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All questions of Differentiation- Continuity and Differentiability for A Level Exam
The area of triangle whose adjacent sides are
is :- a)√70/2 sq. units
- b)9√2 /2 sq. units
- c)3√3 /2 sq. units
- d)2√3 /2 sq. units
Correct answer is option 'A'. Can you explain this answer?
The area of triangle whose adjacent sides are is :
is :a)
√70/2 sq. units
b)
9√2 /2 sq. units
c)
3√3 /2 sq. units
d)
2√3 /2 sq. units
|
Suresh Iyer answered |
Area of triangle = ½(a * b)
a = (1, 0, -2) b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)½
|a * b| = (70)½
Area of triangle = ½(a * b)
= [(70)½]/2
a = (1, 0, -2) b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)½
|a * b| = (70)½
Area of triangle = ½(a * b)
= [(70)½]/2
Therefore, f(x) continuous everywhere.The function f(x) = ax, 0 < a < 1 is- a)increasing
- b)strictly decreasing on R
- c)neither increasing or decreasing
- d)decreasing
Correct answer is option 'D'. Can you explain this answer?
The function f(x) = ax, 0 < a < 1 is
a)
increasing
b)
strictly decreasing on R
c)
neither increasing or decreasing
d)
decreasing
|
Krishna Iyer answered |
f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Using approximation find the value of 
- a)2.025
- b)2.001
- c)2.01
- d)2.0025
Correct answer is option 'D'. Can you explain this answer?
Using approximation find the value of
a)
2.025
b)
2.001
c)
2.01
d)
2.0025
|
Gunjan Lakhani answered |
Let x=4, Δx=0.01
y=x^½ = 2
y+Δy = (x+ Δx)^½ = (4.01)^½
Δy = (dy/dx) * Δx
Δy = (x^(-1/2))/2 * Δx
Δy = (½)*(½) * 0.01
Δy = 0.25 * 0.01
Δy = 0.0025
So, (4.01)^½ = 2 + 0.0025 = 2.0025
- a)cos(sin-1x) + c
- b)sin-1x + c
- c)sin(cos-1x) + c
- d)x + c
Correct answer is option 'D'. Can you explain this answer?
a)
cos(sin-1x) + c
b)
sin-1x + c
c)
sin(cos-1x) + c
d)
x + c
|
Divey Sethi answered |
cos(sin-1x)/(1-x2)½……………….(1)
t = sin-1 x
dt = dx/(1-x2)½
Put the value of dt in eq(1)
= ∫cost dt
= sint + c
= sin(sin-1 x) + c
⇒ x + c
t = sin-1 x
dt = dx/(1-x2)½
Put the value of dt in eq(1)
= ∫cost dt
= sint + c
= sin(sin-1 x) + c
⇒ x + c
Integral of sin5x.cos2x is: - a)
- b)
- c)
- d)
Correct answer is 'A'. Can you explain this answer?
Integral of sin5x.cos2x is:
a)
b)
c)
d)
|
Learners Habitat answered |
So Looking at this integral, we have
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0- a)6
- b)4
- c)8
- d)2
Correct answer is option 'B'. Can you explain this answer?
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0
a)
6
b)
4
c)
8
d)
2
|
Dr Manju Sen answered |
f (x) = x2 - 8x + 12
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
A vector of magnitude 14 units, which is parallel to the vector
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A vector of magnitude 14 units, which is parallel to the vector
a)
b)
c)
d)
|
EduRev JEE answered |
Given vector = i + 2j - 3k
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14 unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14 unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)
If xy = 2, then dy/dx is- a)-x2/2
- b)y2
- c)-2/x2
- d)-y/x
Correct answer is option 'D'. Can you explain this answer?
If xy = 2, then dy/dx is
a)
-x2/2
b)
y2
c)
-2/x2
d)
-y/x
|
Hansa Sharma answered |
xy = 2
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
- a)– 1
- b)0
- c)1
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
a)
– 1
b)
0
c)
1
d)
1/2
|
Angad Gupta answered |
We have to use L'Hopital Rule It is in the form 0/0 So first we have to differentiate it After differentiating we get sinx/2x Then again differentiate it We get cosx/2 and now we get the answer as 1/2
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that- a)c ε [a, b] such that f'(c) = 0
- b)c ε (a, b) such that f'(c) = 0
- c)c ε (a, b] such that f'(c) = 0
- d)c ε [a, b) such that f'(c) = 0
Correct answer is option 'B'. Can you explain this answer?
