Questions: Continuity and Differentiability
Continuity and Differentiability26 min readLesson 5 of 6
Q1. Define continuity of a function at a point.
Answer (English): A function f is continuous at a point c when the limit of f as x approaches c exists, equals f(c), and the one-sided limits from both sides agree with that value. In symbols, the left limit, the right limit and the value at c must all be the same number. If any one of the three differs, the function is discontinuous at c.
Hinglish: Continuous at c tab hai jab left limit, right limit aur f(c) teeno barabar hon. Koi ek alag ho to discontinuity hai.
Q2. Is the function f(x) = |x| continuous and differentiable at x = 0?
Answer (English): The absolute value function is continuous at zero because the left limit is 0, the right limit is 0 and the value is 0. It is not differentiable at zero because the left derivative is minus 1 while the right derivative is plus 1. The graph has a corner there, which is exactly what a mismatch of one-sided slopes means.
Hinglish: |x| zero par continuous hai, kyunki dono limits aur value 0 hain. Par differentiable nahi, kyunki left derivative -1 aur right derivative +1 hai. Graph par corner hai.
Q3. Describe where the staircase function [x] fails to be continuous on the real line.
Answer (English): The greatest integer function jumps at every integer: just below an integer its value is one less than at the integer itself. So the left limit and the right limit differ at every integer point. The function is left continuous at integers but not right continuous, and it is continuous at every non-integer point.
Hinglish: Greatest integer function har integer par jump karta hai, isliye left limit aur right limit alag hote hain. Non-integer points par continuous hai.
Q4. Differentiate sin of the quantity tan inverse of e to the power x with respect to x.
Answer (English): Let u be tan inverse of e to the power x, so the function is sin u. The derivative is cos u times du by dx. Now du by dx equals e to the power x divided by 1 plus e to the power 2x. Substituting back gives cos of tan inverse of e to the power x times that fraction. At x equal to 0 the cosine becomes root 2 by 2 and the fraction becomes 1 by 2, so the value is root 2 by 4.
Hinglish: Chain rule lagao: outer sin, inner tan inverse, phir e^x. Derivative me cos(u) multiply e^x/(1+e^(2x)) aata hai. x = 0 par value root 2 by 4 hoti hai.
Q5. Find dy by dx when y equals x to the power x.
Answer (English): Take logarithms: log y equals x log x. Differentiate both sides with respect to x, so y' divided by y equals log x plus 1. Multiplying by y gives the derivative as x to the power x times the quantity log x plus 1. The logarithm step is needed because the power rule does not apply when the exponent is a variable.
Hinglish: Dono taraf log lo: log y = x log x. Differentiate karke y'/y = log x + 1, isliye y' = x^x (log x + 1).
Q6. Find dy by dx for the parametric curve x = a cos t, y = a sin t.
Answer (English): Differentiate each coordinate with respect to t: dx by dt equals minus a sin t and dy by dt equals a cos t. Dividing gives dy by dx equal to minus cot t. This matches the slope of a circle at the parameter t, which is expected because the curve is a circle.
Hinglish: dx/dt = -a sin t, dy/dt = a cos t, isliye dy/dx = -cot t. Circle ka slope bhi yahi hota hai.
Q7. Find the second derivative of y = x cubed.
Answer (English): The first derivative is 3x squared. Differentiating again gives 6x, which is the second derivative. It measures how the slope changes: positive for positive x and negative for negative x. For a polynomial, each differentiation drops the degree by one.
Hinglish: y' = 3x², phir y'' = 6x. Second derivative slope ka rate of change batata hai.
Q8. Check the continuity of f(x) = (x squared minus 1) divided by (x minus 1) at x = 1.
Answer (English): The function is not defined at x equal to 1, so it cannot be continuous there. However the limit exists: factor the numerator as (x minus 1)(x plus 1), cancel the common factor for x not equal to 1, and the limit is 2. The graph is a straight line with a hole at (1, 2).
Hinglish: x = 1 par function defined nahi hai, isliye continuous nahi. Par limit 2 hai, kyunki numerator factor hoke cancel ho jaata hai. Graph me (1, 2) par hole hai.
Q9. If a function is differentiable at a point, what can you conclude about its continuity there?
Answer (English): Differentiability implies continuity. The existence of a finite derivative forces the difference quotient to settle down, which in turn forces the function values to approach the value at the point. The converse is false, as the absolute value function shows at zero.
Hinglish: Differentiable implies continuous, par continuous se differentiability nahi milti. |x| zero par continuous hai par differentiable nahi.
Q10. Differentiate e to the power sin inverse x.
Answer (English): This is a chain rule problem with two layers. The outer exponential keeps the same exponential when differentiated, and the inner sin inverse has derivative one over the square root of one minus x squared. Multiplying gives e to the power sin inverse x over the square root of one minus x squared.
Hinglish: Outer e^u, inner sin⁻¹x. Derivative = e^(sin⁻¹x) / √(1 - x²).
Marks pattern on this chapter
Objective items ask for continuity at a point, differentiability of a standard function or a direct derivative. Two and three mark items ask for a constant, a logarithmic derivative or a parametric derivative. Five mark items combine a second derivative with a chain of substitutions. Write the rule used at every stage and finish with the check the question asks for.
A final check habit: for a continuity constant, substitute the value back and verify both pieces agree; for a derivative, differentiate the answer once mentally to see if it returns the integrand shape. These checks cost ten seconds and catch most sign errors.
Quiz below: five multiple choice questions on continuity, differentiability, chain rule, logarithmic differentiation and second derivatives. Attempt them after the formula sheet and the two topic pages.