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CO-ORDINATE GEOMETRY
To specify the position of a point in space, we use rectangular
axes coordinate system. This system consists of ( i ) origin ( ii ) axis
or axes. If point is known to be on a given line or in a particular
direction, only one coordinate is necessary to specify its position.
If it is in a plane, two coordinates are required. If it is in space
three coordinates are needed.
Origin
This is any fixed point which is convenient to you. All measurement
are taken w.r.t. this fixed point.
Axis or Axes
Any fixed direction passing through origin and convenient to you
can be taken as an axis. If the position of a point or position of all
the points under consideration always happen to be in a particular
direction, then only one axis is required. This is generally called
the x -axis. If the positions of all the points under consideration
are always in a plane, two perpendicular axes are required. These
are generally called x ansd y -axis. If the points are distributed in
a space, three perpendicular axes are taken which are called x , y
and z -axis.
Position of a point in xy plane
The position of a point is specified by its distances from origin
along (or parallel to) x and y -axis as shown in figure. Here
x -coordinate and y -coordinate is called abscissa and ordinate
respectively.
x
y
y
x x
( x,y )
origin
y
Distance Formula
The distance between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by
2 2 d = x 2 (^) − x 1 (^) + y 2 (^) − y 1
Example 1: Find value of a if distance between the points
(–9 cm, a cm) and (3 cm, 3 cm) is 13 cm.
Sol. By using distance formula ( ) ( )
2 2 d = x 2 (^) − x 1 (^) + y 2 (^) − y 1
⇒ (^) ( ) [ ]
(^2 ) 13 = 3 − − 9 + 3 − a
⇒ 132 = 12^2 + (3 – a )^2 ⇒ (3 – a )^2 = 13^2 – 12^2 = 5^2
⇒ (3 – a ) = ± 5 ⇒ a = –2 cm or 8 cm
Example 2: A dog wants to catch a cat. The dog follows
the path whose equation is y – x = 0 while the cat follows
the path whose equation is x^2 + y^2 = 8. The coordinates of
possible points of catching the cat are:
Sol. Let catching point be ( x 1 , y 1 ) then, y 1 – x 1 = 0 and
x 12 + y 12 = 8
Therefore, 2 x 12 = 8 ⇒ x 12 = 4 ⇒ x 1 = ±2 ; So possible
points are (2, 2) and (–2, –2).
Example 3: Distance between two points ( 8, – 4) and
(0, a ) is 10. All the values are in the same unit of length.
Find the positive value of a.
Sol. From distance formula (8 – 0)^2 + (–4 – a )^2 = 100
⇒ (4 + a ) 2 = 36 ⇒ a = 2 or –
Train Your Brain
1. Find the distance of point (–12, 5) from the origin. 2. Find the distance between points
( i ) (5, 4) and (6, 7)
( ii ) (–7, 3) and (7, –6)
3. Find the value of a if distance between ( a , 3) cm and (–2, 6) cm is 5 cm.
Concept Application
Mathematical Tools
CHAPTER
Mathematical Tools 41
TRIGONOMETRY
Angle
it is measure of change in direction.
Radius
arc Radius
O
r
r
s
q
Angle (θ) =
Arc
Radius
s
r
Angles measured in anticlockwise
and clockwise direction are
usually taken positive and negative
respectively.
System of measurement of an angle
(A) Sexagesimal system:
In this system, angle is measured in degrees.
In this system, 1 right angle = 90°, 1° = 60′ (arc minutes),
1 ′ = 60′′ (arc seconds)
(B) Circular system:
In this system, angle is measured in radian. If arc = radius
then θ = 1 rad
Relation between degrees and radian
2π rad = 360° ⇒ π rad = 180° ⇒ 1 rad =
π
To convert from degree to radian multiply by 180
π
°
To convert from radian to degree multiply by
π
Example 4: A circular arc of length π cm. Find angle
subtended by it at the centre in radian and degree.
6 cm
6 cm
q^ p^ cm
Sol.
cm rad = 30° 6 cm 6
s
r
π π θ = = =
Example 5: The moon’s distance from the earth is 360000 km
and its diameter subtends an angle of 42′ at the eye of the
observer. The diameter of the moon in kilometers is
(1) 4400 (2) 1000
(3) 3600 (4) 8800
Sol. Here angle is very small so diameter ≈ arc
42 42 42 rad 60 60 180 1800
^ π^ π θ = ′= (^) × (^) = × × =
Diameter =
360000 4400 km 1800 7
R θ = × × =
Train Your Brain
4. Convert 15° into radians 5. Convert
π radians into degrees
Concept Application
Trigonometric Ratios (T-ratios)
Following ratios of the sides of a right angled triangle are known as trigonometrical ratios.
Base ( B )
Perpendicular (
P
Hypotenuse (
H
q
sin
P
H
θ =
cos
B
H
θ =
tan
P
B
θ =
cosec sin
H
P
θ = = θ
1 sec cos
H
B
θ = = θ
cot tan
B
P
θ = = θ
Trigonometric Identities
In figure, P^2 + B^2 = H^2 Divide by H^2 ,
2 2 1
P B
H H
+^ =
⇒ sin^2 θ+ cos^2 θ = 1
Divide by B^2 ,
2 2 1
P H
B B
+^ =
2 2 ⇒ 1 + tan θ = sec θ
Divide by P^2 ,
2 2 1
B H
P P
2 2 ⇒ 1 + cot θ = cosecθ
Commonly Used Values of Trigonometric Functions
Angle (θ) 0° 30° 37° 45° 53° 60° 90°
sin θ 0
cos θ 1
tan θ 0
Four Quadrants and ASTC Rule
In first quadrant, all trigonometric ratios are positive. In second quadrant, only sinθ and cosecθ are positive. In third quadrant, only tanθ and cotθ are positive.
