SOLUTION OF PROBLEM: 7

Let z and w are two non-zero compex numbers such that |z| =|w| and arg(z) + arg(w)=180(degree), then z is

SOLUTION: we can have two different ways to approach the solutions

CONVENTIONAL METHOD: 

Let z = r e(ip) , where r is |z| and p is arg(z)

also, w = r e(iq), where r is |w| and q is ar(w)

as given |z|=|w| and p + q =180(degree), we can write w = r e(i(180-p))= r(cos(180-p)+i sin(180-p))

w = r(-cos(p) +i sin(p)) = -r( cos(p) - isin(p)) = -1/z =-z( bar).

Hence, w = -z(bar)

NOTE: e(ip) is e to the power ip.

GRAPHICAL METHOD:

SOLUTION OF PROBLEM: 6

Find the value of k such that x +y = k is normal to the parabola (y)sq =12x. (COR/M)

SOLUTION: We already know that y = mx -2am -a(m(cube)) is normal to the parabola y(sq)=4ax. Here in the given problem a=3 and if we compare x +y = k with the standard equation of the normal we get m=-1 and k = -2am -a(m(cube)).

By putting the values of m and a, we get k =9

NOTE: y(sq) is y square and m(cube) is m cube

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GRAPH OF ONE-TO-ONE FUNCTION

UR QUESTIONS, OUR SOLUTION: 3

How a one-to-one function looks like on cartesian plane ? ( DC/M)
( asked by RAVI JHA)

SOLUTION: One-to-one functions has unique mapping, that is for a unique value of x we will have unique value of y. All functions which are strictly increasing or strictly decreasing in their domain will be one-one functions.

Otherwise, they will be many-one.

For eg: y = x, y = log(x), y = e(x) are all strictly increasing function they will be one-to-one.

y = -x , y =e(-x) are strictly decreasing functions they will also be one-to-one.

whereas, y= sin(x) , y = |x|, y = x(sq), y= cos(x), y = tan(x) will all be many-one functions because their graph is decreasing for some values of x and increasing for others.

Note: e(x) is e raise to x, e(-x) is e raise to -x, x(sq) is x sqaure

FIND THE VALUES OF a, b, c and d

PROBLEM : 12

Find the values of a, b, c and d such that   ( DC/M)

     


     

MEET THE TOPPER

GRAPH OF |tanx|

PROBLEM: 11

Plot the curve given by the function y = |tanx|  (DC/M)

SOLUTION OF PROBLEM: 5

Find the area enclosed by the curve |x| +|y| =1 (IC/M)

SOLUTION: The equation will have four different cases

x + y =1, when x>0 and y>0 ( we will draw x + y=1line & will take its part which is lying in Ist quadrant)

 x-y =1, when x>0 and y<0 ( we will draw x - y=1line & will take its part which is lying in IVth quadrant)

-x +y =1,when x<0 and y>0 ( we will draw -x + y=1line & will take its part which is lying in IInd quadrant)

-x -y =1,when x<0 and y<0 ( we will draw -x - y=1line & will take its part which is lying in IIIrd quadrant)

The enclosed figure will be a square of a side root(2).

Therefore, the area enclose will that of square that is 2 sq units

SOLUTION OF PROBLEM: 4

Find out the quadratic equation having equal and opposite roots as that of equation (x sq -16) =0.                    ( ALG/E)

.
 Note: (x) sq stands for x square

SOLUTION: The given(x)sq-16=0 has its roots as +4 and-4. According to the given condition the required equation has its roots as equal and opposite to the given equation's roots that is the required equation will have its roots  as -4 and +4 .The required equation will therefore be (x+4)(x-4)=0 or (x)sq-16=0, which is nothing but the original equation. 

ONTO AND INTO FUNCTIONS

PROBLEM: 10

Let function f is from R to R be defined by f(x) =2x + sin(x) for x belongs to R. Then f is

a) one-to-one and onto

b) one-to-one but not onto.

c) onto but not one-to-one

d) neither one-to-one nor onto.      ( DC/CH)

AREA UNDER THE CURVE

PROBLEM : 9

Find the area of the region bounded by the curves y=e(x) and the lines x=0 and y=e . (IC/E)
NOTE: e(x) is e to the power x.

INCREASING FUNCTION

PROBLEM: 8

Prove that the f(x) = 2(x)cube +3(x)square +6x +10 will always be an increasing function. (DC/M)

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MAXIMUM DISTANCE OF A LINE FROM A POINT

UR QUESTIONS, OUR SOLUTION: 2

Find the equation of a line passing through (1,2) and having maximum distance from the point (3,1). (COR/M)

( asked by ESHA PRASHAR)

SOL: 

We can solve this problem in two different ways one is using Graphs and other by Conventional method.

GRAPHICAL METHOD

we already know we will have infinetly many lines passing through (1,2), so let's examine all those lines and picked the one having maximum distance from point (3,1).Clearly the line perpendicular to line joining (1,2) and (3,1) is the required line having maximum distance from (3,1).( shown by bold line).

