12th Grade > Mathematics
DETERMINANTS MCQs
Total Questions : 60
| Page 2 of 6 pages
Answer: Option B. -> 0
:
B
dndxn[Δ(x)]=∣∣
∣
∣
∣∣dndxnxndndxnsinxdndxncosxn!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∣∣
∣
∣
∣∣=01SinceR1≡R2}.
:
B
dndxn[Δ(x)]=∣∣
∣
∣
∣∣dndxnxndndxnsinxdndxncosxn!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∣∣
∣
∣
∣∣=01SinceR1≡R2}.
Answer: Option B. -> 18
:
B
Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0in the given equation, we have
t=∣∣
∣∣0−1312−4−340∣∣
∣∣=−12+30=18.
:
B
Since it is an identity in λ so satisfied by every value of
λ. Now put λ=0in the given equation, we have
t=∣∣
∣∣0−1312−4−340∣∣
∣∣=−12+30=18.
Answer: Option C. -> 0
:
C
∣∣
∣∣0a−b−a0cb−c0∣∣
∣∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).
:
C
∣∣
∣∣0a−b−a0cb−c0∣∣
∣∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).
Answer: Option D. -> None of these
:
D
Applying C1→C1+C2+C3
Then, ∣∣
∣
∣
∣∣0loga(yz)loga(zx)0logb(zx)logb(xy)0logc(xy)logc(yz)∣∣
∣
∣
∣∣=0
:
D
Applying C1→C1+C2+C3
Then, ∣∣
∣
∣
∣∣0loga(yz)loga(zx)0logb(zx)logb(xy)0logc(xy)logc(yz)∣∣
∣
∣
∣∣=0
Answer: Option B. -> 0
:
B
f(X)=x(x+1)∣∣
∣∣1112xx−1x3x(x−1)(x−1)(x−2)x(x−1)∣∣
∣∣=x(x+1)(x−1)∣∣
∣∣1112xx−1x3xx−2x∣∣
∣∣
Applying C2→C2−C1 and C3→C3−C1 then
f(x)=x(x+1)(x−1)∣∣
∣∣1002x−x−1−x3x−2x−2−2x∣∣
∣∣=x2(x+1)2(x−1)∣∣
∣∣1002x−1−13x−2−2∣∣
∣∣=0∴f(200)=0
:
B
f(X)=x(x+1)∣∣
∣∣1112xx−1x3x(x−1)(x−1)(x−2)x(x−1)∣∣
∣∣=x(x+1)(x−1)∣∣
∣∣1112xx−1x3xx−2x∣∣
∣∣
Applying C2→C2−C1 and C3→C3−C1 then
f(x)=x(x+1)(x−1)∣∣
∣∣1002x−x−1−x3x−2x−2−2x∣∣
∣∣=x2(x+1)2(x−1)∣∣
∣∣1002x−1−13x−2−2∣∣
∣∣=0∴f(200)=0
Answer: Option C. -> 1
:
C
It has a non-zero solution if
∣∣
∣∣1k−13−k−11−31∣∣
∣∣=0⇒−6k+6=0⇒k=1.
:
C
It has a non-zero solution if
∣∣
∣∣1k−13−k−11−31∣∣
∣∣=0⇒−6k+6=0⇒k=1.
Answer: Option B. -> 1
:
B
For infinitely many solutions, the two equations must be identical
⇒k+1k=8k+3=4k3k−1⇒(k+1)(k+3)=8k and 8(3k-1) = 4k(k+3)
⇒k2−4k+3=0 and k2−3k+2=0.
By cross multiplication, k2−8+9=k3−2=1−3+4
k2=1 and k=1 ;∴ k=1.
:
B
For infinitely many solutions, the two equations must be identical
⇒k+1k=8k+3=4k3k−1⇒(k+1)(k+3)=8k and 8(3k-1) = 4k(k+3)
⇒k2−4k+3=0 and k2−3k+2=0.
By cross multiplication, k2−8+9=k3−2=1−3+4
k2=1 and k=1 ;∴ k=1.
Answer: Option C. -> ΔABC is an isosceles triangle
:
C
Δ=∣∣
∣
∣∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R2→R2−R3
=4∣∣
∣
∣∣a2b2c2abc(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R3→R3−R1+2R2
∴Δ=4∣∣
∣∣a2b2c2abc111∣∣
∣∣=−4(a−b)(b−c)(c−a)=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.
:
C
Δ=∣∣
∣
∣∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R2→R2−R3
=4∣∣
∣
∣∣a2b2c2abc(a−1)2(b−1)2(c−1)2∣∣
∣
∣∣
Applying R3→R3−R1+2R2
∴Δ=4∣∣
∣∣a2b2c2abc111∣∣
∣∣=−4(a−b)(b−c)(c−a)=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.
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