Class 11th
Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
3 months agoContributor-Level 9
S? = ∑ tan? ¹ (6? / (2²? ¹ + 3²? ¹) from r=1 to k. (Assuming n in image is r)
t? = tan? ¹ (6? / (2²? ¹ + 3²? ¹)
= tan? ¹ ( (3/2) * (3/2)^ (2r) / ( (9/4) + (3/2)^ (2r+2) (This seems overly complex. Let's re-examine the image's simplification).
t? = tan? ¹ (6? / (2 * 4? + 3 * 9? ). The image simplifies the denominator to 2²? ¹ + 3²? ¹, which is different. Following the image's next step:
t? = tan? ¹ [ 6? / ( 1 + (3/2)^ (2r+1) ] (This denominator is incorrect).
The image seems to simplify t? into:
t? = tan? ¹ (3/2)? ¹) - tan? ¹ (3/2)? )
S? = [tan? ¹ (3/2)²) - tan? ¹ (3/2)] + [tan? ¹ (3/2)³) - tan? ¹ (3/2)²)] + . + [t
New answer posted
3 months agoContributor-Level 10
Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.
New answer posted
3 months agoContributor-Level 9
(1 - x + x²)³? = ∑ a? x? (from j=0 to 3n)
= a? + a? x + a? x² + . + a? x³? (I)
Let A = a? + a? + a? + .
Let B = a? + a? + a? + .
In (I) put x = 1: (1 - 1 + 1)³? = 1.
1 = a? + a? + a? + a? + . (A + B = 1)
In (I) put x = -1: (1 - (-1) + (-1)²)³? = 3³?
3³? = a? - a? + a? - a? + . (A - B = 3³? )
(This seems incorrect based on the provided solution. Following the image:)
In (I) put x = -1, (1+1+1)^n = 1. (There must be a typo in the original problem, probably (1-x+x²)^n).
Assuming (1-x+x²)^n. Put x=-1 gives 3^n.
The provided text says putting x=-1 gives 1.
1 = a? - a? + a? - a? + .
Adding the two equations: 2 = 2 (a? + a? + a? + .) = 2A
New answer posted
3 months agoContributor-Level 9
Equation of chord of x² + y² = 25 with mid point (h, k) is xh + yk = h² + k².
Or, y = (-h/k)x + (h² + k²)/k.
If this touches the ellipse x²/9 + y²/16 = 1, then the condition for tangency c² = a²m² + b² must be satisfied.
Here, m = -h/k, c = (h²+k²)/k, a²=9, b²=16.
(h² + k²)/k)² = 9 (-h/k)² + 16
(h² + k²)²/k² = 9h²/k² + 16
⇒ (h² + k²)² = 9h² + 16k²
∴ Required locus (x² + y²)² = 9x² + 16y².
New answer posted
3 months agoThe number of roots of the equation, (81)sin²? + (81)cos²? = 30 in the interval [0, π] is equal to :
Contributor-Level 9
(81)^sin²x + (81)^cos²x = 30.
(81)^sin²x + (81)^ (1-sin²x) = 30.
Let y = 81^sin²x.
y + 81/y = 30
y² - 30y + 81 = 0
(y - 3) (y - 27) = 0
⇒ y = 3 or y = 27.
Either 81^sin²x = 3 ⇒ 3^ (4sin²x) = 3¹ ⇒ sin²x = 1/4 ⇒ sin x = ±1/2. x = π/6, 5π/6.
OR, 81^sin²x = 27 ⇒ 3^ (4sin²x) = 3³ ⇒ sin²x = 3/4 ⇒ sin x = ±√3/2. x = π/3, 2π/3.
(as 0 ≤ x ≤ π)
Total possible solutions = 4.
New answer posted
3 months agoContributor-Level 10
For the quadratic equation (k+1)tan²x - √2λ tanx + (k-1) = 0, the roots are tanα and tanβ.
Sum of roots: tanα + tanβ = √2λ / (k+1).
Product of roots: tanα tanβ = (k-1) / (k+1).
tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)
tan (α + β) = [√2λ / (k+1)] / [1 - (k-1)/ (k+1)]
tan (α + β) = [√2λ / (k+1)] / [ (k+1 - k + 1)/ (k+1)] = (√2λ) / 2 = λ/√2.
Given tan² (α+β) = 50.
(λ/√2)² = 50
λ²/2 = 50 ⇒ λ² = 100 ⇒ λ = ±10.
New answer posted
3 months agoContributor-Level 10
Truth table for (p → q) ∧ (q → ~p).
| p | q | p → q | ~p | q → ~p | (p → q) ∧ (q → ~p) |
|-|-|-|-|-|-|
| T | T | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
The final column is F, T, which is the truth table for ~p.
Therefore, (p → q) ∧ (q → ~p) is equivalent to ~p.
New answer posted
3 months agoContributor-Level 9
y² - 2x = 4√ (y² - 2x + 1)
Let y² - 2x = k.
k = 4√ (k+1) ⇒ k² = 16 (k+1) ⇒ k²-16k-16=0
(This appears to be from another problem. The provided solution text says:)
y² = 2x = 4 (1/2 x)
a > 2 * (1/2) ⇒ a > 1.
New answer posted
3 months agoContributor-Level 10
The tangent to the parabola y² = 4ax is y = mx + a/m.
For y² = 4x, a=1. So, the tangent is y = mx + 1/m.
The given line is y = mx + 4.
Comparing the two, 1/m = 4 ⇒ m = 1/4.
The line is y = (1/4)x + 4.
This line is also tangent to x² = 2by.
Substitute y into the parabola equation:
x² = 2b (1/4)x + 4)
x² = ( b/2 )x + 8b
x² - ( b/2 )x - 8b = 0.
For tangency, the discriminant (D) is zero.
D = (-b/2)² - 4 (1) (-8b) = 0.
b²/4 + 32b = 0.
b ( b/4 + 32) = 0.
b = 0 (not possible) or b/4 = -32 ⇒ b = -128.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 681k Reviews
- 1800k Answers