Let a `in R` and let f: `R to R` be given by f(x) `=x^(5) -5x+a.` then

Options

A. f(x) has three real roots, if `a gt 4`
B. f(x) has only one realroot, if `a gt 4`
C. f(x0 has three real roots, if `a lt -4`
D. f(x) has three real roots, if `-4lt a lt4`

Answer

B, D

Explanation

PLAN
(i) Concepts of curve tracing are used in this question.
(ii)n Number of roots are taken out from the cuirve traced.
Let `y=x^(5)-5x`
`(i) As x to oo, y to oo and as x to-oo, y to-oo`
(ii) Also, at `x=0, y=0,` thus the curve passes through the origin.
(iii) `(dy)/(dx)=5x^(4)-5=5(x^(4)-1)=5(x^(2)-1)(x^(2)+1)=5(x-1)(x+1)(x^(2)+1)`

Now, `(dy)/(dx)gt0 "in"(-oo,-1)uu(1,oo), thus f(x) ` is increasing in these intervals. Also, `(dy)/(dx)lt0"in"(-1,1),"thus decreasing in"(-1,1).`
(iv) Also, at `x=-1,dy//dx` changes its sign from + be to -ve.
`therefore x=-1` is point of local mixima. Similarly, x=1 is point of local minima. Local maximum value, `y=(-1)^(5)-5(-1)=4` Local minimum value , `y=(1)^(5)-5(1)=-4`

Now, let `y=-a`
As evident from the graph, if `-a in (-4,4)`
`i.e. alpha in (-4,+4)`
Then, f(x) has three real roots and if `-a gt4or -alt-4,` then f(x) has one real root. i.e. for `a lt-4gt4, f(x)`n has one real root.