MHT CET202615 April 2026Evening ShiftMathematicsContinuity and DifferentiabilityActual
The value of f(0) so that the function f(x) = (256 - 8x)^ 1 4 - 4 16 - 4(64 + 3x)^ 1 3 , x 0 is continuous at x = 0 , is
Options
- A- 1 8
- B1 8
- C1 64
- D8
Correct answer
B. 1 8
Step-by-step solution
For f(x) to be continuous at x = 0 , we must have f(0) = _ x 0 f(x) . _ x 0 (256 - 8x)^ 1 4 - 4 16 - 4(64 + 3x)^ 1 3 This is a 0 0 indeterminate form. Applying L'Hospital's rule: _ x 0 1 4 (256 - 8x)^ - 3 4 (-8) -4 1 3 (64 + 3x)^ - 2 3 3 = _ x 0 -2(256 - 8x)^ - 3 4 -4(64 + 3x)^ - 2 3 Substituting x = 0 : = -2(256)^ - 3 4 -4(64)^ - 2 3 = -2(4^4)^ - 3 4 -4(4^3)^ - 2 3 = -2(4⁻³) -4(4⁻²) = -2 1 64 -4 1 16 = - 1 32 - 1 4 = 1 8 Answer: 1 8