One Mark Questions
Q.1 Is
a solution of the equation 


[AI 2008]
Sol.
We have 
When




Therefore,
is not a solution of the given equation.

When





Therefore,

Q.2 Is
a solution of the equation 


[AI 2008]
Sol.
We have 
When




Therefore,
is a solution of the given equation.

When





Therefore,

Q.3 Show that
is a solution of equation


[Foreign 2008]
Sol. We have 
When




Therefore ,
is a solution of the given equation.

When





Therefore ,

Q.4 For what value of k are the roots of the quadratic equation
real and equal.

[Delhi 2008]
Sol. We have 
Here
Since


For real and equal roots, D = 0




So, at
the equation has real and equal roots.

Here

Since



For real and equal roots, D = 0




So, at

Q.5 For what value of k are roots of the quadratic equation
equals and reals.

[AI 2008C]
Sol. We have 
Here
Since


For real and equal roots, D = 0



So, at
equation has real and equal roots.

Here

Since



For real and equal roots, D = 0



So, at

Q.6 For what value of k does
have equal roots.
[AI 2008C]

[AI 2008C]
Sol. Here 
Since
![\Rightarrow D = {\left[ {2\left( {k - 12} \right)} \right]^2} - 4 \times 2 \times \left( {k - 12} \right)](http://dronstudy.com/wp-content/plugins/latex/cache/tex_e4b30721132d4027a8f55b39c9a14f7a.gif)

![\Rightarrow D = 4\left( {k - 12} \right)[k - 12 - 2]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_53c7be2567c7b97f25330b86bf4a8b02.gif)

Now for equal and real roots D = 0
Then

Because for
the given equation
becomes a linear equation. So,
is rejected.
Therefore, only
is solution for the equation having real and equal roots.

Since

![\Rightarrow D = {\left[ {2\left( {k - 12} \right)} \right]^2} - 4 \times 2 \times \left( {k - 12} \right)](http://dronstudy.com/wp-content/plugins/latex/cache/tex_e4b30721132d4027a8f55b39c9a14f7a.gif)

![\Rightarrow D = 4\left( {k - 12} \right)[k - 12 - 2]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_53c7be2567c7b97f25330b86bf4a8b02.gif)

Now for equal and real roots D = 0
Then


Because for



Therefore, only

Q.7 Find the discriminant of the quadratic equation 

[AI 2009]
Sol. We have 
Here
Since



So, the discriminant of quadratic equation is

Here

Since




So, the discriminant of quadratic equation is

Q.8. Write the nature of roots of quadratic equation 

[Foreign 2009]
Sol. We have 
Here
, b =
, 
Since



As
, the equation has real and equal roots.

Here



Since




As

Two Marks Questions
Q.1 Solve for 

[Foreign 2005; Delhi 2006 C]
Sol. We have 
Here
,
, 
Since
![\Rightarrow D = {\left[ { - 2\left( {{a^2} + {b^2}} \right)} \right]^2} - 4 \times 1 \times {\left[ {\left( {{a^2} - {b^2}} \right)} \right]^2}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_fc777665523064df60c3344128edc433.gif)
![\Rightarrow D = 4\left[ {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {{a^2} - {b^2}} \right)}^2}} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_0cb5fd54f12b4e2884bcd777fce04718.gif)
![\Rightarrow D = 4\left[ {\left( {{a^2} + {b^2} + {a^2} - {b^2}} \right)\left( {{a^2} + {b^2} - a + {b^2}} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_f49a44b96d0167352d110036d16bfa95.gif)
[Because
]


Now


Then values of x are
or 

Here



Since

![\Rightarrow D = {\left[ { - 2\left( {{a^2} + {b^2}} \right)} \right]^2} - 4 \times 1 \times {\left[ {\left( {{a^2} - {b^2}} \right)} \right]^2}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_fc777665523064df60c3344128edc433.gif)
![\Rightarrow D = 4\left[ {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {{a^2} - {b^2}} \right)}^2}} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_0cb5fd54f12b4e2884bcd777fce04718.gif)
![\Rightarrow D = 4\left[ {\left( {{a^2} + {b^2} + {a^2} - {b^2}} \right)\left( {{a^2} + {b^2} - a + {b^2}} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_f49a44b96d0167352d110036d16bfa95.gif)
[Because



Now



Then values of x are


Q.2 Solve for 

[Delhi 2006]
Sol. We have 







Q.3 A two digit number is such that the product of its digits is 35. When 18 is added to the number, the digits interchange their places. Find the number.
[Delhi 2006]
Sol.Let digit at ten’s place 
Digit at unit’s place
Therefore, Number
Now
,
………..(1)
A.T.Q.




