Bank Guides Notes Mocks Board Questions / Maharashtra State Board Class 11 / Mathematics Continuity 87 Maharashtra State Board Class 11 Mathematics practice questions with answers and worked solutions.
Difficulty: 17 easy · 44 moderate · 26 hard. Most-asked subtopics: Continuity of a Function at a Point (37), Types of Discontinuity (19), Continuity from the Left and from the Right (14). Updated 16 August 2026. Balbharati textbook · 8.1 SolvedEx.1 Easy Add #1 · Continuity of a Function at a Point
Discuss the continuity of the function f ( x ) = ∣ x − 3 ∣ f(x) = |x - 3| f ( x ) = ∣ x − 3∣ at x = 3 x = 3 x = 3 . Balbharati textbook · 8.1 SolvedEx.2 Moderate Add #2 · Types of Discontinuity
Determine whether the function f f f is continuous on the set of real numbers where
f ( x ) = 3 x + 1 f(x) = 3x + 1 f ( x ) = 3 x + 1 , for x < 2 x < 2 x < 2
= 7 = 7 = 7 , for 2 ≤ x < 4 2 \le x < 4 2 ≤ x < 4
= x 2 − 8 = x^2 - 8 = x 2 − 8 for x ≥ 4 x \ge 4 x ≥ 4 .
If it is discontinuous, state the type of discontinuity. Balbharati textbook · 8.1 SolvedEx.3 Moderate Add #3 · Continuity of a Function at a Point
Test whether the function f ( x ) f(x) f ( x ) is continuous at x = − 4 x = -4 x = − 4 , where
f ( x ) = x 2 + 16 x + 48 x + 4 f(x) = \frac{x^2 + 16x + 48}{x + 4} f ( x ) = x + 4 x 2 + 16 x + 48 , for x ≠ − 4 x \ne -4 x = − 4
= 8 = 8 = 8 , for x = − 4 x = -4 x = − 4 . Balbharati textbook · 8.1 SolvedEx.4 Moderate Add #4 · Continuity Over an Interval
Discuss the continuity of f ( x ) = 9 − a 2 f(x) = \sqrt{9 - a^2} f ( x ) = 9 − a 2 , on the interval [ − 3 , 3 ] [-3, 3] [ − 3 , 3 ] . Balbharati textbook · 8.1 SolvedEx.5 Moderate Add #5 · Continuity from the Left and from the Right
Show that the function f ( x ) = ⌊ x ⌋ f(x) = \lfloor x \rfloor f ( x ) = ⌊ x ⌋ is not continuous at x = 0 , 1 x = 0, 1 x = 0 , 1 in the interval [ − 1 , 2 ) [-1, 2) [ − 1 , 2 ) . Done a few? Here’s your next move
Balbharati textbook · 8.1 SolvedEx.6 Hard Add #6 · Continuity of a Function at a Point
Discuss the continuity of the following function at x = 0 x = 0 x = 0 , where
f ( x ) = x 2 sin ( 1 x ) f(x) = x^2 \sin\left(\frac{1}{x}\right) f ( x ) = x 2 sin ( x 1 ) , for x ≠ 0 x \ne 0 x = 0
= 0 = 0 = 0 , for x = 0 x = 0 x = 0 . Balbharati textbook · 8.1 SolvedEx.7 Hard Add #7 · Continuity of a Function at a Point
Find k k k if f ( x ) f(x) f ( x ) is continuous at x = 0 x = 0 x = 0 , where
f ( x ) = x e x + tan x sin 3 x f(x) = \frac{x e^x + \tan x}{\sin 3x} f ( x ) = s i n 3 x x e x + t a n x , for x ≠ 0 x \ne 0 x = 0
= k = k = k , for x = 0 x = 0 x = 0 . Balbharati textbook · 8.1 SolvedEx.8 Hard Add #8 · Continuity of a Function at a Point
If f f f is continuous at x = 1 x = 1 x = 1 , where
f ( x ) = sin ( π x ) x − 1 + a f(x) = \frac{\sin(\pi x)}{x - 1} + a f ( x ) = x − 1 s i n ( π x ) + a , for x < 1 x < 1 x < 1
= 2 π = 2\pi = 2 π , for x = 1 x = 1 x = 1
= 1 + cos ( π x ) π ( 1 − x ) 2 + b = \frac{1 + \cos(\pi x)}{\pi(1 - x)^2} + b = π ( 1 − x ) 2 1 + c o s ( π x ) + b , for x > 1 x > 1 x > 1 ,
then find the values of a a a and b b b . Identify discontinuities for the following functions as either a jump or a removable discontinuity on R.
