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 has

Many 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

N²

C. Zero

SOLUTION

Solution : C

Question 4

$If f, g and h are differentiable functions of x and △ ⎛ ⎜ ⎜ ⎝ x ⎞ ⎟ ⎟ ⎠ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h (x f)' & (x g)' & (x h)' (x^{2} f)'' & (x^{2} g)'' & (x^{2} h)'' \end{matrix} ∣ ∣ ∣ ∣ ∣, then △' ⎛ ⎜ ⎜ ⎝ x ⎞ ⎟ ⎟ ⎠ is$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')' & (x^{3} h'')' & (x^{3} h'')' \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'') & (x^{3} h'') & (x^{3} h'') \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')'' & (x^{3} h'')'' & (x^{3} h'')'' \end{matrix} ∣ ∣ ∣ ∣ ∣$

0

SOLUTION

Solution : A

$∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h x f' + f & x g' + g & x h' + h 4 x f' + 2 f + x^{2} f'' & 4 x g' + 2 g + x^{2} g'' & 4 x h' + 2 h + x^{2} h'' \end{matrix} ∣ ∣ ∣ ∣ ∣ Operating R_{2} \to R_{1} - R_{1}; R_{3} \to R_{3} - 4 R_{2} + 2 R_{1} and shifting x of R_{2} to R_{3} △ ⎛ ⎜ ⎜ ⎝ x ⎞ ⎟ ⎟ ⎠ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' x^{3} f'' & x^{3} g'' & x^{3} h'' \end{matrix} ∣ ∣ ∣ ∣ ∣ \Rightarrow △' ⎛ ⎜ ⎜ ⎝ x ⎞ ⎟ ⎟ ⎠ = 0 + 0 + ∣ ∣ ∣ ∣ ∣ \begin{matrix} f & g & h f' & g' & h' (x^{3} f'')' & (x^{3} g'')' & (x^{3} h'')' \end{matrix} ∣ ∣ ∣ ∣ ∣$

Question 5

If $ω$ is a cube root of unity and $Δ = ∣ ∣ ∣ \begin{matrix} 1 & 2 ω ω & ω^{2} \end{matrix} ∣ ∣ ∣$ , then $Δ^{2}$ is equal to

- ω

ω^{2}

C. 1

ω

SOLUTION

Solution : D

Since $Δ = ω^{2} - 2 ω^{2} = - ω^{2} .$ Therefore $Δ^{2} = ω^{4} = ω .$

Question 6

If $Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣$ , then $Δ_{2} Δ_{1}$ is equal to

a c

b d

(b - a) (d - c)

a b c

SOLUTION

Solution : B

$Δ_{2} Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 1 & 0 c + a d & b d \end{matrix} ∣ ∣ ∣ = b d .$

Question 7

If $A_{1}, B_{1}, C_{1} . . . .$ are respectively the co-factors of the elements $a_{1}, b_{1}, c_{1} . . . .$ of the determinant $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣,$ then $∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ =$

a_{1} Δ

a_{1} a_{3} Δ

(a_{1} + b_{1}) Δ

D. None of these

SOLUTION

Solution : A

$B_{2} = ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{3} & c_{3} \end{matrix} ∣ ∣ ∣ = a_{1} c_{3} - c_{1} a_{3} C_{2} = - ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{3} & b_{3} \end{matrix} ∣ ∣ ∣ = - (a_{1} b_{3} - a_{3} b_{1}) B_{3} = - ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{2} & c_{2} \end{matrix} ∣ ∣ ∣ = - (a_{1} c_{2} - a_{2} c_{1}) C_{3} = ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣ = (a_{1} b_{2} - a_{2} b_{1}) ∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} - a_{3} c_{1} & - (a_{1} b_{3} - a_{3} b_{1}) - (a_{1} c_{2} - a_{2} c_{1}) & a_{1} b_{2} - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & - a_{1} b_{3} - a_{1} c_{2} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & a_{3} b_{1} - a_{1} c_{2} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & - a_{1} b_{3} a_{2} c_{1} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & a_{3} b_{1} a_{2} c_{1} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = {a_{1}}^{2} (b_{2} c_{3} - b_{3} c_{2}) + a_{1} b_{1} (- c_{3} a_{2} + a_{3} c_{2}) + a_{1} c_{1} (- a_{3} b_{2} + a_{2} b_{3}) + c_{1} b_{1} (a_{3} a_{2} - a_{2} a_{3}) = a_{1} Δ .$

Question 8

If A = $∣ ∣ ∣ ∣ \begin{matrix} 5 & 6 & 3 - 4 & 3 & 2 - 4 & - 7 & 3 \end{matrix} ∣ ∣ ∣ ∣,$ then cofactors of the elements of $2^{n d}$ row are

A. 39, -3, 11

B. -39, 3, 11

C. -39, 27, 11

D. -39, -3, 11

SOLUTION

Solution : C

$C_{21} = (- 1)^{2 + 1} (18 + 21) = - 39 C_{22} = (- 1)^{2 + 2} (15 + 12) = 27 C_{23} = (- 1)^{2 + 3} (- 35 + 24) = 11 .$