[using (1)]






Because
, so it can be rejected
Putting
in eq(1) , we get

Therefore, number

Digit at unit’s place

Therefore, Number

Now


A.T.Q.











Because

Putting


Therefore, number

Q.4 Using the quadratic formula; solve the equation 

[AI 2006]
Sol. We have 
Here
Since
![\Rightarrow D = \left[ { - \left( {4{b^4} - 3{a^4}} \right) - 4\left( {{a^2}{b^2}} \right) \times \left( { - 12{a^2}{b^2}} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_31f2cf028a661b60dc3cf03fbdfc4081.gif)




Therefore,![x = {{ - B \pm \sqrt D } \over {2A}} = {{ - \left[ { - \left( {4{b^4} - 3{a^4}} \right)} \right] \pm \sqrt {{{\left( {4{b^4} + 3{a^4}} \right)}^2}} } \over {2{a^2}{b^2}}}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_6aedebda61906f3953014d1edc1cabc6.gif)




Here

Since

![\Rightarrow D = \left[ { - \left( {4{b^4} - 3{a^4}} \right) - 4\left( {{a^2}{b^2}} \right) \times \left( { - 12{a^2}{b^2}} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_31f2cf028a661b60dc3cf03fbdfc4081.gif)




Therefore,
![x = {{ - B \pm \sqrt D } \over {2A}} = {{ - \left[ { - \left( {4{b^4} - 3{a^4}} \right)} \right] \pm \sqrt {{{\left( {4{b^4} + 3{a^4}} \right)}^2}} } \over {2{a^2}{b^2}}}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_6aedebda61906f3953014d1edc1cabc6.gif)



Q.5 The sum of two natural numbers is 8. Determine the numbers if sum of their reciprocals is 8/15
[AI 2006]
Sol.Let one number
Therefore, other natural number
A.T.Q.
…………..(1)
So
(By (1))
A.T.Q.







When
, other number
When
, other number
Therefore, Numbers are 3 and 5

Therefore, other natural number

A.T.Q.

So

A.T.Q.







When


When


Therefore, Numbers are 3 and 5
Q.6 Solve for 

[Foreign 2006]
Sol. 
A =
, B =
, C =
Since
![\Rightarrow D = {\left[ {8\left( {{a^2} - {b^2}} \right)} \right]^2} - 4 \times {\left( {a + b} \right)^2}16{\left( {a - b} \right)^2}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_5b9f140e845ca7049121d2b90f41d454.gif)

Therefore,![x = {{ - B} \over {2A}} = {{ - 8\left( {{a^2} - {b^2}} \right)} \over {2{{\left( {a + b} \right)}^2}}} = {{ - 4\left[ {\left( {a - b} \right)} \right]} \over {a + b}}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_cede7b7bb46296d554471581f028c8bd.gif)


A =



Since

![\Rightarrow D = {\left[ {8\left( {{a^2} - {b^2}} \right)} \right]^2} - 4 \times {\left( {a + b} \right)^2}16{\left( {a - b} \right)^2}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_5b9f140e845ca7049121d2b90f41d454.gif)

Therefore,
![x = {{ - B} \over {2A}} = {{ - 8\left( {{a^2} - {b^2}} \right)} \over {2{{\left( {a + b} \right)}^2}}} = {{ - 4\left[ {\left( {a - b} \right)} \right]} \over {a + b}}](http://dronstudy.com/wp-content/plugins/latex/cache/tex_cede7b7bb46296d554471581f028c8bd.gif)

Q.7 Two numbers differ by 3 and their product is 504. Find the numbers.
[Foreign 2006]
Sol. Let one number = x
Therefore, other number = x + 3
A.T.Q.





When
, other number
When
, other number
Therefore, Numbers are 21, 24 or – 24 , – 21
Therefore, other number = x + 3
A.T.Q.






When


When


Therefore, Numbers are 21, 24 or – 24 , – 21
Q.8 Rewrite the following as a quadratic equation in x and then solve for x. 

[AI 2006 C]
Sol. We have 















Q.9 Find the value of p so that the quadratic equation
has two equal roots.

