Exercise 5.1 — Continuity and Differentiability

1. 5.1.12 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Prove that the function f(x)=5x-3 is continuous at x=0, at x=-3 and at x=5.

   Hint. A polynomial is continuous at every real number, so just confirm the limit matches the function value at each point.

   Since f is a polynomial, lim(x->a) f(x) = f(a) at every a, since polynomials are built from sums and products of continuous functions. At a=0: f(0)=-3, limit=-3. At a=-3: f(-3)=-18, limit=-18. At a=5: f(5)=22, limit=22. All limits match the function values.

   ✦ f is continuous at x=0, x=-3, and x=5 (and in fact everywhere, being a polynomial).

2. 5.1.22 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Examine the continuity of the function f(x)=2x^2-1 at x=3.

   Hint. Polynomials are continuous everywhere, so just verify the limit equals f(3).

   f(3) = 2(9)-1 = 17. Since f is a polynomial, lim(x->3) f(x) = f(3) = 17, so the limit matches the function value.

   ✦ f is continuous at x=3.

3. 5.1.36 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Examine the following functions for continuity: (a) f(x)=x-5 (b) f(x)=1/(x-5), x!=5 (c) f(x)=(x^2-25)/(x+5), x!=-5 (d) f(x)=|x-5|.

   Hint. Each is built from continuous pieces (polynomials, quotients with nonzero denominator, or the modulus function), so continuity follows from the algebra of continuous functions on the stated domain.

   (a) A polynomial, continuous everywhere. (b) A rational function; since the domain excludes x=5, the denominator never vanishes on the domain, so it is continuous throughout. (c) A rational function with denominator x+5, which never vanishes since x!=-5 is excluded from the domain; continuous throughout. (d) The composition of the continuous modulus function with the continuous linear function x-5; continuous everywhere.

   ✦ All four functions are continuous on their stated domains.

4. 5.1.43 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.

   Hint. x^n is a polynomial (a power function), so it is continuous everywhere, in particular at the specific point x=n.

   Since f(x)=x^n is a polynomial, lim(x->n) f(x) = n^n = f(n), because polynomials are continuous at every real number, including the point x=n itself.

   ✦ f is continuous at x=n.

5. 5.1.54 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Is the function f defined by f(x)=x if x<=1, 5 if x>1, continuous at x=0? At x=1? At x=2?

   Hint. At x=0 and x=2, only one branch is relevant nearby, so check each branch's own continuity; at x=1, compare the left-hand limit, right-hand limit, and f(1).

   At x=0: only the x<=1 branch matters nearby, and x is continuous there, so f is continuous at x=0. At x=1: LHL = lim(x->1-) x = 1. RHL = lim(x->1+) 5 = 5. f(1)=1 (from the x<=1 branch). Since LHL is not equal to RHL, f is discontinuous at x=1. At x=2: only the x>1 branch matters nearby, and the constant function 5 is continuous, so f is continuous at x=2.

   ✦ Continuous at x=0 and x=2; discontinuous at x=1.

6. 5.1.63 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Find all points of discontinuity of f, where f(x)=2x+3 if x<=2, 2x-3 if x>2.

   Hint. Both pieces are polynomials, so check only the boundary point x=2.

   At x=2: LHL = 2(2)+3 = 7. RHL = 2(2)-3 = 1. Since LHL is not equal to RHL, f is discontinuous at x=2. Both pieces are polynomials elsewhere, hence continuous there.

   ✦ f is discontinuous only at x=2.

7. 5.1.74 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Find all points of discontinuity of f, where f(x)=|x|+3 if x<=-3, -2x if -3<x<3, 6x+2 if x>=3.

   Hint. Check both boundary points, x=-3 and x=3, comparing LHL, RHL, and the function value at each.

   At x=-3: LHL (from |x|+3) = |-3|+3 = 6. RHL (from -2x) = -2(-3) = 6. f(-3) = |-3|+3 = 6 (first branch). All three match, so f is continuous at x=-3. At x=3: LHL (from -2x) = -2(3) = -6. RHL/f(3) (from 6x+2, since x>=3 includes 3) = 6(3)+2 = 20. Since LHL is not equal to RHL, f is discontinuous at x=3.