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that
a)
c ε [a, b] such that f'(c) = 0
b)
c ε (a, b) such that f'(c) = 0
c)
c ε (a, b] such that f'(c) = 0
d)
c ε [a, b) such that f'(c) = 0
|
Abhinay Tripathi answered |
Answer is
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
Whta is the derivatve of y = log5 (x)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Whta is the derivatve of y = log5 (x)
a)
b)
c)
d)
|
Tanuja Kapoor answered |
y = log5 x = ln x/ln 5 → change of base
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
Infinity Academy answered |
On differentiating both sides with respect to x
The maximum value of f (x) = sin x in the interval [π,2π] is
a) 6
b) 0
c) -2
d) -4
Correct answer is option 'B'. Can you explain this answer?
|
|
Kiran Mehta answered |
f(x) = sin x
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.
- a)3 cos2 x. cos 4x
- b)cos 3x. cos3x
- c)-9 cos 3x. cos2x sinx
- d)3 sin2x.sin 4x
Correct answer is option 'D'. Can you explain this answer?
a)
3 cos2 x. cos 4x
b)
cos 3x. cos3x
c)
-9 cos 3x. cos2x sinx
d)
3 sin2x.sin 4x
|
Deepak Kapoor answered |
Given: y = sin 3x . sin3
= 3 sin2x.sin 4x
= 3 sin2x.sin 4x
Evaluate: 
- a)
- b)1/√3 arc tan[(x-2)/√5] + C
- c)
- d)
%7D%7D%7D%2B%7BC%7D)
Correct answer is option 'D'. Can you explain this answer?
Evaluate:
a)
b)
1/√3 arc tan[(x-2)/√5] + C
c)
d)
|
|
Deepak Kapoor answered |
Let's apply the integral substitution,
substitute 
Now use the standard integral :}}})
substitute back u=(x-2) and add a constant C to the solution,
Evaluate: 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Evaluate:
a)
b)
c)
d)
|
|
Leelu Bhai answered |
I = ∫√(x² + 5x)dx= ∫√(x² + 5x + 25/4 - 25/4)= ∫√{(x + 5/2)² - (5/2)²}={1/2(x+5/2)(√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}= {(2x + 5)/4 (√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}Thus, option D is correct...
If a, b, c and d are the position vectors of the points A, B, C and D such that a + c = b + d, then ABCD is a- a)Trapezium
- b)Rectangle
- c)Square
- d)Parallelogram
Correct answer is option 'D'. Can you explain this answer?
If a, b, c and d are the position vectors of the points A, B, C and D such that a + c = b + d, then ABCD is a
a)
Trapezium
b)
Rectangle
c)
Square
d)
Parallelogram
|
|
Tejas Verma answered |
Given:
If we divide both sides with 2 we get,
If we divide both sides with 2 we get,
⇒ Mid pt. of AC = Mid. pt. of BD
∴ ABCD is a parallelogram.
∴ ABCD is a parallelogram.
Differentiate sin2(θ2 + 1) with respect to θ2- a)sin(2θ2 + 1)
- b)cos(2θ2 + 2)
- c)sin(2θ2 + 2)
- d)cos(2θ2 + 1)
Correct answer is option 'C'. Can you explain this answer?
Differentiate sin2(θ2 + 1) with respect to θ2
a)
sin(2θ2 + 1)
b)
cos(2θ2 + 2)
c)
sin(2θ2 + 2)
d)
cos(2θ2 + 1)
|
|
Gunjan Lakhani answered |
y = sin2(θ2+1)
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
A real function f is said to be continuous if it is continuous at every point in …… .- a)[-∞,∞]
- b)The range of f
- c)The domain of f
- d)Any interval of real numbers
Correct answer is option 'A'. Can you explain this answer?
A real function f is said to be continuous if it is continuous at every point in …… .
a)
[-∞,∞]
b)
The range of f
c)
The domain of f
d)
Any interval of real numbers
|
Saranya Choudhury answered |
Its domain. This means that for any point x in the domain of f, as x approaches a certain value a, the value of f(x) approaches f(a). In other words, there are no sudden jumps or gaps in the graph of f.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is- a)Parallel to the x axis
- b)Parallel to the y axis
- c)Parallel to the line joining the end points of the curve
- d)Parallel to the line y = x
Correct answer is option 'C'. Can you explain this answer?
