CBSE > Class 11 > Mathematics | CBSE Board | Homework Help

Find the critical points, here they are 1 and 3.
solve for x<1,
-(x-1)-(-(x-3))≥0
-x+1+x-3≥0
-2≥0 (not possible)
Thus, x should be greater than 1
For 1 ≤ x ≤ 3
(x-1)-(-(x-3))≥0
2x-4≥0
x≥2
So one solution set is 2≤x≤3
For x>3
(x-1)-(x-3)≥0
2≥0
Therefore from above two results,
x≥2

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ANSWER

Find the sum to n terms of the sequence 5+55+555... to n terms

Posted by Jasleen Kaur 4 years, 2 months ago

CBSE > Class 11 > Mathematics

2 answers

Raunak Chauhan 4 years, 2 months ago

Answer : 5+ 55+555+......n term 5(1+11+111+........n term) Dividing and multliply by 9 5/9(9+99+999.....n term) 5/9{(10-1)+(100-1)+1000-1)... n term} 5/9{(10+10²+10³+... n term)-(1+1+1+...n term)} 5/9[{10(10*n-1)/10-1)}-n] 5/81{10(10*n-1)-n

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Gaurav Seth 4 years, 2 months ago

Answer:

Sum = 5 + 55 + 555 + …. n terms.
= 5/9 [9 + 99 + 999 + …. n terms]
= 5/9 [(10 – 1) + (100 – 1) + (1000 – 1) + … n terms]
= 5/9 [10 + 100 + 1000 ….. – (1 + 1 + … 1)]
= 5/9 [10(10ⁿ – 1)/(10 – 1) + (1 + 1 + … n times)]
= 50/81(10ⁿ – 1) – 5n/9

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ANSWER

5542-111

Posted by Mansi Pareek 4 years, 2 months ago

CBSE > Class 11 > Mathematics

4 answers

Naveen Kumar 4 years, 2 months ago

5432

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Anju Ramesh Kumar 4 years, 2 months ago

Calculator nhi tha kya apke pass ?

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Gaur Saab?? 4 years, 2 months ago

5431??

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Śěřãj The Cute? 4 years, 2 months ago

5431 How simple ???

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ANSWER

Find the lines through (2,-1) which cut the axes so that the intercepts are equal in magnitude

Posted by Prabh Prabh 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Sruti Mohanty 4 years, 2 months ago

As the intercepts are equal therefore x+y=a. As the equation passes through points(2,-1), a=2-1=1 Therefore the required equation is x+y=1

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ANSWER

Flower

Posted by Swaradhya E.Enandan Verma 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Śěřãj The Cute? 4 years, 2 months ago

Variety ????

2Thank You

ANSWER

Which term of the A. P. 15, 20, 25, ... is 115?

Posted by Himanshu Sharma 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Anmol Agarwal 4 years, 2 months ago

Given, a=25 d=20−25=−5 Tn<0 a+(n−1)d<0 25+(n−1)(−5)<0 25<5(n−1) 5<(n−1) n>6 ⇒n=7 7th term is negative. a7=25+(7−1)(−5) =25−5×6 =−5

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ANSWER

Find the domain of f(x)=√x+2 +1/√10-x

Posted by Harshita Bansal❤️ 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Deepanshu Tyagi 4 years, 2 months ago

In this type of equations ,if we have to find out domain we have to solve the denominator- So, => √10-x=0 => [x=√10] Domain=set of real no.s except √10.

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ANSWER

What is property of modolus function. With example

Posted by Sadique Ansari 3 years, 10 months ago

CBSE > Class 11 > Mathematics

1 answers

Sia ? 3 years, 10 months ago

The graph of modulus function is continuous having a corner at x=0. It means that graph is differentiable at all points except at x=0 (we can not draw a tangent at a corner). Since graph is symmetric about y-axis, modulus function is even function. We also see that there are pair of x values for non-zero values of y.

