Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If p, q, r are in A.P. and are positive, the roots of the quadratic

equation p $x^{2}$ + qx + r = 0 are all real for ___.

| $\frac{r}{p}$ - 7| ≥ 4 $\sqrt{3}$

| $\frac{p}{r}$ - 7| < 4 $\sqrt{3}$

All p and r

No p and r

SOLUTION

Solution : A

p,q,r are positive and are in A.P.

∴ q = $\frac{p + r}{2}$ .........(i)

∵ The roots of p $x^{2}$ + qx + r = 0 are real

⇒ $q^{2}$ ≥ 4pr ⇒ ${[\frac{p + r}{2}]}^{2}$ ≥ 4pr [using (i)]

⇒ $p^{2}$ + $r^{2}$ - 14pr ≥ 0

⇒ ${(\frac{r}{p})}^{2}$ - 14 ( $\frac{r}{p}$ ) + 1 ≥ 0 ( ∵ p>0 and p ≠ 0)

${(\frac{r}{p} - 7)}^{2}$ - $({4 \sqrt{3})}^{2}$ ≥ 0

⇒| $\frac{r}{p}$ - 7| ≥ 4 $\sqrt{3}$ .

Question 2

Sum of first n terms in the following series

${c o t}^{- 1}$ 3 + ${c o t}^{- 1}$ 7 + ${c o t}^{- 1}$ 13 + ${c o t}^{- 1}$ 21 + .........n terms.is given by

${t a n}^{- 1}$ ( $\frac{n}{n + 2}$ )

${c o t}^{- 1}$ ( $\frac{n + 2}{n}$ )

${t a n}^{- 1}$ (n+1) - ${t a n}^{- 1}$ 1

All of these

SOLUTION

Solution : D

Let S = 3 + 7 + 13 + 21 + ...... + $T_{π}$ ⇒ $T_{π}$ = $n^{2}$ + n + 1.

Let $T_{r}$ = ${c o t}^{- 1}$ ( $r^{2}$ + r + 1) = ${t a n}^{- 1}$ (r + 1) - ${t a n}^{- 1}$ r.

Put r = 1, 2,........, n and add, we get the required sum

${t a n}^{- 1}$ (n+1) - ${t a n}^{- 1}$ 1 = ${t a n}^{- 1}$ ( $\frac{n}{n + 2}$ ) = ${c o t}^{- 1}$ ( $\frac{n + 2}{n}$ )

Question 3

The sum of the series $1^{2}$ .2 + $2^{2}$ .3 + $3^{2}$ .4 + ........ to n terms is

$\frac{n^{3} {(n + 1)}^{3} (2 n + 1)}{24}$

$\frac{n (n + 1) (3 n^{2} + 7 n + 2)}{12}$

$\frac{n (n + 1)}{6}$ [n(n+1) + (2n + 1)]

$\frac{n (n + 1)}{12}$ [6n(n+1) + 2(2n + 1)]

SOLUTION

Solution : B

$T_{π}$ = $n^{2}$ (n + 1) = $n^{3}$ + $n^{2}$

$S_{π}$ = $\sum$ $T_{π}$ = $\sum$ $n^{3}$ + $\sum$ $n^{2}$ = ${[\frac{n (n + 1)}{2}]}^{2}$ + $\frac{n (n + 1) (2 n + 1)}{6}$

= $\frac{n (n + 1)}{2}$ [ $\frac{n (n + 1)}{2}$ + $\frac{2 n + 1}{3}$ ] = $\frac{n (n + 1) (3 n^{2} + 7 n + 2)}{12}$

Question 4

1 . The sum of the series 1 + 3x + 6 $x^{2}$ + 10 $x^{3}$ + ....... ∞ will be

$\frac{1}{(1 - x)^{2}}$

$\frac{1}{1 - x}$

$\frac{1}{(1 + x)^{2}}$

$\frac{1}{(1 - x)^{3}}$

SOLUTION

Solution : D

Let S = 1 + 3x + 6 $x^{2}$ + 10 $x^{3}$ + ....... ∞

⇒ x.S = x + 3 $x^{2}$ + 6 $x^{3}$ + ..........∞

Subtracting S(1-x) = 1 + 2x + 3 $x^{2}$ + 4 $x^{3}$ + ........∞

⇒ x(1 - x)S = x + 2 $x^{2}$ + 3 $x^{3}$ + .........∞

Again subtracting,

⇒ S[(1 - x) - x(1 - x)] = 1 + x + $x^{2}$ + $x^{3}$ + ........... ∞

⇒ S[(1 - x)(1 - x)] = $\frac{1}{1 - x}$ ⇒ S = $\frac{1}{(1 - x)^{3}}$

Question 5

The sum of infinite terms of the following series

1 + $\frac{4}{5}$ + $\frac{7}{5^{2}}$ + $\frac{10}{5^{3}}$ + ...........∞ will be

$\frac{3}{16}$

$\frac{35}{8}$

$\frac{35}{4}$

$\frac{35}{16}$

SOLUTION

Solution : D

Let the sum to infinity of the arithmetic-geometric series be

S = 1 + 4 . $\frac{1}{5}$ + 7. $\frac{1}{5^{2}}$ + 10. $\frac{1}{5^{3}}$ +........

