PYQs Homework Notes Mocks Board Daily Homework JEE Mains Maths Question type Constants that make a piecewise function continuous at a point This type of question, asked 16 times: 2021 Paper 10, 2021 Paper 14, 2021 Paper 18, 2021 Paper 22, 2021 Paper 26, 26 Jul 2022, 29 Jul 2022, 27 Jan 2024, 31 Jan 2024, 4 Apr 2024, 23 Jan 2025, 3 Apr 2025, 23 Jan 2026 S1, 23 Jan 2026 S2, 24 Jan 2026 S2, 8 Apr 2026 S2. All 16 questions below, oldest first.
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Question 1 2021 Paper 10
80 Let f : R → R f:\mathbb{R}\to\mathbb{R} f : R → R be a function defined as f ( x ) = { sin ( a + 1 ) x + sin 2 x 2 x , x < 0 b , x = 0 x + b x 3 − x b x 5 / 2 , x > 0 f(x)=\begin{cases}\dfrac{\sin(a+1)x+\sin 2x}{2x}, & x<0 \\ b, & x=0 \\ \dfrac{\sqrt{x+bx^{3}}-\sqrt{x}}{bx^{5/2}}, & x>0\end{cases} f ( x ) = ⎩ ⎨ ⎧ 2 x sin ( a + 1 ) x + sin 2 x , b , b x 5/2 x + b x 3 − x , x < 0 x = 0 x > 0 . If f f f is continuous at x = 0 x=0 x = 0 , then the value of a + b a+b a + b is equal to: A. B. -2
C. -3
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Question 2 2021 Paper 14
62 Let f ( x ) = { μ , x = 2 e tan ( x − 2 ) x − [ x ] , x < 2 ∣ x 2 − 5 x + 6 ∣ λ ( − x 2 + 5 x − 6 ) μ , x > 2 f\left( x \right)=\left\{ \begin{matrix} \mu, & x= 2 \\ e^{\frac{\tan\left( x- 2 \right)}{x-\left\lbrack x \right\rbrack}}, & x< 2 \\ \frac{\left| x^{2}- 5x+ 6 \right|\lambda}{\left( -x^{2}+ 5x- 6 \right)\mu}, & x> 2 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ μ , e x − [ x ] t a n ( x − 2 ) , ( − x 2 + 5 x − 6 ) μ ∣ x 2 − 5 x + 6 ∣ λ , x = 2 x < 2 x > 2 If f f f is continuous at x = 2 x = 2 x = 2 , then μ + λ \mu+\lambda μ + λ is equal to
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Question 3 2021 Paper 18
90 Let a , b ∈ R , b ≠ 0 a,b \in R,b \neq 0 a , b ∈ R , b = 0 , Define a function f ( x ) = { a s i n π 2 ( x − 1 ) , for x ≤ 0 tan 2 x − sin 2 x b x 3 , for x > 0. f(x) =\left\{ \begin{matrix} asin\frac{\pi}{2}(x- 1), & \text{~for~}x\leq 0 \\ \frac{\tan2x- \sin2x}{bx^{3}}, & \text{~for~}x> 0. \end{matrix} \right. f ( x ) = { a s in 2 π ( x − 1 ) , b x 3 t a n 2 x − s i n 2 x , for x ≤ 0 for x > 0. If f f f is continuous at x = 0 x = 0 x = 0 , then 10 − a b 10 - ab 10 − ab is equal to
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Question 4 2021 Paper 22
75 If the function f ( x ) = { 1 x log e ( 1 + x a 1 − x b ) , x < 0 k , x = 0 cos 2 x − sin 2 x − 1 x 2 + 1 − 1 , x > 0 f(x)=\left\{\begin{matrix}\frac{1}{x}\log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right) & ,\ x<0 \\ k & ,\ x=0 \\ \frac{\cos^{2}x-\sin^{2}x-1}{\sqrt{x^{2}+1}-1} & ,\ x>0\end{matrix}\right. f ( x ) = ⎩ ⎨ ⎧ x 1 log e ( 1 − b x 1 + a x ) k x 2 + 1 − 1 c o s 2 x − s i n 2 x − 1 , x < 0 , x = 0 , x > 0 is continuous at x = 0 x=0 x = 0 , then 1 a + 1 b + 4 k \frac{1}{a}+\frac{1}{b}+\frac{4}{k} a 1 + b 1 + k 4 is equal to: A. -5