Mathematical Tools 43
6. Evaluate
( i ) sin(210°) ( ii ) cos120°
( iii ) sin(330°) ( iv ) tan(300°)
7. If tanθ is
find
( i ) sinθ ( ii ) cosθ
Concept Application
A few important trigonometric formulae:
1. Addition Formulae:
( a ) sin ( A + B ) = sin A cos B + cos A sin B
( b ) cos ( A + B ) = cos A cos B – sin A sin B
( c ) tan ( A + B ) =
tan tan
1 tan tan
A B
A B
2. Subtraction Formulae:
( a ) sin ( A – B ) = sin A cos B – cos A sin B
( b ) cos ( A – B ) = cos A cos B + sin A sin B
( c ) tan ( A – B ) =
tan tan
1 tan tan
A B
A B
3. Multiplication Formulae:
( a ) sin 2 A = 2 sin A cos A
sin A = (^) 2 sin cos 2 2
A A
( b ) cos 2 A = cos^2 A – sin^2 A = 2 cos^2 A – 1 = 1 – 2 sin^2 A
cos A = cos^2 2
A
A
( c ) tan 2 A = 2
2 tan
1 tan
A
− A
tan A = 2
2 tan 2
1 tan 2
A
A
Small Angle Approximation
If θ is small in radians then sin θ ≈ θ, cos θ ≈ 1 and tanθ ≈ θ
Note: here θ must be in radian.
Smaller the value of θ, better will be approximation
Maximum and Minimum Values of Some useful Trigonometric
Functions
( i ) − 1 ≤ sin θ ≤ 1
( ii ) − 1 ≤ cos θ ≤ 1
( iii ) 2 2 2 2 − a + b ≤ a cos θ ± b sinθ ≤ a + b
Example 9: Find the value of
( i ) sin 74° ( ii ) cos 106°
( iii ) sin 15° ( iv ) cos 75°
Sol. ( i ) sin 74° = sin (2 × 37°) = 2 sin 37° cos 37°
( ii ) cos 106° = cos (2 × 53°) = cos 2 53° – sin 2 53° 2 2 3 4 9 16 7
5 5 25 25
^ −
( iii ) sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45°
sin 30 2 2 2 2 2 2
° = × − × =
( iv ) cos 75° = cos (45° + 30°) = cos 45° cos30° – sin 45°
sin 30 2 2 2 2 2 2
° = × − × =
Example 10: Find the approximate values of ( i ) sin 10°
( ii ) tan 20° ( iii ) cos 10°.
Sol. ( i ) sin10 sin 10 sin 180 18 18
π π π ° = (^) °× (^) = ≈ °
( ii ) tan 20 tan 20 tan 180 9 9
π^ π^ π ° = (^) °× (^) = ≈ °
( iii ) cos10 cos 10 cos 1 180 18
π^ π ° = (^) °× (^) = = °
Example 11: Find maximum and minimum values of y :
( i ) y = 2 sin x ( ii ) y = 4 – cos x
( iii ) y = 3sin x + 4 cos x
Sol. ( i ) y max = 2(1) = 2 and y min = 2(–1) = –
( ii ) y max = 4 – (–1) = 4 + 1 = 5 and y min = 4 – (1) = 3
( iii ) y max = (^3 2) + 42 = 5 and y min
Example 12: What is the value of x for which y is maximum
y = k sin 2 x.
(1) (^) rad 2
π (2) (^) rad 4
π
(3) rad 3
π (4) (^) rad 6
π
Sol. As sin 2 x ≤ 1 so y will be maximum if sin 2 x = 1.
Therefore 2 rad. 2 4
x x
π π = ⇒ =
Train Your Brain
44 PW NEET (XI) Module-1 P H Y S I C S
Example 13: The position of a particle moving along x -axis
varies with time t according to equation x = 3 sinωt
- cosωt where ω is constant. Find the region in which the particle is confined.
Sol. (^) x = 3 sin ω − t cosω t
( ) (^ )
2 x max (^) 3 1 2
2 ∴ = + − = and
( ) (^ )
2 x min (^) 3 1 2
2 = − + − = −
Thus, the particle is confined in the region − 2 ≤ x ≤ 2
8. Find maximum and minimum value of
( i ) 8 – 6 cos x
( ii ) 3 sin x – 4 cos x
( iii ) 5 sin( x ) + 12 cos x + 4
9. Find approximate value of
( i ) sin (1°)
( ii ) cos(5°)
( iii ) tan (3°)
10. Find the value of
sin 37 5
( i ) cos(74°) ( ii ) sin(106°)
Concept Application
ALGEBRA
Quadratic Equation
An algebraic equation of second order (highest power of the
variable is equal to 2) is called a quadratic equation.
The equation ax^2 + bx + c = 0,
is the general form of quadratic equation where a ≠ 0.
