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Asked on 15 Apr CBSE/Class 10/Mathematics
I want to get tution as a tutors in maths,sst
Lesson Posted on 13 Apr CBSE/Class 10/Mathematics
Manibhushan Kumar
I have 15 years of experience in Maths and Stats.I have started teaching 1 year after completing my graduate...
Lesson Posted on 13 Apr CBSE/Class 10/Mathematics
Sudhansu Bhushan R.
I am teaching Physics, Chemistry and Mathematics for Class  IX and class X in respect of I.C.S.E.,...
The log function
Background: Any positive number, y, can be written as 10 raised to some power, x. We can write this relationship in equation form:
y = 10^{ x}
For example it is obvious that 1000 can be written as 10^{ 3}. It is not so obvious that 16 can be written as 10^{ 1.2}. How do we know that this is the correct power of 10? Because we get it from the graph shown below.
To make this graph we made a table of a few obvious values of y = 10^{ x} as shown below, left. Then we plotted the values in the graph (they are the red dots) and drew a smooth curve through them. Then we observed that the curve went through y = 16 and x = 1.2 (the black dot). This means that 16 = 10^{ 1.2}.
We next define a function called the logarithm that takes a number like 16 as input, calculates that it can be written as 10^{ 1.2}, and returns the exponent 1.2 as its output value. Here is the formal definition.

Example 1: Evaluate log ( 10^{ 5.7} ). In this example the argument of the log function is already expressed as 10 raised to an exponent, so the log function simply returns the exponent:
log ( 10^{ 5.7} ) = 5.7
Example 2: Evaluate log ( 1000 ). The argument is a number that is easily expressed as 10 raised to an exponent. We do this and the log function then returns the exponent:
log ( 1000 ) = log ( 10^{ 3} ) = 3
Example 3: Evaluate log ( 16 ). The argument is a number which we don't know how to express as 10 raised to an exponent (unless we remember the above discussion which said that 16 = 10^{ 1.2} ). Therefore we use a calculator or the Algebra Coach to evaluate it:
log ( 16 ) = 1.2
Example 4: log ( x ). The argument is an expression. Until we can evaluate that expression we must leave this logarithm as is.
Graph: The blue curve shown to the right is the graph of the log function. Notice that for any positive x it returns a single value. For any negative x it is undefined. If you compare this graph of the log function to thegraph of y = 10^{ x} then you see that one can be gotten from the other by interchanging the x and y axes.
For comparison we have also shown the graph of the ln function (the natural log function) in red. The ln graph has the same shape as the log graph but is 2.3 times as tall. The ln function is described below.
An important feature of the log function is that it increases very slowly as x becomes very large. It describes nicely how the human ear percieves loud sounds and the way the human eye percieves bright lights.
Domain and range: The domain of the log function is all positive real numbers and the range is all real numbers.
The log function can be extended to the complex numbers, in which case the domain is all complex numbers except zero. The logarithm of zero is always undefined.
Some special values of the log function
Solving the Equation 10^{ x} = y for x by using the log function
Suppose that x is unknown but that 10^{ x} equals a known value y. Then finding x requires solving the following equation for x:
10^{ x} = y
The solution is:
x = log (y)
This is because finding log (y) means expressing y as 10 to an exponent, and then returning that exponent. But the original equation says that that exponent is x. Note that if y is negative then there is no real solution. However there is a complex solution. Furthermore if y = 0 then there is no solution at all.
log (x) and 10^{ x} are inverse functions
Consider the 10^{ x} function which takes x and returns 10^{ x}, like this:
The log function is defined to do exactly the opposite, namely:
Therefore these are inverse functions.
Note the following:
10^{ log( y)} = y
and if we replace y in the second equation by the y of the first equation we get this identity:
x = log (10^{ x })
These identities are useful for showing how the log and antilog functions cancel each other.
How to use the log function in the Algebra Coach
Algorithm for the log function
Click here to see the algorithm that computers use to evaluate the log function.