Balbharati textbook · 8.1 SolvedEx.9(1) Moderate Add #9 · Types of Discontinuity
f ( x ) = x 2 − 3 x − 18 x − 6 f(x) = \frac{x^2 - 3x - 18}{x - 6} f ( x ) = x − 6 x 2 − 3 x − 18 Balbharati textbook · 8.1 SolvedEx.9(2) Moderate Add #10 · Types of Discontinuity
g ( x ) = 3 x + 1 g(x) = 3x + 1 g ( x ) = 3 x + 1 , for x < 3 x < 3 x < 3
= 2 − 3 x = 2 - 3x = 2 − 3 x , for x ≥ 3 x \ge 3 x ≥ 3 Balbharati textbook · 8.1 SolvedEx.9(3) Moderate Add #11 · Types of Discontinuity
h ( x ) = 13 − x 2 h(x) = 13 - x^2 h ( x ) = 13 − x 2 , for x < 5 x < 5 x < 5
= 13 − 5 x = 13 - 5x = 13 − 5 x , for x > 5 x > 5 x > 5 Balbharati textbook · 8.1 SolvedEx.10 Hard Add #12 · Types of Discontinuity
Show that the function
f ( x ) = 5 cos x − e ( π 2 − x ) cot x f(x) = \dfrac{5^{\cos x} - e^{\left(\frac{\pi}{2} - x\right)}}{\cot x} f ( x ) = cot x 5 c o s x − e ( 2 π − x ) , for x ≠ π 2 x \ne \dfrac{\pi}{2} x = 2 π
= log 5 − e = \log 5 - e = log 5 − e , for x = π 2 x = \dfrac{\pi}{2} x = 2 π
has a removable discontinuity at x = π 2 x = \dfrac{\pi}{2} x = 2 π . Redefine the function so that it becomes continuous at x = π 2 x = \dfrac{\pi}{2} x = 2 π . Balbharati textbook · 8.1 SolvedEx.11 Moderate Add #13 · Continuity of a Function at a Point
If f ( x ) = ( 3 x + 2 2 − 5 x ) 1 x f(x) = \left( \dfrac{3x + 2}{2 - 5x} \right)^{\frac{1}{x}} f ( x ) = ( 2 − 5 x 3 x + 2 ) x 1 , for x ≠ 0 x \ne 0 x = 0 , is continuous at x = 0 x = 0 x = 0 then find f ( 0 ) f(0) f ( 0 ) . Balbharati textbook · 8.1 SolvedEx.12 Hard Add #14 · Continuity of a Function at a Point
If f ( x ) f(x) f ( x ) is defined on R R R , discuss the continuity of f f f at x = π 2 x = \dfrac{\pi}{2} x = 2 π , where
f ( x ) = 5 cos x + 5 − cos x − 2 ( 3 cot x ) . log ( 2 + π − 2 x 2 ) f(x) = \dfrac{5^{\cos x} + 5^{-\cos x} - 2}{(3\cot x).\log\left(\dfrac{2 + \pi - 2x}{2}\right)} f ( x ) = ( 3 cot x ) . log ( 2 2 + π − 2 x ) 5 c o s x + 5 − c o s x − 2 , for x ≠ π 2 x \ne \dfrac{\pi}{2} x = 2 π
= 2 log 5 3 = \dfrac{2\log 5}{3} = 3 2 log 5 , for x = π 2 x = \dfrac{\pi}{2} x = 2 π . Examine the continuity of
Balbharati textbook · Ex 8.1 Q1(i) Easy Add #15 · Continuity of a Function at a Point