The general solution of above equation is
2 4
2
b b ac x a
If values of x be x 1 and x 2 then
2
1
b b ac x a
and
2
2
b b ac x a
Here x 1 and x 2 are called roots of equation ( i ). We can easily see that
sum of roots (^1 )
b x x a
and
product of roots (^1 )
c x x a
Example 14: Find roots of equation 2 x^2 – x – 3 = 0.
Sol. Compare this equation with standard quadratic equation ax^2 + bx + c = 0, we have a = 2, b = –1, c = –3.
Now from
2 2 4 ( 1)^ ( 1)^ 4(2)( 3) ; 2 2(2)
b b ac x x a
− ± −^ − −^ ±^ −^ −^ −
x
x x x
or 1 2
x = x = −
Train Your Brain
11. Find the root of quadratic equation
( i )
x + x − =
( ii ) 4 x^2 + 6 x – 12 = 0
12. Find the sum and product of roots of equation
( i )
2 5 2 3 0 3 3
x − x + =
( ii )
x − x + =
Concept Application
MATHEMATICAL SERIES
Arithmetic Progression (AP)
General form : a , a + d , a +2 d , ..., a + ( n – 1) d
Here a = first term, d = common difference
Sum of n terms [^ (^1 ) ]^2 ( 1 )
n
n n S = a + a + n − d = a + n − d
Geometrical Progression (GP)
General form : a , ar , ar^2 ,..., ar n –1^ Here a = first term,
r = common ratio
Sum of n terms
( 1 )
1
n
n
a r S r
, r < 1
46 PW NEET (XI) Module-1 P H Y S I C S
necessary for a concise and precise description of the phenomena.
These mathematical formulae are expressed in form of equations
and known as function.
Thus, a function describing a physical process expresses an
unknown physical quantity in terms of one or more known physical
quantities. We call the unknown physical quantity as dependent
variable and the known physical quantities as independent
variables. For the sake of simplicity, we consider a function
that involves a dependent variable y and only one independent
variable x. It is denoted y = f ( x ) and is read as y equals to f of
x. Here f ( x ) is the value of y for a given x. Following are some
examples of functions.
y = 2 x + 1, y = 2 x 2
- 3 x + 1, y = sin x , y = ln (2 x + 1)
Knowledge of the dependent variable for different values of the
independent variable, and how it changes when the independent
variable varies in an interval is collectively known as behavior of
the function.
Graph of a Function
Graph of a function is the diagrammatic representation of
a function and allows us to visualize it. To plot a graph the
dependent variable (here y ) is usually taken on the ordinate and
the independent variable (here x ) on the abscissa. Graph being an
alternative way to represent a function does not require elaborate
calculations and explicitly shows behavior of the function in a
concerned interval.
Graphs of Some Commonly Used Functions
Linear, parabolic, trigonometric and exponential functions are the
most common in use.
( i ) Straight line Equation and its Graph
When the dependent variable y varies linearly with the
independent variable x , the relationship between them is
represented by a linear equation of the type given below
y = mx + c. The equation is also shown in graph by an
arbitrary line.
y
x
c
q
–x
y^
mx
c
Here m & c are known as slope of the line and intercept on the
y -axis, respectively.
Slope:
Slope of a line is a quantitative measure to express the inclination
of the line. It is expressed by ratio of change in ordinate to change
in abscissa.
y
x
c q –x
q
D y
D x
y 2
y 1
x 1 x 2
2 2 = ∆ x + ∆ y
Slope of a line
2 1
2 1
y y y m x x x
= slope of tangent
When the x and the y axes are scaled identically, slope equals to tangent of the angle, which line makes with the positive x -axis.
m = tanθ
It is positive if y increases with increase in x , negative if y decreases with increase in x , zero if y remains unchanged with change in x and infinite if y changes but x remains unchanged. For these cases the line is inclined up, inclined down, parallel to x -axis and parallel to y -axis respectively as shown in the adjoining figure by lines A , B , C and D respectively.
B^ A
C
D
y
–x x
–y
Intercept
It is equals to the value of ordinate y , where the line cuts the y -axis. It may be positive, negative or zero for lines crossing the positive y -axis, negative y -axis and passing through the origin respectively.
Example 19: A parallelogram ABCD is shown in figure.
A
B
C
D (0, –1)
y
x
Column-I Column-II
(A) Equation of side AB (P) 2 y + x = 2
(B) Equation of side BC (Q) 2 y – x = 2
(C) Equation of side CD (R) 2 y + x = –
(D) Equation of side DA (S) 2 y – x = –
(T) y + 2 x = 2
Train Your Brain
Mathematical Tools 47
Sol.
For side AB : ( )
m c
y = x +
For side BC :
m c
y x
For side CD :
m c
y = x −
For side DA :
m c y x
Example 20: Frequency f of a simple pendulum depends on
its length l and acceleration g due to gravity according to the
following equation
g f l
π
Graph between which of the following quantities is a
straight line? (1) f on the ordinate and l on the abscissa
(2) f on the ordinate and l on the abscissa
(3) f^2 on the ordinate and l on the abscissa
(4) f^2 on the ordinate and
l
on the abscissa
Sol.^2
f l
18. Four lines are drawn in the figure below. Match the entries in Column-I with entries in Column-II
A
B
C
D
y
x
( –2,0)
Column-I Column-II
A. Equation of line AB P. 3 y – x + 3 = 0
B. Equation of line BC Q. 2 y + x + 2 = 0
C. Equation of line CD R. y = x + 2
D. Equation of line AD S. 3 y + x = 2
T. 3 y + 2 x = 6
Concept Application
( ii ) Quadratic equation and its graph
A function of the form y = ax^2 + bx + c is known as quadratic
equation. The graph of the quadratic function is a U-shaped
curve is called a parabola. The simplest parabola has the form
y = ax^2. Its graph is shown in the following figure.
