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Here f(x) = x
Function f is defined for all real numbers
Let c be any real number.
∴ f is continuous at x = c
But c is any real number.
∴ f is continuous at every real number.
g(x) = x – [x]
Let a be any integer.
.
∴ f is discontinuous at x = a
But a is any integral point.
∴ f is discontinuous at all integral points.
Let f(x) = [ x ]. Df = R
Let a be any real number ∈ Df.
Two cases arise:
Case I. If a is not an integer, then
⇒ f is continuous at x = a
Case II. If a ∈ 1, then f(a) = [ a ] = a and
∴ f is not continuous at x = a, a ∈ I.
∴ function is discontinuous at every integral point.
Tips: -
1. Domain of continuity for the function [x] is R – I.
2. From the graph of [x], done in earlier class, it is also clear that [x] is discontinuous at integral points.
, when x = 3, then f will be continuous at x = 3.
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Discuss continuity of the function f given by
f(x) = | x – 1| + | x – 2 ] at x = 1 and x = 2.
Also f(0) = value of (x +1) at x = 0
= 0 + 1 = 1
∴ f (x) is continuous at x = 0.
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is not defined at x = 0. Find the value of/(x) so that f is continuous at x = 0.
∴ f is right continuous at x = 3 and left continuous at x = 5
Subtracting (2) from (1), we get,
– 2 a = – 6, ⇒ a = 3
Putting a = 3 in (1) we get,
3 x 3 + b = 1, ⇒ b = – 8
∴ a = 3, b = –8.
Also f(0) = a
Since f(x) is continuous at x = 0
f(x) = |x|.Df=R
Let a be any real number ∈ Df
But a is any real number ∈ Df
∴ f is continuous function
⇒ | x | is continuous at every point of domain.
Here f(x) = x3 + x2 –1
Function f is defined for all real numbers.
Let c be any real number
∴ f is continuous at c
But c is any real number
∴ f is continuous at every real number
∴ f is continuous function.
Here f(x) = x – 5
Function f is defined for all real numbers.
Let c be any real number.
∴ f(c) = c – 5
∴ f is continuous at x = c
But c is any real number.
∴ f is continuous at every real number.
f(x) = | x – 5 |
Here f(x) = | x – 5 |
Function f is defined for all real numbers.
Let c be any real number.
∴ f is continuous at.x = c.
But c is any real number
∴ f is continuous at every real number.
Let f(x) = sin x Df = R
Let a be any real number ∈ Df
∴ f is continuous at x = a
But a is any real number ∈ Df
∴ f is continuous function
∴ sin x is continuous at every point of its domain,
(i) Let f(x) = cos x. Df = R
Let a be any real number ∈ Df
∴ f is continuous as x = a
But a is any real number ∈ Df
∴ f is continuous at every point of its domain.
i.e., cos x is continuous at every point of its domain.
∴ f is continuous at x = a
But a is any real number ∈ Df
∴ f is continuous function at every point of domain,
∴ cosec x is continuous at every point of domain.
∴ f is continuous at x = a
But a is any real number ∈ Df
∴ f is continuous at every point of domain,
∴ sec x is continuous at every point of domain.
∴ f is continuous at x = a.
But a is any real number ∈ Df.
∴ f is continuous at every point of domain,
∴ cot x is continuous at every point of domain.
f(x) = sin x – cos x
Df = R
Let a be any real number.
f(x) = sin x cos.x
Df = R
Let a be any real number.
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Here f(x + y) = f(x) + f(y) ∀ x, y∈ R ..(1)
Let f be continuous at x = 0
⇒ f is continuous at x = c
But c is any real number
∴ f is continuous ∀ .x ∈ R.
if possible, let (f + g ) be continuous, and as f is continuous then
(f + g ) – f is a continuous function (∵ difference of two continuous function is continuous)
⇒ g is continuous, which is a contradiction
Hence (f + g ) is discontinuous.
Note : If f and g are discontinuous, then f + g ,fg need not be discontinuous.
For example
But (fg) (x) = (tan x) (cos x) = sin x, is continuous
(ii) 
but (f + g) (x) = 0, is continuous at x = 0
Case II : a = 0
∴ f is continuous at x = a = 0
Thus, f is continuous at all points of its domain.
