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1

MCQ (Single Correct Answer)

If the coefficients of r^{th}, (r+1)^{th}, and (r + 2)^{th} terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation

A

$${m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0$$

B

$${m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0$$

C

$${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$

D

$${m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0$$

Let r = 2

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T_{2} = $${}^m{C_1}.y$$

Coefficient of T_{2} = $${}^m{C_1}$$

3rd term = T_{3} = $${}^m{C_2}.{y^2}$$

Coefficient of T_{3} = $${}^m{C_2}$$

4th term = T_{4} = $${}^m{C_3}.{y^3}$$

Coefficient of T_{2} = $${}^m{C_3}$$

$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$

$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$

$$ \Rightarrow $$ 6m^{2} - 6m = 6m +m(m^{2} - 3m + 2)

$$ \Rightarrow $$ 6m^{2} - 6m = 6m + m^{3} - 3m^{2} + 2m

$$ \Rightarrow $$ 6m - 6 = 6 + m^{2} - 3m + 2

$$ \Rightarrow $$ m^{2} - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get

m^{2} - 9m + 14 = 0. So Option C is correct.

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T

Coefficient of T

3rd term = T

Coefficient of T

4th term = T

Coefficient of T

$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$

$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$

$$ \Rightarrow $$ 6m

$$ \Rightarrow $$ 6m

$$ \Rightarrow $$ 6m - 6 = 6 + m

$$ \Rightarrow $$ m

Now put r = 2 at each option and find answer.

In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get

m

2

MCQ (Single Correct Answer)

If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as

A

$$1 - {3 \over 8}{x^2}$$

B

$$3x + {3 \over 8}{x^2}$$

C

$$ - {3 \over 8}{x^2}$$

D

$${x \over 2} - {3 \over 8}{x^2}$$

$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$

= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)

$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$

= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$

= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$

= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$

[ As x^{3} is so small we can ignore $$-{3 \over {16}}{x^3}$$]

= $$ - {3 \over 8}{x^2}$$

= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)

$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$

= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$

= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$

= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$

[ As x

= $$ - {3 \over 8}{x^2}$$

3

MCQ (Single Correct Answer)

If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation

A

$$a - b = 1$$

B

$$a + b = 1$$

C

$${a \over b} = 1$$

D

$$ab = 1$$

General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is T_{r+1}.

T_{r+1} = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x^{7},

$$ \Rightarrow $$ 22 - 3r = 7

$$ \Rightarrow $$ r = 5

So coefficient of x^{7} = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is T_{r+1}.

T_{r+1} = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x^{-7},

11 - 3r = -7

$$ \Rightarrow $$ r = 6

$$\therefore$$ Coefficient of x^{-7} = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

According to question,

Coefficient of x^{7} = Coefficient of x^{-7}

$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$ \Rightarrow $$ $$ab = 1$$

T

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x

$$ \Rightarrow $$ 22 - 3r = 7

$$ \Rightarrow $$ r = 5

So coefficient of x

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is T

T

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x

11 - 3r = -7

$$ \Rightarrow $$ r = 6

$$\therefore$$ Coefficient of x

According to question,

Coefficient of x

$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$ \Rightarrow $$ $$ab = 1$$

4

MCQ (Single Correct Answer)

If $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$ and $$I = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by the principle of mathematical induction

A

$${A^n} = nA - \left( {n - 1} \right){\rm I}$$

B

$${A^n} = {2^{n - 1}}A - \left( {n - 1} \right){\rm I}$$

C

$${A^n} = nA + \left( {n - 1} \right){\rm I}$$

D

$${A^n} = {2^{n - 1}}A + \left( {n - 1} \right){\rm I}$$

Given $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$

$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$$

and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$$

So we can say $${A^n}$$ = $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$

Now $$nA - \left( {n - 1} \right){\rm I}$$

= $$\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$$ - $$\left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right]$$

= $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$ = $${A^n}$$

$$\therefore$$ $${A^n} = nA - \left( {n - 1} \right){\rm I}$$

$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$$

and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$$

So we can say $${A^n}$$ = $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$

Now $$nA - \left( {n - 1} \right){\rm I}$$

= $$\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$$ - $$\left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right]$$

= $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$ = $${A^n}$$

$$\therefore$$ $${A^n} = nA - \left( {n - 1} \right){\rm I}$$

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