# Free Objective Test 01 Practice Test - 11th and 12th

A.

B.

C.

D.

#### SOLUTION

Solution : D

The cofactor of the element '4' in the determinant ∣ ∣ ∣ ∣1351234280110211∣ ∣ ∣ ∣ is

A. 4
B. -10
C. 10
D. -4

#### SOLUTION

Solution : C

The cofactor of element 4, in the 2nd row and 3rd column is
=(1)2+3 ∣ ∣131801021∣ ∣ = - {1(-2)-3(8-0)+1.16}
=10.

x + ky  z = 0, 3x  ky z = 0 and x  3y + z = 0 has non-zero solution for k =

A. -1
B. 0
C. 1
D. 2

#### SOLUTION

Solution : C

It has a non-zero solution if
∣ ∣1k13k1131∣ ∣ = 0  6k + 6 = 0  k = 1 .

The number of values of k for which the system of equations (k+1)x + 8y = 4k, kx + (k+3)y = 3k1 has infinitely many solutions, is

A. 0
B. 1
C. 2
D. Infinite

#### SOLUTION

Solution : B

For infinitely many solutions, the two equations must be identical
k+1k = 8k+3 = 4k3k1 (k+1)(k+3) = 8k and 8(3k-1) = 4k(k+3)
k24k+3 = 0 and k23k+2 = 0 .
By cross multiplication, k28+9 = k32 = 13+4
k2 = 1 and k=1 ; k=1.

If Δ(x) = ∣ ∣ ∣xnsin xcos xn!sinnπ2cosnπ2aa2a3∣ ∣ ∣, then the value of dndxn[Δ(x)] at x=0 is

A. -1
B. 0
C. 1
D. Dependent of a

#### SOLUTION

Solution : B

dndxn[Δ(x)] = ∣ ∣ ∣ ∣dndxnxndndxnsin xdndxncos xn!sin(nπ2)cos(nπ2)aa2a3∣ ∣ ∣ ∣=∣ ∣ ∣ ∣n!sin(x+nπ2)cos(x+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣ ∣ ∣ ∣ [Δn(x)]x=0 = ∣ ∣ ∣ ∣n!sin(0+nπ2)cos(0+nπ2)n!sin(nπ2)cos(nπ2)aa2a3∣ ∣ ∣ ∣ = 0              1Since R1  R2}.

If f(n)=αn+βn and ∣ ∣ ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣ ∣ ∣=k(1α)2(1β)2(αβ)2, then k is equal to

A. 1
B. -1
C. αβ
D. αβγ

#### SOLUTION

Solution : A

Δ=∣ ∣ ∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣ ∣ ∣   =∣ ∣1111αβ1α2β2∣ ∣×∣ ∣1111αβ1α2β2∣ ∣
Applying C2C2C3C3C1,
∣ ∣1001α1β11α21β21∣ ∣2    =(α1)2(β1)2(βα)2    =(1α)2(1β)2(αβ)2
Hence, k=1

If Δ1=∣ ∣111abca2b2c2∣ ∣,Δ2=∣ ∣1bca1cab1abc∣ ∣, then

A. Δ1+Δ2=0
B. Δ1+2Δ2=0
C. Δ1=Δ2
D. Δ1=2Δ2

#### SOLUTION

Solution : A

Δ1=∣ ∣111abca2b2c2∣ ∣, & Δ2=∣ ∣1bca1cab1abc∣ ∣
In Δ2,Transposing the determinant
Δ2=∣ ∣111bccaabaac∣ ∣
C1aC1,C2bC2,C3CC3
Δ2=1abc∣ ∣abcabcabcabca2b2c2∣ ∣
Taking abc common from R2
Δ2=∣ ∣abc111a2b2c2∣ ∣
R1R2
Δ2=∣ ∣111abca2b2c2∣ ∣
Δ2=Δ1Δ2+Δ1=0

If α,β are non real numbers satisfying x31=0 then the value of ∣ ∣λ+1αβαλ+β1β1λ+α∣ ∣ is equal to