   ✦ f is discontinuous only at x=3.

8. 5.1.83 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Find all points of discontinuity of f, where f(x)=|x|/x if x!=0, 0 if x=0.

   Hint. For x<0, |x|/x simplifies to -1; for x>0, it simplifies to 1 — so check the jump at x=0.

   For x<0, |x|=-x, so |x|/x = -1. For x>0, |x|=x, so |x|/x = 1. So LHL = -1 and RHL = 1 as x approaches 0. Since LHL is not equal to RHL, f is discontinuous at x=0. Elsewhere the function is a constant (-1 or 1) on each side, hence continuous.

   ✦ f is discontinuous only at x=0.

9. 5.1.94 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

   Find all points of discontinuity of f, where f(x)=x/|x| if x<0, -1 if x>=0.

   Hint. Simplify x/|x| for x<0 first — it may turn out to match the other branch exactly.

   For x<0, |x|=-x, so x/|x| = x/(-x) = -1. So the first branch also equals -1 for every x<0, exactly matching the second branch's constant value of -1 for x>=0. This means f(x)=-1 for every real x.

   ✦ f is the constant function -1 everywhere, so it has no points of discontinuity.

10. 5.1.104 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find all points of discontinuity of f, where f(x)=x+1 if x>=1, x^2+1 if x<1.

    Hint. Check the boundary point x=1; both pieces are polynomials elsewhere.

    At x=1: LHL (from x^2+1) = 1+1 = 2. RHL/f(1) (from x+1, since x>=1 includes 1) = 1+1 = 2. Since LHL, RHL, and f(1) all equal 2, f is continuous at x=1.

    ✦ f has no points of discontinuity; it is continuous everywhere.

11. 5.1.114 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find all points of discontinuity of f, where f(x)=x^3-3 if x<=2, x^2+1 if x>2.

    Hint. Check the boundary point x=2; both pieces are polynomials elsewhere.

    At x=2: LHL/f(2) (from x^3-3, since x<=2 includes 2) = 8-3 = 5. RHL (from x^2+1) = 4+1 = 5. Since LHL, RHL, and f(2) all equal 5, f is continuous at x=2.

    ✦ f has no points of discontinuity; it is continuous everywhere.

12. 5.1.124 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find all points of discontinuity of f, where f(x)=x^10-1 if x<=1, x^2 if x>1.

    Hint. Check the boundary point x=1.

    At x=1: LHL/f(1) (from x^10-1, since x<=1 includes 1) = 1-1 = 0. RHL (from x^2) = 1. Since LHL is not equal to RHL, f is discontinuous at x=1.

    ✦ f is discontinuous only at x=1.

13. 5.1.134 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Is the function defined by f(x)=x+5 if x<=1, x-5 if x>1, a continuous function?

    Hint. Check the boundary point x=1.

    At x=1: LHL/f(1) (from x+5, since x<=1 includes 1) = 1+5 = 6. RHL (from x-5) = 1-5 = -4. Since LHL is not equal to RHL, f is not continuous at x=1.

    ✦ No, f is not a continuous function (it is discontinuous at x=1).

14. 5.1.145 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Discuss the continuity of f, where f(x)=3 if 0<=x<=1, 4 if 1<x<3, 5 if 3<=x<=10.

    Hint. Check both boundary points, x=1 and x=3.

    At x=1: LHL/f(1) (from the first branch, since 0<=x<=1 includes 1) = 3. RHL (from the second branch) = 4. Since LHL is not equal to RHL, f is discontinuous at x=1. At x=3: LHL (from the second branch) = 4. RHL/f(3) (from the third branch, since 3<=x<=10 includes 3) = 5. Since LHL is not equal to RHL, f is discontinuous at x=3.

    ✦ f is discontinuous at x=1 and x=3, and continuous at every other point of [0,10].

15. 5.1.155 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Discuss the continuity of f, where f(x)=2x if x<0, 0 if 0<=x<=1, 4x if x>1.

    Hint. Check both boundary points, x=0 and x=1.

    At x=0: LHL (from 2x) = 0. RHL/f(0) (from the middle branch, since 0<=x<=1 includes 0) = 0. All match, so f is continuous at x=0. At x=1: LHL/f(1) (from the middle branch, since 0<=x<=1 includes 1) = 0. RHL (from 4x) = 4. Since LHL is not equal to RHL, f is discontinuous at x=1.

