# Free Objective Test 02 Practice Test - 11th and 12th

The system of equations

X+2y+3z=4

2x+3y+4z=5

3x+4y+5z=6   has
A.

Many solutions

B.

No solution

C.

Unique solution

D.

Atmost two solutions

#### SOLUTION

Solution : A  A.

f=3 and g=-5

B.

f=-3 and g=-5

C.

f=-3 and g=-9

D.

f=-5 and g=9

#### SOLUTION

Solution : D  A.

N

B.

N2

C. Zero
D.

1

#### SOLUTION

Solution : C If f, g and h are differentiable functions of x and ⎜ ⎜x⎟ ⎟=∣ ∣ ∣fgh(xf)'(xg)'(xh)'(x2f)''(x2g)''(x2h)''∣ ∣ ∣,then '⎜ ⎜x⎟ ⎟ is

A.

∣ ∣ ∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣ ∣ ∣

B.

∣ ∣ ∣fghf'g'h'(x3f'')(x3h'')(x3h'')∣ ∣ ∣

C.

∣ ∣ ∣fghf'g'h'(x3f'')''(x3h'')''(x3h'')''∣ ∣ ∣

D. 0

#### SOLUTION

Solution : A

∣ ∣ ∣fghxf'+fxg'+gxh'+h4xf'+2f+x2f''4xg'+2g+x2g''4xh'+2h+x2h''∣ ∣ ∣Operating R2R1R1; R3R34R2+2R1and shifting x of R2 to R3⎜ ⎜x⎟ ⎟=∣ ∣ ∣fghf'g'h'x3f''x3g''x3h''∣ ∣ ∣'⎜ ⎜x⎟ ⎟=0+0+∣ ∣ ∣fghf'g'h'(x3f'')'(x3g'')'(x3h'')'∣ ∣ ∣

If ω is a cube root of unity and Δ = 12ωωω2, then Δ2 is equal to

A. ω
B. ω2
C. 1
D. ω

#### SOLUTION

Solution : D

Since Δ = ω22ω2 = ω2. Therefore Δ2 = ω4 = ω.

If Δ1 = 10ab and Δ2 = 10cd, then Δ2Δ1 is equal to

A. ac
B. bd
C. (b  a)(d  c)
D. abc

#### SOLUTION

Solution : B

Δ2Δ1 = 10cd 10ab = 10c+adbd = bd.

If A1, B1, C1.... are respectively the co-factors of the elements a1, b1, c1.... of the determinant Δ = ∣ ∣a1b1c1a2b2c2a3b3c3∣ ∣, then B2C2B3C3 =

A. a1Δ
B. a1a3Δ
C. (a1+b1)Δ
D. None of these

#### SOLUTION

Solution : A

B2 = a1c1a3c3 = a1c3  c1a3C2 = a1b1a3b3 = (a1b3  a3b1)B3 = a1c1a2c2 = (a1c2  a2c1)C3 = a1b1a2b2 = (a1b2  a2b1)B2C2B3C3 = a1c3  a3c1(a1b3  a3b1)(a1c2  a2c1)a1b2  a2b1                  =a1c3a1b3a1c2a1b2+a1c3a3b1a1c2a2b1+a3c1a1b3a2c1a1b2+a3c1a3b1a2c1a2b1=a12(b2c3b3c2)+a1b1(c3a2+a3c2)+a1c1(a3b2+a2b3)+c1b1(a3a2a2a3) = a1Δ .

If A = ∣ ∣563432473∣ ∣, then cofactors of the elements of 2nd row are

A. 39, -3, 11
B. -39, 3, 11
C. -39, 27, 11
D. -39, -3, 11

#### SOLUTION

Solution : C

C21 = (1)2+1(18+21) = 39C22 = (1)2+2(15+12) = 27C23 = (1)2+3(35+24) = 11 .