Therefore,the slope of the required line is -(3-1)/(1-2)=2. Hence the line will be (y-2)=2 (x-1) or y=2x.

CONVENTIONAL METHOD

The general equation of line passing through (1,2) is y=mx+2-m. Its distance from (3,1) will be

d=|2m+1|/sqrt(msq+1) where sqrt is square root and m sq is m square.

differentiating d with respect to m, we will get m=2 as the point of maxima.

Therefore, the required will be y=2x.

ARGUMENT OF z

PROBLEM: 7

Let z and w be two non- zero complex numbers such 

that |z|=|w| and arg(z) +arg(w) =180(degree), then z is

                                                                                 (ALG/M)

NORMAL TO PARABOLA

PROBLEM: 6

Find the value of k such that the line x +y =k is normal to the 

 parabola (y)sq=12x.  (COR/M)

NOTE: (y)sq stands for y square.

AREA UNDER THE CURVES

PROBLEM: 5

 Find the area enclosed by the curve |x| +|y|=1 
( IC/M)

OUR SOLUTION OF PROBLEM: 3

Find out the distance between the parallel lines given by x + 3y =7 and 5x + 15y = 25. ( COR/E)

Solution: We already know the formula to get the distance between two parallele lines.

ax +by+c=0 and ax+by+d=0 are parallel lines , the distance between them is |c-d|/sqrt( asq +bsq)

where sqrt stands for square root and sq stands for square.

 lines x + 3y =7 and 5x + 15y = 25 are || but the coefficient of x and y of both the equation are not equal.

So, let's make them equal by dividing 2nd equation by 5

We have  x + 3y -7=0 and   x + 3y -5=0 by applying the above formula

distance=|5-7|/sqrt(10)= 2/sqrt(10)

OUR SOLUTION OF PROBLEM: 2

Find the interval in which 
f(x)=2log(x-2)-xsq+4x+1 increases
a) (1,2)
b) (2,3)
c) (5/2,3)
d) (2,4)

Note: xsq term stands for x square.                         (DC/M)

SOLUTION:

Before starting up the problem, let’s first of all find out the domain of the given function

Since, we have (x-2) is inside the log it should be positive, it will happen only when x>2.

Hence the domain of given function is x > 2 therefore option a) is ruled out.

Now differentiate the given function, we get f ‘(x)= -(x-3)(x-1)/ (x-2)

For f(x) to increasing, f ‘ (x) > 0

Since, x> 2, the denominator of f ‘ (x ) is always positive, which means for f (x) to increasing -(x-3)(x-1) >0

which means x will lie from ( 1,3), but since the function is defined for x >2, we will have (2,3 )as the interval where function is increasing.

Hence the correct choices are (2,3) and (2.5 ,3)

 

GRAPHS OF |sinx| and (sinx) square

UR QUES, OUR SOL: 1

Q. What is the difference between the curves of |sinx| and (sinx) sq.    ( DC/ M)

(asked by RAVI JHA)

SOL: The curve of y = |sinx| can easily be drawn from the sinx as the portion of the curve below x- axis will get inverted.

For the y =(sinx) sq, we can write it as (1-cos(2x))/ 2 , which we can easily plot by plotting cos2x, then inverting and halving it so as to get -cos2x/2 and later shifting the origin, we will get the shape of   

y= (sinx) sq

EQUAL AND OPPOSITE ROOTS

PROBLEM: 4

Find out the quadratic equation having equal and opposite roots as that of equation (x sq -16) =0.       ( ALG/E)

Note: xsq stands for x square.

DISTANCE BETWEEN PARALLEL LINES

PROBLEM: 3

Find out the distance between the parallel lines given by x + 3y =7 and 5x + 15y = 25.   ( COR/E)

OUR SOLUTION OF PROBLEM: 1

QUESTION : Find the equation of the circle touching the axes and having radius of 1 unit.

SOLUTION:

Since the required circle is touching both the axes, the coordinate of the centre of circle will be in the form of (r,r) where r is the radius of circle.
So the eq. of circle will be (x-r)sq +(y-r)sq = r sq
but here we can have in total 4 such circles
Having centre at (+-r, +-r)

Each of these circle will be 4 different quadrant.
In the given problem r=1
So the eq of circles will be
(x-1)sq +(y-1)sq = r sq Ist quadrant circle
(x+1)sq +(y-1)sq = r sq IInd quadrant circle
(x+1)sq +(y+1)sq = r sq IIird quadrant circle
(x-1)sq +(y+1)sq = r sq IVth quadrant circle

DIFFERENTIAL CALCULUS

PROBLEM: 2

Find the interval in which f(x)=2log(x-2)-xsq+4x+1 increases
a) (1,2)
b) (2,3)
c) (5/2,3)
d) (2,4)                                                   ( DC/M)


Note: xsq term stands for x square.

CIRCLE TOUCHING AXES

PROBLEM: 1

Find the equation of the circle touching the axes and having radius of 1 unit.   ( COR/M)

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