    ✦ f is discontinuous only at x=1; continuous everywhere else, including x=0.

16. 5.1.165 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Discuss the continuity of f, where f(x)=-2 if x<=-1, 2x if -1<x<=1, 2 if x>1.

    Hint. Check both boundary points, x=-1 and x=1.

    At x=-1: LHL/f(-1) (from the first branch, since x<=-1 includes -1) = -2. RHL (from the middle branch) = 2(-1) = -2. All match, so f is continuous at x=-1. At x=1: LHL/f(1) (from the middle branch, since -1<x<=1 includes 1) = 2(1) = 2. RHL (from the third branch) = 2. All match, so f is continuous at x=1.

    ✦ f is continuous everywhere; it has no points of discontinuity.

17. 5.1.174 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the relationship between a and b so that the function f defined by f(x)=ax+1 if x<=3, bx+3 if x>3, is continuous at x=3.

    Hint. Set the left-hand and right-hand limits at x=3 equal to each other.

    LHL/f(3) (from ax+1, since x<=3 includes 3) = 3a+1. RHL (from bx+3) = 3b+3. For continuity, these must be equal: 3a+1 = 3b+3, which simplifies to 3a-3b = 2.

    ✦ 3a - 3b = 2 (equivalently, a - b = 2/3)

18. 5.1.185 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    For what value of lambda is the function defined by f(x)=lambda(x^2-2x) if x<=0, 4x+1 if x>0, continuous at x=0? What about continuity at x=1?

    Hint. Evaluate the x<=0 branch at x=0 for any lambda — it may not depend on lambda at all, which would settle the question immediately.

    LHL/f(0) (from lambda(x^2-2x), since x<=0 includes 0) = lambda(0-0) = 0, for every value of lambda, since both x^2 and 2x vanish at x=0 regardless of lambda. RHL (from 4x+1) = 1. Since 0 is never equal to 1, no value of lambda makes f continuous at x=0. At x=1, only the x>0 branch (4x+1, a polynomial) is relevant nearby, so f is continuous there for every value of lambda.

    ✦ No value of lambda makes f continuous at x=0; f is continuous at x=1 for every value of lambda.

19. 5.1.194 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Show that the function defined by g(x)=x-[x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

    Hint. At an integer n, the greatest integer function itself jumps from n-1 to n, so track how that jump affects x-[x] from each side.

    Let n be any integer. As x approaches n from the left, [x]=n-1, so g(x)=x-(n-1) approaches n-(n-1)=1. As x approaches n from the right, [x]=n, so g(x)=x-n approaches n-n=0. Also g(n)=n-[n]=n-n=0. Since the left-hand limit (1) is not equal to the right-hand limit (0), g is discontinuous at every integer n.

    ✦ g is discontinuous at every integer, since the left-hand limit is always 1 while the right-hand limit and function value are always 0 there.

20. 5.1.203 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Is the function defined by f(x)=x^2-sin(x)+5 continuous at x=pi?

    Hint. This is a sum/difference of a polynomial and sine, both continuous everywhere, so continuity follows immediately.

    Since x^2, sin(x), and the constant 5 are each continuous everywhere, their sum/difference f(x)=x^2-sin(x)+5 is continuous everywhere, by the algebra of continuous functions, and in particular at x=pi.

    ✦ Yes, f is continuous at x=pi.

21. 5.1.213 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Discuss the continuity of the following functions: (a) f(x)=sin(x)+cos(x) (b) f(x)=sin(x)-cos(x) (c) f(x)=sin(x).cos(x).

    Hint. Each is a sum, difference, or product of the continuous functions sin(x) and cos(x).

    Since sin(x) and cos(x) are continuous everywhere, their sum (a), difference (b), and product (c) are each continuous everywhere, by the algebra of continuous functions.

    ✦ All three functions are continuous everywhere.

