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1

WB JEE 2009

MCQ (Single Correct Answer)

If $$i = \sqrt { - 1} $$ and n is positive integer, then $${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}$$ is equal to

A
1
B
i
C
in
D
0

Explanation

$${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}$$

$$ = {i^n}(1 + i + {i^2} + {i^3}) = {i^n}(1 + i - 1 - i) = 0$$

2

WB JEE 2008

MCQ (Single Correct Answer)

If 1, $$\omega$$, $$\omega$$2 are cube roots of unity, then $$\left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr } } \right|$$ has value

A
0
B
$$\omega$$
C
$$\omega$$2
D
$$\omega$$ + $$\omega$$2

Explanation

Property of $$\omega$$ is $$\omega$$3 = 1 and 1 + $$\omega$$ + $$\omega$$2 = 0

$$\therefore$$ $$\left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr } } \right|$$

$$ = 1(1 - {\omega ^n} \times {\omega ^{2n}}) - {\omega ^n}({\omega ^{2n}} - {\omega ^n} \times {\omega ^n}) + {\omega ^{2n}}({\omega ^{2n}} \times {\omega ^{2n}} - {\omega ^n})$$

$$ = 1 - {\omega ^{3n}} - {\omega ^{3n}} + {\omega ^{3n}} + {\omega ^{6n}} - {\omega ^{3n}} = 1 - 2{({\omega ^3})^n} + {({\omega ^3})^{2n}}$$

$$ = 1 - 2{(1)^n} + {(1)^{2n}} = 1 - 2 + 1 = 0$$.

3

WB JEE 2008

MCQ (Single Correct Answer)

For two complex numbers z1, z2 the relation $$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$ holds if

A
$$\arg ({z_1}) = \arg ({z_2})$$
B
$$\arg ({z_1}) + \arg ({z_2}) = {\pi \over 2}$$
C
$${z_1}{z_2} = 1$$
D
$$\left| {{z_1}} \right| = \left| {{z_2}} \right|$$

Explanation

Let $${z_1} = {r_1}(\cos {\theta _1} + i\sin {\theta _1})$$ and $${z_2} = {r_2}(\cos {\theta _2} + i\sin {\theta _2})$$

$$\therefore$$ $${z_1} + {z_2} = ({r_1}\cos {\theta _1} + {r_2}\cos {\theta _2}) + i({r_1}\sin {\theta _1} + {r_2}\sin {\theta _2})$$

$$\left| {{z_1} + {z_2}} \right| = \sqrt {{{({r_1}\cos {\theta _1} + {r_2}\cos {\theta _2})}^2} + {{({r_1}\sin {\theta _1} + {r_2}\sin {\theta _2})}^2}} $$

$$ = \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2})} $$

$$\left| {{z_1}} \right| = {r_1},\,\left| {{z_2}} \right| = {r_2}$$

$$ \Rightarrow \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2})} = {r_1} + {r_2}$$ (given $$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$)

(squaring both sides)

$$ \Rightarrow r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2}) = r_1^2 + r_2^2 + 2{r_1}{r_2}$$

$$ \Rightarrow \cos ({\theta _1} - {\theta _2}) = \cos 0^\circ \Rightarrow {\theta _1} = {\theta _2}$$

$$ \Rightarrow \arg ({z_1}) = \arg ({z_2})$$.

4

WB JEE 2008

MCQ (Single Correct Answer)

A and B are two points on the Argand plane such that the segment AB is bisected at the point (0, 0). If the point A, which is in the third quadrant has principal amplitude $$\theta$$, then the principal amplitude of the point B is

A
$$-$$ $$\theta$$
B
$$\pi$$ $$-$$ $$\theta$$
C
$$\theta$$ $$-$$ $$\pi$$
D
$$\pi$$ + $$\theta$$

Explanation

A lies in 3rd quadrant, let $$\theta$$ is amplitude of A

$$\therefore$$ $$-$$180$$^\circ$$ < $$\theta$$ < $$-$$90$$^\circ$$ (principal amplitude)

B will lie in Ist quadrant, so principal amplitude $$\phi$$ will lie in 0 < $$\phi$$ < 90$$^\circ$$.

From figure

$$-$$ $$\theta$$ + $$\phi$$ = 180$$^\circ$$

$$\phi$$ = 180$$^\circ$$ $$-$$ ($$-$$$$\theta$$) ($$\because$$ $$\theta$$ is $$-$$ ve)

= 180$$^\circ$$ + $$\theta$$ = $$\pi$$ + $$\theta$$

e.g. if $$\theta$$ = $$-$$120$$^\circ$$

then $$\phi$$ = 180$$^\circ$$ + ($$-$$120$$^\circ$$)

= 60$$^\circ$$ (principal amplitude of B).

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