A. 0
B. λ3
C. λ3+1
D. λ31

#### SOLUTION

Solution : B

x31=0x=1,ω,ω2
Here, α=ω,β=ω2∣ ∣ ∣λ+1ωω2ωλ+ω21ω21λ+ω∣ ∣ ∣
Applying C1C1+C2+C3, then
∣ ∣ ∣λωω2λλ+ω21λ1λ+ω∣ ∣ ∣
Applying R2R2R1 and R3R3R1, then we get
∣ ∣ ∣λωω20λ+ω2ω1ω201ωλ+ωω2∣ ∣ ∣=λ((λ+ω2ω)(λ+ωω2)(1ω)(1ω2))=λ(λ2)=λ3

If Dk=∣ ∣ ∣1nn2kn2+n+1n2+n2k1n2n2+n+1∣ ∣ ∣ and nk=1Dk=56, then n equals

A. 4
B. 6
C. 8
D. None of these

#### SOLUTION

Solution : D

nk=1Dk=∣ ∣ ∣nk=11nn2nk=1kn2+n+1n2+n2nk=1knk=11n2n2+n+1∣ ∣ ∣
=∣ ∣ ∣nnnn2+nn2+n+1n2+nn2n2n2+n+1∣ ∣ ∣=56
=Applying C2C2C1C3C3C1, then
∣ ∣n00n2+n10n0n+1∣ ∣=56n(n+1)=56=7×8n=7

If sin 2x = 1, then ∣ ∣0cosxsinxsinx0cosxcosxsinx0∣ ∣2 equals

A. 3
B. 0
C. 1
D. None of these

#### SOLUTION

Solution : B

sin2x=1thenX=π4
Then, ∣ ∣0cosxsinxsinx0cosxcosxsinx0∣ ∣2=∣ ∣ ∣ ∣ ∣012121201212120∣ ∣ ∣ ∣ ∣2
=(12×12×12)∣ ∣011101110∣ ∣2
=18{01(01)1(1)}2
=0

If A+B+C=π, then ∣ ∣ ∣sin(A+B+C)sinBcosCsinB0tanAcos(A+B)tnaA0∣ ∣ ∣ is equal to

A. 1
B. 0
C. -1
D. 2

#### SOLUTION

Solution : B

Δ=∣ ∣sinπsinBcosCsinB0tanAcos(πC)tnaA0∣ ∣=∣ ∣0sinBcosCsinB0tanAcosCtanA0∣ ∣
=0 (Δ is skew symmetric)

Let f(x)=∣ ∣cosxsinxcosxcox2xsin2x2cos2xcos3xsin3x3cos3x∣ ∣, then f(π2) is equal to

A. 8
B. 6
C. 4
D. 2

#### SOLUTION

Solution : C

f(x)=∣ ∣sinxsinxcosx2sin2xsin2x2cos2x3sin3xsin3x3cos3x∣ ∣+∣ ∣cosxcosxcosxcos2x2cos2x2cos2xcos3x3cos3x3cos3x∣ ∣+∣ ∣cosxsinxsinxcos2xsin2x4sin2xcos3xsin3x9sin3x∣ ∣
f(π2)=∣ ∣110002310∣ ∣+0+∣ ∣011100019∣ ∣=2(13)+0+1(91)=4+8=4

If a,b,c are  non – zero real numbers and if the equations (a-1) x = y + z, (b -1)y = z + x, (c - 1)z = x + y has a non trivial solution, then ab + bc + ca is equal to

A. a +b + c
B. abc
C. 1
D. None of these

#### SOLUTION

Solution : B

For non trivial solution
∣ ∣a11111b1111c∣ ∣=0
Applying C1C1C3 and C2C2C3, then
∣ ∣a010b1cc1c∣ ∣=0a(b+bcc)0+c(0b)=0ab+bc+ca=abc

∣ ∣1aba1cbc1∣ ∣=

A. 1+a2+b2+c2
B. 1a2+b2+c2
C. 1+a2+b2c2
D. 1+a2b2+c2

#### SOLUTION

Solution : A

∣ ∣1aba1cbc1∣ ∣=1(1+c2)a(a+bc)+b(ac+b)=1+a2+b2+c2.

If pλ4+qλ3+rλ+5λ+t=∣ ∣λ2+3λλ1λ+3λ+12λλ4λ3λ+43λ∣ ∣, the value of t is

A. 16
B. 18
C. 17
D. 19

#### SOLUTION

Solution : B

Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0 in the given equation, we have
t=∣ ∣013124340∣ ∣=12+30=18.