Question 9

If the system of linear equation $x + 2 a y + a z = 0, x + 3 b y + b z = 0, x + 4 c y + c z = 0$ 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

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 a & a 1 & 3 b & b 1 & 4 c & c \end{matrix} ∣ ∣ ∣ ∣ = 0, [C_{2} \to C_{2} - 2 C_{3}] \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & a 1 & b & b 1 & 2 c & c \end{matrix} ∣ ∣ ∣ ∣ = 0, [R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}] \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & a 0 & b & b - a 0 & 2 c - b & c - b \end{matrix} ∣ ∣ ∣ ∣ = 0; b (c - b) - (b - a) (2 c - b) = 0$
On simplification, $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$
$∴$ a, b, c are in Harmonic progression.

Question 10

The system of equations $x + y + z = 2, 3 x - y + 2 z = 6$ and $3 x + y + z = - 18$ 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 $⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 3 & - 1 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 26 - 18 \end{matrix} ⎤ ⎥ ⎦$
On solving the above system we get the unique solution x = -10, y = -4, z = 16.

Question 11

The value of the determinant $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} l o g_{a} (\frac{x}{y}) & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) l o g_{b} (\frac{y}{z}) & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) l o g_{c} (\frac{z}{x}) & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is

A. 1

B. -1

l o g_{c} x y z

D. None of these

SOLUTION

Solution : D

Applying $C_{1} \to C_{1} + C_{2} + C_{3}$
Then, $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) 0 & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) 0 & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0$

Question 12

If $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x + 1 2 x & x (x - 1) & x (x + 1) 3 x (x - 1) & x (x - 1) (x - 2) & x (x^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then f(200) is equal to

A. 1

B. 0

C. 200

D. -200

SOLUTION

Solution : B

$f (X) = x (x + 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 x & x - 1 & x 3 x (x - 1) & (x - 1) (x - 2) & x (x - 1) \end{matrix} ∣ ∣ ∣ ∣ = x (x + 1) (x - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 x & x - 1 & x 3 x & x - 2 & x \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ then
$f (x) = x (x + 1) (x - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 2 x & - x - 1 & - x 3 x & - 2 x - 2 & - 2 x \end{matrix} ∣ ∣ ∣ ∣ = x^{2} (x + 1)^{2} (x - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 2 x & - 1 & - 1 3 x & - 2 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 0 ∴ f (200) = 0$

Question 13

If a, b, c are sides of a triangle and $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} (a + 1)^{2} & (b + 1)^{2} & (c + 1)^{2} (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ , then

Δ A B C

s an equilateral triangle

Δ A B C

is right angled isosceles triangle

Δ A B C

is an isosceles triangle

Δ A B C

None of the above

SOLUTION

Solution : C

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} (a + 1)^{2} & (b + 1)^{2} & (c + 1)^{2} (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{2} \to R_{2} - R_{3}$
$= 4 ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} a & b & c (a - 1)^{2} & (b - 1)^{2} & (c - 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{3} \to R_{3} - R_{1} + 2 R_{2}$
$∴ Δ = 4 ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b^{2} & c^{2} a & b & c 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = - 4 (a - b) (b - c) (c - a) = 0$
If a – b = 0 or b – c = 0 or c – a = 0
$∴$ a = b or b = c or c = a
$∴ Δ A B C$ is an isosceles triangle.

Question 14

$∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & c b & - c & 0 \end{matrix} ∣ ∣ ∣ ∣ =$

A. -2abc

B. abc

C. 0

a^{2} + b^{2} + c^{2}

SOLUTION

Solution : C

$∣ ∣ ∣ ∣ \begin{matrix} 0 & a & - b - a & 0 & c b & - c & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ (Since value of determinant of skew-symmetric matrix of odd orders is 0).

Question 15

If a,b and care non zero numbers, then $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

A. abc

a^{2} b^{2} c^{2}

C. ab+bc+ca

D. None of these

SOLUTION

Solution : D

Multiplying $R_{1}$ by $a, R_{2}$ by b and $R_{3}$ by c, we have
$Δ = \frac{1}{a b c} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a b^{2} c^{2} & a b c & a b + a c a^{2} b c^{2} & a b c & b c + a b a^{2} b^{2} c & a b c & a c + b c \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{a^{2} b^{2} c^{2}}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & a b + a c a c & 1 & b c + a b a b & 1 & a c + b c \end{matrix} ∣ ∣ ∣ ∣ = a b c ∣ ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & \sum a b a c & 1 & \sum a b a b & 1 & \sum a b \end{matrix} ∣ ∣ ∣ ∣ ∣ {b y C_{3} \to C_{3} + C_{1}} = a b c . \sum a b ∣ ∣ ∣ ∣ \begin{matrix} b c & 1 & 1 c a & 1 & 1 a b & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0, [S i n c e C_{2} \equiv C_{3}]$ .
Trick : Put a=1, b=2, c=3 and check it.

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsObjective Test 02Objective Test 02Objective Test 02

Related Exams
Objective Test 02
Objective Test 02
Objective Test 02