Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.0k views
in Mathematics by (90.2k points)
closed by
For what real values of `a` do the roots of the equation `x^2-2x-(a^2-1)=0` lie between the roots of the equation `x^2-2(a+1)x+a(a-1)=0.`

1 Answer

0 votes
by (95.0k points)
selected by
 
Best answer
`" "x^(2)-2x- (a^(2)-1)=0" "` (i)
`" "x^(2)-2(a+1)x+a(a-1)=0" "` (ii)
For equation (i), `x= 2pm sqrt(4+4(a^(2)+1))/(2)= 1 pm a`
Now roots of (i) lies between the roots of equation (ii).
`" "f_(1)(x)= x^(2)-2x-(a^(2)-1)`
`" "f_(2)(x)= x^(2)-2(a+1)x+a(a-1)`
Hence the graphs of expressions for the equation are as shown in the following figure :
image
`" "` If the roots `f_(1)(x)=0` are real.
`" "D ge 0`
`rArr" "4+4(a^(2)-1) ge 0`
`rArr" "a^(2) ge 0`, which is always true
Also `" "f_(2)(1-a) lt 0 and f_(2)(1+a) lt 0`
`rArr" "(1-a^(2))-2(a+1)(1-a)+a(a-1) lt 0`
`rArr" "(1-a)[(1-a)-2a-2-a] lt 0`
`rArr" "(1-a)(-4a-1) lt 0`
`rArr" "(a-1)(4a+1) lt 0`
`rArr" "-(1)/(4) lt a lt 1" "` (iii)
and `" "(1+a)^(2)-2(a+1)(a+1)+a(a-1) lt 0`
`rArr" "-(a+1)^(2)+ a(a-1) lt 0`
`rArr" "-a^(2)-2a-1+a^(2)-a lt 0`
`rArr " "3a+1 gt 0`
`rArr" "a gt - (1)/(3)" "` (iv)
From (iii) and (iv), common values
`" "- (1)/(4) lt a lt 1`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...