B. 5
C. -4
D. 4
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Question 5 2021 Paper 26
74 Let f : R → R f:\mathbb{R} \rightarrow \mathbb{R} f : R → R be defined as f ( x ) = { 2 sin ( − π x 2 ) , if x < − 1 ∣ a x 2 + x + b ∣ , if − 1 ≤ x ≤ 1 sin ( π x ) , if x > 1 f(x) =\left\{ \begin{matrix} 2\sin\left( -\frac{\pi x}{2} \right), & \text{if } x < -1 \\ \left| ax^{2}+ x + b \right|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ 2 sin ( − 2 π x ) , a x 2 + x + b , sin ( π x ) , if x < − 1 if − 1 ≤ x ≤ 1 if x > 1 If f ( x ) f(x) f ( x ) is continuous on R \mathbb{R} R , then a + b a + b a + b equals: A. 3
B. -1
C. -3
D. 1
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Question 6 26 Jul 2022
67 If the function f ( x ) = { log e ( 1 − x + x 2 ) + log e ( 1 + x + x 2 ) s e c x − cos x , x ∈ ( − π 2 , π 2 ) − { 0 } k f(x) =\left\{ \begin{matrix} \frac{\log_{e}\left( 1 - x +x^{2} \right)+\log_{e}\left( 1 + x +x^{2} \right)}{secx - \cos x},x \in\left( \frac{- \pi}{2},\frac{\pi}{2} \right)\mathbf{- \{ 0\}} \\ k \end{matrix} \right. f ( x ) = { sec x − c o s x l o g e ( 1 − x + x 2 ) + l o g e ( 1 + x + x 2 ) , x ∈ ( 2 − π , 2 π ) − { 0 } k is continuous at x = 0 x = 0 x = 0 , then k k k is equal to : A. 1
B. -1
C. e
D. 0
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Question 7 29 Jul 2022
154 Let the function f ( x ) = { log e ( 1 + 5 x ) − log e ( 1 + α x ) x ;if x ≠ 0 10 if x = 0 f(x) =\left\{ \begin{matrix} \frac{\log_{e}(1 + 5x) -\log_{e}(1 + \alpha x)}{x} & \text{;if~}x \neq 0 \\ 10 & \text{~if~}x = 0 \end{matrix} \right. f ( x ) = { x l o g e ( 1 + 5 x ) − l o g e ( 1 + α x ) 10 ;if x = 0 if x = 0 be continuous at x = 0 x = 0 x = 0 . The α \alpha α is equal to: A. 10
B. -10
C. 5
D. -5
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Question 8 27 Jan 2024
73 Consider the function. f ( x ) = { a ( 7 x − 12 − x 2 ) b ∣ x 2 − 7 x + 12 ∣ , x < 3 2 sin ( x − 3 ) x − [ x ] , x > 3 b , x = 3 f(x) =\left\{ \begin{matrix} \frac{a\left( 7x - 12 -x^{2} \right)}{b\left| x^{2}- 7x + 12 \right|} & ,x < 3 \\ 2^{\frac{\sin(x - 3)}{x - \lbrack x\rbrack}} & ,x > 3 \\ b & ,\ x = 3 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ b ∣ x 2 − 7 x + 12 ∣ a ( 7 x − 12 − x 2 ) 2 x − [ x ] s i n ( x − 3 ) b , x < 3 , x > 3 , x = 3 Where [ x ] \lbrack x\rbrack [ x ] denotes the greatest integer less than or equal to x x x . If S S S denotes the set of all ordered pairs ( a , b ) (a,b) ( a , b ) such that f ( x ) f(x) f ( x ) is continuous at x = 3 x = 3 x = 3 , then the number of elements in S S S is : A. 2
