Surds And Indices Questions For IBPS Clerk PDF

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Question 1: If $lo{g}_{3}2,lo{g}_3(2^x–5),lo{g}_3(2^x–7/2)$ are in arithmetic progression, then the value of x is equal to

a) 5

b) 4

c) 2

d) 3

Question 2: Let $u=({\log }_{2}x{)}^2–6{\log }_{2}x+12$ where x is a real number. Then the equation $x^u=256$, has

a) no solution for x

b) exactly one solution for x

c) exactly two distinct solutions for x

d) exactly three distinct solutions for x

Question 3: If x = -0.5, then which of the following has the smallest value?

a) $2^{1/x}$

b) $1/x$

c) $1/x^2$

d) $2^x$

e) $1/\sqrt{-x}$

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Question 4: Which among $2^{1/2},3^{1/3},4^{1/4},6^{1/6}$, and ${12}^{1/12}$ is the largest?

a) $2^{1/2}$

b) $3^{1/3}$

c) $4^{1/4}$

d) $6^{1/6}$

e) ${12}^{1/12}$

Question 5: If $lo{g}_{y}x=(a\ast lo{g}_{z}y)=(b\ast lo{g}_{x}z)=ab$, then which of the following pairs of values for (a, b) is not possible?

a) (-2, 1/2)

b) (1,1)

c) (0.4, 2.5)

d) ($\pi$, 1/ $\pi$)

e) (2,2)

Question 6: If x >= y and y > 1, then the value of the expression $lo{g}_x(x/y)+lo{g}_y(y/x)$ can never be

a) -1

b) -0.5

c) 0

d) 1

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Question 7: If $f(x)=\log \frac{(1+x)}{(1-x)}$, then f(x) + f(y) is

a) $f(x+y)$

b) $f\frac{(x+y)}{(1+xy)}$

c) $(x+y)f\frac{1}{(1+xy)}$

d) $\frac{f(x)+f(y)}{(1+xy)}$

Question 8: $2^{73}-2^{72}-2^{71}$ is the same as

a) $2^{69}$

b) $2^{70}$

c) $2^{71}$

d) $2^{72}$

Question 9: Find the value of $\frac{1}{1+\frac{1}{3-\frac{4}{2+\frac{1}{3-\frac{1}{2}}}}}$ + $\frac{3}{3–\frac{4}{3+\frac{1}{2-\frac{1}{2}}}}$

a) $\frac{13}{7}$

b) $\frac{15}{7}$

c) $\frac{11}{21}$

d) $\frac{17}{28}$

Question 10: If ${\log }_2{\log }_7(x^2–x+37)$ = 1, then what could be the value of ‘x’?

a) 3

b) 5

c) 4

d) None of these

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Question 11: Which of the following is true?

a) $7^{(3^2)}=(7^3{)}^2$

b) $7^{(3^2)}>(7^3{)}^2$

c) $7^{(3^2)}<(7^3{)}^2$

d) None of these

Question 12: If ${\log }_{2}x.{\log }_{\frac{x}{64}}2={\log }_{\frac{x}{16}}2$. Then x is

a) 2

b) 4

c) 16

d) 12

Question 13: What is the value of $\sqrt{\frac{a}{b}}$, If ${\log }_4{\log }_4{4}^{a-b}=2{\log }_4(\sqrt{a}-\sqrt{b})+1$

a) -5/3

b) 2

c) 5/3

d) 1

Question 14: ${\log }_{5}2$ is

a) An integer

b) A rational number

c) A prime number

d) An irrational number

Question 15: Find the coefficient of $x^{12}$ in the expansion of $(1–x^6{)}^4(1–x{)}^{-4}$

a) 113

b) 119

c) 125

d) 132

Question 16: $(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})÷\sqrt{\frac{16}{81}}=?$

a) $\frac{5}{16}$

b) $\frac{7}{12}$

c) $\frac{3}{8}$

d) None of these

Question 17: Find the value of x from the following equation:
${\log }_{10}3+{\log }_{10}(4x+1)={\log }_{10}(x+1)+1$

a) 2/7

b) 7/2

c) 9/2

d) None of the above

Question 18: If $\log 3,log(3^x–2)$ and $log(3^x+4)$ are in arithmetic progression, then x is equal to

a) $\frac{8}{3}$

b) ${\log }_{3}8$

c) ${\log }_{2}3$

d) $8$

Question 19: The value of $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\dots ..\mathrm{\infty }}}}}$ is

a) 1

b) 2

c) 3

d) 4

Question 20: The unit digit in the product of $(8267{)}^{153}\times (341{)}^{72}$ is

a) 1

b) 2

c) 7

d) 9

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Answers & Solutions:

