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

Question 1

SOLUTION

Solution : D

Question 2

The cofactor of the element '4' in the determinant $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 5 & 1 2 & 3 & 4 & 2 8 & 0 & 1 & 1 0 & 2 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is

A. 4

B. -10

C. 10

D. -4

SOLUTION

Solution : C

The cofactor of element 4, in the $2^{n d}$ row and $3^{r d}$ column is
$= (- 1)^{2 + 3} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 8 & 0 & 1 0 & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣$ = - {1(-2)-3(8-0)+1.16}
=10.

Question 3

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

A. -1

B. 0

C. 1

D. 2

SOLUTION

Solution : C

It has a non-zero solution if
$∣ ∣ ∣ ∣ \begin{matrix} 1 & k & - 1 3 & - k & - 1 1 & - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow - 6 k + 6 = 0 \Rightarrow k = 1 .$

Question 4

The number of values of k for which the system of equations $(k + 1) x + 8 y = 4 k, k x + (k + 3) y = 3 k - 1$ 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
$\Rightarrow \frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4 k}{3 k - 1} \Rightarrow (k + 1) (k + 3) = 8 k$ and 8(3k-1) = 4k(k+3)
$\Rightarrow k^{2} - 4 k + 3 = 0$ and $k^{2} - 3 k + 2 = 0 .$
By cross multiplication, $\frac{k^{2}}{- 8 + 9} = \frac{k}{3 - 2} = \frac{1}{- 3 + 4}$
$k^{2} = 1$ and k=1 ; $∴$ k=1.

Question 5

If $Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{n} & s i n x & c o s x n! & s i n \frac{n π}{2} & c o s \frac{n π}{2} a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then the value of $\frac{d^{n}}{d x^{n}} [Δ (x)] a t x = 0$ is

A. -1

B. 0

C. 1

D. Dependent of a

SOLUTION

Solution : B

$\frac{d^{n}}{d x^{n}} [Δ (x)] = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{d^{n}}{d x^{n}} x^{n} & \frac{d^{n}}{d x^{n}} s i n x & \frac{d^{n}}{d x^{n}} c o s x n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} n! & s i n (x + \frac{n π}{2}) & c o s (x + \frac{n π}{2}) n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \Rightarrow [Δ^{n} (x)]_{x = 0} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} n! & s i n (0 + \frac{n π}{2}) & c o s (0 + \frac{n π}{2}) n! & s i n (\frac{n π}{2}) & c o s (\frac{n π}{2}) a & a^{2} & a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0 1 S i n c e R_{1} \equiv R_{2}} .$

Question 6

If $f (n) = α^{n} + β^{n}$ and $∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$ , then k is equal to

A. 1

B. -1

α β

α β γ

SOLUTION

Solution : A

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + α + β & 1 + α^{2} + β^{2} 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} & 1 + α^{4} + β^{4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & α & β 1 & α^{2} & β^{2} \end{matrix} ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & α & β 1 & α^{2} & β^{2} \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{2} \to - C_{2} - C_{3} \to C_{3} - C_{1}$ ,
${∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & α - 1 & β - 1 1 & α^{2} - 1 & β^{2} - 1 \end{matrix} ∣ ∣ ∣ ∣}^{2} = (α - 1)^{2} (β - 1)^{2} (β - α)^{2} = (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$
Hence, k=1

Question 7

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$ , then

Δ_{1} + Δ_{2} = 0

Δ_{1} + 2 Δ_{2} = 0

Δ_{1} = Δ_{2}

Δ_{1} = 2 Δ_{2}

SOLUTION

Solution : A

$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣,$ & $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$
In $Δ_{2}$ ,Transposing the determinant
$Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b c & c a & a b a & a & c \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to a C_{1}, C_{2} \to b C_{2}, C_{3} \to C C_{3}$
$Δ_{2} = \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} a & b & c a b c & a b c & a b c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$
Taking abc common from $R_{2}$
$Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c 1 & 1 & 1 a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$
$R_{1} \leftrightarrow R_{2}$
$Δ_{2} = - ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow Δ_{2} = - Δ_{1} \Rightarrow Δ_{2} + Δ_{1} = 0$

Question 8

If $α, β$ are non real numbers satisfying $x^{3} - 1 = 0$ then the value of $∣ ∣ ∣ ∣ \begin{matrix} λ + 1 & α & β α & λ + β & 1 β & 1 & λ + α \end{matrix} ∣ ∣ ∣ ∣$ is equal to

A. 0

λ^{3}

λ^{3} + 1

λ^{3} - 1

SOLUTION

Solution : B

$x^{3} - 1 = 0 ∴ x = 1, ω, ω^{2}$
Here, $α = ω, β = ω^{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} λ + 1 & ω & ω^{2} ω & λ + ω^{2} & 1 ω^{2} & 1 & λ + ω \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $C_{1} \to C_{1} + C_{2} + C_{3}$ , then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} λ & ω & ω^{2} λ & λ + ω^{2} & 1 λ & 1 & λ + ω \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R - 2 \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , then we get
$∣ ∣ ∣ ∣ ∣ \begin{matrix} λ & ω & ω^{2} 0 & λ + ω^{2} - ω & 1 - ω^{2} 0 & 1 - ω & λ + ω - ω^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = λ ((λ + ω^{2} - ω) (λ + ω - ω^{2}) - (1 - ω) (1 - ω^{2})) = λ (λ^{2}) = λ^{3}$