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ꜰɪɴᴅ ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏꜰ ꜱᴜʙꜱᴇᴛꜱ ᴏꜰ ᴀ×ʙ ᴡʜᴇʀᴇ ᴀ ={3,6,9} ᴀɴᴅ ʙ={x,ʏ,ᴢ}

Posted by .... .... 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Nisha Kashyap 4 years, 2 months ago

No of relation=3×3=9,and no of subsets of A×B is 9^2=81

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Value of tan pi/8

Posted by सत्य सनातन? 4 years, 2 months ago

CBSE > Class 11 > Mathematics

3 answers

Nisha Kashyap 4 years, 2 months ago

Pi=180 so ans is180/8

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Sruti Mohanty 4 years, 2 months ago

(Square root of 2)-1

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Sruti Mohanty 4 years, 2 months ago

_/2-1

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ANSWER

Find the domain of f(x)=√x+2+1/√10-x

Posted by Harshita Bansal❤️ 4 years, 2 months ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Evaluate cosec (-1200°)

Posted by Harshita Bansal❤️ 4 years, 2 months ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Value of iota

Posted by Ishika Dixit 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Gaurav Seth 4 years, 2 months ago

he square root of a negative real number is called an imaginary quantity or imaginary number. e.g., √-3, √-7/2

The quantity √-1 is an imaginary number, denoted by ‘i’, called iota.

Integral Powers of Iota (i)

i=√-1, i² = -1, i³ = -i, i⁴=1

So, i⁴ⁿ⁺¹= i, i⁴ⁿ⁺² = -1, i⁴ⁿ⁺³ = -i, i⁴ⁿ⁺⁴ = i⁴ⁿ = 1

In other words,

iⁿ = (-1)^n/2, if n is an even integer
iⁿ = (-1)^(n-1)/2.i, if is an odd integer

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ANSWER

. Sn denotes the sum of the first n terms of AP. If S2n = 5 Sn, then prove that ??? ??? = ??/4

Posted by Nik . 4 years, 2 months ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

in a survey of 25 students, it was found that 15 students bad taken Mathematics, 12 had taken Physics, I had taken chemistry, 9 had taken Mathematics and Physics, 3had taken all the three subjects, 4 had taken Physics and Chemistry. find the number of students that had ci, Only chemistry Only Mathematics. cii) Only Physics. cin physics and Chemistry but not Mathematics. > Mathematics and Physics but not Chemistry vi) Only one of the subjects. vil) At least one of the three subjects.

Posted by Aleena Philip Shaji 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Gaurav Seth 4 years, 2 months ago

Total no. of student=25

Let n(M)=student who studying mathematics

n(C)=student who studying chemistry

n(P)=student who studying Physics

∴n(M)=15,n(P)=12,n(C)=11,n(M∩P)=9,n(M∩C)=5,n(P∩C)=4,n(M∩P∩C)=3

Number of student who studying mathematics and physic but not chemistry

⇒n(M∩P)−n(M∩P∩C)

⇒9−3=6

Number of student who studying chemistry and physic but not mathematics

⇒n(P∩C)−n(M∩P∩C)

⇒4−3=1

Number of student who studying mathematics and chemistry but not physics

⇒n(M∩C)−n(M∩P∩C)

⇒5−3=2

∴No.of student who had taken two of the three subjects=6+1+2=9

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ANSWER

show that the locus of a point which moves in such a way that the square of its distance from base of an isosceles triangle is equal to the rectangle under its distance from the other side is a circle

Posted by Komal Yadav 4 years, 2 months ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Write the given sets in the set builder form 4,8,12,16,20

Posted by Tilak Gupta 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Tec Om 4 years, 2 months ago

A={x: 4 * n where n belongs to N; 1<=x<6}

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ANSWER

Express 2cos4x sin2x

Posted by Tilak Gupta 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

First Name 4 years, 2 months ago

=sin(6x) + sin(2x)

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ANSWER

Can i find maths exampler questions here?