The ln function
Background: You might find it useful to read the previous section on the log function before reading this section. Recall that the log function takes a number like 16 as input, calculates that it can be written as 10^{ 1.2}, and then returns the exponent 1.2 as its output value. But why use base 10? After all, probably the only reason that the number 10 is important to humans is that they have 10 fingers with which they first learned to count. Maybe on some other planet they use base 8!
In fact probably the most important number in all of mathematics (click here to see why) is the number 2.7182818284590…, which we give the name e, in honor of Leonard Euler, who first discovered it. It will be important to be able to take any positive number, y, and express it as e raised to some power, x. We can write this relationship in equation form:
y = e^{ x}
For example 5 can be written as e^{ 1.6} (the exponent is approximate). How do we know that this is the correct power of e? Because we get it from the graph shown below.
To make this graph we made a table of a few obvious values of y = e^{ x} as shown below, left. Then we plotted the values in the graph (they are the red dots) and drew a smooth curve through them. Then we observed that the curve went through y = 5 and x = 1.6 (the black dot). This means that 5 = e^{ 1.6}.
If you compare this graph to that of y = 10^{ x} you see that both have the same socalled exponential growth shape but that this graph grows more slowly.
We next define a function called the natural logarithm that takes a number like 5 as input, calculates that it can be written as e^{ 1.6}, and returns the exponent 1.6 as its output value. Here is the formal definition.

From now on, to avoid confusion, we will call ln(x) the natural logarithm function and log(x) the base 10 logarithm function.
Example 1: Evaluate ln ( e^{ 4.7} ). In this example the argument of the ln function is already expressed as e raised to an exponent, so the ln function simply returns the exponent:
ln ( e^{ 4.7} ) = 4.7
Example 2: Evaluate ln ( 8.6 ). The argument is a number which we don't know how to express as e raised to an exponent . Therefore we use a calculator or the Algebra Coach to evaluate it:
ln ( 8.6 ) = 2.15
Example 3: ln ( x ). The argument is an expression. Until we can evaluate that expression we must leave this natural logarithm as is.
Example 4: Evaluate ln ( e ). Express the argument as e raised to the exponent 1 and return the exponent:
ln ( e ) = ln ( e^{ 1} ) = 1
Example 5: Evaluate ln ( 1 ). Express the argument as e raised to the exponent 0 and return the exponent:
ln ( 1 ) = ln ( e^{ 0} ) = 0
Graph: The red curve shown to the right is the graph of the ln function. Notice that for any positive x it returns a single value. For any negative x it is undefined. If you compare this graph of the ln function to thegraph of y = e^{ x} then you see that one can be gotten from the other by interchanging the x and y axes.
For comparison we have also shown the graph of the base 10 log function in blue. The log graph has the same shape as the ln graph but is only 43% as tall.
Domain and range: The domain of the ln function is all positive real numbers and the range is all real numbers.
The ln function can be extended to the complex numbers, in which case the domain is all complex numbers except zero. The natural logarithm of zero is always undefined.
Solving the Equation e^{ x} = y for x by using the ln function
Suppose that x is unknown but that e^{ x} equals a known value y. Then finding x requires solving this equation for x:
e^{ x} = y
The solution is:
x = ln (y)
because finding ln (y) means expressing y as e to an exponent, and then returning that exponent. But the original equation says that that exponent is x. Note that if y is negative then there is no real solution. However there is a complex solution. Furthermore if y = 0 then there is no solution at all.
ln (x) and e^{ x} are inverse functions
Consider the e^{ x} function which takes x and returns e^{ x}, like this:
The ln function is defined to do exactly the opposite, namely:
Therefore these are inverse functions.
Note the following:
e^{ ln (y)} = y
and if we replace y in the second equation by the y of the first equation we get this identity:
x = ln (e^{ x})
These identities are useful for showing how the ln and e^{ x} functions cancel each other.
How to use the ln function in the Algebra Coach
Algorithm for the ln function
Click here to see the algorithm that computers use to evaluate the natural logarithm function.
