Question
If f(x)= \ array l x^2-1,0 < x < 2 \\ 2 x+3,2 x < 3 array ., then the quadratic equation whose roots are _ x 2^ - f(x) and _ x 2^ + f(x) is
BITSAT
Mathematics
Limits
2023
BITSAT 2023 (Memory Based Paper 2)
BITSAT Mathematics Question (2023) — Solution
Options
- A. x^2-6 x+9=0
- B. x^2-7 x+8=0
- C. x^2-14 x+49=0
- D. x^2-10 x+21=0
Answer
D. x^2-10 x+21=0
Step-by-step solution
Since, f(x)= cases x^2-1, & 0 Now, _ x 2^ - f(x)= _ x 2^ - (x^2-1 ) aligned & = _ h 0 [(2-h)^2-1 ]= _ h 0 [4+h^2-4 h-1 ] \\ & = _ h 0 [h^2-4 h+3 ]=0-0+3=3 aligned and _ x 2^ + f(x)= _ x 2^ + (2 x+3) = _ h 0 [2(2+h)+3]= _ h 0 [4+2 h+3]=7 Hence the quadratic equation whose roots are 3 and 7 is given by x^2-(3+7) x+(3+7)=0 x^2 -10 x+21=0
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