[Delhi 2011]
Sol. We have 

Here
For equal roots
Since





But
[Because for p=0 quadratic equation becomes linear equation]
So,
is the root of the equation.


Here

For equal roots

Since






But

So,

Q.10 Find the value of m so that the quadratic equation
has two equal roots.

[AI 2011]
Sol. We have 

Here
For equal roots D = 0
Since




But
[Because for m = 0 quadratic equation becomes linear equation]
So,
is the root of the equation.


Here

For equal roots D = 0
Since





But

So,

Q.11 For what value of k does the quadratic equation
has two equal roots.

[AI 2011]
Sol. We have 
Here
For equal roots

![\Rightarrow {\left( 2 \right)^2}{\left[ {\left( {k - 5} \right)} \right]^2} - 4 \times \left( {k - 5} \right) \times 2 = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_75e9a8c1db23ed0a130673e72b16e85b.gif)






But
[Because for k = 5 quadratic equation becomes linear equation]
So
is the root of the equation.

Here

For equal roots


![\Rightarrow {\left( 2 \right)^2}{\left[ {\left( {k - 5} \right)} \right]^2} - 4 \times \left( {k - 5} \right) \times 2 = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_75e9a8c1db23ed0a130673e72b16e85b.gif)






But

So

Q.12 Find the value of k for which the roots of the quadratic equation
are equal.

[Delhi 2013]
Sol. We have 
Here
for real and equal roots

![\Rightarrow {\left[ {2\left( {k - 4} \right)} \right]^2} - 4x\left( {k - 4} \right) \times 2 = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_ef6c43962dc5c48cc749b174c9483d6d.gif)


But
[Because for k = 4 quadratic equation becomes linear equation]
So,
is the root of equation.

Here



![\Rightarrow {\left[ {2\left( {k - 4} \right)} \right]^2} - 4x\left( {k - 4} \right) \times 2 = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_ef6c43962dc5c48cc749b174c9483d6d.gif)


But

So,

Q.13 Solve for 

[Delhi 2013]
Sol. We have 



, 
or 








Q.14 Solve for 

[AI 2013]
Sol. We have 




or 







Q.15 Solve for 

[Foreign 2013]
Sol. We have 



or 






Three Marks Questions
Q.1 A passenger train takes 2 hours less for a journey of 300 km. It its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
[Delhi 2005 C, 2006]
Sol. Let usual speed of the train = x km / hr
Distance = 300 km
Time taken =
[Because
]
It speed = (x + 5) km/hr
Then time taken =
A.T.Q.








[reject x=-30 because speed cannot negative)
Therefore, usual speed = 25 km / hr
Distance = 300 km
Time taken =


It speed = (x + 5) km/hr
Then time taken =

A.T.Q.









Therefore, usual speed = 25 km / hr
Q.2 A speed of a boat in still water is 11 km / hr. It can go 12 km upstream and return downstream to the original point in 2 hr 45 minutes. Find the speed of the stream.
[AI 2006]
Sol. Let speed of the stream = x km / hr
Speed of the boat in still water = 11 km / hr
Therefore, up stream speed = (1 – x) km / hr
and downstream speed =(11 + x) km / hr
Distance = 12 km
Time taken for downstream direction =
Time taken for upstream direction =
A.T.Q.
2 hr 45 min








[Reject x=-5 because speed cannot be negative]
So x = 5 km/hr is the speed of the stream.
Speed of the boat in still water = 11 km / hr
Therefore, up stream speed = (1 – x) km / hr
and downstream speed =(11 + x) km / hr
Distance = 12 km
Time taken for downstream direction =

Time taken for upstream direction =

A.T.Q.










So x = 5 km/hr is the speed of the stream.
Q.3 A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km / hr less than that of the fast train. Find the speeds of the two trains..
[Foreign 2006]
Sol. Let speed of fast train = x km / hr
Speed of slow train = 
A.T.Q.








[rejected]
So, speed so of the that train = 50 km / hr
And speed of the slow train = 40 km / hr.


A.T.Q.









So, speed so of the that train = 50 km / hr
And speed of the slow train = 40 km / hr.
Q.4 Seven years ago Varun’s age was five times the square of Swati’s age. three years hence Swati’s age will be two-fifth of Varun’s age. Find their present ages.
[Delhi 2006 C]
Sol. Let Varun’s present age = x years.
And Swati’s age = y years
Case 1 : Seven years ago
Varun’s age was (x – 7) years and
A.T.Q.

