JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 17)
Let a $$\in$$ R and f : R $$\to$$ R be given by f(x) = x5 $$-$$ 5x + a. Then,
f(x) has three real roots , if a > 4
f(x) has only one real root, if a > 4
f(x) has three real roots, if a < $$-$$4
f(x) has three real roots, if $$-$$4 < a < 4
Explanation
In the equation $$f(x) = {x^5} - 5x + a$$, there are different polynomials depending on the parameter a. Now, for the roots of each of these, in general, f(x) = 0.
That is, $$a = 5x - {x^5} = x(5 - {x^5})$$
Hence, the parameter a is a function of x. That is,
$$a(x) = x(5 - {x^5})$$
Now, $$a'(x) = 5 - 5{x^4}$$
Therefore, extrema occurs at a'(x) = 0
That is, when x4 = 1 or x = 1 and x = $$-$$1 (only real roots considered)
Here, $$a''(x) = - 20{x^3}$$
$$a''(1) < 0$$ (maximum)
$$a''( - 1) > 0$$ (minimum)
Hence, the maximum value is
a(1) = 4
The minimum value is
a($$-$$1) = $$-$$4
Hence, when $$-$$4 < a < 4, there are three points that is, x values where f(x) = 0, that is, three roots of f(x) for any value of a lying in ($$-$$4, 4)}. .... (1)
When | a | > 4, there is only one x for which f(x) = 0 ...... (2)
Hence, from statements (1) and (2), we can conclude that options (B) and (D) are correct.
That is, $$a = 5x - {x^5} = x(5 - {x^5})$$
Hence, the parameter a is a function of x. That is,
$$a(x) = x(5 - {x^5})$$
Now, $$a'(x) = 5 - 5{x^4}$$
Therefore, extrema occurs at a'(x) = 0
That is, when x4 = 1 or x = 1 and x = $$-$$1 (only real roots considered)
Here, $$a''(x) = - 20{x^3}$$
$$a''(1) < 0$$ (maximum)
$$a''( - 1) > 0$$ (minimum)
Hence, the maximum value is
a(1) = 4
The minimum value is
a($$-$$1) = $$-$$4
Hence, when $$-$$4 < a < 4, there are three points that is, x values where f(x) = 0, that is, three roots of f(x) for any value of a lying in ($$-$$4, 4)}. .... (1)
When | a | > 4, there is only one x for which f(x) = 0 ...... (2)
Hence, from statements (1) and (2), we can conclude that options (B) and (D) are correct.
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