Function f is defined for all points of the real line.
Let c be any real number.
Three cases arise :
Case I: c < 0
Case II:c > 1
Case III: c = 0
∴ f is continuous at x = 0
Thus f has no point of discontinuity.
Case II: If c > 1, then f(c) = c – 2.
∴f is continuous at all points x > 1.
Case III : If c = 1, then![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/98/62/04434da5848e25812395fa2c81cd.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/98/62/04434da5848e25812395fa2c81cd.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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∴ f is not continuous at x = 1
∴ x = 1 is the only point of discontinuity of f.
∴ f is not a continuous function.
Function f is defined for all real numbers except 0. Therefore domain of f is D1 ∪ D2 where
D1 = {.x∈R : r < 0 }, D2 = {x ∈ R : x > 0}
Now two cases arise :
Case 1: Let c ∈ D1. In this case f(x) = x + 2.
∴ f (x) is continuous at x = c
But c is any point of D1.
∴ f is continuous in D1.
Case II : Let c ∈ D2. In this case f(x) = –x + 2.
∴ f(x) is continuous at x = c.
But c is any point of D2.
∴ f is continuous in D2.
Now f is continuous at all points in the domain of f. f
∴ f is continuous.
Case II : c > 2
∴ f is continuous at all points x > 2.
Case III : x = 2
∴ f is not continuous at x = 2
∴ x = 2 is the only point of discontinuity of f.
∴ f is continuous at all point x < – 3
Case II: c = –3
∴ f is continuous at x = – 3
Case III: – 3 < c < 3
f(x) = – 2 x is a continuous function as it is a polynomial.
Case IV : c = 3
∴ f is continuous at all points x > 3.
Find all points of discontinuity of f, where f is defined by
Find all points of discontinuity of f, where f is defined by
Find all points of discontinuity of f, where f is defined by
Function f is defined at all points of the real line.
When x < – 1, we have f(x) = – 2; which is constant and so it is continuous.
∴ f is continuous at x = – 1
In the interval –1 < x < 1, we have f(x) = 2 x, which being a linear polynomial, is continuous.
∴ f is continuous at x = 1
When x > 1, we have f (x) = 2. which is constant and so it is continuous.
Sponsor Area
Find the values of a and b such that the function defined by
is a continous function.
Here f(x) = sin (x2)
Let g(x) =sinx and h(x) = x2
∴(g o h) (x) = g (h(x)) = g(x2 ) = sin x2
Now f(x) = (g o h)(x) and g, h are both continuous functions.
∴ f is continuous function.
Here f(x)=cos(x2)
Let g(x)= cosx and h(x)=x2
∴ (g o h)(x)=g(h(x))=g(x2) =cos x2
Now f(x)=(g o h)(x) and g.h are both continuous functions
∴ f is continuous function.
Let f(x) =| x |, g (x) =sin x
Df = R, Rf =[ 0, ∞ )
Dg =R, Rg=[–1,1]
∵ Rf. C Dg
∵ g o f is defined
and ( g o f) (x) =g(f(x)) =g(| x |)=sin| x |
Now f and g are both continuous for every x ∈ R.
∴ g o f i.e., sin | x | is continuous for every x ∈ R.
Here f (x) = | cos x |
Let g(x) = | x | and h(x) = cos x
∴ (g o h) (x) = g (h(x)) = g(cos x) = | cos x |
Now f(x) = (g o h)(x) and g,h are both continuous functions.
∴ f is continuous function.
Here f(x) = |1–x+|x||
Let g(x)=1–x+|x| and h(x) = |x|
∴ (h o g)(x) = h (g (x)) = h(1–x+|x|) = |1–x+|x||
Now polynomial function 1 – x is a continuous function.
Also |x| is a continuous function
We know that sum of two continuous function is a continuous function.
∴ 1–x+|x| is a continuous function.
Now (h o g)(x) = |1–x+|x || is the composite of two continuous functions h and g.
∴ f(x) = |1–x+|x|| is a continuous function.
Find all the points of discontinuity of f defined by
f(x) =|x|–|x+1|.
