If {a}=x²{i}+x{j}+3{k} and {b}=x{i}-4{j}+2{k} and {a}·{b}>6 then ______

Correct Answer is : x∈(−2,0)∪(2,∞)x(-2,0)(2,∞)x∈(−2,0)∪(2,∞)

Correct Answer is : x∈(−2,0)∪(2,∞)x(-2,0)(2,∞)x∈(−2,0)∪(2,∞)

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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 32 of 160

Marks: +1, -0

\vec{a}=x^{2}\hat{i}+x\hat{j}+3\hat{k}

\vec{b}=x\hat{i}-4\hat{j}+2\hat{k}

\vec{a}\cdot\vec{b}>6

then ______

Solution:

Given,

\vec{a}=x^{2}\hat{i}+x\hat{j}+3\hat{k}

\vec{b}=x\hat{i}-4\hat{j}+2\hat{k}

a . b > 6

\Rightarrow (x^{2}\hat{i}+x\hat{j}+3\hat{k})(x\hat{i}-4\hat{j}+2\hat{k})>6

\Rightarrow x^{2}\cdot x+(x)(-4)+3(2>6

[a\cdot b=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}

, where

a=a_{1}\hat{i}+b_{1}j+c_{1}k

and

b=a_{2}\hat{i}+b_{2}j+c_{2}\hat{k}

]

\Rightarrow x^{3}-4x+6>6

\Rightarrow x^{3}-4x+6-6>0

\Rightarrow x^{3}-4x>0

\Rightarrow x(x^{2}-4)>0

\Rightarrow x(x-2)(x+2)>0

Using wavy curve method critical points are

x=0,2,-2

∴ Solution is

x\in(-2,0)\cup(2,\infty)

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