MHT CET questions
29293 published questions with worked solutions.
- Let f(x) = cases 5^ 1 x , & x < 0 [x], & x 0, R cases , then at x = 0
- The function f(x) = x^2 - 10x + 26 x^4(x^2 - 9)(1 + 27x^2) is continuous for
- If f(x) = cases x + 2, & when x < 1 4x - 1, & when 1 x 3 x^2 + 5, & when x > 3 cases then the correct statement is
- If f(x) = cases ( x^2 a ) - a, & when x < a 0, & when x = a a - ( x^2 a ), & when x > a cases , then
- If f(x) = cases 1 + px - 1 - px x , & -1 x < 0 [2mm] 2x + 1 x - 2 , & 0 x 1 cases is continuous in [-1, 1], then p is equal to
- If f(x) is continuous at x = 2, where f(x) = ax + b for 0 x < 2 = bx + 11 for 2 x 4 and f(3) = 2, then value of a and b is
- If f(x) = cases x 2 - 1, & 0 x < 1 [1mm] 1 2 , & 1 x < 2 cases g(x) = (2x + 1)(x - k) + 3, 0 x then g[f(x)], will be continuous...
- If function f(x) = 1 + x - [3] 1 + x x is continuous at x = 0, then f(0) is equal to
- If f(x) = 1 - x 4x - , x 4 , x [0, 2 ] is a continuous functions, then f ( 4 ) is equal to
- If f(x) = cases x + 3, & x < 1 5x - 1, & 1 x 3 x^2 + 5, & x > 3 cases , then correct statement is
- If f(x) = cases 1, & x 2 ax + b, & 2 < x < 4 7, & x 4 cases is continuous at x = 2 and x = 4, then
- Let f(x) = cases x^4 - 5x^2 + 4 |(x - 1)(x - 2)| , & x 1, 2 6, & x = 1 12, & x = 2 cases then f(x) is continuous on the set
- Let f(x) = (1 + x + x^2) + (1 - x + x^2) x - x Then the value of f(0) so that f is continuous at x = 0 is
- The value of f(0) so that the function f(x) = 1 + x - (1 + x)^ 1 3 x becomes continuous is equal to
- If f(x) = _e(1 + x^2 x) x^3 , x 0 is continuous at x = 0, then f(0) must be defined as
- If f(x) = _ (1 - 3x) (1 + 3x), x 0 = k, x = 0 Is continuous at x = 0, then the value of k is
- If f(x) = cases 1 - x x , & x 0 [1mm] k, & x = 0 cases is continuous at x = 0 then k =
- The function f(x) = (1 + ax) - (1 - bx) x is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is...
- Function f(x) = 1 - 4x 8x^2 , where x 0 and f(x) = k, where x = 0 is a continuous function at x = 0, then the value of k will be
- The value of p and q for which the function f(x) = cases (p + 1)x + x x , & x < 0 [1mm] q, & x = 0 [1mm] x + x^2 - x x^ 3/2 , & x...