Here f(x) =|x|–|x+1|
Let g(x) = |x| and h(x) = x+1
∴ (g o h)(x) = g(h(x)) = g(x+1) = |x+1|
Now g and h are both continuous functions
∴ (g o h) is a continuous function.
∴ |x+1| is continuous function
Also | x | is continuous function.
Now difference of two continuous functions is a continuous function.
∴ f(x) = |x|–|x+1| is continuous.
Consider the function f given by
f(x) = |x - 1| + |x - 2| + |x - 3|
This function is continuous everywhere
Differentiability at x = 1
Similarly f is not derivable at x = 2, 3. Also f is differentiable at any other point.
Hence the result.
Consider the function f given by
f(x) = |x - 1| + |x - 2|
This function is continuous everywhere
Differentiability at x = 1
Similarly f is not derivable at x = 2. Also f is differentiable at any other point.
Hence the result.
Let function f be defined by
f(x) = |x - 1| + |x - 2| + |x - 3| + |x - 4| + |x - 5|
This function is continuous everywhere.
Also it is differentiable everywhere except at x = 1, 2, 3, 4, 5.
Hence the result.
Find the equation of the normal to the curve x2 = 4 y which passes through the point (1, 2).
The equation of the curve is x2 = 4y ...(1)
Let normal at (h, k) pass through (1, 2).
Since (h, k) lies on (1)
...(2)
Slope of tangent at (h, k) = 
Equation of normal at (h, k) is 
Let y=x2 + 3x + 2
Differentiating both sides w.r.t. x, we get,
(1 - x2). 2 y1 y2 + y12 (-2 x) = 2 m2 y yi
Dividing both sides by 2 y1 , we get,
(1 - x2) y2 - x y1 = m2 y
Differentiating both sides w.r.t. x, we get
(1-x2)·2 y1 y2+y12 (-2 x)=a2·2 y y1
Dividing both sides by 2 y1 , we get,
f(x)=x2 + 2 x - 8
This is polynomial in x.
(i) Sisce every polynomial in x is continuous function for all x.
∴ f is continuous in [-2, 2]
(ii) f'(x) = 2 x + 2, which exists in (-4, 2)
∴ f is differentiable in (-4, 2)
(iii) f(-4) = (-4)2 + 2(-4)-8 = 16 - 8 - 8 = 0
f(2) = (2)2 + 2 (2) - 8 = 4 + 4 - 8 = 0
∴ f(-4) = f(2).
∴ f satisfies all the conditions of Rolle's Theorem.
∴ there exists at least one value c of x such that f'(c) = 0, where - 4 < c < 2.
Now f'(c) = 0 gives us 2 c + 2 = 0 or 2 c = -2
∴ c = - 1 ∈ (- 4, 2)
∴ Rolle's Theorem is verified.
Here f(x) = x(x - 1)2 = x3 - 2 x2 + x
It is a polynomial in x.
(i) Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [0, 1].
(ii) f'(x) = 3 x2 - 4 x + 1, which existsyn (0, 1)
∴ f(x) is derivable in (0, 1).
(iii) f(0) = 0, f(1) = 0
∴ f(0) = f(1)
∴ f(x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that
f'(c) = 0 where 0 < c 1.
It is a polynomial in x.
(i) since every polynomial in x is a continuous function for every value of x
∴ f(x) is continuous in [0, 8]
(ii) f'(x) = 8 - 2 x, which exists in (0, 8)
∴ f(x) is derivable in (0, 8)
(iii) f(0) = 0 - 0 = 0
f(8) = 8 (8) - (8)2 = 64 - 64 = 0
∴ f(0) = f(8)
∴ f(x) satisfies all the conditions of Rolle's theorem.
∴ there must exist at least one value c of x such that f'(c) = 0 where 0 < c < 8.
Now f'(c) = 0 gives 8 - 2 c = 0 ⇒ c = 4 ∈ (0, 8)
∴ Rolle's theorem is verified.
Here f(x) = x2
This is a polynomial in x
(a) Since every polynomial in x is continuous for all x
∴ f(x) is continuous in [- 1, 1].
(b) f'(x) = 2 x, which exists in (- 1, 1)
∴ f(x) is derivable in (- 1, 1).