1) Answer (D)

$2log(2^x–5)=log2+log(2^x–7/2)$
Let $2^x=t$
=> $(t-5{)}^2=2(t-7/2)$
=> $t^2+25–10t=2t–7$
=> $t^2–12t+32=0$
=> t = 8, 4
Therefore, x = 2 or 3, but $2^x$ > 5, so x = 3

2) Answer (B)

$x^u=256$

Taking log to the base 2 on both the sides,

$u\ast {\log }_{2}x={\log }_{2}256$

=>$[({\log }_{2}x{)}^2–6{\log }_{2}x+12]\ast {\log }_{2}x=8$

$(lo{g}_{2}x{)}^3–6(lo{g}_{2}x{)}^2+12lo{g}_{2}x=8$

Let $lo{g}_{2}x=t$

$t^3–6t^2+12t–8=0$

$(t-2{)}^3=0$

Therefore, $lo{g}_{2}x=2$

=> $x=4$ is the only solution

Hence, option B is the correct answer.

3) Answer (B)

$2^p$ is always positive

$x^2$ is always non negative.

$1/\sqrt{-x}$ is always positive.

$\frac{1}{x}$ is negative when x is negative.

In this case, x is negative => $\frac{1}{x}$ is smallest.

4) Answer (B)

Make the power equal and compare the denominators.

$2^{1/2}$ can be written as ${64}^{1/12}$

$3^{1/3}$ can be written as ${81}^{1/12}$

$4^{1/4}$ can be written as ${64}^{1/12}$

$6^{1/6}$ can be written as ${36}^{1/12}$

Among these, ${81}^{1/12}$ is the greatest => $3^{1/3}$ is the greatest.

5) Answer (E)

$lo{g}_{y}x=ab$
$a\ast lo{g}_{z}y=ab$ => $lo{g}_{z}y=b$
$b\ast lo{g}_{x}z=ab$ => $lo{g}_{x}z=a$
$lo{g}_{y}x$ = $lo{g}_{z}y\ast lo{g}_{x}z$ => $logx/logy$ = $logy/logz\ast logz/logx$
=> $\frac{logx}{logy}=\frac{logy}{logx}$
=> $(logx{)}^2=(logy{)}^2$
=> $logx=logy$ or $logx=-logy$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

6) Answer (D)

$lo{g}_x(x/y)+lo{g}_y(y/x)$ = $1–lo{g}_x(y)+1–lo{g}_y(x)$
= $2–(lo{g}_{x}y+1/lo{g}_{x}y)$ <= 0 (Since $lo{g}_{x}y+1/lo{g}_{x}y$ >= 2)
So, the value of the expression cannot be 1.

7) Answer (B)

If $f(x)=\log \frac{(1+x)}{(1-x)}$ then $f(y)=\log \frac{(1+y)}{(1-y)}$

Also Log (A*B)= Log A + Log B

f(x)+f(y) = $\log \frac{(1+x)(1+y)}{(1-x)(1-y)}$ solving we get $\log \frac{1+\frac{(x+y)}{(1+xy)}}{1-\frac{(x+y)}{(1+xy)}}$

Hence option B.

8) Answer (C)

$2^{71}(2^2–2^1–1)$
$2^{71}(4-2-1)$
$2^{71}$

9) Answer (B)

First term can be reduced to $\frac{33}{21}$ or $\frac{11}{7}$

And second term can be reduced to $\frac{16}{28}$ or $\frac{4}{7}$

Sum will be = $\frac{11}{7}$ + $\frac{4}{7}$ = $\frac{15}{7}$

10) Answer (C)

${\log }_2{\log }_7(x^2–x+37)$ = 1

${\log }_7(x^2–x+37)$ = $2$

$(x^2–x+37)$ = $7^2$

Given eq. can be reduced to $x^2–x+37=49$

So x can be either -3 or 4.

