Let A and B be two points with position vector 2 - - - - 3 k- and - - -2 - respectively such that -AB- - 3 then limx - x - -2 1x-1 may be equal to-

A(2,1, 3), B(β, α^2 , 0) |AB| nbsp; = 3 = √((β 2)^2+(α^2 1)^2 + 9) Squaring both sides ⇒ 9 = nbsp;(β 2)^2+(α^2 1)^2 + 9 ⇒ 0 = (β 2)^2+(α^2 1)^2 ∴β = ...

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Standard XII

Mathematics

Local Minima

Let A and B b...

Question

Let $A$ and $B$ be two points with position vector $2^i +^j - 3^k$ and $β^i + α^{2}^j$ respectively such that $| - - \to A B | = 3$ then $lim x \to β (x - α^{2})^{\frac{1}{x - (α + 1)}}$ may be equal to:

1

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e

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e^{2}

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Solution

The correct option is B $e$
$A (2, 1, - 3), B (β, α^{2}, 0)$
$| - - \to A B | = 3 = \sqrt{(β - 2)^{2} + (α^{2} - 1)^{2} + 9}$
Squaring both sides
$\Rightarrow 9 = (β - 2)^{2} + (α^{2} - 1)^{2} + 9$
$\Rightarrow 0 = (β - 2)^{2} + (α^{2} - 1)^{2}$
$∴ β = 2, α^{2} = 1$
Now when $α = 1$ ,
$lim x \to β (x - α^{2})^{\frac{1}{x - (α + 1)}} = lim x \to 2 (x - 1)^{\frac{1}{x - 2}} = e^{λ}$
where $λ = lim x \to 2 (x - 1 - 1) \times \frac{1}{x - 2} = lim x \to 2 \frac{x - 2}{x - 2} = 1$
$lim x \to β (x - α^{2})^{\frac{1}{x - (α + 1)}} = e^{1} = e$
Now when $α = - 1$ ,
$lim x \to β (x - α^{2})^{\frac{1}{x - (α + 1)}} = lim x \to 2 (x - 1)^{\frac{1}{x}} = 1^{\frac{1}{2}} = 1$

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Local Minima

Standard XII Mathematics

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Let A and B be two points with position vector 2^i+^j−3^k and β^i+α2^j respectively such that |−−→AB|=3 then limx→β(x−α2)1x−(α+1) may be equal to:

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