f ( x ) = x 3 + 2 x 2 − x − 2 f(x) = x^{3} + 2x^{2} - x - 2 f ( x ) = x 3 + 2 x 2 − x − 2 at x = − 2 x = -2 x = − 2 . Balbharati textbook · Ex 8.1 Q1(ii) Easy Add #16 · Continuity from the Left and from the Right
f ( x ) = sin x f(x) = \sin x f ( x ) = sin x , for x ≤ π 4 x \le \dfrac{\pi}{4} x ≤ 4 π
= cos x = \cos x = cos x , for x > π 4 x > \dfrac{\pi}{4} x > 4 π , at x = π 4 x = \dfrac{\pi}{4} x = 4 π Balbharati textbook · Ex 8.1 Q1(iii) Easy Add #17 · Continuity of a Function at a Point
f ( x ) = x 2 − 9 x − 3 f(x) = \dfrac{x^{2} - 9}{x - 3} f ( x ) = x − 3 x 2 − 9 , for x ≠ 3 x \ne 3 x = 3
= 8 = 8 = 8 for x = 3 x = 3 x = 3 Examine whether the function is continuous at the points indicated against them.
Balbharati textbook · Ex 8.1 Q2(i) Easy Add #18 · Continuity from the Left and from the Right
f ( x ) = x 3 − 2 x + 1 f(x) = x^{3} - 2x + 1 f ( x ) = x 3 − 2 x + 1 , if x ≤ 2 x \le 2 x ≤ 2
= 3 x − 2 = 3x - 2 = 3 x − 2 , if x > 2 x > 2 x > 2 , at x = 2 x = 2 x = 2 . Balbharati textbook · Ex 8.1 Q2(ii) Moderate Add #19 · Continuity of a Function at a Point
f ( x ) = x 2 + 18 x − 19 x − 1 f(x) = \dfrac{x^{2} + 18x - 19}{x - 1} f ( x ) = x − 1 x 2 + 18 x − 19 , for x ≠ 1 x \ne 1 x = 1
= 20 = 20 = 20 for x = 1 x = 1 x = 1 , at x = 1 x = 1 x = 1 Balbharati textbook · Ex 8.1 Q2(iii) Moderate Add #20 · Continuity from the Left and from the Right
f ( x ) = x tan 3 x + 2 f(x) = \dfrac{x}{\tan 3x} + 2 f ( x ) = tan 3 x x + 2 , for x < 0 x < 0 x < 0
= 7 3 = \dfrac{7}{3} = 3 7 , for x ≥ 0 x \ge 0 x ≥ 0 , at x = 0 x = 0 x = 0 . Balbharati textbook · Ex 8.1 Q3 Moderate Add #21 · Continuity Over an Interval
Find all the points of discontinuities of f ( x ) = ⌊ x ⌋ f(x) = \lfloor x \rfloor f ( x ) = ⌊ x ⌋ on the interval ( − 3 , 2 ) (-3, 2) ( − 3 , 2 ) . Balbharati textbook · Ex 8.1 Q4 Easy Add #22 · Continuity of a Function at a Point
Discuss the continuity of the function f ( x ) = ∣ 2 x + 3 ∣ f(x) = |2x + 3| f ( x ) = ∣2 x + 3∣ , at x = − 3 / 2 x = -3/2 x = − 3/2 Test the continuity of the following functions at the points or interval indicated against them.