11) Answer (B)

$7^{(3^2)}=7^9$

$(7^3{)}^2=7^6$

So $7^{(3^2)}>(7^3{)}^2$

12) Answer (B)

${\log }_{2}x.{\log }_{\frac{x}{64}}2={\log }_{\frac{x}{16}}2$

i.e. $\frac{logx}{log2}\ast \frac{lo{g}_2}{logx-log64}=\frac{log2}{logx-log16}$

i.e. $\frac{logx\ast (logx-log16)}{logx-log64}$ = $\log 2$

let t = log x

Therefore,  $\frac{t\ast (t-log16)}{t-log64}$ = $\log 2$

$t^2-4\ast log2\ast t=t\ast log2-6\ast (log2{)}^2$

I.e. $t^2-5\ast log2\ast t-6\ast (log2{)}^2$ = 0

I.e. $t^2-3\ast log2\ast t-2\ast log2\ast t-6\ast (log2{)}^2$ = 0

i.e. $t\ast (t-3\ast log2)-2\ast log2\ast (t-3\ast log2)$ = 0

i.e $t=2\ast log2$ or $t=3\ast log2$

i.e $logx=log4$ or $logx=log8$

therefore $x=4$ or $8$

therefore our answer is option ‘B’

13) Answer (C)

$\sqrt{\frac{a}{b}}$, If ${\log }_4{\log }_4{4}^{a-b}=2{\log }_4(\sqrt{a}-\sqrt{b})+{\log }_{4}4$

i.e. ${\log }_4{\log }_4{4}^{a-b}={\log }_4((\sqrt{a}-\sqrt{b}{)}^2)\ast 4$

i.e. ${\log }_4{4}^{a-b}=((\sqrt{a}-\sqrt{b}{)}^2)\ast 4$

i.e. (a-b)*${\log }_{4}4=((\sqrt{a}-\sqrt{b}{)}^2)\ast 4$

i.e. a-b = 4a+4b-8$\sqrt{ab}$

i.e. 3a + 5b – 8$\sqrt{ab}$ = 0

i.e. $3{\sqrt{\frac{a}{b}}}^2$ – 8$\sqrt{\frac{a}{b}}$+5 = 0

put $\sqrt{\frac{a}{b}}$ = t

therefore 3$t^2$ – 8t + 5 = 0

solving we get t = 1 or t = $\frac{5}{3}$

i.e. $\sqrt{\frac{a}{b}}$ = 1 or $\frac{5}{3}$

but if $\sqrt{\frac{a}{b}}$ = 1 then a=b then ${\log }_4(\sqrt{a}-\sqrt{b})$ will become indefinite

Therefore  $\sqrt{\frac{a}{b}}$ = $\frac{5}{3}$

Therefore our answer is option ‘C’

14) Answer (D)

Let ${\log }_{5}2$ = y

Let us assume  ${\log }_{5}2$ is a rational number.

${\log }_{5}2$ = p/q, where p and q are co primes.

5^(p/q)=2 => 5^p=2^q.

5^p=5*5*5*5*5*5*5………………p times

2^p=2*2*2*2*2*2*2………………q times

No value of p and q can satisfy the equation. Hence y is an irrational number.

15) Answer (C)

We can write $(1–x^6{)}^4$ = $4C0(1{)}^4(x^6{)}^0$ – $4C1(1{)}^3(x^6{)}^1$ + $4C2(1{)}^2(x^6{)}^2$ – $4C3(1{)}^1(x^6{)}^3$+ $4C4(1{)}^0(x^6{)}^4$

$\Rightarrow$ $(1–x^6{)}^4=(1-4x^6+6x^{12}-4x^{18}+x^{24})$

Therefore, we can say

$(1–x^6{)}^4(1–x{)}^{-4}=(1-4x^6+6x^{12}-4x^{18}+x^{24})\ast (1–x{)}^{-4}$

We have to find out coefficient of $x^{12}$, $x^6$, $x^0$ in $(1–x{)}^{-4}$.

We can use binomial expansion for negative coefficients. Therefore, coefficient of $x^{12}$ in $(1–x{)}^{-4}$

$\Rightarrow$ ${\displaystyle \frac{(-4)\ast (-4-1)\ast (-4-2)\ast \dots \ast (-4-11)}{12!}}$

$\Rightarrow$ ${\displaystyle \frac{15!}{12!\ast 3!}}$

$\Rightarrow$ ${\displaystyle \frac{15\ast 14\ast 13}{3\ast 2\ast 1}}$

$\Rightarrow$ $455$

Similarly, coefficient of $x^6$ in $(1–x{)}^{-4}$

$\Rightarrow$ ${\displaystyle \frac{(-4)\ast (-4-1)\ast (-4-2)\ast \dots \ast (-4-5)}{6!}}$

$\Rightarrow$ ${\displaystyle \frac{9!}{6!\ast 3!}}$

$\Rightarrow$ ${\displaystyle \frac{7\ast 8\ast 9}{3\ast 2\ast 1}}$

$\Rightarrow$ $84$

Coefficient of $x^0$ in $(1–x{)}^{-4}$ is 1.

