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1

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

Let a, b and c be three real numbers satisfying

$$[\matrix{ a & b & c \cr } ]\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right] = [\matrix{ 0 & 0 & 0 \cr } ]$$ ........(E)

Let $$\omega$$ be a solution of $${x^3} - 1 = 0$$ with $${\mathop{\rm Im}\nolimits} (\omega ) > 0$$. If a = 2 with b and c satisfying (E), then the value of $${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$$ is equal to

A
$$-$$2
B
2
C
3
D
$$-$$3

Explanation

Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$ \Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$ \Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$ \Rightarrow a + b + c = 0$$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ b = 6a and c = $$-$$ 7a

If a = 2, b = 12 and c = $$-$$14

$$\therefore$$ $${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$$

$$ \Rightarrow {3 \over {{\omega ^2}}} + {1 \over {{\omega ^{12}}}} + {3 \over {{\omega ^{ - 14}}}} = {3 \over {{\omega ^2}}} + 1 + 3{\omega ^2}$$

$$ = 3\omega + 1 + 3{\omega ^2}$$

$$ = 1 + 3(\omega + {\omega ^2})$$

$$ = 1 - 3 = - 2$$

2

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

Let a, b and c be three real numbers satisfying

$$[\matrix{ a & b & c \cr } ]\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right] = [\matrix{ 0 & 0 & 0 \cr } ]$$ .......(E)

If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is

A
0
B
12
C
7
D
6

Explanation

Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$ \Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$ \Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$ \Rightarrow a + b + c = 0$$ ....... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ $$b = 6a$$ and $$c = - 7a$$

As (a, b, c) lies on $$2x + y + z = 1$$

$$ \Rightarrow 2a + b + c = 1$$

$$ \Rightarrow 2a + 6a - 7a = 1$$

$$\Rightarrow$$ a = 1, b = 6 and c = $$-$$7

$$\therefore$$ $$7a + b + c = 7 + 6 - 7 = 6$$

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