# IPMAT Question Paper 2019 | IPM Indore Quants

###### IPMAT Sample Paper | IPMAT Question Paper | Question 39

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Question 39: The function f(x) = $$frac{x^{3} - 5x^{2} - 8x}{3}\\$ is 1. positive and monotonically increasing for x $\in$-$infty, \frac{5-\sqrt{57}}{2}\\$) and x $\in$$frac{5+\sqrt{57}}{2}, +\infty\\$) 2. negative and monotonically decreasing for x $\in$-$infty, \frac{5-\sqrt{57}}{2}\\$ and x $\in$$frac{5+\sqrt{57}}{2},+\infty\\$) 3. negative and monotonically increasing for x $\in$-$infty, \frac{5-\sqrt{57}}{2}\\$) and positive and monotonically increasing for x $\in$$frac{5+\sqrt{57}}{2},+\infty\\$) 4. positive and monotonically increasing for x $\in$-$infty, \frac{5-\sqrt{57}}{2}\\$) and negative and monotonically decreasing for x $\in$$frac{5+\sqrt{57}}{2},+\infty\\$) ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer Let’s find the roots of the function, That is where the function takes the value 0. f$x) = 0.
$$frac{x^{3} - 5x^{2} - 8x}{3}\\$ = 0 x3- 5x2- 8x = 0 x$x2−5x−8) = 0

x = 0 or x2−5x−8 = 0

x2−5x−8 = 0

x2−5x + $$frac{25}{4}\\$ −$\frac{25}{4}\\$ −8 = 0$x - $$frac{5}{2}\\$ )2- $\frac{57}{4}\\$ = 0$x - $$frac{5}{2}\\$ )2= $\frac{57}{4}\\$ x - $\frac{5}{2}\\$ = +$\frac{\sqrt{57}}{2}\\$$or) x - $$frac{5}{2}\\$ = - $\frac{\sqrt{57}}{2}\\$ x = $\frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$ or x = $\frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ Therefore, the roots of the equation $\frac{x^{3} - 5x^{2} - 8x}{3}\\$ are 0, $\frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$, $\frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ That means, when plotted f$x) will be 0 at 3 points when x = 0 or $$frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$ or $\frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ Arranged in increasing order the roots are $\frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ , 0 , $\frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$. Check f$x) at a point which is less than $$frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ and at a point greater than $\frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$ f$-2) = $$frac{$-2$^{3} - 5(-2)^{2} - 8(-2)}{3}$ = $\frac{-8-20-16}{3}\\$ = $\frac{-44}{3}\\$ = -ve. f$10) = $$frac{$10$^{3} - 5(10)^{2} - 8(10)}{3}$ = $\frac{1000-500-80}{3}\\$ = $\frac{420}{3}\\$ = 140 =+ve Since f$x) is negative below $$frac{5}{2}\\$ - $\frac{\sqrt{57}}{2}\\$ and positive above $\frac{5}{2}\\$ + $\frac{\sqrt{57}}{2}\\$. And since it is a continuous function, the function looks something like this. From the graph, only Option C is correct. f$x) is negative and monotonically increasing for x $$in$-$infty, \frac{5-\sqrt{57}}{2}\\$) and positive and monotonically increasing for x $\in$$frac{5+\sqrt{57}}{2},+\infty\\$) The question is"The function f$x) = $$frac{x^{3} - 5x^{2} - 8x}{3}\\$ is" ##### Hence, the answer is negative and monotonically increasing for x $\in$-$infty, \frac{5-\sqrt{57}}{2}\\$) and positive and monotonically increasing for x $\in$$frac{5+\sqrt{57}}{2},+\infty\\$) Choice C is the correct answer ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ###### Already have an Account? ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Mobile:$91) 99626 48484 / 94459 38484
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