Class 12th

Let $A$ and $B$ be two independent events such that $P (A) = \frac{1}{3}$ $\frac{}{}$ and $P (B) = \frac{1}{6}$ $\frac{}{}$ . Then, which of the following is TRUE?

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Class 12th Inverse Trigonometric Functions

Contributor-Level 9

A and B are independent events so P(A/B) = 1/3

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7 months ago

The inverse function of $f (x) = \frac{8^{2 x} - 8^{- 2 x}}{8^{2 x} + 8^{- 2 x}}, x \in (- 1,1)$ $\frac{^{}^{}}{^{}^{}}$ , is

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Contributor-Level 9

y = (8²?-8?²?)/(8²?+8?²?) ⇒ (1+y)/(1-y) = 8? ⇒ 4x = log?((1+y)/(1-y))
x = (1/4)log?((1+y)/(1-y)), f?¹(x) = (1/4)log?((1+x)/(1-x))

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7 months ago

If $\frac{}{^{} {(^{})}^{}} {(^{})}^{}$ $\int \frac{c o s ? x d x}{{s i n}^{3} ? x {(1 + {s i n}^{6} ? x)}^{2} 3} = f (x) {(1 + {s i n}^{6} ? x)}^{1 / λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ $(\frac{}{})$ is equal to:

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Class 12th Maths Determinants

Contributor-Level 9

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

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7 months ago

For which of the following ordered pairs $(μ, δ)$ , the system of linear equations $x + 2 y + 3 z = 1$ $3 x + 4 y + 5 z = μ$ is inconsistent?

$4 x + 4 y + 4 z = δ$

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Contributor-Level 9

Note D = |3, 4, 5; 1, 2, 3; 4, 4| = 0 (R? → R? - 2R? + 3R? )
Now let P? = 4x+4y+4z-δ=0. If the system has solutions it will have infinite solution, so P? = αP? + βP?
Hence 3α+β=4 and 4α+2β=4 ⇒ α=2 and β=-2
So for infinite solution 2µ-2=δ ⇒ for 2µ ≠ δ+2 System inconsistent

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7 months ago

If ∫(cosθ/(5+7sinθ-2cos²θ))dθ = A logₑ|B(θ)| + C, where C is a constant of integration, then B(θ)/A can be:

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Class 12th Application of Integrals

Contributor-Level 10

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

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7 months ago

The area (in sq. units) of the region A = {(x,y): (x-1)[x] ≤ y ≤ 2√x, 0 ≤ x ≤ 2}, where [t] denotes the greatest integer function, is:

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Contributor-Level 10

y = |x| (x-1)
= { 0, 0 ≤ x < 1
{ x-1, 1 ≤ x < 2
Area = ∫? ² 2√x dx - ∫? ² (1) (1)
= [ (4x³/²)/3]? ² - 1/2 = (8√2)/3 - 1/2

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7 months ago

Which of the following point lies on the tangent to the curve x⁴eʸ + 2√(y+1) = 3 at the point (1,0)?

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Class 12th Differential Equations

Contributor-Level 10

e? y'x? + 4x³e? + 2y' / (2√ (y+1) = 0 at (1,0)
y' + 4 + y' = 0 ⇒ y' = -2
equation of tangent at (1,0) is 2x + y - 2 = 0
So option (C) is correct.

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7 months ago

Let y = y(x) be the solution of the differential equation cosx(dy/dx) + 2ysinx = sin2x, x ∈ (0, π/2). If y(π/3) = 0, then y(π/4) is equal to:

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Contributor-Level 10

dy/dx + 2tanx · y = 2sinx
I.F. = e^ (∫2tanxdx) = sec²x
Solution is y·sec²x = ∫2sinx·sec²xdx + C
ysec²x = 2secx + C
0 = 2·2 + c ⇒ c = -4
ysec²x = 2secx - 4
y (π/4) = √2 - 2

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7 months ago

If for some α ∈ R, the lines L₁: (x+1)/2 = (y-2)/(-1) = (z-1)/1 and L₂: (x+2)/α = (y+1)/(5-α) = (z+1)/1 are coplanar, then the line L₂ passes through the point:

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Contributor-Level 10

Line are coplanar
so | [α, 5-α, 1], [2, -1, 1], | = 0
−5α + (α – 5)3 + 7 = 0
-2α = 8 ⇒ α = −4
⇒ L? : (x+2)/-4 = (y+1)/9 = (z+1)/1
Now by cross checking option (A) is correct.

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7 months ago

Class 12th Determinants

If the system of linear equations
x + y + 3z = 0
x + 3y + k²z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k ∈ R, then x + (k/y)z is equal to:

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