………..(1)
Case II : Three years hence
Varun’s age will be (x + 3) years and
Swati’s age will be (y + 3) years
A.T.Q.

![\Rightarrow y + 3 = {2 \over 5}\left[ {5{{\left( {y - 7} \right)}^2} + 7 + 3} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_76cdd9c14d802e258a3bb7a09a4787e0.gif)








, 
y = 9 [Rejecting y=11/2 because age cannot be in fraction]
Putting y=9 in (1),We get


Therefore, Present age of Swati = 9 years and Varun = 27 years.
And Swati’s age = y years
Case 1 : Seven years ago
Varun’s age was (x – 7) years and
A.T.Q.


Case II : Three years hence
Varun’s age will be (x + 3) years and
Swati’s age will be (y + 3) years
A.T.Q.

![\Rightarrow y + 3 = {2 \over 5}\left[ {5{{\left( {y - 7} \right)}^2} + 7 + 3} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_76cdd9c14d802e258a3bb7a09a4787e0.gif)










y = 9 [Rejecting y=11/2 because age cannot be in fraction]
Putting y=9 in (1),We get


Therefore, Present age of Swati = 9 years and Varun = 27 years.
Q.5 A 2-digit number is such that product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
[AI 2006 C]
Sol. Let digit at unit’s place = x
And digit at ten’s place = y
Therefore, number = 10y + x.
A.T.Q.

And



[at using (1)]





[Reject x=-9]
When x = 2,
Therefore, number =92
And digit at ten’s place = y
Therefore, number = 10y + x.
A.T.Q.

And











When x = 2,

Therefore, number =92
Q.6 A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km / hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
[AI 2006 C]
Sol.Let uniform speed of the train = x km/hr
Distance = 90 km
Therefore, Time taken =
hr
If speed was (x + 15) km / hr
Then time taken =
A.T.Q.








[Rejecting x=-60 because speed cannot be negative]
Therefore, x = 45 km / hr
Hence uniform speed of the train = 45 km/hr
Distance = 90 km
Therefore, Time taken =

If speed was (x + 15) km / hr
Then time taken =

A.T.Q.









Therefore, x = 45 km / hr
Hence uniform speed of the train = 45 km/hr
Q.7 The difference of two numbers is 5 and the difference of their reciprocals is
Find the numbers.

[Delhi 2007]
Sol. Let one number is x
Therefore, other number = x + 5
A.T.Q.








Hence, numbers are -10, -5 or 5, 10
Therefore, other number = x + 5
A.T.Q.








Hence, numbers are -10, -5 or 5, 10
Q.8 By increasing the list price of a book by Rs. 10. A person can buy 10 less books for Rs. 1200. Find the original list price of the book.
[Delhi 2007]
Sol. Let list price of the book = Rs. X
Total cost = Rs. 1200
Therefore, number of books =
If list price of the book = Rs. (x+10)
Then number of books =
A.T.Q.









[Because price cannot be negative so rejected]
Therefore, list price of the book = Rs. 30
Total cost = Rs. 1200
Therefore, number of books =

If list price of the book = Rs. (x+10)
Then number of books =

A.T.Q.










Therefore, list price of the book = Rs. 30
Q.9 The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by
. Find the fraction.

[AI 2007]
Sol. Let Denominator = x
Therefore, numerator = x – 1
Fraction =
A.T.Q.








or
[Rejecting x = -7]

Fraction is

Therefore, numerator = x – 1
Fraction =

A.T.Q.











Fraction is


Q.10 The difference of squares of two natural number is 45. The square of the smaller number is four times the larger number. Find the numbers.
[AI 2007]
Sol.Let larger number be = x
Smaller number be = y
A.T.Q.

………….(1)
Also
[using (1)]




So larger numbers is 9.
Putting x=9 in (1),We get

[Rejecting y=-6]

So Numbers are: 9 and 6
Smaller number be = y
A.T.Q.


