SATHEE: Theory of Equations 5 Question 10

Theory of Equations 5 Question 10

10. Let $a \in R$ and $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a$ . Then,

(a) $f (x)$ has three real roots, if $a > 4$

(b) $f (x)$ has only one real root, if $a > 4$

(d) $f (x)$ has three real roots, if $- 4 < a < 4$

Passage Based Problems

Read the following passage and answer the questions.

Passage I

Consider the polynomial $f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3}$ . Let $s$ be the sum of all distinct real roots of $f (x)$ and let $t = | s |$ .

(2010)

Show Answer

Answer:

Correct Answer: 10. (b,d)

Solution:

PLAN

(i) Concepts of curve tracing are used in this question.

(ii) Number of roots are taken out from the curve traced.

Let $y = x^{5} - 5 x$

(i) As $x \to \infty, y \to \infty$ and as $x \to - \infty, y \to - \infty$

(ii) Also, at $x = 0, y = 0$ , thus the curve passes through the origin.

(iii) $\frac{d y}{d x} = 5 x^{4} - 5 = 5 (x^{4} - 1) = 5 (x^{2} - 1) (x^{2} + 1)$

$= 5 (x - 1) (x + 1) (x^{2} + 1)$

Now, $\frac{d y}{d x} > 0$ in $(- \infty, - 1) \cup (1, \infty)$ , thus $f (x)$ is increasing in these intervals.

Also, $\frac{d y}{d x} < 0$ in $(- 1, 1)$ , thus decreasing in $(- 1, 1)$ .

(iv) Also, at $x = - 1, d y / d x$ changes its sign from + ve to - ve.

$∴ x = - 1$ is point of local maxima.

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. $a \in (- 4, + 4)$

Then, $f (x)$ has three real roots and if $- a > 4$

or $- a < - 4$ , then $f (x)$ has one real root.

i.e. for $a < - 4$ or $a > 4, f (x)$ has one real root.

Theory of Equations 5 Question 10

10. Let $a \in R$ and $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a$ . Then,

Passage Based Problems

Passage I

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Theory of Equations 5 Question 10

10. Let a∈R and f:R→R be given by f(x)=x5−5x+a. Then,

Passage Based Problems

Passage I

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

10. Let $a \in R$ and $f : R \to R$ be given by $f (x) = x^{5} - 5 x + a$ . Then,