PYQs Homework Notes Mocks Board Daily Homework JEE Mains Maths Question type Function defined as the max or min of f(t) for t up to x: continuity and differentiability This type of question, asked 3 times: 2021 Paper 17, 29 Jun 2022, 27 Jan 2024. All 3 questions below, oldest first.
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Question 1 2021 Paper 17
74 Let f : [ 0 , ∞ ) → [ 0 , 3 ] f:[0,\infty)\rightarrow[0,3] f : [ 0 , ∞ ) → [ 0 , 3 ] be a function defined by f ( x ) = { max { sin t : 0 ≤ t ≤ x } , 0 ≤ x ≤ π 2 + cos x , x > π f(x)=\begin{cases} \max\{\sin t : 0\leq t\leq x\}, & 0\leq x\leq\pi \\ 2+\cos x, & x>\pi \end{cases} f ( x ) = { max { sin t : 0 ≤ t ≤ x } , 2 + cos x , 0 ≤ x ≤ π x > π Then which of the following is true? A. f f f is continuous everywhere but not differentiable exactly at one point in ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) B. f f f is differentiable everywhere in ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) C. f f f is not continuous exactly at two points in ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) D. f f f is continuous everywhere but not differentiable exactly at two points in ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) Tap an option to check your answer.
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Question 2 29 Jun 2022
64 Let f : R → R f:R \rightarrow R f : R → R be a function defined by: f ( x ) = { max { t 3 − 3 t : t ≤ x } , x ≤ 2 x 2 + 2 x − 6 , 2 < x < 3 [ x − 3 ] + 9 , 3 ≤ x ≤ 5 2 x + 1 , x > 5 f(x) =\begin{cases} \max\left\{ t^{3}- 3t : t \leq x \right\}, & x \leq 2 \\ x^{2}+ 2x - 6, & 2 < x < 3 \\ \lbrack x - 3\rbrack + 9, & 3 \leq x \leq 5 \\ 2x + 1, & x > 5 \end{cases} f ( x ) = ⎩ ⎨ ⎧ max { t 3 − 3 t : t ≤ x } , x 2 + 2 x − 6 , [ x − 3 ] + 9 , 2 x + 1 , x ≤ 2 2 < x < 3 3 ≤ x ≤ 5 x > 5 Where [ t ] \lbrack t\rbrack [ t ] is the greatest integer less than or equal to t t t . Let m m m be the number of points where f f f is not differentiable and I = ∫ − 2 2 f ( x ) d x I =\int_{- 2}^{2} f(x)dx I = ∫ − 2 2 f ( x ) d x . Then the ordered pair (m, I) is equal to: A. ( 3 , 27 4 ) \left( 3,\frac{27}{4} \right) ( 3 , 4 27 ) B. ( 3 , 23 4 ) \left( 3,\frac{23}{4} \right) ( 3 , 4 23 ) C. ( 4 , 27 4 ) \left( 4,\frac{27}{4} \right) ( 4 , 4 27 ) D. ( 4 , 23 4 ) \left( 4,\frac{23}{4} \right) ( 4 , 4 23 ) Tap an option to check your answer.
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Question 3 27 Jan 2024
152 Consider the function f : ( 0 , 2 ) → R f:(0,2) \rightarrow R f : ( 0 , 2 ) → R defined by f ( x ) = x 2 + 2 x f(x) =\frac{x}{2}+\frac{2}{x} f ( x ) = 2 x + x 2 and the function g ( x ) g(x) g ( x ) defined by g ( x ) = { min { f ( t ) : 0 < t ≤ x } , 0 < x ≤ 1 3 2 + x , 1 < x < 2 g(x) =\left\{ \begin{matrix} \min\{ f(t) : 0 < t \leq x\}, & 0 < x \leq 1 \\ \frac{3}{2}+ x, & 1 < x < 2 \end{matrix} \right. g ( x ) = { min { f ( t ) : 0 < t ≤ x } , 2 3 + x , 0 < x ≤ 1 1 < x < 2 . Then A. g g g is continuous but not differentiable at x = 1 x = 1 x = 1 B. g g g is not continuous for all x ∈ ( 0 , 2 ) x \in (0,2) x ∈ ( 0 , 2 ) C. g g g is neither continuous nor differentiable at x = 1 x = 1 x = 1 D. g g g is continuous and differentiable for all x ∈ ( 0 , 2 ) x \in (0,2) x ∈ ( 0 , 2 ) Tap an option to check your answer.
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