The exp or e^{ x} function
Background: You might find it useful to read the previous section on the ln function before reading this section. There we saw that it is possible to use the number e (which is approximately 2.7182818284590…) as a base and to raise it to any power, x, and produce any positive number y. We can write this relationship in equation form:
y = e^{ x}
Here is a graph of y = e^{ x} (the blue curve). For comparison we also show graphs of y = 2^{ x} and y = 4^{ x}. Because the number e is between 2 and 4 the curve y = e^{ x} lies between the curves y = 2^{ x} and y = 4^{ x}.
All three of these curves are called exponential functions because the independent variable x is in the exponent. All three have the property that the higher up the curve you go the steeper they get. However y = e^{ x} has the special property that at every point along the curve the slope equals the height. This property concerning slopes makes it a very important function in calculus.

Graph: The blue curve is the graph of y = e^{ x} (i.e. of the exp function). It has the property that its slope equals its height everywhere. The dotted red lines show the slope of the curve at various points along the curve. Notice that the slope is 5 when the height is 5, and so on.
If you compare this graph of the exp function to the graph of the ln function then you see that one can be gotten from the other by interchanging the x and y axes.
Domain and range: The domain of the exp function is all real numbers and the range is all positive real numbers.
The exp function can be extended to the complex numbers, in which case the domain and the range is all complex numbers.
Solving the Equation ln (x) = y for x by using the e^{ x} or exp function
Suppose that x is unknown but that ln (x) equals a known value y. Then finding x requires solving this equation for x:
ln (x) = y.
The solution is:
x = e^{ y}, or x = exp (y).
This was explained in the previous section on the ln function.
ln (x) and e^{ x} are inverse functions
This was explained in the previous section on the ln function.
How to use the exp function in the Algebra Coach
Note: You never need to use the functional notation exp (x) in the Algebra Coach. You can always use the notation e^{ x} instead. In fact the default setting for the exponential function option is exp (x) → e^{ x}.
Algorithm for the e^{ x} or exp function
Click here to see the algorithm that computers use to evaluate e^{ x} or exp (x).
Lesson Posted on 03 Apr CBSE/Class 10/Mathematics
Lino P Thomas
I can teach each and every topics very easily and make it stronger for the students. I have an experience...
Hello Students,
Here I am for you to provide some simple and basic lessons on how to deal with trigonometry
Hopefully, You all would be aware of Sine, Cosine, Tangent Functions and their reciprocals.
We can easily understand these functions through drawing a triangle ABC and marking an angle (say at A) inside it.
Then,
Sin A = Opposite side / Hypotenuse
Cos A = Adjacent Side / Hypotenuse
Tan A = Opposite side / Adjacent side
Cosec A, Sec A, Cot A would be their reciprocals respectively.
Hope you guys are clear in this session.
Stay tuned for more .. If you like this, please provide your valuable feedback.
Thanks!!
read lessAnswered on 02 Apr CBSE/Class 10/Mathematics
How to explain trigonometry application
Ayush Ranjan
Facilitator
In physics, trigonometry is used to find the components of vectors, model the mechanics of waves (both physical and electromagnetic) and oscillations, sum the strength of fields, and use dot and cross products. Even in projectile motion you have a lot of application of trigonometry.
read lessLesson Posted on 26 Mar CBSE/Class 10/Mathematics
CHAPTER10 CIRCLES NCERT SOLUTIONS
Rahul S
I am an experienced, qualified teacher and tutor with over 4 years of experience in teaching Maths ,...
Exercise: 10.1
Answer
A circle can have infinite tangents.
(i) A tangent to a circle intersects it in ............... point(s).
(ii) A line intersecting a circle in two points is called a .............
(iii) A circle can have ............... parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ............
Answer
(i) one
(ii) secant
(iii) two
(iv) point of contact
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm
Answer
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
By Pythagoras theorem in ΔOPQ,
OQ^{2} = OP^{2} +^{ }PQ^{2}
⇒ (12)^{2 }= 5^{2} + PQ^{2}
⇒PQ^{2} = 144  25
⇒PQ^{2} = 119
⇒PQ = √119 cm
(D) is the correct option.
other, a secant to the circle.
Answer
AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.
Exercise: 10.2
In Q.1 to 3, choose the correct option and give justification.