Question 9

If $D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $\sum_{k = 1}^{n} D_{k} = 56$ , then n equals

A. 4

B. 6

C. 8

D. None of these

SOLUTION

Solution : D

$∴ \sum_{k = 1}^{n} D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \sum_{k = 1}^{n} 1 & n & n 2 \sum_{k = 1}^{n} k & n^{2} + n + 1 & n^{2} + n 2 \sum_{k = 1}^{n} k - \sum_{k = 1}^{n} 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n & n n^{2} + n & n^{2} + n + 1 & n^{2} + n n^{2} & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 56$
=Applying $C_{2} \to C_{2} \to C_{1} \to C_{3} \to C_{3} - C_{1}$ , then
$∣ ∣ ∣ ∣ \begin{matrix} n & 0 & 0 n^{2} + n & 1 & 0 n & 0 & n + 1 \end{matrix} ∣ ∣ ∣ ∣ = 56 \Rightarrow n (n + 1) = 56 = 7 \times 8 \Rightarrow n = 7$

Question 10

If sin 2x = 1, then ${∣ ∣ ∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2}$ equals

A. 3

B. 0

C. 1

D. None of these

SOLUTION

Solution : B

$∴ s i n 2 x = 1 t h e n X = \frac{π}{4}$
Then, ${∣ ∣ ∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2} = {∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣}^{2}$
$= (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) {∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & - 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2}$
$= \frac{1}{8} {0 - 1 (0 - 1) - 1 (1)}^{2}$
=0

Question 11

If $A + B + C = π$ , then $∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t n a A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

A. 1

B. 0

C. -1

D. 2

SOLUTION

Solution : B

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} s i n π & s i n B & c o s C - s i n B & 0 & t a n A c o s (π - C) & - t n a A & 0 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 0 & s i n B & c o s C - s i n B & 0 & t a n A - c o s C & - t a n A & 0 \end{matrix} ∣ ∣ ∣ ∣$
=0 ( $∵ Δ$ is skew symmetric)

Question 12

Let $f (x) = ∣ ∣ ∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o x 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣$ , then $f^{'} (\frac{π}{2})$ is equal to

A. 8

B. 6

C. 4

D. 2

SOLUTION

Solution : C

$f^{'} (x) = ∣ ∣ ∣ ∣ \begin{matrix} - s i n x & s i n x & c o s x - 2 s i n 2 x & s i n 2 x & 2 c o s 2 x - 3 s i n 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} c o s x & c o s x & c o s x c o s 2 x & 2 c o s 2 x & 2 c o s 2 x c o s 3 x & 3 c o s 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} c o s x & s i n x & - s i n x c o s 2 x & s i n 2 x & - 4 s i n 2 x c o s 3 x & s i n 3 x & - 9 s i n 3 x \end{matrix} ∣ ∣ ∣ ∣$
$f^{'} (\frac{π}{2}) = ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 1 & 0 0 & 0 & - 2 3 & - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ + 0 + ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & - 1 - 1 & 0 & 0 0 & - 1 & 9 \end{matrix} ∣ ∣ ∣ ∣ = 2 (1 - 3) + 0 + 1 (9 - 1) = - 4 + 8 = 4$

Question 13

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
$∣ ∣ ∣ ∣ \begin{matrix} a - 1 & - 1 & - 1 1 & 1 - b & 1 1 & 1 & 1 - c \end{matrix} ∣ ∣ ∣ ∣ = 0$
Applying $C_{1} \to C_{1} - C_{3}$ and $C_{2} \to C_{2} - C_{3},$ then
$∣ ∣ ∣ ∣ \begin{matrix} a & 0 & 1 0 & - b & 1 c & c & 1 - c \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow a (- b + b c - c) - 0 + c (0 - b) = 0 \Rightarrow a b + b c + c a = a b c$

Question 14

$∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣ ∣ ∣ =$

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

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

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

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

SOLUTION

Solution : A

$∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣ ∣ ∣ = 1 (1 + c^{2}) - a (- a + b c) + b (a c + b) = 1 + a^{2} + b^{2} + c^{2}$ .

Question 15

If $p λ^{4} + q λ^{3} + r λ + 5 λ + t = ∣ ∣ ∣ ∣ \begin{matrix} λ^{2} + 3 λ & λ - 1 & λ + 3 λ + 1 & 2 - λ & λ - 4 λ - 3 & λ + 4 & 3 λ \end{matrix} ∣ ∣ ∣ ∣$ , 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
$t = ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & 3 1 & 2 & - 4 - 3 & 4 & 0 \end{matrix} ∣ ∣ ∣ ∣ = - 12 + 30 = 18$ .

Free Objective Test 01 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

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