y
–x x
In general
y = ax^2 + bx + c
2
b D a x a a
+^ +
where D = b^2 – 4 ac
a > 0 ( U shaped)
a <0 (inverted U -shaped)
, vertex 2 4
b D
a a
æ - ö ç -^ ÷= è ø
a > 0 Þ smiling parabola. (parabola opens upwards)
a < 0 Þ sad parabola. (parabola opens downwards)
Vertex can lie in any quadrant depending on value of a , b and c.
Graph of some trigonometric functions
Among all the trigonometric functions, sinusoidal function,which includes both sine and cosine is most commonly in use. Sine Function y = a sin x y
a
x
–a
y = a sin x
y 2
x
a
Here, a is known as the amplitude and equals to the maximum magnitude of y. In the adjoining figure graph of a sine function
is shown, which has amplitude a units.
Mathematical Tools 49
Logarithm
Logarithm is the exponent or power to which a base must be
raised to yield a given number. Expressed mathematically, x
is the logarithm of n to the base b is bx^ = n in which case one
writes x = log b n. For example 2^5 = 32, therefore 5 is the logarithm
of 32 to base 2 or 5 = log 2 32. Similarity since 10^3 = 1000 then
3 = log 10
Logarithms with base 10 are called common and are written
simply as log
Logarithm laws
log a a = 1
log mn = log m + log n
log log log
m m n n
=^ −
log mn^ = n log m
Logarithms with base e denoted by ln are called natural logarithms
where e (Euler’s number) is an irrational number approximates
equal to 2.71828. The relation between natural logarithms and
common logarithm is given by
log e m = 2.303 log 10 m
ln 2 = 0.
log 2 = 0.
ln e = 1
ln 1 = 0
Example 21: Calculate the value of log 10
(Given log 10 2 = 0.301, log 10
Sol. log 10 12 = log 10 (4 × 3) = log 10 4 + log 10
= log 10 22 + log 10 3 = 2log 10 2 + log 10
= 2 × 0.301 + 0.477 = 1.
Train Your Brain
19. Calculate the value of
( i ) log 1016 ( ii ) log 10 4 + log 1012
( iii ) log 1064 ( iv ) log 10 3 – log 1024
Concept Application
DIFFERENTIATION & ITS APPLICATIONS
( i ) Finite difference: The finite difference between two values
of a physical quantity is represented by ∆ notation.
For example: Difference in two values of y is written as ∆ y
as given in the table below.
y 2 100 100 100
y 1
∆ y = y 2 – y 1 50 1 0.
( ii ) Infinitely small difference: The infinitely small difference
means very-very small difference. And this difference is
represented by ‘ d ’ notation instead of ‘∆’.
For example infinitely small difference in the values of y is
written as ‘ dy ’ if y 2 = 100 and y 1 = 99.99999999........
then dy = 0.000000...................
Definition of Differentiation
Another name for differentiation is derivative. Suppose y is a
function of x or y = f ( x )
Differentiation of y with respect to x is denoted by symbol
f ’( x ) where f ’( x ) =
dy
dx
dx is very small change in x and dy is
corresponding very small change in y.
x
y
O a
Circle
x y a
2 2 2
Ellipse
y
O
a x
a = semi major axis b = semi minor axis
2 2
2 2
x y
a b
b y = |x |
y
x
50 PW NEET (XI) Module-1 P H Y S I C S
Notation: There are many ways to denote the derivative of a
function y = f ( x ). Besides f ’( x ), the most common notations are
these:
y ′
“ y prime” or “ y dash”
Nice and brief but does not name the independent variable
dy
dx
“ dy by dx ”
Names the variables and used d for derivative
df
dx
“ df by dx ” Emphasizes the function’s name.
d f x dx
“ d by dx of f ”
Emphasizes the idea that differentiation is an operation performed on f.
d ( f ) “ d of f ” A common operator notation.
y “ y dot”
One of Newton’s notations, now common
for time derivatives i.e.
dy
dt
.
Slope of a Line
It is the tan of angle made by a line with the positive direction of
x -axis, measured in anticlockwise direction.
Slope = tan θ
(In 1 st quadrant tan θ is +ve and 2 nd quadrant tan θ is –ve)
In figure ( a )slope is positive
In figure ( b )slope is negative
θ < 90° ( st quadrant)
θ > 90° ( nd quadrant)
figure ( ) a figure ( ) b
Average Rates of Change
Given an arbitrary function y = f ( x ) we calculate the average rate
of change of y with respect to x over the interval ( x , x + ∆ x ) by
dividing the change in value of y , i.e. ∆ y = f ( x + ∆ x ) – f ( x ), by
length of interval ∆ x over which the change occurred.
Q
R
D y
P y
x + D x x
y + D y
D x
q
The average rate of change of y with respect to x over the
interval [ x , x + ∆ x ] =
y f ( x x ) f ( ) x
x x
Geometrically,
y
x
QR
PR
= tan θ = Slope of the line PQ
therefore we can say that average rate of change of y with respect to x is equal to slope of the line joining P and Q.
