If $\to a = x^i + (x - 1)^j +^k$ and $\to b = (x + 1)^i +^j + a^k$ always make an acute angle with each other for every value of $x ϵ R$ , then

$a ϵ (- \infty, 2)$

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$a ϵ (2, \infty)$

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$a ϵ (- \infty, 1)$

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$a ϵ (1, \infty)$

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Solution

The correct option is B
$a ϵ (2, \infty)$

$\to a . \to b = (x^i + (x - 1)^j +^k) . ((x + 1)^i +^j + a^k) = x (x + 1) + x - 1 + a = x^{2} + 2 x + a - 1$
If the angle between $\to a$ and $\to b$ is acute
$\Rightarrow \to a . \to b > 0 \forall x ϵ R \Rightarrow x^{2} + 2 x + a - 1 > 0 \forall x ϵ R$
The discriminant of the quadratic expression must be negative
$\Rightarrow 4 - 4 (a - 1) < 0 \Rightarrow a > 2$

Suggest Corrections

Similar questions

If $\to a = x^i + (x - 1)^j +^k$ and $\to b = (x + 1)^i +^j + a^k$ always make an acute angle with each other for every value of $x ϵ R$ , then

\to a = x^i + (x - 1)^j +^k

\to b = (x + 1)^i +^j + a^k

x \in R

\to a = x^i + (x - 1)^j +^k

\to b = (x + 1)^i +^j + a^k

\forall x \in R

\to a = -^i + 2^j +^k

\to b = x^i +^j + (x + 1)^k

\to a = x^i + x^{2}^j + 2^k, \to b = - 3^i +^j +^k, \to c = (3 x + 11)^i + (x - 9)^j - 3^k

\to a

\to b

\to c

\to a

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