Balbharati textbook · Ex 8.1 Q5(i) Hard Add #23 · Continuity of a Function at a Point
f ( x ) = x − 1 − ( x − 1 ) 1 3 x − 2 f(x) = \dfrac{\sqrt{x - 1} - (x - 1)^{\frac{1}{3}}}{x - 2} f ( x ) = x − 2 x − 1 − ( x − 1 ) 3 1 , for x ≠ 2 x \ne 2 x = 2
= 1 5 = \dfrac{1}{5} = 5 1 , for x = 2 x = 2 x = 2
at x = 2 x = 2 x = 2 Balbharati textbook · Ex 8.1 Q5(ii) Hard Add #24 · Continuity of a Function at a Point
f ( x ) = x 3 − 8 x + 2 − 3 x − 2 f(x) = \dfrac{x^{3} - 8}{\sqrt{x + 2} - \sqrt{3x - 2}} f ( x ) = x + 2 − 3 x − 2 x 3 − 8 for x ≠ 2 x \ne 2 x = 2
= − 24 = -24 = − 24 for x = 2 x = 2 x = 2 , at x = 2 x = 2 x = 2 Balbharati textbook · Ex 8.1 Q5(iii) Easy Add #25 · Continuity from the Left and from the Right
f ( x ) = 4 x + 1 f(x) = 4x + 1 f ( x ) = 4 x + 1 , for x ≤ 8 3 x \le \dfrac{8}{3} x ≤ 3 8
= 59 − 9 x 3 = \dfrac{59 - 9x}{3} = 3 59 − 9 x , for x > 8 3 x > \dfrac{8}{3} x > 3 8 , at x = 8 3 x = \dfrac{8}{3} x = 3 8 . Balbharati textbook · Ex 8.1 Q5(iv) Hard Add #26 · Continuity of a Function at a Point
f ( x ) = ( 27 − 2 x ) 1 3 − 3 9 − 3 ( 243 + 5 x ) 1 5 f(x) = \dfrac{(27 - 2x)^{\frac{1}{3}} - 3}{9 - 3(243 + 5x)^{\frac{1}{5}}} f ( x ) = 9 − 3 ( 243 + 5 x ) 5 1 ( 27 − 2 x ) 3 1 − 3 , for x ≠ 0 x \ne 0 x = 0
= 2 = 2 = 2 for x = 0 x = 0 x = 0 , at x = 0 x = 0 x = 0 Balbharati textbook · Ex 8.1 Q5(v) Moderate Add #27 · Continuity from the Left and from the Right
f ( x ) = x 2 + 8 x − 20 2 x 2 − 9 x + 10 f(x) = \dfrac{x^{2} + 8x - 20}{2x^{2} - 9x + 10} f ( x ) = 2 x 2 − 9 x + 10 x 2 + 8 x − 20 for 0 < x < 3 0 < x < 3 0 < x < 3 ; x ≠ 2 x \ne 2 x = 2
= 12 = 12 = 12 , for x = 2 x = 2 x = 2
= 2 − 2 x − x 2 x − 4 = \dfrac{2 - 2x - x^{2}}{x - 4} = x − 4 2 − 2 x − x 2 for 3 ≤ x < 4 3 \le x < 4 3 ≤ x < 4
at x = 2 x = 2 x = 2 Identify discontinuities for the following functions as either a jump or a removable discontinuity.