The Derivative of a Function
We know that, average rate of change of y w.r.t. x is
y
x
f ( x x ) – f ( ) x
x
If the limit of this ratio exists as ∆ x → 0, then it is called the derivative of given function f ( x ) and is denoted as
f ’( x ) =
dy
dx
0
lim x
f x x f x
∆ → x
Derivatives with Respect to Time
In physics, we are often looking at how things change over time:
- Velocity is the derivative of position with respect to time
( ) ( ( )).
d v t x t dt
- Acceleration is the derivative of velocity with respect to time:
( ) ( ( )) ( ( ))
2
2
d d a t v t x t dt (^) dt
- Momentum (usually denoted p ) is mass times velocity, and force ( F ) is mass times acceleration, so the derivative of
momentum is ( ).
dp d dv mv m ma F dt dt dt
Geometrical Meaning of Differentiation
The geometrical meaning of differentiation is very much useful in the analysis of graphs in physics. To understand the geometrical meaning of derivatives we should have knowledge of secant and tangent to a curve
Secant and Tangent to a Curve
A secant to a curve is a straight line, which intersects the curve at
any two points. y
x
P
q Secant
Tangent
A tangent is a straight line, which touches the curve at a particular point. Tangent is a limiting case of secant which intersects the curve at two overlapping points.
In the figure ( a ) shown, if value of ∆ x is gradually reduced then the point Q will move nearer to the point P. If the process is continuously repeated (figure ( b )) value of ∆ x will be infinitely small and secant PQ to the given curve will become a tangent at point P.
52 PW NEET (XI) Module-1 P H Y S I C S
20. If acceleration =
dv
dt
. Find acceleration at t = 1 sec
from v = 3 t^2 – 1
2 = − 3 , find
dy y x x dx
y = − 2 x x
, find
dy
dx
2
find
x dy y x dx
= 24. x = 9 y^2 find
dy
dx
Find
dy
dx
for the following
25. y = x 7/2^ 26. y = x – 27. y = x 28. y = x 5 + x 3 + 4 x 1/ + 7 29. y = 5 x^4 + 6 x 3/2^ + 9 x 30. y = ax^2 + bx + c 31. y = 3 x^5 – 3 x –
x
32. Given S = t^2 + 5 t + 3, than
dS
dt
33. Given S = ut +
at^2 , where u and a are constants. Obtain
the value of
dS
dt
34. The area of a blot of ink is growing such that after t seconds, its area is given by A = (3 t^2 + 7) cm^2. Calculate the rate of increase of area at t = 5 seconds. 35. The area of a circle is given by A = π r^2 , where r is the radius. What is the rate of increase of area w.r.t. rate of change of radius.
Obtain the differential coefficient (differentiation) of the following:
36. ( x – 1) (2 x + 5) 37. (9 x 3 - 8 x + 7) (3 x 5 + 5) 38. If t = s − 1 , then the velocity at t = 2 sec is 39. If S = 3 t^2 , then double differentiation of s with. respect to t. is 40. Velocity of a body is given by v = 3 t^2 – 4 t , then rate of
change of velocity w.r.t. to time at t = 1 sec is
41. Acceleration is given by a =
dv
dt
. then acceleration at
t = 10 sec from v = 3 t^2 + t
2 x + 1
x
x
2
3 1
x
x +
Concept Application
Maxima and Minima
Finding maxima and minima of a function using derivatives:
A maximum is a high point and minimum is low point of a function
(see figure below)
Slope = 0 A
Local maximum
C
D
B
–ve slope
+ve slope Slope = 0 +ve slope
local minimum
y = ( ) f x
y
x
In a smoothly changing function a maximum or minimum is always where function flattens out or where slope of tangent line
is zero. We know slope =.
dy
dx
So a function reaches its maximum
or minimum value when 0.
dy
dx
In the neighbourhood of maximum (point A ), slope changes from positive to zero at point A and then becomes negative as x increases
which means ( )
2 slope 0 2 0
d d dy d y
dx dx dx dx
In the neigbourhood of minimum (point B ), slope changes from negative to zero and then becomes positive as x increases which means
2 slope 0 2 0
d d dy d y
dx dx dx dx
Second Derivative Test
When a functions slope 0
dy
dx
at a point and its second derivative at that point is
( i ) less than zero, it is a local maximum
( ii ) greater than zero, it is a local minimum
Example 25: What is the minimum value of y for the curve y = – 8 x^3 + x^4. Sol. y = –8 x^2 + x^4
dy
dx
= –16 x + 4 x^3 = – x (16 – 4 x^2 )
The function will have a maximum or minimum value
when
dy
dx
⇒ x (16 – 4 x^2 ) = 0 ⇒ x = 0 or x = ± 2
Now,
2
2
d y
dx
= 16 – 12 x^2
At x = 0,
2
2
d y
dx
= –16 (maximum)
At x = ±2,
2
2
d y
dx
= –16 + 48 = +32 (minimum)
So, function has minimum value at x = ± 2
y min = –8 × 4 + 16 = –
Train Your Brain
Mathematical Tools 53
Example 26: A ball is thrown vertically upward in the air. Its height y at any time t is given by y = 10 t – 5 t^2 where y is in meters and t is in seconds. What is the maximum height attained by the ball?
