Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The coefficient of $${x^n}$$ in expansion of $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$ is

A

$${\left( { - 1} \right)^{n - 1}}n$$

B

$${\left( { - 1} \right)^n}\left( {1 - n} \right)$$

C

$${\left( { - 1} \right)^{n - 1}}{\left( {n - 1} \right)^2}$$

D

$$\left( {n - 1} \right)$$

Given $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$

= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$

$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$$

So coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}$$

General term of $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$$

$$\therefore$$ Term containing $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$$

So coefficient of $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

$$\therefore$$ coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

= $${}^n{C_n}.{\left( { - 1} \right)^n}$$ + $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$$

= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$

= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$

$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$$

So coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}$$

General term of $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$$

$$\therefore$$ Term containing $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$$

So coefficient of $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

$$\therefore$$ coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

= $${}^n{C_n}.{\left( { - 1} \right)^n}$$ + $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$$

= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$

2

MCQ (Single Correct Answer)

The coefficient of the middle term in the binomial expansion in powers of $$x$$ of $${\left( {1 + \alpha x} \right)^4}$$ and $${\left( {1 - \alpha x} \right)^6}$$ is the same if $$\alpha $$ equals

A

$${3 \over 5}$$

B

$${10 \over 3}$$

C

$${{ - 3} \over {10}}$$

D

$${{ - 5} \over {3}}$$

For $${\left( {1 + \alpha x} \right)^4}$$ the middle term $${T_{{4 \over 2} + 1}}$$ = $${}^4{C_2}.{\alpha ^2}{x^2}$$

$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$

For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$

$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$

$$\therefore$$ According to question,

$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$

$$ \Rightarrow 6 = 20 \times - \alpha $$

$$ \Rightarrow \alpha = - {3 \over {10}}$$

$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$

For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$

$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$

$$\therefore$$ According to question,

$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$

$$ \Rightarrow 6 = 20 \times - \alpha $$

$$ \Rightarrow \alpha = - {3 \over {10}}$$

3

MCQ (Single Correct Answer)

Let $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.$$ Then which of the following is true

A

Principle of mathematical induction can be used to prove the formula

B

$$S\left( K \right) \Rightarrow S\left( {K + 1} \right)$$

C

$$S\left( K \right) \ne S\left( {K + 1} \right)$$

D

$$S\left( 1 \right)$$ is correct

Given $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $$ \ne $$ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)^{2}

Now let S(k) is true

$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k^{2}

$$ \Rightarrow $$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k^{2} + 2k +1

= 3 + (k + 1)^{2}

$$ \Rightarrow $$ S(k + 1) is true.

$$\therefore$$ S(k) $$ \Rightarrow $$ S(k + 1)

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $$ \ne $$ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)

Now let S(k) is true

$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k

$$ \Rightarrow $$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k

= 3 + (k + 1)

$$ \Rightarrow $$ S(k + 1) is true.

$$\therefore$$ S(k) $$ \Rightarrow $$ S(k + 1)

4

MCQ (Single Correct Answer)

The number of integral terms in the expansion of $${\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}$$ is

A

35

B

32

C

33

D

34

General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$

= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$

When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.

And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.

Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.

$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256

$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, t_{n} = a + (n - 1)d

$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33

= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$

When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.

And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.

Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.

$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256

$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, t

$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33

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Complex Numbers

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