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`        Sir i can not able to solve this question Please give me detailed solution of it.`
2 years ago

Dharma Teja
33 Points
```							 There is no formula for the roots of a ﬁfth degree polynomial. Nor can they be identiﬁed by inspection in the present case. At any rate, we are not asked to identify the real roots but only to test some statements about the numbers of real roots (as opposed to the roots themselves). So methods from calculus have to be used and often the answers are obtained by merely looking at a (well-drawn) graph. Note that f(x) is a diﬀerent function of x for diﬀerent values of a. But their graphs are all qualitatively same and can be obtained from each other by suitable vertical shifts. All the four alternatives given can be paraphrased in terms of this function. Hence it suﬃces to draw only the graph of the function y = f(x) = x5 −5x. f′(x) = 5(x4−1) = 5(x2+1)(x2−1) has two distinct, simple real roots, viz. −1 and 1. So, f′ is positive on (−∞,−1), negative on (−1,1) and positive on (1,∞). Since f is an odd function and f(−1) = 4 and f(1) = −4, Now, coming to the various alternatives, for any a ∈ IR, the roots of the equation x5−5x+a = 0 correspond precisely the points of intersection of the graph above with the horizontal line y = −a. draw the graph, it is clear that a horizontal line y = b intersects the graph in three distinct points if b ∈ (−4,4) and only in one point if b  4. This shows that (B) and (D) are correct while (A) and (C) are false. A simple, straightforward question, testing elementary curve sketching.
```
2 years ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions