Free Objective Test 03 Practice Test - 11th and 12th

Question 1

The first 3 terms in the expansion of $(1 + a x)^{n}$ (n ≠ 0) are 1, 6x and 16 $x^{2}$ . Then the value of a and n are respectively

2 and 9

3 and 2

$\frac{2}{3}$ and 9

$\frac{3}{2}$ and 6

SOLUTION

Solution : C

$n_{c_{1}} . a x = 6 x; n_{c_{2}} (a x)^{2} = 16 x^{2} \Rightarrow a n = 6; \frac{n (n - 1)}{2} a^{2} = 16 \Rightarrow a = \frac{2}{3} \Rightarrow n = 9$

Question 2

If the third term in the binomial expansion of $(1 + x)^{m}$ is - $\frac{1}{8} x^{2}$ , then the rational value of m is

$\frac{1}{2}$

SOLUTION

Solution : B

$m_{c_{2}} = - \frac{1}{8} \Rightarrow \frac{m (m - 1)}{2} = - \frac{1}{8} \Rightarrow 4 m^{2} - 4 m + 1 = 0 \Rightarrow (2 m - 1)^{2} = 0 \Rightarrow m = \frac{1}{2}$

Question 3

If the $(r + 1)^{t h}$ term in the expansion of ${(\sqrt[3]{\frac{a}{\sqrt{b}}} + \sqrt{\frac{b}{\sqrt[3]{a}}})}^{21}$ has the same power of a and b, then value of r is

SOLUTION

Solution : A

We have $T_{r + 1}$ = $^{21} C_{r} {(\sqrt[3]{\frac{a}{\sqrt{b}}})}^{21 - r} {(\sqrt{\frac{b}{\sqrt[3]{a}}})}^{r}$

= $^{21} C_{r} a^{7 - (\frac{r}{2})} b^{\frac{2}{3} r - (\frac{7}{2})}$

Since the powers of a and b are the same,

therefore

7 - $\frac{r}{2}$ = $\frac{2}{3}$ r - $\frac{7}{2}$ ⇒ r = 9

Question 4

If $x^{4}$ occurs in the $r^{t h}$ term in the expansion of $(x^{4} + \frac{1}{x^{3}})^{16}$ , then r =

SOLUTION

Solution : C

$T_{r}$ = $^{16} C_{r - 1}$ $(x^{4})^{16 - r}$ $(\frac{1}{x^{3}})^{r - 1}$ = $^{16} C_{r - 1}$ $x^{67 - 7 r}$

$\to 67 - 7 r = 4 \to r = 9$

Question 5

If coefficient of $(2 r + 3)^{t h}$ and $(r - 1)^{t h}$ terms in the expansion of $(1 + x)^{15}$ are equal, then value of r is

SOLUTION

Solution : A

$^{15} C_{2 r + 2}$ = $^{15} C_{r - 2}$

But $^{15} C_{2 r + 2}$ = $^{15} C_{15 - (2 r + 2)}$ = $^{15} C_{13 - 2 r}$

⇒ $^{15} C_{13 - 2 r}$ = $^{15} C_{r - 2}$ ⇒ r = 5.

Question 6

In $(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})^{n}$ if the ratio of $7^{t h}$ term from the beginning to the $7^{t h}$ term from the end is $\frac{1}{6}$ , then n =

None of these

SOLUTION

Solution : C

$\frac{1}{6} = \frac{^{n} C_{6} {(2^{\frac{1}{3}})}^{n - 6} {(3^{- \frac{1}{3}})}^{6}}{^{n} C_{n - 6} {(2^{\frac{1}{3}})}^{6} {(3^{- \frac{1}{3}})}^{n - 6}}$

$6^{- 1} = 6^{\frac{1}{3} (n - 12)}$

$n - 12 = - 3$

$n = 9$

Question 7

$r^{t h}$ term in the expansion of $(a + 2 x)^{n}$ is

$\frac{n (n + 1) . . . . . (n - r + 1)}{r!} a^{n - r + 1} (2 x)^{r}$

$\frac{n (n - 1) . . . . . (n - r + 2)}{(r - 1)!} a^{n - r + 1} (2 x)^{r - 1}$

$\frac{n (n + 1) . . . . . (n - r)}{(r + 1)!} a^{n - r} (x) r$

None of these

SOLUTION

Solution : B

$r^{t h}$ term of $(a + 2 x)^{n}$ is $^{n} C_{r - 1} (a)^{n - r + 1} (2 x)^{r - 1}$