22. 5.1.224 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

    Hint. cos(x) is continuous everywhere like sin(x); the other three are quotients involving sin(x) or cos(x), so they're continuous wherever their denominator is nonzero.

    cos(x) is continuous everywhere, since it is built the same way as sin(x). Since cosec(x) = 1/sin(x) is a quotient, it is continuous wherever sin(x) is not 0, i.e. everywhere except x = n.pi (n an integer). Since sec(x) = 1/cos(x) is a quotient, it is continuous wherever cos(x) is not 0, i.e. everywhere except x = (2n+1).pi/2. Since cot(x) = cos(x)/sin(x) is a quotient, it is continuous wherever sin(x) is not 0, i.e. everywhere except x = n.pi.

    ✦ cos(x) is continuous everywhere; cosec(x) and cot(x) are continuous except at x=n.pi; sec(x) is continuous except at x=(2n+1).pi/2.

23. 5.1.234 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find all points of discontinuity of f, where f(x)=sin(x)/x if x<0, x+1 if x>=0.

    Hint. Use the standard limit lim(x->0) sin(x)/x = 1 to check the boundary point x=0.

    At x=0: LHL = lim(x->0-) sin(x)/x = 1 (the standard trigonometric limit). RHL/f(0) (from x+1, since x>=0 includes 0) = 0+1 = 1. All match, so f is continuous at x=0. Elsewhere, sin(x)/x is continuous for x<0 (a quotient of continuous functions with nonzero denominator), and x+1 is continuous for x>=0.

    ✦ f has no points of discontinuity; it is continuous everywhere.

24. 5.1.244 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Determine if f defined by f(x)=x^2.sin(1/x) if x!=0, 0 if x=0, is a continuous function.

    Hint. Use the squeeze theorem: since sin(1/x) is bounded between -1 and 1, x^2.sin(1/x) is squeezed between -x^2 and x^2.

    For x!=0, -x^2 <= x^2.sin(1/x) <= x^2, since |sin(1/x)| <= 1 always. As x approaches 0, both -x^2 and x^2 approach 0, so by the squeeze theorem, lim(x->0) x^2.sin(1/x) = 0 = f(0). So f is continuous at x=0. Elsewhere, x^2.sin(1/x) is a product/composition of continuous functions (since x!=0 there), hence continuous.

    ✦ Yes, f is continuous everywhere, including at x=0.

25. 5.1.254 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Examine the continuity of f, where f(x)=sin(x)-cos(x) if x!=0, -1 if x=0.

    Hint. Evaluate the limit of the formula piece directly at x=0 and compare to f(0).

    lim(x->0) [sin(x)-cos(x)] = sin(0)-cos(0) = 0-1 = -1 = f(0), since sin(x)-cos(x) is continuous everywhere as a difference of continuous functions, so its limit at 0 equals its value there.

    ✦ f is continuous everywhere, including at x=0.

26. 5.1.264 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the value of k so that the function f defined by f(x)=k.cos(x)/(pi-2x) if x != pi/2, 3 if x=pi/2, is continuous at x=pi/2.

    Hint. Substitute x=pi/2+h and let h approach 0, rewriting cos(x) and pi-2x in terms of h to evaluate the limit.

    Let x = pi/2+h, so as x approaches pi/2, h approaches 0. Then cos(x) = cos(pi/2+h) = -sin(h), and pi-2x = pi-2(pi/2+h) = -2h. So k.cos(x)/(pi-2x) = k.(-sin(h))/(-2h) = (k/2).(sin(h)/h), which approaches k/2 as h approaches 0, since sin(h)/h approaches 1. Setting this equal to f(pi/2)=3: k/2 = 3, so k = 6.

    ✦ k = 6

27. 5.1.273 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the value of k so that the function f defined by f(x)=k.x^2 if x<=2, 3 if x>2, is continuous at x=2.

    Hint. Set the left-hand and right-hand limits at x=2 equal.

    LHL/f(2) (from k.x^2, since x<=2 includes 2) = 4k. RHL = 3. Setting 4k = 3 gives k = 3/4.

    ✦ k = 3/4

28. 5.1.283 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the value of k so that the function f defined by f(x)=k.x+1 if x<=pi, cos(x) if x>pi, is continuous at x=pi.

    Hint. Set the left-hand and right-hand limits at x=pi equal.

    LHL/f(pi) (from k.x+1, since x<=pi includes pi) = k.pi+1. RHL (from cos(x)) = cos(pi) = -1. Setting k.pi+1 = -1 gives k.pi = -2, so k = -2/pi.

