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New answer posted

a year ago

0 Follower 27 Views

A
alok kumar singh

Contributor-Level 10

? P? =? P? ⇒ n!/ (n-r)! = n!/ (n-r-1)! ⇒ n-r=1.
? C? =? C? ⇒ r + (r-1) = n ⇒ n = 2r-1.
Substitute n: (2r-1)-r=1 ⇒ r=2.

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

(2? -2) is multiple of 3.
If n is odd, 2? ≡ (-1)? =-1 (mod 3). 2? -2 ≡ -1-2 = -3 ≡ 0 (mod 3).
If n is even, 2? ≡ (-1)? =1 (mod 3). 2? -2 ≡ 1-2 = -1 (mod 3).
So n must be odd.
2-digit numbers are 10-99 (90 numbers).
Odd numbers are 11,13, .,99. Number of terms = (99-11)/2 + 1 = 45.
Probability = 45/90 = 1/2.

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

For x>2, f (x) = ∫? ¹ (5+1-t)dt + ∫? ² (5+t-1)dt + ∫? (5+t-1)dt
= ∫? ¹ (6-t)dt + ∫? ² (4+t)dt + ∫? (4+t)dt
= [6t-t²/2]? ¹ + [4t+t²/2]? ² + [4t+t²/2]?
= (6-1/2) + (8+2 - (4+1/2) + (4x+x²/2 - (8+2)
= 5.5 + 5.5 + 4x+x²/2 - 10 = 4x+x²/2 + 1.
f (2? ) = 8+2+1 = 11. f (2? ) = 5 (2)+1 = 11. Continuous.
f' (x) = 4+x for x>2. f' (2? ) = 6.
For x<2, f' (x)=5. f' (2? )=5.
Not differentiable at x=2.

New answer posted

a year ago

0 Follower 50 Views

V
Vishal Baghel

Contributor-Level 10

Circle: (x-1)² + (y-3)² = 4. C= (1,3), r=2.
Length of tangent from P (-1,1) is L = √ (-1)²+1²-2 (-1)-6 (1)+6) = √ (1+1+2-6+6) = √4 = 2.

Area of quadrilateral PACB = 2 * Area (PAC) = 2 * (1/2 * L * r) = 2*2=4.
AB is chord of contact. T=0 => -x+y- (x-1)-3 (y-1)+6=0 => -2x-2y+10=0 => x+y=5.
Distance of C from AB = |1+3-5|/√2 = 1/√2.
Length of AB = 2√ (r²-d²) = 2√ (4-1/2) = 2√ (7/2) = √14.
Area of ABD =?

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

T? = ¹? C? (xsinα)¹? (acosα/x)? = ¹? C? x¹? ²? sin¹? α a? cos? α
For term independent of x, 10-2r=0 ⇒ r=5.
T? = ¹? C? sin? α a? cos? α = ¹? C? (sin2α/2)? a?
This is the greatest when sin2α=1.
the greatest value = ¹? C? (a/2)? = 10!/ (5!)².
¹? C? = 252.
252 (a/2)? = 252.
(a/2)? = 1 ⇒ a=2.

New answer posted

a year ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

Coeff of x? in (x²+1/bx)¹¹: T? = ¹¹C? (x²)¹¹? (1/bx)? = ¹¹C? x²²? ³? b?
22-3r=7 ⇒ 3r=15 ⇒ r=5. Coeff is ¹¹C? /b?
Coeff of x? in (x-1/bx²)¹¹: T? = ¹¹C? (x)¹¹? (-1/bx²)? = ¹¹C? x¹¹? ³? (-1)? b?
11-3r=-7 ⇒ 3r=18 ⇒ r=6. Coeff is ¹¹C? (-1)? /b? = ¹¹C? /b?
¹¹C? /b? = ¹¹C? /b? ⇒ b = ¹¹C? /¹¹C? = (11-5+1)/5 = 7/5. This differs from the solution.
Let's check the exponents again.
x? : 22-2r-r=7 => 22-3r=7 => 3r=15 => r=5. Correct.
x? : 11-r-2r=-7 => 11-3r=-7 => 3r=18 => r=6. Correct.
The given solution has b=1.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Area = ∫? ² (2? - logx)dx = [2? /ln2 - (xlnx-x)]? ²
= (4/ln2 - (2ln2-2) - (2/ln2 - (0-1) = 2/ln2 - 2ln2 + 1.
α=2, β=-2, γ=1.
(α+β-2γ)² = (2-2-2)² = 4.

New answer posted

a year ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

log (x + √ (x²+1) is an odd function.
f (-x) = log (-x + √ (x²+1) = log (1/ (x+√ (x²+1) = -log (x+√ (x²+1) = -f (x).
Integral of an odd function over a symmetric interval is 0.

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

xdy - ydx = x³cosxdx
(xdy-ydx)/x² = xcosxdx
d (y/x) = xcosxdx
y/x = ∫xcosxdx = xsinx - ∫sinxdx = xsinx + cosx + c
y = x²sinx + xcosx + cx
y (π) = 0 + π (-1) + cπ = 0 ⇒ c = 1
y = x²sinx + xcosx + x
y (π/2) = (π/2)² (1) + 0 + π/2 = π²/4 + π/2

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).

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