(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Answer
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
By Pythagoras theorem in ΔOPQ,
OQ^{2} = OP^{2} +^{ }PQ^{2}
⇒ (25)^{2 }= OP^{2} + (24)^{2}
⇒OP^{2} = 625  576
⇒OP^{2} = 49
⇒OP = 7 cm
The radius of the circle is option (A) 7 cm.
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60° (B) 70° (C) 80° (D) 90°
Answer
OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
⇒ ∠PTQ + 90° + 110° + 90° = 360°
⇒ ∠PTQ = 70°
∠PTQ is equal to option (B) 70°.
Answer
OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option (A) 50°
Answer
Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B
respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,
∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer
Let AB be the tangent to the circle at point P with centre O.
We have to prove that PQ passes through the point O.
Suppose that PQ doesn't passes through point O. Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
PQ intersect CD at R and also intersect AB at P.
AS, CD // AB PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90° (PQ ⊥ AB)
⇒ ∠ORP = 90°
∠ROP + ∠OPA = 180° (Cointerior angles)
⇒∠ROP + 90° = 180°
⇒∠ROP = 90°
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.
Answer
AB is a tangent drawn on this circle from point A.
∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA^{2} =^{ }AB^{2 }+ BO^{2}
⇒ 5^{2 }= 4^{2 }+ BO^{2}
⇒ BO^{2} = 25  16
⇒ BO^{2} = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.
7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
Answer
Let the two concentric circles with centre O.
AB be the chord of the larger circle which touches the smaller circle at point P.
∴ AB is tangent to the smaller circle to the point P.
⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA^{2} = AP^{2} + OP^{2}
⇒ 5^{2} = AP^{2} + 3^{2}
⇒ AP^{2} = 25  9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Answer
From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC
Answer
We joined O and C
A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
Answer
Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the linesegment joining the points of contact at the centre.
Answer
ABCD is a parallelogram,
∴ AB = CD ... (i)
∴ BC = AD ... (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) =
(DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.
DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Answer
In ΔABC,
Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Now semi perimeter of circle s,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
Area of ΔABC = √s (s  a)(s  b)(s  c)
= √(14 + x) (14 + x  14)(14 + x  x  6)(14 + x  x  8)
= √(14 + x) (x)(8)(6)
= √(14 + x) 48 x ... (i)
also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)
Equating equation (i) and (ii) we get,
√(14 + x) 48 x = 56 + 4x
Squaring both sides,
48x (14 + x) = (56 + 4x)^{2}
⇒ 48x = [4(14 + x)]^{2}/(14 + x)
⇒ 48x = 16 (14 + x)
⇒ 48x = 224 + 16x
⇒ 32x = 224
⇒ x = 7 cm
Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm
Answer
Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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Lesson Posted on 08 Mar CBSE/Class 10/Mathematics CBSE/Class 12/Science/Mathematics
Honesty and Dedication helps in education and success in life.
Jagdish S.
Mathematics teaching for CBSE ICSE,PUC state CET,JEE IIT and Engineering maths ,CAT,MAT,International...
Honesty and Dedication helps in education and success in life.
As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel happiness. Get less mark but with honesty that will make you feel happiness.
read lessAnswered on 02 Apr CBSE/Class 10/Mathematics Tuition/Class IXX Tuition
Sushil K.
Work hard till your signature becomes an autograph
Let the marks of mathematics be x
Than marks of science be 28x
now according to question
(x+3),
(28x4)
The product of new marks will be 180
So,
(x+3) (28x4)=180
24xx^2+723x=180
x^221x+108=0
x^212x9x+108=0
x(x12)9(x12)=0
(x9)(x12)=0
(x9)=0 than x=9
(x12)=0than x=12
If 9 marks got in math than 19 marks got in science
If 12 marks got in math than 16 marks got in science
read lessAnswered on 25 Feb CBSE/Class 10/Mathematics Tuition/Class IXX Tuition
Salman Ahmad Siddiqui
Maths Tutor
Answered on 08 Mar CBSE/Class 10/Mathematics Tuition/Class IXX Tuition
Aravind Kumar
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