If the system of linear equation x+2ay+az = 0, x+3by+bz = 0, x+4cy+cz = 0 has a non zero solution, then a, b, c

A. Are in A.P.
B. Are in G. P.
C. Are in H. P.
D. Satisfy a +2b + 3c = 0

#### SOLUTION

Solution : C

∣ ∣12aa13bb14cc∣ ∣ = 0,     [C2  C22C3] ∣ ∣10a1bb12cc∣ ∣ = 0,     [R3  R3R2, R2  R2R1] ∣ ∣10a0bba02cbcb∣ ∣ = 0 ; b(cb)(ba)(2cb) = 0
On simplification, 2b = 1a+1c
a, b, c are in Harmonic progression.

The system of equations x + y + z =2, 3x  y + 2z =6 and 3x + y + z =18 has

A. A unique solution
B. No solutions
C. An infinite number of solutions
D. Zero solution as the only solution

#### SOLUTION

Solution : A

Given system of equation can be written as 111312311xyz = 2618
On solving the above system we get the unique solution x = -10, y = -4, z = 16.

The value of the determinant  ∣ ∣ ∣ ∣loga(xy)loga(yz)loga(zx)logb(yz)logb(zx)logb(xy)logc(zx)logc(xy)logc(yz)∣ ∣ ∣ ∣ is

A. 1
B. -1
C. logcxyz
D. None of these

#### SOLUTION

Solution : D

Applying C1C1+C2+C3
Then, ∣ ∣ ∣ ∣0loga(yz)loga(zx)0logb(zx)logb(xy)0logc(xy)logc(yz)∣ ∣ ∣ ∣=0

If f(x)=∣ ∣ ∣1xx+12xx(x1)x(x+1)3x(x1)x(x1)(x2)x(x21)∣ ∣ ∣, then f(200) is equal to

A. 1
B. 0
C. 200
D. -200

#### SOLUTION

Solution : B

f(X)=x(x+1)∣ ∣1112xx1x3x(x1)(x1)(x2)x(x1)∣ ∣=x(x+1)(x1)∣ ∣1112xx1x3xx2x∣ ∣
Applying C2C2C1 and C3C3C1 then
f(x)=x(x+1)(x1)∣ ∣1002xx1x3x2x22x∣ ∣=x2(x+1)2(x1)∣ ∣1002x113x22∣ ∣=0f(200)=0

If a, b, c are sides of a triangle and ∣ ∣ ∣a2b2c2(a+1)2(b+1)2(c+1)2(a1)2(b1)2(c1)2∣ ∣ ∣=0, then

A. ΔABC s an equilateral triangle
B. ΔABC is right angled isosceles triangle
C. ΔABC is an isosceles triangle
D. ΔABC None of the above

#### SOLUTION

Solution : C

Δ=∣ ∣ ∣a2b2c2(a+1)2(b+1)2(c+1)2(a1)2(b1)2(c1)2∣ ∣ ∣
Applying R2R2R3
=4∣ ∣ ∣a2b2c2abc(a1)2(b1)2(c1)2∣ ∣ ∣
Applying R3R3R1+2R2
Δ=4∣ ∣a2b2c2abc111∣ ∣=4(ab)(bc)(ca)=0
If   a – b = 0 or b – c = 0 or c – a = 0
a = b or b = c or c = a
ΔABC is an isosceles triangle.

∣ ∣0aba0cbc0∣ ∣=

A. -2abc
B. abc
C. 0
D. a2+b2+c2

#### SOLUTION

Solution : C

∣ ∣0aba0cbc0∣ ∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).

If a,b and care non zero numbers, then Δ=∣ ∣ ∣b2c2bcb+cc2a2cac+aa2b2aba+b∣ ∣ ∣ is equal to

A. abc
B. a2b2c2
C. ab+bc+ca
D. None of these

#### SOLUTION

Solution : D

Multiplying R1 by a,R2 by b and R3 by c, we have
Δ=1abc∣ ∣ ∣ab2c2abcab+aca2bc2abcbc+aba2b2cabcac+bc∣ ∣ ∣=a2b2c2abc∣ ∣bc1ab+acac1bc+abab1ac+bc∣ ∣=abc∣ ∣ ∣bc1abac1abab1ab∣ ∣ ∣{by C3C3+C1}=abc.ab∣ ∣bc11ca11ab11∣ ∣=0, [Since C2C3].
Trick : Put a=1, b=2, c=3 and check it.