⇒ $\frac{1}{5}$ S = $\frac{1}{5}$ + 4. $\frac{1}{5^{2}}$ + 7. $\frac{1}{5^{3}}$ +...........

Subtracting (1 - $\frac{1}{5}$ )S = 1 + 3. $\frac{1}{5}$ + 3. $\frac{1}{5^{2}}$ + 3. $\frac{1}{5^{3}}$ + .......

= 1 + 3( $\frac{1}{5}$ + $\frac{1}{5^{2}}$ + ..............)

⇒ $\frac{4}{5}$ S = 1 + 3. $\frac{1}{5}$ ( $\frac{1}{1 - \frac{1}{5}}$ ) = 1 + $\frac{3}{4}$ = $\frac{7}{4}$ ⇒ S = $\frac{35}{16}$ .

Aliter :  Use direct formula $S_{\infty}$ =   $\frac{a b}{1 - r}$ +   $\frac{d b r}{(1 - r)^{2}}$

Here a = 1, b = 1, d = 3, r =   $\frac{1}{5}$ , therefore

$S_{\infty}$ =   $\frac{1}{1 - \frac{1}{5}}$ +   $\frac{3 \times 1 \times \frac{1}{5}}{(1 - {\frac{1}{5}}^{2})}$ = $\frac{5}{4}$ + $\frac{\frac{3}{5}}{\frac{16}{25}}$ = $\frac{5}{4}$ + $\frac{15}{16}$ = $\frac{35}{16}$ .

Aliter: Use S = [1 + $\frac{r}{1 - r}$ × diff. of A.P. ] $\frac{1}{1 - r}$

Question 6

If a and b are two different positive real numbers, then which of the following relations is true

$2 \sqrt{a b} > (a + b)$

$2 \sqrt{a b} < (a + b)$

$2 \sqrt{a b}$ = $(a + b)$

None of these

SOLUTION

Solution : B

We know that A > G > H

Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G

$\Rightarrow \frac{a + b}{2} > \sqrt{a b}$ or (a + b) > 2 $\sqrt{a b}$

Question 7

If $\frac{1}{b - a} + \frac{1}{b - c}$ = $\frac{1}{a} + \frac{1}{c}$ , then a,b,c are in

A.P

G.P

H.P

In G.P and H.P both

SOLUTION

Solution : C

Since the reciprocals of a and c occur on RHS, let us first assume that a,b,c are in H.P.

So that $\frac{1}{a}$ , $\frac{1}{b}$ , $\frac{1}{c}$ are in A.P.

$\Rightarrow \frac{1}{b} - \frac{1}{a}$ = $\frac{1}{c} - \frac{1}{b}$ = d,say

$\Rightarrow \frac{a - b}{a b}$ = d = $\frac{b - c}{b c} \Rightarrow a - b$ = abd and b - c =bcd

Now LHS = $- \frac{1}{a - b} + \frac{1}{b + c}$ = $- \frac{1}{a b d} + \frac{1}{b c d}$

$= \frac{1}{b d} (\frac{1}{c} - \frac{1}{a})$ = $\frac{1}{b d} (2 d) ∴ \frac{2}{b}$ = $\frac{1}{a} + \frac{1}{c}$ = RHS

$∴$ a, b, c are in H.P. is verified.

Question 8

If sixth term of a H.P. is $\frac{1}{61}$ and its tenth term is $\frac{1}{105}$ , then first term of that H.P. is

$\frac{1}{28}$

$\frac{1}{39}$

$\frac{1}{6}$

$\frac{1}{17}$

SOLUTION

Solution : C

$T_{6}$ of H.P. = $\frac{1}{61} \Rightarrow T_{6}$ of A.P. = 61

and $T_{10}$ of H.P. = $\frac{1}{105} \Rightarrow T_{10}$ of A.P. = 105,

so, a + 5d = 61 ........(i) and a + 9d = 105 .....(ii)

From (i) and (ii), a= 6

Therefore first term of H.P. = $\frac{1}{a}$ = $\frac{1}{6}$ .