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is
a)
Parallel to the x axis
b)
Parallel to the y axis
c)
Parallel to the line joining the end points of the curve
d)
Parallel to the line y = x
|
Ujjwal Gupta answered |
No, option C is correct answer.
Find the maximum and minimum values of f (x) = 2x3 – 24x + 107 in the interval [1, 3].- a)89, 69
- b)89, 75
- c)59, 56
- d)89, -9
Correct answer is option 'B'. Can you explain this answer?
Find the maximum and minimum values of f (x) = 2x3 – 24x + 107 in the interval [1, 3].
a)
89, 69
b)
89, 75
c)
59, 56
d)
89, -9
|
Anjana Sharma answered |
Toolbox:d/dx(x^n) = nx^n−1Maxima & Minima = f′(x) = 0Step 1:f(x) = 2x^3−24x+107Differentiating with
, where C is a constant
, where C is a constant
, where C is a constant
, where C is a constant
If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be- a)3b – 2a
- b)3a – b
- c)3a – 2b
- d)3b – a
Correct answer is option 'A'. Can you explain this answer?
If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be
a)
3b – 2a
b)
3a – b
c)
3a – 2b
d)
3b – a
|
Snehal Choudhary answered |
AC = 3AB
(c - a)= 3(b - a). (c - a)
= 3b -3a
c= 3b--2a
If 
and 
, then the value of scalars x and y are:- a)x = 1 and y = -2
- b)x = -2 and y = 1
- c)x = 2 and y = -1
- d)x = 2 and y = 1
Correct answer is option 'C'. Can you explain this answer?
If and , then the value of scalars x and y are:
and , then the value of scalars x and y are:a)
x = 1 and y = -2
b)
x = -2 and y = 1
c)
x = 2 and y = -1
d)
x = 2 and y = 1
|
Sushil Kumar answered |
Given, a = i + 2j
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,
x = 2
y = -1
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,
x = 2
y = -1
Which of the following functions are not continuous.- a)
- b)
- c)ex
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions are not continuous.
a)
b)
c)
ex
d)
|
Anu answered |
b,c,and d are continuous and [x] is discontinuous at integer points.
If 3 sin(xy) + 4 cos (xy) = 5, then 
= .....- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If 3 sin(xy) + 4 cos (xy) = 5, then = .....
= .....a)
b)
c)
d)
|
|
Dr Manju Sen answered |
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
Derivatve of f(x) 
is given by
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Derivatve of f(x) is given by
is given bya)
b)
c)
d)
|
|
Neha Sharma answered |
y
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
Unit vectors along the axes OX, OY and OZ are denoted by - a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Unit vectors along the axes OX, OY and OZ are denoted by
a)
b)
c)
d)
|
|
Krishna Iyer answered |
represents the unit vectors along the co ordinate axis i.e. OX ,OY and OZ respectively.The unit vector in the direction of 
, where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The unit vector in the direction of , where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:
, where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:a)
b)
c)
d)
|
|
Sushil Kumar answered |
Given, Point A (2,-3,7)
Point B (1,3,-4)
Let vector in the direction of AB be C.
∴ C = B - A
⇒ (1,3,-4) - (2,-3,7)
⇒ ( 1-2 , 3+3 , -4-7 )
⇒ (-1,6,-11)
⇒ -1i + 6j -11k
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158
Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector) = (-1i + 6j -11k)/√158
Unit vector C = (C vector)/(Magnitude of C vector) = (-1i + 6j -11k)/√158
can be equal to
The points with position vectors 
are collinear vectors, Value of a =- a)-20
- b)20
- c)-40
- d)40
Correct answer is option 'C'. Can you explain this answer?
The points with position vectors are collinear vectors, Value of a =
are collinear vectors, Value of a =a)
-20
b)
20
c)
-40
d)
40
|
Om Desai answered |
Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.
The value of 
is:- a)0
- b)3
- c)1/3
- d)1
Correct answer is option 'B'. Can you explain this answer?
