KENDRIYA VIDYALAYA SANGATHAN th
13 KVS JMO 2010 Solution
B.S.Manral
Question1.Let a, b, c ∈ R, a ≠ 0, such that a and 4a+3b+2c have the 2
same sign. Show that the equation ax +bx+c=0 cannot have both the roots in the interval (1,2). Answer1.Let us assume that both the roots lie in (1,2). Case1:
When a >0 the shape of the graph will be like figure (i) and Putting x=1 and x=2 will give a+b+c > 0
…(1)
4a+2b+ c> 0
…(2)
From (1) we get b+c > -a
…(3)
adding corresponding sides of (2) and (3) 4a+3b+2c > -a
as a > 0, 4a+3b+2c,may not be strictly > 0 but it is
given that a and 4a+3b+2c have the same sign, which is a contradiction. Case2:
When a < 0, the shape of the graph will be like figure (ii) and the value of 4a+3b+2c < -a, may have have value > 0, as –ve of a negative number is a positive number , again a contradiction. Hence both the roots of the equation cannot be in the interval (1,2).
Question 2.(a) Find all the integers which are equal to 11 times the
sum of their digits. (b) Prove that
+
is rational number.
Answer 2(a)
It can be easily proved that such numbers can be only of 3-digits. Let the number be 100H+10T+U
∴
11(H+T+U) = 100H+10T+U
⇒ T=89H-10U
…. (1)
As H, T, U are all ≥ 0 and ≤ 9 Hence the only solution of the above equation is H=1, T=9, U=8. Therefore the only number is 198. Answer 2.(b)
Let
+
=p
Cubing we get
2+ +2- +3 ( 4-3p= p
+
3
3
p +3p=4 2
p (p +3)=4 => p=1 which is rational. OR
3
) =p
3 3 We can easily show that = } and = } Hence the value of the given expression is 1, which is a rational number, proved. Question 3. A circle centered at A with radius 1 unit and another
circle centered at B with radius 4 units touch each other externally. A third circle is drawn to touch the first two circles and one of their common external tangents. What is the radius of the third circle? Answer 3.Join AB,AE,BE and draw AL,EM,BN perpendicular to the
tangent, through E and through A draw CD and AK parallel to the tangent to meet AL,BN at C,D and to BN at K. Let the radius of the third circle be r then AC=1-r, AE=1+r,BD=4-r,BE=4+r, BK=3,AB=5 by applying Pythagoras theorem in ∆ACD we get CE=2 r
r ,
ED=4
r
hence AK=6
again applying Pythagoras theorem in ∆AKB we get
r= 4/9.
Question4. Given three non collinear points A,H and G . Construct a
triangle with A as vertex, H as orthocenter and G as centroid.
Answer 4.A,G and H are the three given points Step1. Join AH and AG Step2. Produce AG up to M
such that GM=1/2 AG Step3. Draw a line l
through M and perpendicular to AH. Step4. Draw a line n,
perpendicular to
l at
M,
which meet GH produced at O. O is the circum centre of the required triangle.(Euler’s line). Step5. Taking O as centre and radius OA, draw a circle which
intersect line
l at
B and C, join AB and AC.
∆ABC is the required triangle whose H is the orthocenter and G is
the centroid. 2
Question5. In∆ABC, ∠ A is twice the ∠B. Prove that a =b(b+c). Answer 5. Draw the bisector of ∠BAC to meet BC at L.
Now ∆ABC~∆LAC by AA corollary BC AC
=
AC LC
a b
=
b LC
LC=
b2
……(1)
a
Now as AL is the angle bisector of ∠CAB BL
=
AB
LC AC
Adding 1 to both the sides BC LC a LC
=
AB AC
=
cb
AC
….(2)
b
Multiplying the corresponding sides of equation 1 and equation 2 we get a=
cb b
×
b2 a
2
a =b(b+c) Proved. 2
Question6. The equation x +px+q=0, where p and q are integers , has rational roots . Prove that the roots are integers. Answer 6. ∵ p, q are integers and the roots of the equation are
rational hence 2
Discriminator=P -4q will be a perfect square 2
Let p -4q=r 2
2
⇒ p - r =4q
2
r∈Z
⇒ p and r are either both even or both odd Now the solution of the equation X=
p r
2
Clearly when p and r are both even or both odd
p r
is even
Hence the roots are integers.
Question7. ∆ABC is right angled with ∠ ACB as its right angle, 0
m∠ABC=60 , and AB=10 units. Let P be a point chosen randomly in side ∆ ABC. Extend BP to meet AC at D. What is the probability that BD > 5√ 2.
Answer 7. Take a point E on 0
AC such that∠BEC=45 , applying trigonometry we can easily show that BE=5√ 2 and BC=5 Now for BD to be greater than 5√2, P must move within the ∆AEB
Hence the probability of BD>5 √ 2 =ar(∆AEB)/ar(∆ABC) =
=
AE AC
5 3 5 5 3
Question8. Prove that the sum of the hypotenuse and the altitude of the right angled triangle dropped on the hypotenuse exceed the half perimeter of the triangle. Answer 8.Let ∆ABC is right angled at B, Draw BL perpendicular to
AC. Now in∆ALB
AL+BL>AB
(1)
Triangle inequality In ∆BLC
LC+BL>BC (2)
–do-
Adding (1) and (2) we get AC+2BL>AB+BC Adding AC to both the sides we get 2AC+2BL>AB+BC+AC =>AC+BL>1/2(AB+BC+AC) Proved.
Question 9.How many times is digit zero written when
listing all numbers from 1 to 3333? Answer 9.
From - to
No of Zeros
1-100
11
101-990
178
991-1000
3
1001-2000
300
2001-3000
300
3001-3300
159
3301-3333
12
Total
963 9
25
49
81
Question10. Find out the remainder when x+x +x +x +x
is
3
divided by x -x. Answer 10. Using remainder theorem and putting x3=x
and repeating the process until the degree of remainder < the degree of the divisor We get the remainder=5x
--x---