Also






So larger numbers is 9.
Putting x=9 in (1),We get



So Numbers are: 9 and 6
Q.11 Find the roots of the following equation


[Delhi 2008]
Sol. We have 






Therefore, required roots are 2 , 1







Therefore, required roots are 2 , 1
Q.12 Find the roots of the following equation 

[Delhi 2011]
Sol. We have 
Here a =
, b = -5 , c = 
Since





Or

So required roots are
or 

Here a =


Since






Or


So required roots are


Q.13 Solve the following quadratic equation for 

[AI 2012]
Sol. We have 
Here A = 1, B = - 4a , C =
Since D =



Therefore,


Therefore, required roots are

Here A = 1, B = - 4a , C =

Since D =




Therefore,



Therefore, required roots are

Q.14 If the sum of two natural number is 8 and their product is 15. Find the numbers
[AI 2012]
Sol. Let the first natural number = x
A.T.Q.
x + y = 8 ……………..(1)
Where, y is another natural number
So,First natural number = x
Second natural number =8 – x [By equation (1)]
A.T.Q.






or 
Now if x = 3 then another number is 5
If x = 5 then another number is 3
A.T.Q.
x + y = 8 ……………..(1)
Where, y is another natural number
So,First natural number = x
Second natural number =8 – x [By equation (1)]
A.T.Q.








Now if x = 3 then another number is 5
If x = 5 then another number is 3
Q.15 for what value of k, are the roots of the quadratic equation
equal ?

[Foreign 2013]
Sol. We have 
Here
For equal roots D = 0
Since


![\Rightarrow \left( {k - 4} \right)\left[ {k - 4 - 16} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_5a1017fbd93f33c6c154949e04c07646.gif)
or 
[ If k=4 then quadratic equation becomes linear Equation ]
is the required root.

Here

For equal roots D = 0
Since



![\Rightarrow \left( {k - 4} \right)\left[ {k - 4 - 16} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_5a1017fbd93f33c6c154949e04c07646.gif)




Five Marks Questions
Q.1 In a class test, the sum of the marks obtained by p in Mathematics and science is 28. Had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.
[Delhi 2008]
Sol. Let marks obtained in mathematics be x and marks obtained in science be y.
A.T.Q.
…………….(1)
Also
[using (1)]




A.T.Q.

Also








Therefore, marks obtained in mathematics = 12 or 9
If marks obtained in mathematics = 12


If marks obtained in mathematics = 9


Q.2 The sum of the areas of two squares is 640m2.If the difference in their perimeters be 64m. Find the sides of the two squares.
[Delhi 2008C, Delhi 2008]
Sol. Let side of bigger square = x m
And side of smaller square = y m
A.T.Q.
…………….(1)
And



Substituting in equation (1), we get





or 
If y = 8, then
Therefore, sides of squares are 8 m and 24 m
And side of smaller square = y m
A.T.Q.

And




Substituting in equation (1), we get







If y = 8, then

Therefore, sides of squares are 8 m and 24 m
Q.3 A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream and than to return downstream to the same spot. Find the speed of stream .
[AI 2008]
Sol. Speed of boat = 18 km / hr
Let speed of stream = x km/hr
Upstream speed
km/hr
Downstream speed
km/hr
Time taken to cover 24 km upstream =
hrs.
Time taken to cover 24 km downstream =
hrs.
A.T.Q.








x = 6, x =- 54
[ Because speed cannot be negative]
Therefore, speed of the stream = 6 km / hr
Let speed of stream = x km/hr
Upstream speed

Downstream speed

Time taken to cover 24 km upstream =

Time taken to cover 24 km downstream =

A.T.Q.










Therefore, speed of the stream = 6 km / hr
Q.4 Two water taps together can fill the tank in
hrs. The tap of larger diameter takes 10 hrs. less than smaller one to fill the tank. Find the time in which each tap can separately fill the tank.

[AI 2008]
Sol. Let the two water taps fill the tank separately in (x) and (x-10) hrs
Since, both the taps together fill the tank in
hr
Therefore,








Therefore,
, 25
Rejecting
, we get x = 25
Therefore, two taps separately fill the tank in (x) and (x – 10) i.e. 25 hrs and 15 hrs.
Since, both the taps together fill the tank in

Therefore,









Therefore,

Rejecting

Therefore, two taps separately fill the tank in (x) and (x – 10) i.e. 25 hrs and 15 hrs.
Q.5 A peacock is sitting on the top of a pillar which is 9 m high. From a point 27 m away from the bottom of the pillar a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal at what distance from the hole is the snake caught?
Let the distance covered by peacock 
Therefore, Distance covered by snake
In right angle





And
Therefore, the snake is caught at 12 m from the hole.