(c) f(-1) = (-1)2 =1, f(1) = (1)2 = 1
∴ f(-1) = f(1)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there exists at least one value c of x such that f'(c) = 0 , where - 1 < c < 1
Now f'(c) = 0 gives us 2 c = 0
∴ c = 0 ∈ (-1, 1) ⇒ Rolle's Theorem is verified.
Let f(x) = x2 - 1
It is a polynomial in x
(a) Since every polynomial is a continuous function for every value of x
∴ f is continuous in [-1, 1]
(b) f'(x) = 2 x, which exists on (-1, 1)
∴ f is derivable in (-1, 1)
(c) f(-1) = (-1)2 - 1 = 1 - 1 = 0
f(1) = (1)2 - 1 = 1 - 1 = 0
∴ f(-1) = f(1)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where - 1 < c 1.
Now f'(c) = 0 gives us 2 c = 0 or c = 0
Now c = 0 lies in the open interval (-1, 1)
∴ Rolle's Theorem is verified
Here f(x) = x2 - 4 x + 3
It is a polynomial in x.
(a) Since every polynomial is a continuous function for every value of x
∴ f(x) is continuous in 1 ≤ x ≤ 3.
(b) f'(x) = 2 x, - 4, which exists in 1 < y < 3
∴ f(x) is derivable in 1 < x < 3.
(c) f(i) = (1)2 - 4 × 1 + 3 = 1 - 4 + 3 = 0
f(3) = (3)2 - 4 × 3 + 3 = 9 - 12 + 3 = 0
∴ f(1) = f(3)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there must exist at least one value c of x such that f'(c) = 0, where 1 < c < 3.
Now f'(c) = 0 gives us 2 c - 4 = 0 or, c = 2
Now c = 2 lies in the open interval (1, 3)
∴ Rolle's Theorem is verified
Let f(x) = x2 - 5 x + 4
(a) Since every polynomial is a continuous function for every value of x
∴ f is continuous in [1, 4]
(b) f'(x) = 2 x - 5, which exists in (1, 4)
∴ f is derivable in (1, 4)
(c) f(1) = 1 - 5 + 4 = 0
f(4) = 16 - 20 + 4 = 0
∴ f(1) = f(4)
∴ f satisfies all the condition of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where 1 < c < 4
∴ Rolle's theorem is verified.
Here f(x) = x (x2-4) = x3-4 x
It is a polynomial in x
(i) Since every polynomial in x is a continuous function for every value of x
∴ f(x) is continuous in [-2, 2]
(ii) f'(x) = 3 x2-4, which exists in (-2, 2)
∴ f(x) is derivable in (-2, 2)
(iii) f(-2) = 0, f(2) = 0
∴ f(-2) = f(2)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there must exist at least one real value of c such that f'(c) = 0 where - 2 < c < 2
∴ Rolle's Theorem is verified.
f(x) = x (x - 3)2 = x (x2 - 6x + 9) = x3 - 6 x2 + 9 x.
It is a polynomial in x.
(a) Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [0, 3].
(b) f'(x) = 3 x2 - 12 x + 9, which exists in (0, 3)
∴ f(x) is derivable in (0, 3).
(c) f(0) = 0, f(3) = 0
∴ f(0) = f(3)
∴ f(x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where 0 < c < 3.
f'(c) = 0 gives us 3 c2 - 12 c + 9 = 0 or c2 - 4 c + 3 = 0, i.e., (c - 1) (c - 3) = 0 i.e.. c = 1, 3
Now 3 ∉ (0, 3), but 1 ∈ (0, 3)
∴ Rolle's Theorem is verified.
Let f (x) = (x2 - 1) (x - 2) = x3 - 2 x2 - x + 2
It is a polynomial in x
(a) Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [-1, 2].
(b) f'(x) = 3 x2 - 4 x - 1, which exists in (-1, 2)
∴ f is derivable in (-1, 2).
(c) f(-1) = (1 - 1) (-1 -2) = 0
f(2) = (4 - 1) (2 - 1) (2 - 2) = 0
∴ f(-1) = f(2)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c ofx such that f'(c) = 0 where - 1 < c < 2.
Now f'(c) = 0 gives us 3 c2 - 4 c - 1 = 0
∴ Rolle's Theorem is verified.