Balbharati textbook · Ex 8.1 Q6(i) Easy Add #28 · Types of Discontinuity
f ( x ) = x 2 − 10 x + 21 x − 7 f(x) = \dfrac{x^{2} - 10x + 21}{x - 7} f ( x ) = x − 7 x 2 − 10 x + 21 . Balbharati textbook · Ex 8.1 Q6(ii) Easy Add #29 · Types of Discontinuity
f ( x ) = x 2 + 3 x − 2 f(x) = x^{2} + 3x - 2 f ( x ) = x 2 + 3 x − 2 , for x ≤ 4 x \le 4 x ≤ 4
= 5 x + 3 = 5x + 3 = 5 x + 3 , for x > 4 x > 4 x > 4 . Balbharati textbook · Ex 8.1 Q6(iii) Easy Add #30 · Types of Discontinuity
f ( x ) = x 2 − 3 x − 2 f(x) = x^{2} - 3x - 2 f ( x ) = x 2 − 3 x − 2 , for x < − 3 x < -3 x < − 3
= 3 + 8 x = 3 + 8x = 3 + 8 x , for x > − 3 x > -3 x > − 3 . Balbharati textbook · Ex 8.1 Q6(iv) Easy Add #31 · Types of Discontinuity
f ( x ) = 4 + sin x f(x) = 4 + \sin x f ( x ) = 4 + sin x , for x < π x < \pi x < π
= 3 − cos x = 3 - \cos x = 3 − cos x for x > π x > \pi x > π Show that following functions have continuous extension to the point where f ( x ) f(x) f ( x ) is not defined. Also find the extension. Balbharati textbook · Ex 8.1 Q7(i) Moderate Add #32 · Types of Discontinuity
f ( x ) = 1 − cos 2 x sin x f(x) = \dfrac{1 - \cos 2x}{\sin x} f ( x ) = sin x 1 − cos 2 x , for x ≠ 0 x \ne 0 x = 0 . Balbharati textbook · Ex 8.1 Q7(ii) Moderate Add #33 · Types of Discontinuity
f ( x ) = 3 sin 2 x + 2 cos x ( 1 − cos 2 x ) 2 ( 1 − cos 2 x ) f(x) = \dfrac{3\sin^{2} x + 2\cos x(1 - \cos 2x)}{2\left(1 - \cos^{2} x\right)} f ( x ) = 2 ( 1 − cos 2 x ) 3 sin 2 x + 2 cos x ( 1 − cos 2 x ) , for x ≠ 0 x \ne 0 x = 0 . Balbharati textbook · Ex 8.1 Q7(iii) Moderate Add #34 · Types of Discontinuity
f ( x ) = x 2 − 1 x 3 + 1 f(x) = \dfrac{x^{2} - 1}{x^{3} + 1} f ( x ) = x 3 + 1 x 2 − 1 for x ≠ − 1 x \ne -1 x = − 1 . Discuss the continuity of the following functions at the points indicated against them.
Balbharati textbook · Ex 8.1 Q8(i) Hard Add #35 · Continuity of a Function at a Point
f ( x ) = 3 − tan x π − 3 x f(x) = \dfrac{\sqrt{3} - \tan x}{\pi - 3x} f ( x ) = π − 3 x 3 − tan x , x ≠ π 3 x \ne \dfrac{\pi}{3} x = 3 π
= 3 4 = \dfrac{3}{4} = 4 3 , for x = π 3 x = \dfrac{\pi}{3} x = 3 π , at x = π 3 x = \dfrac{\pi}{3} x = 3 π . Balbharati textbook · Ex 8.1 Q8(ii) Hard Add #36 · Continuity from the Left and from the Right
f ( x ) = e 1 / x − 1 e 1 / x + 1 f(x) = \dfrac{e^{1/x} - 1}{e^{1/x} + 1} f ( x ) = e 1/ x + 1 e 1/ x − 1 , for x ≠ 0 x \ne 0 x = 0
= 1 = 1 = 1 , for x = 0 x = 0 x = 0 , at x = 0 x = 0 x = 0 . Balbharati textbook · Ex 8.1 Q8(iii) Hard Add #37 · Continuity of a Function at a Point
f ( x ) = 4 x − 2 x + 1 + 1 1 − cos 2 x f(x) = \dfrac{4^{x} - 2^{x + 1} + 1}{1 - \cos 2x} f ( x ) = 1 − cos 2 x 4 x − 2 x + 1 + 1 , for x ≠ 0 x \ne 0 x = 0
= ( log 2 ) 2 2 = \dfrac{(\log 2)^{2}}{2} = 2 ( log 2 ) 2 , for x = 0 x = 0 x = 0 , at x = 0 x = 0 x = 0 . Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.
