Algebra

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Algebra

  1. 1. • Disclaimer:This presentation is prepared by trainees of baabtra as a part of mentoring program. This is not official document of baabtra –Mentoring Partner• Baabtra-Mentoring Partner is the mentoring division of baabte System Technologies Pvt . Ltd
  2. 2. ALGEBRAPROBLEMS Ashwin Anand V Email:ashwinanand99@gmail.com Facebook id:ashwinanand99@gmail.com
  3. 3. 1) If f(x) = 5 - 2x, then f -1(-3) = ?a) 3 b) 4c) 6 d) 8
  4. 4. 2) If f(x) = x - 4, then (f o f)(3) = ?a) 4 b) -3c) -5 d) -11
  5. 5. 3) The number of solutions of (x2 + 1)2 +2(x2 + 1) - 3 = 0 is equal to ?a) 1 b) 3c) 0 d) 2
  6. 6. 4) Find the distance between the points (-4 , -5)and (-1 , -1).a) 16 b) 5c) 3 d) 6
  7. 7. 5) Evaluate f(2) - f(1)f(x) = 6x + 1a) 3 b) 4c) 5 d) 6
  8. 8. Solution 1:Find f -1(x) and then Find f -1(- 3)y = 5 - 2x , givenx = 5 - 2y , interchange x and y2y = 5 - x , y = log2(5 - x) , solve for yf -1(x) = log2(5 - x) , inverse functionf -1(- 3) = log2(5 + 3)= log2(23) = 3
  9. 9. Solution 2 :If f(x) = x - 4, then (f o f)(3) =(f o f)(3) = f(f(3)) = f(3 - 4) = f(-1) =-5
  10. 10. Solution 3Let u = x2 + 1 and rewrite the given equation in terms of u as followsu 2 + 2u - 3 = 0Factor and solve the above equation(u + 3)(u - 1) = 0two solutions: u = x2 + 1 = - 3 and u = x2 + 1 = 1Equation x2 + 1 = - 3 has no solutions. Solve the equation x2 + 1 = 1for to getx = 0.The given equation has one solution.
  11. 11. Solution 4The distance d between points (-4 , -5) and (-1 , -1) is given byd = sqrt[ (-1 - -4) 2 + (-1 - -5) 2 ]Simplify.d = sqrt(9 + 16) = 5
  12. 12. Solution 5Given the functionf(x) = 6x + 1f(2) - f(1) is given by.f(2) - f(1) = (6*2 + 1) - (6*1 + 1) = 6
  13. 13. • If this presentation helped you, please visit our page facebook.com/baabtra and like it. Thanks in advance.• www.baabtra.com | www.massbaab.com |ww w.baabte.com
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