The value of is:
is:a)
0
b)
3
c)
1/3
d)
1
|
Sahil Soni answered |
Cross multiply in maths take place in cycle like i》j》k i×j=k j×k=i k×i=j but j×i=-k k×j=-i i×k=-j and dotmultiply takes place as i.i=1j.j=1K.K=1BUTI.J=0J.K=0k.i=0so the correct answer is b
For what values of a and b, f is a continuous function.- a)a=2,b=0
- b)a=1,b=0
- c)a=0,b=2
- d)a=0,b=0
Correct answer is 'A'. Can you explain this answer?
For what values of a and b, f is a continuous function.a)
a=2,b=0
b)
a=1,b=0
c)
a=0,b=2
d)
a=0,b=0
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Tejas Verma answered |
For continuity: LHL=RHL
at x=2,
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
at x = 0,
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
Find the value of x when x is a natural number and 24x< 100.- a){5,6,……..∞}
- b){1,2,3,4}
- c){1,2,3,4,5}
- d){0,1,2,3,4}
Correct answer is option 'B'. Can you explain this answer?
Find the value of x when x is a natural number and 24x< 100.
a)
{5,6,……..∞}
b)
{1,2,3,4}
c)
{1,2,3,4,5}
d)
{0,1,2,3,4}
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Shreya Gupta answered |
We are given: 24x < 100
24x < 100
=> 24x /24 < 100/24 [Dividing both sides by positive number.]
=> x < 25/6
When x is a natural number, in this case, the following values of x make the statement true
x = 1, 2, 3, 4.
The solution set of the inequality is {1, 2, 3, 4}.
What is the region represented by x > 0 and y < 0?- a)2nd quadrant
- b)3rd quadrant
- c)4th quadrant
- d)1st quadrant
Correct answer is option 'C'. Can you explain this answer?
What is the region represented by x > 0 and y < 0?
a)
2nd quadrant
b)
3rd quadrant
c)
4th quadrant
d)
1st quadrant
|
|
Suresh Reddy answered |
If x >0 then x will be positive and y less than 0 means it gets negative value
that means it lies in 4th quadrant.
that means it lies in 4th quadrant.
The solution of the inequality 3(2-x)≥2(1-x) for real x is :- a)x < 4
- b)x > 4
- c)x ≤4
- d)x ≥ 4
Correct answer is option 'C'. Can you explain this answer?
The solution of the inequality 3(2-x)≥2(1-x) for real x is :
a)
x < 4
b)
x > 4
c)
x ≤4
d)
x ≥ 4
|
Mohit Rajpoot answered |
3(2 - x) ≥ 2(1 - x)
6 - 3x ≥ 2 - 2x
4 ≥ x
6 - 3x ≥ 2 - 2x
4 ≥ x
Pairs of consecutive even positive integers, both of which are larger than 7 such that their sum is less than 28, are- a)(8, 9), (9, 10), (10, 11)
- b)(8, 10), (10, 12)
- c)(8, 10), (10, 12), (12, 14)
- d)(8, 10), (10, 12), (12, 14), (14, 16)
Correct answer is option 'C'. Can you explain this answer?
Pairs of consecutive even positive integers, both of which are larger than 7 such that their sum is less than 28, are
a)
(8, 9), (9, 10), (10, 11)
b)
(8, 10), (10, 12)
c)
(8, 10), (10, 12), (12, 14)
d)
(8, 10), (10, 12), (12, 14), (14, 16)
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Neha Joshi answered |
Let x be the smaller of the two consecutive even positive integers .
Then the other integer is x+2.
Since both the integers are larger than 7,x > 7 ....(1)
Also the sum of the two integers is less than 28.
x + (x + 2) < 28
⇒ 2x + 2 < 28
⇒ 2x < 28 − 2
⇒ 2x < 26
=> x < 13…...(2)
From(1) and (2), we get
7 < x <13
Since x is an even number, x can take the values 8, 10 and 12.
Thus the required possible pairs are (8,10),(10,12) and (12,8).
Then the other integer is x+2.
Since both the integers are larger than 7,x > 7 ....(1)
Also the sum of the two integers is less than 28.
x + (x + 2) < 28
⇒ 2x + 2 < 28
⇒ 2x < 28 − 2
⇒ 2x < 26
=> x < 13…...(2)
From(1) and (2), we get
7 < x <13
Since x is an even number, x can take the values 8, 10 and 12.
Thus the required possible pairs are (8,10),(10,12) and (12,8).
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