Sol. y = 10 t – 5 t^2 ⇒
dy
dt
= 10 – 10 t = 0
⇒ t = 1 sec ⇒
2 d y
dt
= – 10 (maximum)
ball attains maximum height at t = 1s
ymax = 10 × 1 – 5 × 1^2 = 5m.
45. Find the minimum value of
( i ) y = 1 + x^2 – 2 x ( ii ) y = 5 x^2 – 2 x + 1
46. The position of a particle moving along the y-axis is given by y = 3 t^2 – t^3 , where y is in meters and t is in sec. The time when particle attains its maximum positive y position is
( a ) 1.5 sec ( b ) 4 sec
( c ) 2 sec ( d ) 3 sec
Concept Application
INTEGRATION
( i ) Integration is the inverse process of differentiation:
Integration is the process of finding the function, whose
derivative is given. For this reason, the process of integration is the inverse process of differentiation.
Consider a function f ( x ), whose derivative w.r.t. x is another
function f ′ ( x ) i.e. (^) ( ( )) ( )
d f x f x dx
If differentiation of f ( x ) w.r.t. x is equal to f′ ( x ), then
f ( x ) + c is called the integration of f ′ ( x ), where c is called
the constant of integration.
Symbolically, it is written as (^) ∫ f^ ′( )^ x dx^^ =^ f^ ( ) x^ + c
Here, f ′ ( x ) dx is called element of integration and ∫ is
called indefinite integral operator. Let us proceed to obtain
integral of x n w.r.t. x, 1 ( ) ( 1)
d (^) n n x n x dx
= +
Since the process of integration is the inverse process of
differentiation,
( 1) or ( 1)
n n ∫^ n^ +^ x dx^ n^ + ∫ x dx
1 or 1
n n x x dx n
=
∫
The above formula holds for all values of n , except n = –1.
It is because, for n = –1,
n 1 1 x dx x dx dx x
− ∫ =^ ∫ =∫
Since 1/ x is differential coefficient of loge x
i. e .,
(log (^) e )
d x dx x
dx log ex x
∫
Similarly, the formulae for integration of some other functions can be obtained if we know the differential
coefficients of various functions.
Basic formulae of integration: Following are a few basic formulae of integration:
1. (^) ∫ 1 dx = x + C 2. (^) ∫ a dx = ax + C 3. (^) (( ) (^ ))
1 / 1 ; 1
n n x dx x n C n
∫ =^ +^ +^ ≠ −
4. (^) sin x dx = − cos x + C ∫ 5. (^) ∫cos x dx = sin x + C
2 ∫ sec^ x dx^ =^ tan x^ + C 7.^
2 ∫csc^ x dx^ = −^ cot x^ + C
8. (^) ∫sec x ( tan x dx ) = sec x + C 9. (^) ∫csc^ x^ (^ cot^ x dx )^ = −^ csc x^ + C 10. (^) ∫( 1/ x dx ) = ln x + C
x x ∫^ e dx^ =^ e^ + C
Example 27: Integrate w.r.t. x :
( i ) x 11/2^ ( ii ) x –7^ ( iii ) x p / q
Sol. ( i )
11 1 2 11/2 2 13/ (^11 ) 1 2
x x dx c x c
= + = +
∫
( ii )
7 1 7 1 6
7 1 6
x x dx c x c
− + − − = + = − + − + ∫
( iii )
1 ( )/
p p (^) q q x^ q p^ q^ q x dx c x c p (^) p q
q
= + = +
∫
Example 28: Find integer w.r.t. x , for (^2)
x
x
Sol. (^) 2
x dx x ∫
1 2 x^2 dx
− + ∫ =
3 x^2 dx
−
∫ =
(^3 ) 2
3 1 2
x
− (^) +
∫
c x
Example 29: ( )
2 x + 1 dx ∫
Sol. 2 x dx + 2 xdx + 1 dx ∫ ∫ ∫
3 2 3 2 2 3 2 3
x x x
Train Your Brain
Mathematical Tools 55
Example 33: A gas expands its volume from V to 3V
as shown in figure. Calculate the work in this process if
W = pdv. ∫ p
P = 6/ V
V
Sol. (^) [ ]
3 3 6 3 6 log 6.6 J
V V V e (^) V V V
W pdv dv v v
∫ ∫
Example 34: Calculate average value of current from t = 0
to t = 4 seconds.
5 A
I
t (sec)
Sol. (^) average
total area 5 2 2.5 amp total time 4
b
a b
a
Idt
I
dt
×
∫
∫
Example 35: Find the shaded area bounded by line y = x
and y = x^2 as shown in figure.
y
x x = 1
y = 1
Sol.
1 1 2
0 0
A = (^) ∫ xdx −∫ x dx
2 1 31
0 0
units 2 3 2 3 6
x x = − = − =
Example 36: Calculate average value of current from t = 0 to t = T seconds ( T = 2π/w).