= $\frac{n!}{(n - r + 1)! (r - 1)!} a^{n - r + 1} (2 x)^{r - 1}$

= $\frac{n (n - 1) . . . . . (n - r + 2)}{(r - 1)!} a^{n - r + 1} (2 x)^{r - 1}$

Question 8

$6^{t h}$ term in expansion of $(2 x^{2} - \frac{1}{3 x^{2}})^{10}$ is

$\frac{4580}{17}$

- $\frac{896}{27}$

$\frac{5580}{17}$

None of these

SOLUTION

Solution : B

Applying $T_{r + 1}$ = $^{n} C_{r}$ $x^{n - r} d$ for $(x + a)^{n}$

Hence $T_{6}$ = $^{10} C_{5}$ $(2 x^{2})^{5}$ (- $\frac{1}{3 x^{2}})^{5}$

= - $(\frac{10!}{5! 5!})$ (32) $(\frac{1}{243})$ = - $\frac{896}{27}$

Question 9

If the ratio of the coefficient of third and fourth term in the expansion of $(x - \frac{1}{2 x})^{n}$ is 1:2, then the value of n will be

-10

SOLUTION

Solution : D

$T_{3}$ = $^{n} C_{2}$ $(x)^{n - 2}$ $(- \frac{1}{2 x})^{2}$ and $T_{4}$ = $^{n} C_{3}$ $(x)^{n - 3}$ $(- \frac{1}{2 x})^{3}$

But according to the condition,

$\frac{- n (n - 1) \times 3 \times 2 \times 1 \times 8}{n (n - 1) (n - 2) \times 2 \times 1 \times 4}$ = $\frac{1}{2}$ ⇒ n = -10

Question 10

Find the value of
$\frac{(18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25)}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 81 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}$

100

SOLUTION

Solution : A

The numerator is of the form

$a^{3}$ + $b^{3} + 3 a b (a + b) = (a + b)^{3}$

∴ N' = $(18 + 7)^{3}$ = $25^{3}$

∴ D' = $3^{6} +^{6} C_{1} \cdot 3^{5} \cdot 2^{1} +^{6} C_{2} \cdot 3^{4} \cdot 2^{2} +^{6} C_{3} \cdot 3^{3} \cdot 2^{3} +^{6} C_{4} \cdot 3^{2} \cdot 2^{4} +^{6} C_{5} \cdot 3^{1} \cdot 2^{5} +^{6} C_{6} \cdot 2^{6}$

This is clearly the expansion of $(3 + 2)^{6} = 5^{6} = (25)^{3}$

$\frac{N^{'}}{D^{'}} = \frac{(25)^{3}}{(25)^{3}} = 1$

Question 11

If number of terms in the expansion of $(x - 2 y + 3 z)^{n}$ are 45, then n =

None of these

SOLUTION

Solution : B

$\frac{(n + 1) (n + 2)}{2}$ = 45 or $n^{2}$ + 3n - 88 = 0 $\Rightarrow$ n = 8 .

Question 12

The greatest integer less than or equal to $(\sqrt{2} + 1)^{6}$ is

196

197

198

199

SOLUTION

Solution : B

Let $(\sqrt{2} + 1)^{6}$ = k + f, where k is integral part and f

the fraction (0 ≤ f < 1).

Let $(\sqrt{2} - 1)^{6}$ = f, (0 < f < 1),

Since 0 < ( $\sqrt{2}$ - 1) < 1

Now, k + f + f' = $(\sqrt{2} + 1)^{6}$ + $(\sqrt{2} - 1)^{6}$

= 2{ $^{6} C_{0} {.2}^{3} +^{6} C_{2} {.2}^{2} +^{6} C_{4} .2 +^{6} C_{6}$ } = 198 ..........(i)

∴ f + f' = 198 - k = an integer

But, 0 ≤ f < 1 and 0 < f' < 1 ⇒ 0 < (f + f') < 2

⇒ f + f' = 1, ( ∵ f + f' is an integer)

∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.