Let f(x) = e1 - x2
(a) Since e1 - x2 is continuous in [-1, 1]
∴ f is continuous in [-1, 1]
(b) f'(x) = e1 - x2 (- 2 x) = - 2 x e1 - x2 , which exists in (-1, 1)
∴ is derivable in (-1, 1)
(c) f(-1) = e1 - 1 = e0 = 1
f(1) = e1 - 1 = e0 = 1
∴ f(-1) = f(1)
∴ satifies all the conditions of Rolle's Theorem.
∴ there must exist at least one value c of x such f' (c) = 0 where - 1 < c < 1
Now f' (c) = 0 gives us - 2 c e1-c2 = 0
⇒ c = 0 ∈ (-1, 1)
∴ Rolle's Theorem is verified.
Let f(x) = log (x2 + 2) - log 3
(a) Since log (x2 + 2) and log 3 are both continuous in [-1, 1]
∴ 1og (x2 + 2) - log 3 is continuous in [-1, 1] ⇒ f is continuous in [-1, 1]
∴ f is derivable in (-1, 1).
(c) f(-1) = log (1 + 2) - log 3 = 0
f(1) = log (1 + 2) - log 3 = 0
∴ f(-1) = f(1)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where - 1 < c < 1
∴ Rolle's Theorem is verified.
].
Here f(x) = sin2 x
(i) We know that sin x is continuous in [0, 
]
Now sin2x, being the product of two continuous function sin x and sin x, is continuous in [0, 
].
(ii) f'(x) = 2 sin x cos x, which exists in (0, 
)
∴ f(x) is derivable in (0, π).
(iii) f(0) = sin2 0 = 0, f(
) = sin2 π = 0
∴ f(0) = f(
)
∴ f{x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that
f'(c) = 0 where 0 < c < 
.
Show that the function 
is continuous but not differentiable at x=3.
f is continuous at x = 3.
For what value of k is the following function continuous at x = 2?
The given function f(x) will be continuous at x = 2, if
Thus, for k = 5, the given function is continuous at x = 2.
Differentiate the following function w.r.t. x:
y =
Now y = u + v
Taking logarithms on both sides, we have,
log u = x log ( sin x )
Differentiating with respect to x, we have,
From (i), (ii) and (iii)
We get,
If y =3 cos ( log x ) + 4 sin ( log x ), then show that
y = 3 cos ( log x ) + 4 sin ( log x )
Differentiating the above function with respect to x, we have,
Again differentiating the above function with respect to x, we have,
Find all points of discontinuity of f, where f is defined as following:
Here, f ( x ) =
The function is defined on all the points and hence continuous
possible points of discontinuity are 3 and -3 . We need to check the
continuity of the function at two points x = 3 and x = - 3 .
Case 1: For x = - 3, f ( - 3 ) = - ( - 3 ) + 3 = 6
So, f is continuous at x = -3
Case 2: For x = 3, f ( 3 ) = 6 ( 3 ) + 2 = 20
Therefore, function f is not continuous at point x = 3
Hence x = 3 is the only point of discontinuity.
Differentiate
Taking log on both sides,
log y = log ( xx cos x )
log y = x cos x log x
Differentiating with respect to x,
Differentiating with respect to x,
Adding (i) and (ii)
The area (in sq. units) of the region
is
B.
If a curve y=f(x) passes through the point (1, −1) and satisfies the differential equation, y(1+xy) dx=x dy, then f(-1/2) is equal to
-2/5
-4/5
2/5
4/5
D.
4/5
Given differential equation is
y( 1+ xy) dx = xdy
⇒ ydx + xy2 dx = xdy
If f and ga re differentiable functions in (0,1) satisfying f(0) =2= g(1), g(0) = 0 and f(1) = 6, then for some c ε] 0,1[
2f'(c) = g'(c)
2f'(c) = 3g'(c)
f'(c) = g'(c)
f'(c) = 2g'(c)
D.
f'(c) = 2g'(c)
Given, f(0) = 2 = g(1), g(0) and f(1) = 6
f and g are differentiable in (0,1)
Let h(x) = f(x)-2g(x) .... (i)
h(0) = f(0)-2g(0)
h(0) = 2-0
h(0) = 2
and h(1) = f(1)-2g(1) = 6-2(2)
h(1) = 2, h(0) = h(1) = 2
Hence, using rolle's theorem
h'(c) = 0, such that cε (0,1)
Differentiating Eq. (i) at c, we get
f'(c) -2g'(c) = 0
f'(c) = 2g'(c)
Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f′(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5).
