(A) 3:5
(B) 4:5
(C) 7:9
(D) Cannot be determined
Ans: A
Solution:
Consider the figure given below
Triangle PQR with Sides PQ, QR and PR Having Lengths
We have, PL and QM are perpendiculars on QR and PR respectively. Hence triangles PLR and QMR are right angled triangles sharing the common acute angle R. Hence they must be similar. Further, we note that right angled triangles MQR and LQX share the common acute angle XQL. Hence triangle XQL must be similar to triangle MQR and thereby also similar to triangle RLP.
Therefore
(1)
Now, PQ, QR, and PR are of lengths 13, 14 and 15 units respectively .Hence , using Heron’s formula, we get that the area of the triangle is 84 square units. But area of triangle
Hence units. By Pythagoras theorem we get square units. Hence .
Thus using (A) and the values of LR and PR, we get, the required ratio as 3:5 as given in option (A).
Q: 9. Given that p, q, x and y are integers such that How may ordered quadruplets (p, q, x, y) satisfy the given equation?
(A) Only one
(B) Eight
(C) Sixteen
(D) Infinitely many
Ans: B
Solution
The given equation is
or
or
From the above equation we can conclusively say that
pair 1
Or
— pair 2
Solving pair 1, we get
Since p, q, x, y are integers, the only possible values are tabulated below.
P | Q | X | Y |
1 | 0 | 1 | 0 |
-1 | 0 | -1 | 0 |
0 | 1 | 0 | -1 |
0 | -1 | 0 | 1 |
Similarly, by solving the 2nd pair, we get
P | Q | X | Y |
0 | 1 | 0 | 1 |
0 | -1 | 0 | -1 |
1 | 0 | -1 | 0 |
-1 | 0 | 1 | 0 |
Hence the number of possible Solutions in integers is 8 as given by option (B).
Q: 10. Vyom, a student of Jago-Re Higher Secondary School calculated the value of as . Then which of the following is true?
(A) He was correct
(B) He made an error of
(C) He made an error of
(D) He made an error of
Ans: B
Solution:
By applying method of induction,
For ,
Thus , we can write .
Hence, option (B).
Q: 11. Let us consider a positive integer N such that the factors of N, 1 and N inclusive, are given by a1, a2 ….. an. Also let a1 + a2 + …. an = p. Then which among is the value of ?
(A)
(B)
(C)
(D) None of these
Ans: C
Solution
The given expression is =
This is because the denominator comprises of the of the factors of N which is N and the numerator comprises of the sum of the complements of in N, which basically boils down to finding the sum of the factors of N .
Hence the given expression has a value as given by option (C).
Q: 12. A quadratic function has two values ‘α’ and ‘β’ between 4 and 7 such that at ‘α’ and ‘β’,. Find the probable value of among the following :
(A)
(B)
(C)
(D)
Ans: B
Solution:
As at ‘α’ and that means they are the root of the quadratic function f(x) so the quadratic function can be rewritten as
Moreover, all are positive quantities. (As α and β lie between 4 and 7). Applying Arithmetic Mean Geometric Mean , we have
...... Eqn (1)
and
...... Eqn (2)
Taking Eqn (1),
or, ),
or,
Similarly from Eqn (2),
Combining both
or,
or,
or,
or,
Clearly options (a), (c), and (d) are greater than 81/16 so the only possible value of
Option (B)
Q: 13. Find the minimum value of the polynomial
(A)
(B)
(C)
(D)
Ans: A
Solution
The given polynomial may be rewritten as
Or,
Thus the given polynomial consists of the sum of three non-negative definite terms and one. Hence the minimum value 1 is attained when each of the other three non negative definite terms become zero at .Hence the answer is option (A).