Balbharati textbook · Ex 8.1 Q9(i) Hard Add #38 · Types of Discontinuity
f ( x ) = e 5 sin x − e 2 x 5 tan x − 3 x f(x) = \dfrac{e^{5\sin x} - e^{2x}}{5\tan x - 3x} f ( x ) = 5 tan x − 3 x e 5 s i n x − e 2 x , for x ≠ 0 x \ne 0 x = 0
= 3 / 4 = 3/4 = 3/4 , for x = 0 x = 0 x = 0 , at x = 0 x = 0 x = 0 . Balbharati textbook · Ex 8.1 Q9(ii) Hard Add #39 · Continuity from the Left and from the Right
f ( x ) = log ( 1 + 3 x ) ( 1 + 5 x ) f(x) = \log_{(1 + 3x)}(1 + 5x) f ( x ) = log ( 1 + 3 x ) ( 1 + 5 x ) for x > 0 x > 0 x > 0
= 32 x − 1 8 x − 1 = \dfrac{32^{x} - 1}{8^{x} - 1} = 8 x − 1 3 2 x − 1 , for x < 0 x < 0 x < 0 , at x = 0 x = 0 x = 0 . Balbharati textbook · Ex 8.1 Q9(iii) Moderate Add #40 · Types of Discontinuity
f ( x ) = ( 3 − 8 x 3 − 2 x ) 1 x f(x) = \left( \dfrac{3 - 8x}{3 - 2x} \right)^{\frac{1}{x}} f ( x ) = ( 3 − 2 x 3 − 8 x ) x 1 , for x ≠ 0 x \ne 0 x = 0 . Balbharati textbook · Ex 8.1 Q9(iv) Easy Add #41 · Continuity from the Left and from the Right
f ( x ) = 3 x + 2 f(x) = 3x + 2 f ( x ) = 3 x + 2 , for − 4 ≤ x ≤ − 2 -4 \le x \le -2 − 4 ≤ x ≤ − 2
= 2 x − 3 = 2x - 3 = 2 x − 3 , for − 2 < x ≤ 6 -2 < x \le 6 − 2 < x ≤ 6 . Balbharati textbook · Ex 8.1 Q9(v) Hard Add #42 · Continuity from the Left and from the Right
f ( x ) = x 3 − 8 x 2 − 4 f(x) = \dfrac{x^{3} - 8}{x^{2} - 4} f ( x ) = x 2 − 4 x 3 − 8 , for x > 2 x > 2 x > 2
= 3 = 3 = 3 , for x = 2 x = 2 x = 2
= e 3 ( x − 2 ) 2 − 1 2 ( x − 2 ) 2 = \dfrac{e^{3(x - 2)^{2}} - 1}{2(x - 2)^{2}} = 2 ( x − 2 ) 2 e 3 ( x − 2 ) 2 − 1 , for x < 2 x < 2 x < 2 Balbharati textbook · Ex 8.1 Q10(i) Hard Add #43 · Continuity of a Function at a Point
If f ( x ) = 2 + sin x − 3 cos 2 x f(x) = \dfrac{\sqrt{2 + \sin x} - \sqrt{3}}{\cos^{2} x} f ( x ) = cos 2 x 2 + sin x − 3 , for x ≠ π 2 x \ne \dfrac{\pi}{2} x = 2 π , is continuous at x = π 2 x = \dfrac{\pi}{2} x = 2 π then find f ( π 2 ) f\left(\dfrac{\pi}{2}\right) f ( 2 π ) . Balbharati textbook · Ex 8.1 Q10(ii) Hard Add #44 · Continuity of a Function at a Point
If f ( x ) = cos 2 x − sin 2 x − 1 3 x 2 + 1 − 1 f(x) = \dfrac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{3x^{2} + 1} - 1} f ( x ) = 3 x 2 + 1 − 1 cos 2 x − sin 2 x − 1 for x ≠ 0 x \ne 0 x = 0 , is continuous at x = 0 x = 0 x = 0 then find f ( 0 ) f(0) f ( 0 ) . Balbharati textbook · Ex 8.1 Q10(iii) Hard Add #45 · Continuity of a Function at a Point