B. Infinitely many
C. 4
D. 1
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Question 9 31 Jan 2024
77 Let g ( x ) g(x) g ( x ) be a linear function and f ( x ) = { g ( x ) , x ≤ 0 ( 1 + x 2 + x ) 1 x , x > 0 f(x) =\left\{ \begin{matrix} g(x) & ,x\leq 0 \\ \left( \frac{1 +x}{2 +x} \right)^{\frac{1}{x}} & ,x> 0 \end{matrix} \right. f ( x ) = { g ( x ) ( 2 + x 1 + x ) x 1 , x ≤ 0 , x > 0 , is continuous at x = 0 x= 0 x = 0 . If f ′ ( 1 ) = f ( − 1 ) f^{'}(1) =f( - 1) f ′ ( 1 ) = f ( − 1 ) , then the value of g ( 3 ) g(3) g ( 3 ) is A. 1 3 log e ( 4 9 e 1 / 3 ) \frac{1}{3}\log_{e}\left( \frac{4}{9e^{1/3}} \right) 3 1 log e ( 9 e 1/3 4 ) B. 1 3 log e ( 4 9 ) + 1 \frac{1}{3}\log_{e}\left( \frac{4}{9} \right) + 1 3 1 log e ( 9 4 ) + 1 C. log e ( 4 9 ) − 1 \log_{e}\left( \frac{4}{9} \right) - 1 log e ( 9 4 ) − 1 D. log e ( 4 9 e 1 / 3 ) \log_{e}\left( \frac{4}{9e^{1/3}} \right) log e ( 9 e 1/3 4 ) Tap an option to check your answer.
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Question 10 4 Apr 2024
61 Let f : R → R f:R \rightarrow R f : R → R be a function given by f ( x ) = { 1 − cos 2 x x 2 , x < 0 α , x = 0 , where α , β ∈ R . If β 1 − cos x x , x > 0 f(x) =\left\{ \begin{matrix} \frac{1 - \cos2x}{x^{2}} & ,x < 0 \\ \alpha & ,x = 0,\text{~where~}\alpha,\beta \in R\text{. If~} \\ \frac{\beta\sqrt{1 - \cos x}}{x} & ,x > 0 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ x 2 1 − c o s 2 x α x β 1 − c o s x , x < 0 , x = 0 , where α , β ∈ R . If , x > 0 f f f is continuous at x = 0 x = 0 x = 0 , then α 2 + β 2 \alpha^{2}+\beta^{2} α 2 + β 2 is equal to: A. 48
B. 12
C. 3
D. 6
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Question 11 23 Jan 2025
53 If the function f ( x ) = { 2 x { sin ( k 1 + 1 ) x + sin ( k 2 − 1 ) x } , x < 0 4 x = 0 2 x log e ( 2 + k 1 x 2 + k 2 x ) x > 0 f(x) =\left\{ \begin{matrix} \frac{2}{x}\left\{ \sin\left( k_{1}+ 1 \right)x + \sin\left( k_{2}- 1 \right)x \right\} & ,x < 0 \\ 4 & x = 0 \\ \frac{2}{x}\log_{e}\left( \frac{2 +k_{1}x}{2 +k_{2}x} \right) & x > 0 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ x 2 { sin ( k 1 + 1 ) x + sin ( k 2 − 1 ) x } 4 x 2 log e ( 2 + k 2 x 2 + k 1 x ) , x < 0 x = 0 x > 0 is continuous at x = 0 x = 0 x = 0 , then k 1 2 + k 2 2 k_{1}\ ^{2}+k_{2}\ ^{2} k 1 2 + k 2 2 is equal to A. 8
B. 20
C. 5
D. 10
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Question 12 3 Apr 2025
69 Let f ( x ) = { ( 1 + a x ) 1 / x , x < 0 1 + b , x = 0 ( x + 4 ) 1 / 2 − 2 ( x + c ) 1 / 3 − 2 , x > 0 f(x) =\left\{ \begin{matrix} (1 + ax)^{1/x} & , & x < 0 \\ 1 + b & , & x = 0 \\ \frac{(x + 4)^{1/2}- 2}{(x + c)^{1/3}- 2} & , & x > 0 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ ( 1 + a x ) 1/ x 1 + b ( x + c ) 1/3 − 2 ( x + 4 ) 1/2 − 2 , , , x < 0 x = 0 x > 0 be continuous at x = 0 x = 0 x = 0 . Then e a b c e^{a}bc e a b c is equal to. A. 64