I
0 T
t (sec)
I = 5 sin wt
T /
Sol.^0 [ ] average (^0)
0
5sin 5 cos
T
T T
t
I t T dt
ω
= = − ω ω
∫
∫
[ ]
cos T cos 0 0 T T
π = − ω − = ω = ω (^)
62. Find the area bounded by the curve y = 4 – x^2 from x = –2 to x = 2 63. Find the shaded area bounded by the curve y = (^) x and y = x as shown in figure?
y
x x = 1
y = 1
64. Find the area bounded by y = 2 x – x^2 and x -axis between x = 0 to x = 2 as shown in figure. y
x x = 2
Concept Application
56 PW NEET (XI) Module-1 P H Y S I C S
TROGNOMETRY
1. Change degree into radian:
(1) 160° (2) 135°
(3) 75° (4) 65°
2. Change radian into degree:
π (2)
π
π (4)
π
π
3. Evaluate:
(1) cos 15° (2) cos 53°
(3) tan 37° (4) sin53° – cos 37°
4. Evaluate:
sin
cos
sin
cos
(3) sin 105° (4) sin300°
(5) cos 240° (6) sin 2 (20°) + sin 2 (70°)
5. Evaluate:
(1) 2 sin 15° cos 15° (2) sin 22.5° cos 22.5°
(3) tan 75° (4) sin 2 22.5°
6. Evaluate:
cos 4
π
sin 3
π
sin 4
π
tan 6
π
7. 1 + sinθ is equal to
(1) (sin θ + cos θ) (2) sin θ – cos θ
(3) (^) sin cos 2 2
θ θ
θ θ −
8. (^) 1 + cosθ is equal to
(1) (^) 2 sin 2
θ (2) (^) 2 cos 2
θ
sin 2 2
θ (4)
cos 2 2
θ
9. If
sin cos 7 , sin cos 3
θ + θ
θ − θ
, then tan θ =
ALGEBRA
10. The equation whose roots are the squares of the roots of the
equation ax 2
(1) a^2 x^2 + b^2 x + c^2 = 0
(2) a^2 x^2 – ( b^2 – 4 ac ) x + c^2 =
(3) a 2 x 2
(4) a^2 x^2 + ( b^2 – ac ) x + c^2 =
11. The value of ' a ' for which one root of quadratic equation ( a^2 – 5 a + 3) x^2 + (3 a – 1) x + 2 = 0 is twice as large as
other is:
AP AND GP SERIES
12. Find the sum of first 20 natural Number.
(1) 210 (2) 200
(3) 220 (4) 230
13. The sum of the first five multiples of 3 is:
(1) 45 (2) 55 (3) 65 (4) 75
14. Find sum of
15. Find sum of
16. Find the sum of the geometric series:
4 – 12 + 36 – 108 + ………….. to 10 terms
(1) 59048 (2) –
(3) –59048 (4) 30421
BINOMIAL APPROXIMATION
17. Find approximate value of the (^) 0.95 :
(1) 1 (2) 0.595 (3) 0.60 (4) 0.
18. Find approximate value of the 104 :
(1) 10.2 (2) 12 (3) 13.5 (4) 15
19. Find approximate value of the (4.04)^3
(1) 60.05 (2) 75.63 (3) 65.92 (4) 55.
20. Find approximate value of the (9.6) 4
PRARAMBH EXERCISE-1 (TOPICWISE)PRARAMBH EXERCISE-1 (TOPICWISE)
58 PW NEET (XI) Module-1 P H Y S I C S
3 4
r = − + θ (^) θ θ
(1) 12 θ–2^ – 12θ–4^ + 4θ–5, 24θ–3^ + 48θ–5^ + 20θ–
(2) –12θ–2^ + 12θ–4^ – 4θ–5, 24θ–3^ – 48θ–5^ + 20θ–
(3) –6θ–2^ + 12θ–4^ – 8θ–5, 12θ–3^ – 24θ–5^ + 10θ–
(4) –8θ
12θ
- 6θ - , 24θ - - 24θ - + 10θ -
7 3 2 ω = 3 z − 7 z + 21 z
(1) 21 z^6 +21 z^2 – 42 z , 126 z^5 + 42 z – 42.
(2) 14 z^6 –28 z^2 + 22 z , 120 z^5 –21 z + 42.
(3) 28 z^6 –14 z^2 + 42 z , 122 z^5 –42 z + 21
(4) 21 z 6 –21 z 2
- 42 z , 126 z 5 –42 z + 42
39. y = sin x +cos x
(1) cos x – cos x , – sin x – sin x
(2) sin x – sin x , – sin x – cos x
(3) cos x – sin x , – sin x – cos x
(4) sin x + cos x , – cos x – cos x
Direction ( No. 40 to 42 ) : Derivative of given functions w.r.t. the
independent variable x is.
40. y = x sin x
(1) sin x + x cos x (2) sin x – x cos x
2 2 cos x – x sin x (4)
2 2 sin x – x cos x
x y = e n x
x x e e nx x
2
x x e e nx x
2
x x e e nx x
x x e e nx x
2 y = ( x −1) ( x + x +1)
dy x dx
2 3
dy x dx
2 2
dy x dx
dy x dx
Direction ( No. 43 to 45 ) : Derivative of given function w.r.t. the
independent variable is
sin
cos
x y x
(1) sec 2 x (2) sec x (3) sec 2 2 x (4) sec 3 2 x
x y x
2
y x
2
y x
y x
2
y x
2
x z x
2
2 2
x x
x
2
2 2
x x
x
2
2
x x
x
2
2
x x
x
Direction ( No. 46 to 49 ) :
dy
dx
for following functions is.