Statement 1 is false, statement 2 is true
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
Statement 1 is true, statement 2 is false
C.
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
f(x) = 7 – 2x; x < 2
= 3; 2 ≤ x ≤ 5
= 2x – 7; x > 5
f(x) is constant function in [2, 5]
f is continuous in [2, 5] and differentiable in (2, 5) and f(2) = f(5)
by Rolle’s theorem f′(4) = 0
∴ Statement 2 and statement 1 both are true and statement 2 is correct explanation for statement 1.
The shortest distance between line y - x = 1 and curve x = y2 is
√3/4
3√2 /8
8/3√2
4/√3
A.
√3/4
B.
3√2 /8
y - x = 1
y2 = x
Let f : R → R be a continuous function defined
by f(x) = 1/ex + 2e-x
Statement - 1: f(c) = 1/3, for some c ∈ R.
Statement-2: 0 < f(x)≤ 
, for all x ∈ R.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
A.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
The equation of the tangent to the curve
, that is parallel to the x-axis, is
y = 0
y = 1
y = 3
y =2
C.
y = 3
Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then
f(6) ≥ 8
f(6) < 8
f(6) < 5
f(6) = 5
A.
f(6) ≥ 8
As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem
If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2 , x, y ∈ R and f(0) = 0, then f(1) equals
-1
0
2
1
B.
0
let 
. If f(x) is continuous in 
then f (π/4) is
1
-1/2
-1
1/2
B.
-1/2
A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is
(x-1)2
(x-1)3
(x+1)3
(x+1)2
B.
(x-1)3
f''(x) = 6(x - 1) f'(x) = 3(x - 1)2 + c ........ (i)
At the point (2, 1) the tangent to graph is y = 3x - 5 Slope of tangent = 3
∴ f'(2) = 3(2 - 1)2 + c = 3
3 + c = 3
⇒ c = 0
∴ From equation (i) f'(x) = 3(x - 1)2
f'(x) = 3(x - 1)2 f(x) = (x - 1)3 + k ...... (ii)
Since graph passes through (2, 1)
∴ 1 = (2 - 1)2 + k k = 0
∴ Equation of function is f(x) = (x - 1)3
The differential equation for the family of curves,x2 +y2 -2ay = 0 where a is an arbitrary constant is
2(x2-y2)y' = xy
(x2+y2)y' = xy
2(x2+y2)y' = xy
(x2- y2)y' = xy
A.
2(x2-y2)y' = xy
2x + 2yy′ - 2ay′ = 0
The solution of the differential equation ydx + (x + x2y) dy = 0 is
-1/ XY =C
-1/XY + log y = C
1/XY + log y = C
log y =Cx
B.
-1/XY + log y = C
y dx + x dy + x2y dy = 0
If then the
standard deviation of the 9 items x1, x2, ...., x9 is
3
9
4
2
D.
2
Standard deviation of xi - 5 is
As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.
so σ of xi is 2
Sponsor Area
Sponsor Area
?
continous at x=1 ?
W.r.t x.
w.r.t x.
w.r.t x.
w.r.t x.
w.r.t x.
w.r.t x
w.r.t x.
w.r.t x.
w.r.t x.
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...')
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#6 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#7 {main}</pre>](/application/zrc/images/qvar/MAEN12078068.png)
, x>0
.
.
w.r.t.x.
and express the result free from trigonometrical symbols.
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
#2 [internal function]: _hx_lambda->execute('Http Error #502')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #502')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #502')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #502')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12079484.png)
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
#2 [internal function]: _hx_lambda->execute('Http Error #404')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12079690-1.png)
is continuous at x = 0.
is equal to
. if p (0) = 850, then the time at which the population becomes zero is
equal to
, where c > 0, is a parameter, is of order and degree as follows:
then the solution of the equation is
