Free Objective Test 02 Practice Test - 11th and 12th
Question 1
The system of equations
X+2y+3z=4
2x+3y+4z=5
3x+4y+5z=6 hasMany solutions
No solution
Unique solution
Atmost two solutions
SOLUTION
Solution : A
Question 2
f=3 and g=-5
f=-3 and g=-5
f=-3 and g=-9
f=-5 and g=9
SOLUTION
Solution : D
Question 3
N
N2
1
SOLUTION
Solution : C
Question 4
If f, g and h are differentiable functions of x and △⎛⎜ ⎜⎝x⎞⎟ ⎟⎠=∣∣ ∣ ∣∣fgh(xf)'(xg)'(xh)'(x2f)''(x2g)''(x2h)''∣∣ ∣ ∣∣,then △'⎛⎜ ⎜⎝x⎞⎟ ⎟⎠ is
∣∣ ∣ ∣∣fghf'g'h'(x3f'')'(x3h'')'(x3h'')'∣∣ ∣ ∣∣
∣∣ ∣ ∣∣fghf'g'h'(x3f'')(x3h'')(x3h'')∣∣ ∣ ∣∣
∣∣ ∣ ∣∣fghf'g'h'(x3f'')''(x3h'')''(x3h'')''∣∣ ∣ ∣∣
SOLUTION
Solution : A
∣∣ ∣ ∣∣fghxf'+fxg'+gxh'+h4xf'+2f+x2f''4xg'+2g+x2g''4xh'+2h+x2h''∣∣ ∣ ∣∣Operating R2→R1−R1; R3→R3−4R2+2R1and shifting x of R2 to R3△⎛⎜ ⎜⎝x⎞⎟ ⎟⎠=∣∣ ∣ ∣∣fghf'g'h'x3f''x3g''x3h''∣∣ ∣ ∣∣⇒△'⎛⎜ ⎜⎝x⎞⎟ ⎟⎠=0+0+∣∣ ∣ ∣∣fghf'g'h'(x3f'')'(x3g'')'(x3h'')'∣∣ ∣ ∣∣
Question 5
If ω is a cube root of unity and Δ = ∣∣∣12ωωω2∣∣∣, then Δ2 is equal to
SOLUTION
Solution : D
Since Δ = ω2−2ω2 = −ω2. Therefore Δ2 = ω4 = ω.
Question 6
If Δ1 = ∣∣∣10ab∣∣∣ and Δ2 = ∣∣∣10cd∣∣∣, then Δ2Δ1 is equal to
SOLUTION
Solution : B
Δ2Δ1 = ∣∣∣10cd∣∣∣ ∣∣∣10ab∣∣∣ = ∣∣∣10c+adbd∣∣∣ = bd.
Question 7
If A1, B1, C1.... are respectively the co-factors of the elements a1, b1, c1.... of the determinant Δ = ∣∣ ∣∣a1b1c1a2b2c2a3b3c3∣∣ ∣∣, then ∣∣∣B2C2B3C3∣∣∣ =
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∣∣∣ =∣∣∣a1c3−a1b3−a1c2a1b2∣∣∣+∣∣∣a1c3a3b1−a1c2−a2b1∣∣∣+∣∣∣−a3c1−a1b3a2c1a1b2∣∣∣+∣∣∣−a3c1a3b1a2c1−a2b1∣∣∣=a12(b2c3−b3c2)+a1b1(−c3a2+a3c2)+a1c1(−a3b2+a2b3)+c1b1(a3a2−a2a3) = a1Δ .
Question 8
If A = ∣∣ ∣∣563−432−4−73∣∣ ∣∣, then cofactors of the elements of 2nd row are
SOLUTION
Solution : C
C21 = (−1)2+1(18+21) = −39C22 = (−1)2+2(15+12) = 27C23 = (−1)2+3(−35+24) = 11 .
Question 9
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
SOLUTION
Solution : C
∣∣ ∣∣12aa13bb14cc∣∣ ∣∣ = 0, [C2 → C2−2C3]⇒ ∣∣ ∣∣10a1bb12cc∣∣ ∣∣ = 0, [R3 → R3−R2, R2 → R2−R1]⇒ ∣∣ ∣∣10a0bb−a02c−bc−b∣∣ ∣∣ = 0 ; b(c−b)−(b−a)(2c−b) = 0
On simplification, 2b = 1a+1c
∴ a, b, c are in Harmonic progression.
Question 10
The system of equations x + y + z =2, 3x − y + 2z =6 and 3x + y + z =−18 has
SOLUTION
Solution : A
Given system of equation can be written as ⎡⎢⎣1113−12311⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦ = ⎡⎢⎣26−18⎤⎥⎦
On solving the above system we get the unique solution x = -10, y = -4, z = 16.
Question 11
The value of the determinant ∣∣ ∣ ∣ ∣∣loga(xy)loga(yz)loga(zx)logb(yz)logb(zx)logb(xy)logc(zx)logc(xy)logc(yz)∣∣ ∣ ∣ ∣∣ is
SOLUTION
Solution : D
Applying C1→C1+C2+C3
Then, ∣∣ ∣ ∣ ∣∣0loga(yz)loga(zx)0logb(zx)logb(xy)0logc(xy)logc(yz)∣∣ ∣ ∣ ∣∣=0
Question 12
If f(x)=∣∣ ∣ ∣∣1xx+12xx(x−1)x(x+1)3x(x−1)x(x−1)(x−2)x(x2−1)∣∣ ∣ ∣∣, then f(200) is equal to
SOLUTION
Solution : 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
Question 13
If a, b, c are sides of a triangle and ∣∣ ∣ ∣∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣∣ ∣ ∣∣=0, then
SOLUTION
Solution : 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.
Question 14
∣∣ ∣∣0a−b−a0cb−c0∣∣ ∣∣=
SOLUTION
Solution : C
∣∣ ∣∣0a−b−a0cb−c0∣∣ ∣∣=0 (Since value of determinant of skew-symmetric matrix of odd orders is 0).
Question 15
If a,b and care non zero numbers, then Δ=∣∣ ∣ ∣∣b2c2bcb+cc2a2cac+aa2b2aba+b∣∣ ∣ ∣∣ is equal to
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∣∣ ∣ ∣∣bc1∑abac1∑abab1∑ab∣∣ ∣ ∣∣{by C3→C3+C1}=abc.∑ab∣∣ ∣∣bc11ca11ab11∣∣ ∣∣=0, [Since C2≡C3].
Trick : Put a=1, b=2, c=3 and check it.