5 y = (2 x +1)
(1)
3 10(2 x + 1) (2)
4 10(2 x +1)
3 10(2 x −1) (4) 4 10(2 x −1)
9 y = (4 −3 ) x
8 − 8(4 − 3 ) x (2) 9 − 27(4 −3 ) x
9 − 27(4 + 3 ) x (4) 8 − 27(4 −3 ) x
7 1 7
x y
− = (^) −
8 1 7
x −
8 1 7
x
− −
5 1 7
x
− −
4 1 7
x
− −
49. y = 2sin( ω x + φ) where w and φ constants
(1) 2 ωcos( ω x + φ) (2) 2 ωcos( ω x – φ)
(3) ωcos( ω x + φ) (4) 2 ωcosec( ω x + φ)
SLOPE OF TANGENT
50. Find the slope of tangent of curve y = 1 + x^2 – 2 x at (3, 3).
(1) 1 (2) 2
(3) 3 (4) 4
51. Find the slope of tangent of curve y = 5 x^2 + 2 x + 1 at (0, 0).
(1) 1 (2) 2 (3) 3 (4) 4
52. If y = a (1 – cosθ) and x = a (θ + sinθ), then find the slope of
tangent of y verses x at 2
π θ =.
53. The slope of the normal to the curve y = x 2 - (^2)
x
at (– 1, 0) is:
DIFFERENTIATION AS A RATE MEASUREMENT
54. Suppose that the radius r and area A = π r^2 of a circle are differentiable functions of t. equation that relates dA / dt to dr / dt is:
dA dr r dt dt
= π (2)
dA (^) 2 dr r dt dt
= π
2 2
dA dr r dt dt
= π (4) 2
dA dr r dt dt
= π
Mathematical Tools 59
3 y = 2 u , u = 8 x − 1. Find
dy
dx
(1) 2 48(8 x − 1) (2) 2 48(8 x +1)
(3) 48(8 x − 1) (4) 48(8 x +1)
56. y = sin u , u = 3 x + 1. Find
dy
dx
(1) 3cos(3 x –1) (2) 3cos(3 x +1)
(3) 3sin(3 x –1) (4) 3sin(3 x +1)
57. (^) sin , cos. Find dy y u u x dx
(1) –cos u. sin x (2) 2 − cos u. 1 + u
2 − sin x. 1 − y (4) All of above
2 2 3 1,. Find
dy y t x t dx
MAXIMA & MINIMA
59. Maximum and minimum values of function 2 x^3 – 15 x^2 + 36 x + 11 is
(1) 39, 38 (2) 93, 83 (3) 45, 42 (4) 59, 58
60. Find out minimum/maximum value of y = 1 – x 2 also find out those points where value is minimum/maximum.
(1) max 2, x = –1 (2) max 1, x = 0
(3) min 1, x = –1 (4) min 2, x = 0
61. For y = ( x –2) 2 , what is the maximum/minimum value and
the point at which y is maximum/minimum?
(1) max 2, x = 0 (2) max 0, x = 0
(3) min 1, x = –1 (4) min 0, x = 2
62. Particle's position as a function of time is given by x = - t^2 + 4 t + 4 find the maximum value of position co-ordinate
of particle.
(1) 2 (2) 4 (3) –8 (4) 8
63. Find out minimum/maximum value of y = 2 x^3 – 15 x^2 + 36 x + 11 also find out those points where value is minimum/maximum.
(1) max = 39 at x = 2, min = 39 at x = –
(2) max = 39 at x = 3, min = 38 at x = 2
(3) max = 39 at x = 2, min = 38 at x = 3
(4) max = 39 at x = 2, min = 38 at x = –
64. Determine the position where potential energy will be minimum if U ( x ) = 100 – 50 x + 1000 x^2 J.
(1) 0.25 × 10
(3) 2.5 × 10–1^ (4) 250 × 10–
65. Find out minimum/maximum value of y = 4 x^2 – 2 x + 3 also find out those points where value is minimum/maximum.
min , 4 2
= x = (2)
max , 4 4
= x =
min , 4 4
= x = (4)^
max , 4 2
= x =
66. x + y = 10 then find the maximum value of f = xy?
(1) 25 (2) 30 (3) 15 (4) 35
67. If + r = 12 here is length of cylinder and r is radius of cylinder then find maximum value of volume of cylinder
(1) 156 π (2) 350 π (3) 256 π (4) 250 π
INTEGRATION
68. (^) ∫ (2 ) x dx will be
(1) x^2 + C (2) 2 x + C (3) 2 x^2 + C (4) – x^2 + C
2 ∫ (^ x ) dx^ will be
(1) x + C (2) 2 x + C (3)
3
x
2
x
70. (^) ∫( x^2 – 2 x + 1) dx will be
3 2
3
x − x x + C (2)
3 2
3
x − x + x + C
3 2
3
x
3 2
3
x
71. (^) ∫(–3 x –4) dx will be
(1) x –3^ + C (2) x^3 + C
(3) –3 x –3^ + C (4) 3 x –3^ + C
72. (^) ∫ 2
x
dx will be
x
− + C (2)
x
+ C
x
x
73. (^) ∫
2 x
dx will be
3 2 x + C (2) 3 x + C
3 x + C (4)
4 x + C
74. (^) ∫ 3
3 x
dx will be
3 4
x
2 3
x
2
x^3 + C (4)
2 3
x
75. (^) ∫ 3 3
x x
dx will be
4 2 3 2 3 4
4 2
x x
4 2 3 2 3 1
4 2
x x
3 3
3 4 3 2
4 2
x x
4 2
3 3 3 3
4 2
x x
