JEE Advance - Mathematics (2021 - Paper 1 Online - No. 17)
For x $$\in$$ R, the number of real roots of the equation $$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$$ is ________.
Answer
4
Explanation
Given,
$$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$$ .... (i)
For $$-$$1 $$\le$$ x $$\le$$ 1 i.e., x$$\in$$[$$-$$1, 1]
From Eq. (i), we get
$$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$$
$$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$$
$$ \Rightarrow 7{x^2} + x - 5 = 0$$
$$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$$
Here, both values of x are acceptable.
For | x | > | i.e. x $$\in$$($$-$$ $$\infty$$, $$-$$1) $$\cup$$ (1, $$\infty$$)
From Eq. (i), we get
$$3{x^2} - 4({x^2} - 1) + x - 1 = 0$$
$$ \Rightarrow {x^2} - x - 3 = 0$$
$$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$$
Again here, both values of x are acceptable.
Hence, total number of solutions is 4.
$$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$$ .... (i)
For $$-$$1 $$\le$$ x $$\le$$ 1 i.e., x$$\in$$[$$-$$1, 1]
From Eq. (i), we get
$$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$$
$$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$$
$$ \Rightarrow 7{x^2} + x - 5 = 0$$
$$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$$
Here, both values of x are acceptable.
For | x | > | i.e. x $$\in$$($$-$$ $$\infty$$, $$-$$1) $$\cup$$ (1, $$\infty$$)
From Eq. (i), we get
$$3{x^2} - 4({x^2} - 1) + x - 1 = 0$$
$$ \Rightarrow {x^2} - x - 3 = 0$$
$$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$$
Again here, both values of x are acceptable.
Hence, total number of solutions is 4.
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