Posted by Siddiqua Rahman 4 years, 2 months ago

CBSE > Class 11 > Mathematics

2 answers

Sruti Mohanty 4 years, 2 months ago

Yes the questions are there only

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Śěřãj The Cute? 4 years, 2 months ago

May be not.

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ANSWER

Deleted topics of maths?please koi points Mai likh ke dedo

Posted by Gaur Saab?? 4 years, 2 months ago

CBSE > Class 11 > Mathematics

4 answers

Gaur Saab?? 4 years, 2 months ago

Yes Gaur saab?

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?Ritesh Gupta? 4 years, 2 months ago

Gaur saab..???

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Gaur Saab?? 4 years, 2 months ago

Thanks Aadya

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Aadya Singh 4 years, 2 months ago

</a>

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ANSWER

What is deleted syllabus of maths???please reply!!

Posted by Gaur Saab?? 4 years, 2 months ago

CBSE > Class 11 > Mathematics

3 answers

Gaur Saab?? 4 years, 2 months ago

Oh Bhai usme deleted topics kha hai

1Thank You

Gaur Saab?? 4 years, 2 months ago

Mujhe toh mila nhi

2Thank You

Gaur Saab?? 4 years, 2 months ago

Where

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ANSWER

If tan(a-b)=sin2b/5-cos2b, then find tan a : tan b

Posted by Soham Biswas 4 years, 2 months ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

In a circle of diameter 40cm, the length of chord is 20cm . Find the length of minor arc of the chord.

Posted by Dikshant Shokeen 4 years, 2 months ago

CBSE > Class 11 > Mathematics

2 answers

Gaurav Seth 4 years, 2 months ago

Given : a circle of diameter 40cm,the length of a chord is 20 cm,

To Find : the length of the minor arc of the chord

Solution:

Chord Length = 20 cm

Diameter = 40 cm

Perpendicular from center on chord bisect the chord

Hence Half of chord = 10 cm

Radius = 40/2 = 20 cm

Sin ( 1/2 of chord angle ) = 10 /20

=> Sin ( 1/2 of chord angle ) = 1/2

=> Sin ( 1/2 of chord angle ) = Sin 30°

=> 1/2 of chord angle = 30°

=> chord angle = 60°

Minor arc angle = 60°

or another way to get angle

as Radius = 20 cm

Chord length = 20 cm

Hence it forms an Equilateral triangle

Hence angle formed by chord at center = 60°

Minor arc angle = 60°

length of the major arc of the chord = (60/360) * 2π * Radius

= ( 1/6 ) * 2π * 10

= 10π/3

= 50 * 3.14/3

= 10.47 cm

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Yogita Ingle 4 years, 2 months ago

Diameter of the circle =40 cm
∴ Radius (r) of the circle = 40/2 =20 cm
Let AB be a chord (length = 20 cm) of the circle.
In △OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, △OAB is an equilateral triangle.
∴θ=60^∘=π/3 radian
We know that in a circle of radius r unit, if an arc of length I unit subtends an angle θ radian at the centre, then θ=l/r
∴π/3 = arc AB/20

⟹arc AB = 20/ 3π cm

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ANSWER

Derivative of secx

Posted by सत्य सनातन? 4 years, 2 months ago

CBSE > Class 11 > Mathematics

3 answers

Korou Khundrakpam 4 years, 2 months ago

You will get tanx*secx

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Korou Khundrakpam 4 years, 2 months ago

Chain rule*

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Korou Khundrakpam 4 years, 2 months ago

Put secx=1/cosx, the apply chai rule

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ANSWER

By using the concept of equation of a line,prove that the three points (3,0) (-2,-2) and (8,2) are collinear.

Posted by Priyanka Gurjar 4 years, 2 months ago

CBSE > Class 11 > Mathematics

1 answers

Gaurav Seth 4 years, 2 months ago

In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (–2, –2) is

It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.

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ANSWER

If x²-2x- 3<0,then X lies in the interval which

Posted by Yashansh😎😇😇 Jii 4 years, 2 months ago

CBSE > Class 11 > Mathematics