Question 13

The number of terms which are free from radical signs in the expansion of $(y^{\frac{1}{5}} + x^{\frac{1}{10}})^{55}$ is

None of these

SOLUTION

Solution : B

In the expansion of $(y^{\frac{1}{5}} + x^{\frac{1}{10}})^{55}$ , the general

term is

$T_{r + 1}$ = $^{55} C_{r} (y^{\frac{1}{5}})^{55 - r} (x^{\frac{1}{10}})^{r}$ = $^{55} C_{r} y^{11 - \frac{r}{5}} x^{\frac{r}{10}}$ .

This $T_{r + 1}$ will be independent of radicals if the exponents $\frac{r}{5}$ and $\frac{r}{10}$ are integers, for

0 ≤ r ≤ 55 which is possible only when

r = 0, 10, 20, 30, 40, 50.

∴ There are six terms viz. $T_{1}, T_{11}, T_{21}, T_{31}, T_{41}, T_{51}$

which are independent of radicals.

Question 14

In the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1024}$ , the number of integral terms is

128

129

130

131

SOLUTION

Solution : B

Here n = 1024 = $2^{10}$ , a power of 2, where as the

power of 7 is $\frac{1}{8}$ = $2^{- 3}$

Now first term $^{1024} C_{0}$ ( $5^{\frac{1}{2}})^{1024}$ = $5^{512}$ (integer)

And after 8 terms, the $9^{t h}$ term =   $^{1024} C_{8}$ ( $5^{\frac{1}{2}})^{1016}$ ( $7^{\frac{1}{8}})^{8}$ = an integer

Again, $17^{t h}$ term =   $^{1024} C_{16}$ ( $5^{\frac{1}{2}})^{1008}$ ( $7^{\frac{1}{8}})^{16}$

= An integer.
Continuing like this, we get an A.P. 1, 9, 17,..........., 1025,

because $1025^{t h}$ term = the last term in the expansion

=   $^{1024} C_{1024}$ ( $7^{\frac{1}{8}})^{1024}$ = $7^{128}$ (an integer)

If n is the number of terms of above A.P. we have

1025 = $T_{n}$ = 1 + (n - 1)8 ⇒ n = 129.

Question 15

If n is positive integer and three consecutive coefficients in the expansion of $(1 + x)^{n}$ are in the ratio 6 : 33 : 110, then n =

SOLUTION

Solution : C

Let the consecutive coefficient of $(1 + x)^{n}$ are

$^{n} C_{r - 1}$ , $^{n} C_{r}$ , $^{n} C_{r + 1}$

By condition, $^{n} C_{r - 1}$ : $^{n} C_{r}$ : $^{n} C_{r + 1}$ = 6 : 33 : 110

Now $^{n} C_{r - 1}$ : $^{n} C_{r}$ = 6 : 33

$\Rightarrow$ 2n - 13r + 2 = 0 ..............(i)

and $^{n} C_{r}$ : $^{n} C_{r + 1}$ = 33: 110

$\Rightarrow$ 3n - 13r - 10 = 0 ...............(ii)

Solving (i) and (ii), we get n = 12 and r = 2.

Aliter: We take first n = 4 [By alternate (a)] but

(a) does not hold. Similarity (b).

So alternate (c), n = 12 gives

$(1 + x)^{12}$ $= [1 + 12 x + \frac{12.11}{2.1}$ $x^{2}$ + $\frac{12.11.10}{3.2.1} x^{3} + . . . . . . . . .]$

So coefficient of II, III and IV terms are

$12, 6 \times 11, 2 \times 11 \times 10.$

So, $12 : 6 \times 11 : 2 \times 11 \times 10 = 6 : 33 : 110.$

Free Objective Test 03 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

Related ExamsObjective Test 03Objective Test 03Objective Test 03

Related Exams
Objective Test 03
Objective Test 03
Objective Test 03