det

DET

Read the following passage carefully and answer question No. (1 - 5):

If F (x) = , where gi, fi, hi, (i = 1, 2, 3) are functions of x. To differentiate F (x), we can

differentiate one row (or column) at a time, keeping others unchanged.

F¢ (x) =

or F¢ (x) = .

1. If F (x) = , then the value of F¢ (a) will be

(A) 0 (B) 1(C) 2 (D) none of these

2. If a, b, c are in A.P. and f (x) = , then f¢ (x) is

(A) 0 (B) 1

(C) a + bc (D)

3. If f (x) = , then f¢ (0) is

(A) 1 (B) - 1(C) 2 (D) 1/3

4. If F (x), G (x), H (x) are 3 polynomials of degree 2. Then f (x) = is a polynomial

of degree(A) 2 (B) 3(C) 4 (D) 0

5. If f (x) = , then f² (a) is

(A) 1 (B) 2(C) 3 (D) 0

1. A 2. A 3. C 4. D5. D

1. f¢(X) = + +

F¢(x) = 0

2. f¢(x) =

R1 R1 + R3

f¢(x) = 0 +

(since a + c = 2b) f¢(x) = 0.

3. f¢(x) = +

+

f¢(0) = 0 + 0 +

= 2(1 – 0) = 2.

4. f¢(x) =

f¢(x) = 0 f(x) = cost. deg(f(x)) = 0.

5. f¢(x) = 0 +

f¢(x) = 0 + + 0

f¢¢(a) = = 12 0 = 0.

Passage III

Read the following comprehension carefully:

A determinant is called cyclic if it follows the arrangement symmetrically with a, b and c e.g.

. Now if we increase the degree of any row in this determinant its value is increased

by an expression which is also cyclic and increases the degree of the value of determinant then use this statement to solve the following questions.

Answer the following questions:

1. The value of is equal to

(A) (a - b) (b - c) (c - a) (B) (a - b) (b - c) (c - a) (a + b + c)(C) (a - b) (b - c) (c - a) (ab + bc + ca) (D) (a - b) (b - c) (c - a) abc





1. B 2. C 3. D

1.

C1 C1 – C2 and C2 C1 – C3

=

= 1(a – b) (b – c) [b2 + bc + c2 – a2- – ab – b2]= (a – b) (b – c) (a + b + c)

2.

= 1 (a – b) (b – c) [(a + b) (a2 + ab + b2) – (b + c) (b2 + bc + c2)] = (a – b ) (b – c) (c – a) (ab + bc + ca) = (a – b) (b – c) (c – a) (ab + bc + ca).

3.

= C1 C1 – C2 and C2 C2 – C3

= abc (a – b) (b – c) (c – a).

det

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