If f ( x ) = 4 x − π + 4 π − x − 2 ( x − π ) 2 f(x) = \dfrac{4^{x - \pi} + 4^{\pi - x} - 2}{(x - \pi)^{2}} f ( x ) = ( x − π ) 2 4 x − π + 4 π − x − 2 for x ≠ π x \ne \pi x = π , is continuous at x = π x = \pi x = π , then find f ( π ) f(\pi) f ( π ) . Balbharati textbook · Ex 8.1 Q11(i) Hard Add #46 · Continuity of a Function at a Point
If f ( x ) = 24 x − 8 x − 3 x + 1 12 x − 4 x − 3 x + 1 f(x) = \dfrac{24^{x} - 8^{x} - 3^{x} + 1}{12^{x} - 4^{x} - 3^{x} + 1} f ( x ) = 1 2 x − 4 x − 3 x + 1 2 4 x − 8 x − 3 x + 1 , for x ≠ 0 x \ne 0 x = 0
= k = k = k , for x = 0 x = 0 x = 0
is continuous at x = 0 x = 0 x = 0 , find k k k . Balbharati textbook · Ex 8.1 Q11(ii) Moderate Add #47 · Continuity of a Function at a Point
If f ( x ) = 5 x + 5 − x − 2 x 2 f(x) = \dfrac{5^{x} + 5^{-x} - 2}{x^{2}} f ( x ) = x 2 5 x + 5 − x − 2 , for x ≠ 0 x \ne 0 x = 0
= k = k = k for x = 0 x = 0 x = 0
is continuous at x = 0 x = 0 x = 0 , find k k k . Balbharati textbook · Ex 8.1 Q11(iii) Moderate Add #48 · Continuity from the Left and from the Right
If f ( x ) = sin 2 x 5 x − a f(x) = \dfrac{\sin 2x}{5x} - a f ( x ) = 5 x sin 2 x − a , for x > 0 x > 0 x > 0
= 4 = 4 = 4 for x = 0 x = 0 x = 0
= x 2 + b − 3 = x^{2} + b - 3 = x 2 + b − 3 , for x < 0 x < 0 x < 0
is continuous at x = 0 x = 0 x = 0 , find a a a and b b b . Balbharati textbook · Ex 8.1 Q11(iv) Moderate Add #49 · Continuity Over an Interval
For what values of a a a and b b b is the function
f ( x ) = a x + 2 b + 18 f(x) = ax + 2b + 18 f ( x ) = a x + 2 b + 18 , for x ≤ 0 x \le 0 x ≤ 0
= x 2 + 3 a − b = x^{2} + 3a - b = x 2 + 3 a − b , for 0 < x ≤ 2 0 < x \le 2 0 < x ≤ 2
= 8 x − 2 = 8x - 2 = 8 x − 2 , for x > 2 x > 2 x > 2 ,
continuous for every x x x ? Balbharati textbook · Ex 8.1 Q11(v) Moderate Add #50 · Continuity Over an Interval
For what values of a a a and b b b is the function
f ( x ) = x 2 − 4 x − 2 f(x) = \dfrac{x^{2} - 4}{x - 2} f ( x ) = x − 2 x 2 − 4 , for x < 2 x < 2 x < 2
= a x 2 − b x + 3 = ax^{2} - bx + 3 = a x 2 − b x + 3 , for 2 ≤ x < 3 2 \le x < 3 2 ≤ x < 3
= 2 x − a + b = 2x - a + b = 2 x − a + b , for x ≥ 3 x \ge 3 x ≥ 3
continuous for every x x x on R R R ? Showing the first 50 of 87.
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