in Quantitative Aptitude edited by
37 views
3 votes
3 votes

The number of integers that satisfy the equality $\left( x^{2} – 5x + 7 \right)^{x+1} = 1$ is 

  1. $2$
  2. $3$
  3. $5$
  4. $4$
in Quantitative Aptitude edited by
by
1.3k points 5 6 33
37 views

1 Answer

1 vote
1 vote

Given that, $(x^{2} – 5x + 7)^{x+1} = 1 $

We know that, for $a^{b} = 1 $

  • If $ b = 0 \Rightarrow a^{0} = 1 $
  • If $ a = 1 \Rightarrow 1^{b} = 1 $
  • If $ a = -1, b = \text{even} \Rightarrow (-1)^{\text{even}} = 1 $

$\textsf{Case 1} : \; x+1 = 0 $

$ \Rightarrow \boxed{ x = – 1 \;\textsf{(This case will be accepted)} } $

$ \textsf{Case 2} : \; x^{2} – 5x + 7 = 1 $

$ \Rightarrow x^{2} – 5x + 6 = 0 $

$ \Rightarrow x^{2} – 3x – 2x + 6 = 0 $

$ \Rightarrow x (x – 3) – 2 (x-3) = 0 $

$ \Rightarrow (x-2)(x-3) = 0 $

$ \Rightarrow \boxed{x = 2,3 \;\textsf{(This case will be accepted)} } $

$\textsf{Case 3}: \; x^{2} -5x + 7 = -1 $

$\Rightarrow x^{2} – 5x + 8 = 0 $

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{25-32}}{2} $

$\Rightarrow x = \dfrac{5 \pm \sqrt{-7}}{2} $

$\Rightarrow \boxed{x = \frac{5 \pm 7i}{2} \;\textsf{(This case will be rejected)}} \qquad [\because \sqrt{-1} = i ] $

Because we don’t want value of $x$ is complex number. 

$\therefore$ The number of integers satisfies the equation is $3.$

Correct Answer$: \text{B}$

edited by
by
4.9k points 3 7 28
Answer:

Related questions

Ask
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true