Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1 page covers all 34 questions, solved step-by-step, showing exactly how CBSE awards method marks for continuity proofs — from the formal definition (left hand limit equals right hand limit equals the function's value) to the algebra of continuous functions that lets you build new continuous functions from ones you already know.
This is the longest exercise in the chapter, and it builds in four clear stages. Questions 1–5 get you comfortable with the definition itself — proving a function is continuous at a given point by checking all three conditions directly, and testing continuity of a piecewise function class 12 style at more than one point. Questions 6–19 are the bulk of the exercise: finding all points of discontinuity for a series of piecewise functions, including the classic |x|/x case and the greatest integer function continuity proof that shows up in almost every CBSE sample paper. Questions 20–25 shift to continuity of trigonometric functions — sums, differences and products of sin x and cos x, plus the secant, cosecant and cotangent functions — using the algebra of continuous functions instead of checking limits from scratch every time. The final block, 26–34, is exam-favourite territory: finding the value of k in continuity problems (or a, b and λ) that makes a piecewise function continuous at a boundary point, a question type that appears in nearly every board paper's 2-mark or 3-mark section, closing with proofs that \cos(x^2), |\cos x| and \sin|x| are continuous everywhere.
f(x)=5x-3 is a polynomial function, so it is defined at every real number and its limit at any point equals its value there.
At x=0: \lim_{x\to 0} f(x) = 5(0)-3 = -3 = f(0).
At x=-3: \lim_{x\to -3} f(x) = 5(-3)-3 = -18 = f(-3).
At x=5: \lim_{x\to 5} f(x) = 5(5)-3 = 22 = f(5).
f(3) = 2(3)^2 - 1 = 17.
\lim_{x\to 3} f(x) = 2(3)^2 - 1 = 17.
(a) A polynomial function — continuous at every real number.
(b) A rational function, defined for all x \neq 5 — continuous at every point of its domain.
(c) For x \neq -5, f(x) = \dfrac{(x-5)(x+5)}{x+5} = x-5, a polynomial on its domain — continuous at every point of its domain.
(d) The modulus of a polynomial — composite of two continuous functions, so continuous everywhere.
f(x)=x^n is a polynomial function, so it is defined for every real number and every polynomial function is continuous.
In particular, \lim_{x\to n} x^n = n^n = f(n).
At x=0: this lies in the region x\le 1 where f(x)=x, a polynomial — continuous.
At x=1: LHL =\lim_{x\to 1^-} x = 1. RHL =\lim_{x\to 1^+} 5 = 5. Since LHL ≠ RHL, f is not continuous at x = 1.
At x=2: this lies in the region x \gt 1 where f(x)=5, a constant function — continuous.
Each piece is a polynomial, so f is continuous everywhere except possibly at x=2.
At x=2: LHL =2(2)+3=7. RHL =2(2)-3=1.
At x=-3: LHL =|-3|+3=6. RHL =-2(-3)=6. f(-3)=6. All equal — continuous at x = -3.
At x=3: LHL =-2(3)=-6. RHL =6(3)+2=20. LHL ≠ RHL — discontinuous at x = 3.
For x \gt 0, \frac{|x|}{x}=1; for x \lt 0, \frac{|x|}{x}=-1.
At x=0: LHL =-1, RHL =1.
For x \lt 0, \frac{x}{|x|}=\frac{x}{-x}=-1.
So f(x)=-1 for every real x — for x < 0 it comes from the first rule, and for x ≥ 0 it's given directly as −1. f is really the constant function f(x)=-1.
At x=1: LHL =1^2+1=2. RHL =1+1=2. f(1)=2.
At x=2: LHL =2^3-3=5. RHL =2^2+1=5. f(2)=5.
At x=1: LHL =1^{10}-1=0. RHL =1^2=1.
At x=1: LHL =1+5=6. RHL =1-5=-4.
At x=1: LHL =3, RHL =4 — not equal, discontinuous.
At x=3: LHL =4, RHL =5 — not equal, discontinuous.
On each open interval the function is constant, hence continuous there.
At x=0: LHL =2(0)=0, RHL =0, f(0)=0 — continuous.
