Maths

If two straight lines whose direction cosines are given by the relations l+m-n=0, 3l² +m² +cnl=0 are parallel, then the positive value of c is:

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l + m – n = 0 => n = l + m
3l² + m² + cnl = 0
3l² + m² + cl (l+m) = 0
= (3+c) (l/m)² + c (l/m) + 1 = 0
? Lines are parallel
D = 0
c = 4 (as c > 0)

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6 months ago

Let the eccentricity of an ellipse x²/a² + y²/b² = 1, a>b, be 1/4. If this ellipse passes through the point (-4√(2/5), 3), then a² + b² is equal to:

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Contributor-Level 10

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31

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6 months ago

In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x +y = 4. Let the point B lie on the line x + 3y = 7. If (α, β) is the centroid of ∆ABC, then 15(α + β) is equal to:

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2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

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6 months ago

If dy/dx * 2ʸ - (2ʸ-1)/(2ˣ-1) = 0, x,y > 0, y(1)=1, then y(2) is equal to:

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dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3

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6 months ago

The value of the integral ∫ (x³+x) / (e^(x³)+1) dx is equal to:

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Contributor-Level 10

I = ∫? ² (x³+|x|)/ (e|x|+1) dx . (i)
I = ∫? ² (x³+|x|)/ (e? |x|+1) dx . (ii)
= ∫? ² |x| dx = 2∫? ² x dx
= [x²/2]? ² = (16/4 + 4/2) - 0
= 4+2=6

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6 months ago

If ∫ (x²+1)eˣ / (x+1)² dx = f(x) eˣ + C, where C is a constant, then d³f/dx³ at x=1 is equal

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I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

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6 months ago

If cos⁻¹(y/2) = logₑ(x/5)⁵, |y|<2, then:

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cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0

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6 months ago

The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 – 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to:

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Contributor-Level 10

CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 * CD * AB = 1/2 * 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10

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6 months ago

The number of distinct real roots of x⁴ - 4x + 1 = 0 is:

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Contributor-Level 10

f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution

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New answer posted

6 months ago

Maths Functions

Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:

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