2n-1)d]
S? = 2n [a + (4n-1)d]
We are given S? - S? = 1000.
S? - S? = 2n [a + (4n-1)d] - n [a + (2n-1)d]
= (2na + 8n²d - 2nd) - (na + 2n²d - nd)
= na + 6n²d - and = n [a + (6n-1)d].
So, n [a + (6n-1)d] = 1000.
We need to find S? n. Assuming the pattern S? n = kn [a + (2kn-1)d], then S? n would be 3n [a + (6n-1)d].
S? n = 3 * (n [a + (6n-1)d]) = 3 * 1000 = 3000.

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If 15 sin⁴α + 10 cos⁴α = 6, for some α ∈ R, then the value of 27 sec⁶α + 8 cosec⁶α is equal to :

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Given the equation 15 sin? + 10 cos? = 6.
Divide by cos? : 15 tan? + 10 = 6 sec?
Using sec²? = 1 + tan²? , we get sec? = (1 + tan²? )² = 1 + 2tan²? + tan?
15 tan? + 10 = 6 (1 + 2tan²? + tan? ).
15 tan? + 10 = 6 + 12tan²? + 6tan?
9 tan? - 12 tan²? + 4 = 0.
This is a quadratic in tan²? : (3 tan²? - 2)² = 0.
? 3 tan²? = 2? tan²? = 2/3.
From this, we find sin²? and cos²? If tan²? = 2/3, then sin²? = 2/5 and cos²? = 3/5.
Also, sec²? = 1 + tan²? = 5/3 and cosec²? = 1 + cot²? = 1 + 3/2 = 5/2.
The expression to evaluate is 27 sec? + 8 cosec? = 27 (sec²? )³ + 8 (cosec²? )³.
= 27 (5/3)³ + 8 (5/2)³ = 27 (125/27) + 8 (125/8) =

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Consider a hyperbola H: x² - 2y² = 4. Let the tangent at a point P(4, √6) meet the x-axis at Q and latus rectum at R(x₁, y₁), x₁ > 0. If F is a focus of H which is nearer to the point P, then the area of ∆QFR is equal to :

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