At x=1: LHL =0, RHL =4(1)=4 — not equal, discontinuous.
At x=-1: LHL =-2, RHL =2(-1)=-2, f(-1)=-2 — continuous.
At x=1: LHL =2(1)=2, RHL =2, f(1)=2 — continuous.
LHL at x=3: 3a+1. RHL: 3b+3. f(3)=3a+1.
For continuity, 3a+1=3b+3, i.e. 3a-3b=2.
LHL at x=0: \lambda(0-0)=0. RHL: 4(0)+1=1. f(0)=0.
For continuity we'd need 0=1, which is impossible — so no value of λ makes f continuous at x = 0.
At x=1, f is given by the polynomial 4x+1 for all x near 1 (since x = 1 > 0), so f is continuous at x = 1 for every value of λ.
Let n be any integer. As x\to n^-, [x]=n-1, so g(x)\to n-(n-1)=1.
As x\to n^+, [x]=n, so g(x)\to n-n=0.
LHL (1) ≠ RHL (0) at every integer n.
f is a sum/difference of continuous functions (a polynomial, sine, and a constant), so f is continuous everywhere.
In particular, \lim_{x\to \pi} f(x) = \pi^2-\sin\pi+5 = \pi^2+5 = f(\pi).
Sine and cosine are both continuous functions on \mathbb{R}.
(a), (b) Sum and difference of continuous functions are continuous — both are continuous everywhere.
(c) Product of continuous functions is continuous — continuous everywhere.
Cosine is continuous at every real number (proved analogously to sine).
Cosecant =\dfrac{1}{\sin x} is a quotient of continuous functions, so continuous wherever \sin x\neq 0, i.e. for all x\neq n\pi.
Secant =\dfrac{1}{\cos x} is continuous wherever \cos x\neq 0, i.e. for all x\neq (2n+1)\frac{\pi}{2}.
Cotangent =\dfrac{\cos x}{\sin x} is continuous wherever \sin x\neq 0, i.e. for all x\neq n\pi.
At x=0: LHL =\lim_{x\to 0^-}\dfrac{\sin x}{x}=1 (standard limit). RHL =0+1=1. f(0)=1.
Elsewhere, each piece is continuous on its own domain.
Since -1\le \sin\frac{1}{x}\le 1, we have -x^2 \le x^2\sin\frac{1}{x}\le x^2.
As x\to 0, both bounds \to 0, so by the squeeze principle \lim_{x\to 0} x^2\sin\frac{1}{x}=0.
This equals f(0)=0.
\lim_{x\to 0}(\sin x - \cos x) = \sin 0 - \cos 0 = 0-1=-1.
This equals f(0)=-1.
Put x=\frac{\pi}{2}+h, so h\to 0 as x\to \frac{\pi}{2}. Then \cos x = \cos(\frac{\pi}{2}+h)=-\sin h and \pi-2x = -2h.
\lim_{h\to 0}\dfrac{k(-\sin h)}{-2h} = \dfrac{k}{2}\lim_{h\to 0}\dfrac{\sin h}{h} = \dfrac{k}{2}.
For continuity, \dfrac{k}{2}=3.
LHL =k(2)^2=4k. RHL =3.
Set 4k=3.
LHL =k\pi+1. RHL =\cos\pi=-1.
Set k\pi+1=-1 \Rightarrow k\pi=-2.
LHL =5k+1. RHL =3(5)-5=10.
Set 5k+1=10 \Rightarrow 5k=9.
Continuity at x=2: 5=2a+b …(i)
Continuity at x=10: 10a+b=21 …(ii)
Subtracting (i) from (ii): 8a=16 \Rightarrow a=2. Substituting back: b=5-2(2)=1.
f is the composite g\circ h where g(x)=\cos x and h(x)=x^2, both continuous everywhere.
f is the composite g\circ h where g(x)=|x| and h(x)=\cos x, both continuous everywhere.
f is the composite g\circ h where g(x)=\sin x and h(x)=|x|, both continuous everywhere.
|x| and |x+1| are each composites of continuous functions, hence continuous everywhere.
The difference of two continuous functions is continuous.
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