Question 9

If the 7th term of a H.P is $\frac{1}{10}$ and the 12th term is $\frac{1}{25}$ , then the 20th term is

$\frac{1}{37}$

$\frac{1}{41}$

$\frac{1}{45}$

$\frac{1}{49}$

SOLUTION

Solution : D

Considering corresponding A.P.

a + 6d = 10 and a + 11d = 25 $\Rightarrow d$ = 3, a= -8

$\Rightarrow T_{20}$ = a + 19d = -8 + 57 = 49

Hence 20th term of the corresponding H.P. is $\frac{1}{49}$

Question 10

If a, b, c are three distinct positive real numbers which are in H.P., then $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ is

Greater than or equal to 10

Less than or equal to 10

Only equal to 10

None of these

SOLUTION

Solution : D

we have $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. Let $\frac{1}{a}$ = p - q, $\frac{1}{b}$ = p and

$\frac{1}{c}$ = p + q, where p,q > 0 and p > q. Now, substitute these

values in $\frac{3 a + 2 b}{2 a - b} + \frac{3 c + 2 b}{2 c - b}$ then it reduces to

$10 + \frac{14 q^{2}}{p^{2} - q^{2}}$ which is obviously greater than

10(as p > q > 0).

Trick: Put a = 1, b = $\frac{1}{2}$ , c = $\frac{1}{3}$ .

The expression has the value $\frac{3 + 1}{2 - \frac{1}{2}} + \frac{1 + 1}{\frac{2}{3} - \frac{1}{2}}$ = $\frac{8}{3} + 12 > 0$

Question 11

The first term of a harmonic is $\frac{1}{7}$ and the second term is $\frac{1}{9}$ . The $12^{t h}$ term is

$\frac{1}{19}$

$\frac{1}{29}$

$\frac{1}{17}$

$\frac{1}{27}$

SOLUTION

Solution : B

Here first term of A.P. be 7 and second be 9, then $12^{t h}$ term will be 7 + 11 $\times$ 2 = 29

Hence $12^{t h}$ term of the H.P. be $\frac{1}{29}$ .

Question 12

If x, y, z are in G.P. and $a^{x}$ = $b^{y}$ = $c^{z}$ , then

$l o g_{a} c$ = $l o g_{b} a$

$l o g_{b} a$ = $l o g_{c} b$

$l o g_{c} b$ = $l o g_{a} c$

None of these

SOLUTION

Solution : B

x, y, z are in G.P., then $y^{2}$ = x.z

Now $a^{x}$ = $b^{y}$ = $c^{z}$ = m

$\Rightarrow x l o g_{d} a$ = $y l o g_{d} b$ = $z l o g_{d} c$ = $l o g_{d} m$

$\Rightarrow x$ = $l o g_{a} m$ , y= $l o g_{b} m$ , z = $l o g_{c} m$

Again as x, y, z are in G.P., so $\frac{y}{x}$ = $\frac{z}{y}$

$\Rightarrow \frac{l o g_{b} m}{l o g_{a} m}$ = $\frac{l o g_{c} m}{l o g_{b} m}$ $\Rightarrow l o g_{b} a$ = $l o g_{c} b$

Question 13

If the ratio of the sum of first three terms and the sum of first six terms of a G.P. be 125 : 152, then the common ratio $r$ is ___.

$\frac{3}{5}$

$\frac{5}{3}$

$\frac{2}{3}$

$\frac{3}{2}$

SOLUTION

Solution : A

Given that $\frac{a + a r + a r^{2}}{a + a r + a r^{2} + a r^{3} + a r^{4} + a r^{5}} = \frac{125}{152} .$

$\Rightarrow \frac{1 + r + r^{2}}{1 + r + r^{2} + r^{3} + r^{4} + r^{5}} = \frac{125}{152} i.e., \frac{1 + r + r^{2}}{(1 + r + r^{2}) + r^{3} (1 + r + r^{2})} = \frac{125}{152} \Rightarrow \frac{1}{1 + r^{3}} = \frac{125}{152} \Rightarrow r^{3} = \frac{152}{125} - 1 = \frac{27}{125} \Rightarrow r = \frac{3}{5}$

Question 14

The terms of a G.P. are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is ___.

$\frac{\sqrt{5} - 1}{2}$

$\frac{- 1 - \sqrt{5}}{2}$

$2 \sqrt{5}$

SOLUTION

Solution : A

The given condition can be expressed as
$T_{n}$ = $T_{n + 1} + T_{n + 2},$ where $n \geq 1.$
If $a$ is the first term and $r$ is the common ratio, then by the condition is
$a r^{n - 1} = a r^{n} + a r^{n + 1} . \Rightarrow r^{n - 1} = r^{n} (1 + r)$
$\Rightarrow r^{2} + r - 1 = 0$
$\Rightarrow r = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$

Since each term is positive, the common ratio of the GP is $\frac{\sqrt{5} - 1}{2} .$

Question 15

If the sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle term is 15, then greatest number of the series will be

A. 5

B. 7

C. 9

D. 11

SOLUTION

Solution : B

Let four numbers are $a - 3 d, a - d, a + d, a + 3 d .$
Now $(a - 3 d) + (a + 3 d) = 8 \Rightarrow a = 4$ and $(a - d) (a + d) = 15 \Rightarrow a^{2} - d^{2} = 15 \Rightarrow d = 1$ Thus required numbers are 1,3,5,7. Hence greatest number is 7.

Free Objective Test 02 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

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