    ✦ k = -2/pi

29. 5.1.293 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the value of k so that the function f defined by f(x)=k.x+1 if x<=5, 3x-5 if x>5, is continuous at x=5.

    Hint. Set the left-hand and right-hand limits at x=5 equal.

    LHL/f(5) (from k.x+1, since x<=5 includes 5) = 5k+1. RHL (from 3x-5) = 3(5)-5 = 10. Setting 5k+1 = 10 gives 5k = 9, so k = 9/5.

    ✦ k = 9/5

30. 5.1.305 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find the values of a and b such that the function f defined by f(x)=5 if x<=2, ax+b if 2<x<10, 21 if x>=10, is a continuous function.

    Hint. Set up one equation at x=2 and another at x=10, then solve the resulting pair simultaneously.

    At x=2: LHL/f(2) = 5. RHL (from ax+b) = 2a+b. So 2a+b = 5. At x=10: LHL (from ax+b) = 10a+b. RHL/f(10) = 21. So 10a+b = 21. Subtracting the first equation from the second: 8a = 16, so a = 2. Substituting back: 2(2)+b = 5, so b = 1.

    ✦ a = 2, b = 1

31. 5.1.313 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Show that the function defined by f(x)=cos(x^2) is a continuous function.

    Hint. This is a composition of two continuous functions, cos and the squaring function.

    Since x^2 is continuous everywhere and cos is continuous everywhere, their composition f(x)=cos(x^2) is continuous everywhere, by the composition rule for continuous functions.

    ✦ f is continuous everywhere.

32. 5.1.323 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Show that the function defined by f(x)=|cos(x)| is a continuous function.

    Hint. This is a composition of two continuous functions, the modulus and cos.

    Since cos(x) is continuous everywhere and the modulus function is continuous everywhere, their composition f(x)=|cos(x)| is continuous everywhere, by the composition rule for continuous functions.

    ✦ f is continuous everywhere.

33. 5.1.333 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Examine that sin|x| is a continuous function.

    Hint. This is a composition of two continuous functions, sin and the modulus.

    Since |x| is continuous everywhere and sin is continuous everywhere, their composition f(x)=sin|x| is continuous everywhere, by the composition rule for continuous functions.

    ✦ f is continuous everywhere.

34. 5.1.344 marksNCERT Class 12 Mathematics, Continuity and Differentiability, Reprint 2026-27

    Find all the points of discontinuity of f defined by f(x)=|x|-|x+1|.

    Hint. This is a difference of two modulus (absolute value) functions, each continuous everywhere.

    Both |x| and |x+1| are continuous everywhere (each a composition of the continuous modulus function with a continuous linear function), so their difference f(x)=|x|-|x+1| is continuous everywhere, by the algebra of continuous functions — even though it has corners (non-differentiable points) at x=0 and x=-1, continuity is not affected by those corners.

    ✦ f has no points of discontinuity; it is continuous everywhere.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh105.pdf) — Exercise 5.1 (34 questions), Exercise 5.2 (10 questions), Exercise 5.3 (15 questions), Exercise 5.4 (10 questions), Exercise 5.5 (18 questions), Exercise 5.6 (11 questions), Exercise 5.7 (11 questions), plus the chapter's Miscellaneous Exercise (22 questions), 131 questions total — the largest chapter in the project so far. The old stub taught Rolle's Theorem and the Mean Value Theorem as a full chapter section, but neither appears anywhere in the current book — confirmed by a full-text search finding zero mentions of 'mean value' or 'Rolle' anywhere in the chapter, consistent with the syllabus line itself, which never mentions MVT. The old stub also collapsed all seven real exercises into one invented 45-question group with no solutions file behind it. Every derivative in this file (all 131 questions) was independently computed and verified with sympy, including the second-order-derivative proofs, the parametric-form questions, and the implicit-differentiation identities in the Miscellaneous Exercise; several answers were re-expressed in the clean logarithmic-differentiation form the book itself uses (e.g. dy/dx = y.[...]) rather than sympy's fully-expanded single-fraction form, since that is the technique the exercise is testing.. Questions are referenced from the NCERT textbook for identification.