Graphical Interpretation of Continuity
Trending Questions
- Domain is [−1, 2)
- Product of all integral values of x in the domain is −1
- Sum of all integral values of x in the domain is 0
- Domain is {−1, 0, 1}
Which of the following functions is (are) bijective?
- f:(−∞, 0]→(0, π2]defined by f(x)=sin−1(ex)
- f:[−1, 1]→{−1, 0, 1} defined by f(x)=sgn(sin−1|x|−cos−1|x|)
- f:[−3, 0]→[cos3, 1] defined by f(x)=cosx
- f:R−Z→R defined by f(x)=ln{x}
- a+b+c=0
- |a+b+c|=1
- abc=0
- |abc|=1
Let R be the feasible region (convex polygon) for a linear programming problem and let,
Z = + be the objective function. When Z has an optimal value (maximum or minimum), where the variables and are subject to constraints described by linear inequalities, this optimal value must occur at ____________of the feasible region.
corner point
either in the interior or on the boundary line
any point in the interior
any point on the bounadary line
- R={(1, 1), (2, 2), (3, 3), (2, 1), (1, 2), (2, 3), (3, 2)}
- R−1=R
- Domain of R={1, 2, 3}
- Range of R={5}
- (−1, 1√3)
- (−∞, −1]∪[1√3, ∞)
- [−∞, 1√3)
- (−∞, −1)
Then, select the correct statement(s).
- For x∈[−2π, 2π], y=0 at:
x=−5π4, −π4, 3π4, 7π4 - Domain: R−(8n+1)π/4; n∈Z
- Domain: R−(4n+1)π/4; n∈Z
- For x∈[−2π, 2π], y=0 at:
x=−5π4, −π4, 3π4, 5π4
Among which of these points is f(x)=tanx discontinuous?
−π4
0
π4
π2
Among which of these points is f(x)=tanx discontinuous?
−π4
π4
π2
0
- Onto if f is onto
- One-one if f is one-one
- Continuous if f is continuous
- None of these
- Onto if f is onto
- One-one if f is one-one
- Continuous if f is continuous
- None of these
cosx>0 ∀ x∈(π, 6π)
- (7π2, 9π2)
- (11π2, 6π)
- (3π2, 5π2)∪(7π2, 9π2)∪(11π2, 6π)
- (3π2, 5π2)
The function f(x)=tanx where x∈(−π4, π4).
is continuous
is discontinuous
is increasing
is decreasing
The function f(x)=tanx where x∈(−π4, π4).
is continuous
is discontinuous
is increasing
is decreasing
- f is one-one
- Df=[−1, 2)
- f is many one
- Rf={π4, π2, 3π4}
- many-one onto
- one-one but not onto
- one-one onto
- none of these
- f(f(3))=π
- f(x) is periodic with fundamental period 2π.
- f(x) is neither even nor odd.
- Range of f(x) is [0, 2π].
- (π−1)2
- π2
- (π−2)2
- Maximum area is always infinite
Let x1<x2<x3<...<xn<... be all the points of local maximum of f and y1<y2<y3<...<yn<... be all the points of local minimum of f.
Then which of the following options is/are correct?
- x1<y1
- xn+1−xn>2 for every n
- xn∈(2n, 2n+12) for every n
- |xn−yn|>1 for every n
Find the number of discontinuities of the given function between x = 0 and x =2.
- 2, 1
- 2, 2
- 1, 1
- 1, 2
- (−2π, −π) ∪ (π, 2π)
- (−2π, −3π2) ∪ (−π2, π2) ∪ (3π2, 2π)
- [−2π, −π2] ∪ [π2, 2π]
- [−π2, π2]
where [x] denotes the greatest integer less than or equal to x.
- f(x) is continuous at x=0
- f(x) is continuous in (−1/2, 0)
- f(x) is continuous in (−1/2, 1/2)
- f(x) is continuous in (0, 1/2)
The function f(x)=tanx where x∈(−π4, π4).
is continuous
is discontinuous
is increasing
is decreasing
Let x1<x2<x3<...<xn<... be all the points of local maximum of f and y1<y2<y3<...<yn<... be all the points of local minimum of f.
Then which of the following options is/are correct?
- xn+1−xn>2 for every n
- |xn−yn|>1 for every n
- x1<y1
- xn∈(2n, 2n+12) for every n