Therefore, Distance covered by snake

In right angle






And

Therefore, the snake is caught at 12 m from the hole.
Q.6 A person on tour has Rs. 4200 for his expenses. If he extends his tour for 3 days he has to cut down his daily expenses by Rs. 70. Find the duration of the tour.
[AI 2008]
Sol. Let number of days tour = x
Total expenses = Rs. 4200
Therefore, daily expenses = Rs.
If number of days of tour = x + 3
Then daily expenses = Rs.
A.T.Q.









or
[Rejecting x=-15 because days cannot be negative]

So, number of days of tour is 12 days.
Total expenses = Rs. 4200
Therefore, daily expenses = Rs.

If number of days of tour = x + 3
Then daily expenses = Rs.

A.T.Q.












So, number of days of tour is 12 days.
Q.7 A trader bought a number of articles for Rs. 900,five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought .
[Foreign 2009]
Sol. Number of articles = x
Cost of each articles = y
A.T.Q.

…………….(1)








(rejected), x = 75
Therefore, No of articles = 75
Cost of each articles = y
A.T.Q.











Therefore, No of articles = 75
Q.8 Two years ago a man’s age was three times the square of his son’s age. Three years hence his age will be four times his son’s age. Find their present ages.
[Foreign 2009]
Sol. Age of man = x years
Age of son = y years
A.T.Q.



……………….(1)
Also
………………..(2)
[ Using (1)]




(rejected, age can never be in fraction form)

Putting y=5 in (1),We get


Hence, Present age of Son is 5 years and Present age of Father is 29 years.
Age of son = y years
A.T.Q.




Also









Putting y=5 in (1),We get


Hence, Present age of Son is 5 years and Present age of Father is 29 years.
Multiple choice questions
Q.1 The roots of the equation
where p is constant, are

[Delhi 2011]
(a) 
(b)
(c)
(d)

(b)

(c)

(d)

Sol. (b) We have









Q.2 The roots of the equation
, where m is a constant, are -

[AI 2011]
(a) 
(b)
(c)
(d)

(b)

(c)

(d)

Sol. We have 

![\Rightarrow x\left[ {x - \left( {3 + m} \right)} \right] + m\left[ {x - \left( {m + 3} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_bb4a408b3e05b8bce6f872143b6beb01.gif)
![\Rightarrow \left( {x + m} \right)\left[ {x - \left( {3 + m} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_2193e752888dd8a42835c82755dc36ff.gif)



![\Rightarrow x\left[ {x - \left( {3 + m} \right)} \right] + m\left[ {x - \left( {m + 3} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_bb4a408b3e05b8bce6f872143b6beb01.gif)
![\Rightarrow \left( {x + m} \right)\left[ {x - \left( {3 + m} \right)} \right]](http://dronstudy.com/wp-content/plugins/latex/cache/tex_2193e752888dd8a42835c82755dc36ff.gif)

Q.3 The roots of the quadratic equation
, Where
is a constant, are


[Foreign 2011]
(a) 
(b)
(c)
(d)

(b)

(c)

(d)

Sol. (b)

![\Rightarrow x\left( {x + \alpha+ 6} \right) - \left( {x + 1} \right)\left[ {\left( {x + \alpha + 6} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_ea617581cd9529a356a8a843c69d35d7.gif)
![\Rightarrow \left[ {x - \left( {\alpha + 1} \right)} \right]\left[ {x + \left( {\alpha + 6} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_3e061963b6ee5fa6030c85e75944c95e.gif)



![\Rightarrow x\left( {x + \alpha+ 6} \right) - \left( {x + 1} \right)\left[ {\left( {x + \alpha + 6} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_ea617581cd9529a356a8a843c69d35d7.gif)
![\Rightarrow \left[ {x - \left( {\alpha + 1} \right)} \right]\left[ {x + \left( {\alpha + 6} \right)} \right] = 0](http://dronstudy.com/wp-content/plugins/latex/cache/tex_3e061963b6ee5fa6030c85e75944c95e.gif)

Q.4 If the quadratic equation
has two equal roots, then the values of m are

[Foreign 2012]
(a) 
(b) 0, 2
(c) 0, 1
(d) -1, 0

(b) 0, 2
(c) 0, 1
(d) -1, 0
Sol. We have 
Here


For equal roots D = 0




Here



For equal roots D = 0



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