Therefore, we can say that the coefficient of $x^{12}$ in the expansion of $(1–x^6{)}^4(1–x{)}^{-4}$ = 455+(-4*84)+(1*6) = 125. Hence, option C is the correct answer.

Alternative Solution:

$(1-x^6{)}^4=(1-x{)}^4(1+x+x^2+x^3+x^4+x^5{)}^4$

$\Rightarrow (1-x^6{)}^4(1-x{)}^{-4}=(1+x+x^2+x^3+x^4+x^5{)}^4$

Hence we need to find coeff of  $x^{12}$ in $(1+x+x^2+x^3+x^4+x^5{)}^4$ =  $(1+x+x^2+x^3+x^4+x^5)\times$$(1+x+x^2+x^3+x^4+x^5)\times$$(1+x+x^2+x^3+x^4+x^5)\times$$(1+x+x^2+x^3+x^4+x^5)$

This will be equal to number of integral solutions for a + b + c + d = 12, 0<=a,b,c,d<=4

a is the power of x from the first expression, b is the power of x

Lets find the set of values for (a,b,c,d)

(5,5,2,0) => Number of ways of arranging = 4!/2! = 12

(5,5,1,1) => Number of ways of arranging = 4!/(2!*2!) = 6

(5,4,3,0) => Number of ways of arranging = 4! = 24

(5,4,2,1) => Number of ways of arranging = 4! = 24

(5,3,2,2) => Number of ways of arranging = 4!/2! = 12

(5,3,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,4,0) => Number of ways of arranging = 4!/3! = 4

(4,4,3,1) => Number of ways of arranging = 4!/2! = 12

(4,4,2,2) => Number of ways of arranging = 4!/(2!*2!) = 6

(4,3,3,2) => Number of ways of arranging = 4!/2! = 12

(3,3,3,3) => Number of ways of arranging = 4!/4! = 1

Hence the coeff of $x^{12}$ = 24*2 + 12*5 + 6*2 + 4 + 1 = 125

16) Answer (A)

$(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})÷\sqrt{\frac{16}{81}}=x$ This can be simplified as

$(\frac{15}{27}-\frac{5}{12})÷\frac{4}{9}=x$

$(\frac{5}{36})\ast \frac{9}{4}=x$

x=$\frac{5}{16}$. Hence, option A is the correct answer.

17) Answer (B)

${\log }_{10}3+{\log }_{10}(4x+1)={\log }_{10}(x+1)+1$ can be written as

${\log }_{10}3+{\log }_{10}(4x+1)={\log }_{10}(x+1)+{\log }_{10}10$

We know that ${\log }_{10}a+{\log }_{10}b={\log }_{10}ab$

${\log }_{10}3\ast (4x+1)={\log }_{10}(x+1)\ast 10$

$12x+3=10x+10$

$x=7/2$. Hence, option B is the correct answer.

18) Answer (B)

If $log3,log(3^x–2)$ and $log(3^x+4)$ are in arithmetic progression
Then, $2\ast log(3^x–2)=log3+log(3^x+4)$
Thus, $log(3^x–2{)}^2=log3(3^x+4)$
Thus, $(3^x–2{)}^2=3(3^x+4)$
=> $3^{2x}–4\ast 3^x+4=3\ast 3^x+12$
=> $3^{2x}–7\ast 3^x–8=0$
=> $(3^x+1)\ast (3^x-8)=0$
But $3^x+1\ne 0$
Thus, $3^x=8$
Hence, $x=lo{g}_{3}8$
Hence, option B is the correct answer.

19) Answer (C)

Let, $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\dots ..\mathrm{\infty }}}}}$ = $x$
Thus, $\sqrt{7+\sqrt{7-x}}$ = $x$
=> $7+\sqrt{7-x}=x^2$
=> $7-x=(x^2-7{)}^2$
Putting options we get,
x=1 => 6$\ne (-6{)}^2$
x=2 => 5$\ne (-3{)}^2$
x=3 => 4=$(9-7{)}^2$
x=4 => 3$\ne (9{)}^2$
Hence, option C is the correct answer.

20) Answer (C)

The unit digit in the product of $(8267{)}^{153}\times (341{)}^{72}$ is the same as the unit digit in the product of $(7{)}^{153}\times (1{)}^{72}$
1 raised to anything is 1.
7 has a cyclicity  of 4. Thus, $7^{153}=7^1=7$
Hence, option C is the correct answer.

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