B. 72
C. 48
D. 36
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Question 13 23 Jan 2026 S1
59 Let f ( x ) = { a x 2 + 2 a x + 3 4 x 2 + 4 x − 3 , x ≠ − 3 2 , 1 2 b x = − 3 2 , 1 2 f(x) =\left\{ \begin{matrix} \frac{ax^{2}+ 2ax + 3}{4x^{2}+ 4x - 3}, & x \neq -\frac{3}{2},\frac{1}{2} \\ b & x = -\frac{3}{2},\frac{1}{2} \end{matrix} \right. f ( x ) = { 4 x 2 + 4 x − 3 a x 2 + 2 a x + 3 , b x = − 2 3 , 2 1 x = − 2 3 , 2 1 be continuous at x = − 3 2 x = -\frac{3}{2} x = − 2 3 . If f o f ( x ) = 7 5 fof(x) =\frac{7}{5} f o f ( x ) = 5 7 , then x x x is equal to : A. 2
B. 1
C. 0
D. 1.4
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Question 14 23 Jan 2026 S2
51 If f ( x ) = { a ∣ x ∣ + x 2 − 2 ( sin ∣ x ∣ ) ( cos ∣ x ∣ ) x , x ≠ 0 b , x = 0 f(x) =\left\{ \begin{matrix} \frac{a|x| +x^{2}- 2(\sin|x|)(\cos|x|)}{x} & ,x \neq 0 \\ b & ,x = 0 \end{matrix} \right. f ( x ) = { x a ∣ x ∣ + x 2 − 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ ) b , x = 0 , x = 0 is continuous at x = 0 x = 0 x = 0 , then a + b a + b a + b is equal to : A. 1
B. 2
C. 0
D. 4
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Question 15 24 Jan 2026 S2
70 Let [ t ] \lbrack t\rbrack [ t ] denote the greatest integer less than or equal to t t t . If the function f ( x ) = { b 2 sin ( π 2 [ π 2 ( cos x + sin x ) cos x ] ) , x < 0 sin x − 1 2 sin 2 x x 3 , x > 0 a , x = 0 f\left( x \right)=\left\{ \begin{matrix} b^{2}\sin\left( \frac{\pi}{2}\left\lbrack \frac{\pi}{2}\left( \cos x + \sin x \right)\cos x \right\rbrack \right) & \ ,x < 0 \\ \frac{\sin x -\frac{1}{2}\sin2x}{x^{3}} & \ ,x > 0 \\ a & \ ,x = 0 \end{matrix} \right. f ( x ) = ⎩ ⎨ ⎧ b 2 sin ( 2 π [ 2 π ( cos x + sin x ) cos x ] ) x 3 s i n x − 2 1 s i n 2 x a , x < 0 , x > 0 , x = 0 is continuous at x = 0 x = 0 x = 0 , then a 2 + b 2 a^{2}+b^{2} a 2 + b 2 is equal to A. B. C. D. Tap an option to check your answer.
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Question 16 8 Apr 2026 S2
69 Let f ( x ) = { 1 3 , x ≤ π / 2 b ( 1 − sin x ) ( π − 2 x ) 2 , x > π / 2 f(x) =\left\{ \begin{matrix} \frac{1}{3} & , & x\leq\pi/2 \\ \frac{b(1 - \sin x)}{(\pi- 2x)^{2}} & , & x>\pi/2 \end{matrix} \right. f ( x ) = { 3 1 ( π − 2 x ) 2 b ( 1 − s i n x ) , , x ≤ π /2 x > π /2 , If f f f is continuous at x = π / 2 x =\pi/2 x = π /2 , then the value of ∫ 0 3 b − 6 ∣ x 2 + 2 x − 3 ∣ d x \int_{0}^{3\text{ }b - 6} \left| x^{2}+ 2x - 3 \right|dx ∫ 0 3 b − 6 x 2 + 2 x − 3 d x is : A. 5
B. 2
C. 3
D. 4
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