Surds And Indices Questions For IBPS Clerk PDF

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Question 1: If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7/2)$ 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 $log_y x = (a*log_z y) = (b*log_x z) = ab$, 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 $log_x (x/y) + log_y (y/x)$ 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: $(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})\div\sqrt{\frac{16}{81}}=?$

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 $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-…..\infty}}}}$ 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)

$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
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,

$u * \log_{2}{x} = \log_{2}{256}$

=>$[({\log_2 x})^2 – 6 {\log_2 x} + 12] * \log_{2}{x} = 8$

$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$

Let $log_2 x = t$

$t^3 – 6t^2 +12t – 8 = 0$

$(t-2)^3 = 0$

Therefore, $log_2 x = 2$

=> $x = 4$ 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)

$log_y x = ab$
$a*log_z y = ab$ => $log_z y = b$
$b*log_x z = ab$ => $log_x z = a$
$log_y x$ = $log_z y * log_x z$ => $log x/log y$ = $log y/log z * log z/log x$
=> $\frac{log x}{log y} = \frac{log y}{log x}$
=> $(log x)^2 = (log y)^2$
=> $log x = log y$ or $log x = -log y$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

6) Answer (D)

$log_x (x/y) + log_y (y/x)$ = $1 – log_x (y) + 1 – log_y (x)$
= $2 – (log_x y + 1/log_x y)$ <= 0 (Since $log_x y + 1/log_x y$ >= 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. $\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$

i.e. $\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\log{2}$

let t = log x

Therefore,  $\frac{t * (t-log{16})}{t-log{64}}$ = $\log{2}$

$t^2-4*log 2*t = t*log 2-6*(log 2)^2$

I.e. $t^2-5*log 2*t-6*(log 2)^2$ = 0

I.e. $t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0

i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0

i.e $t=2*log 2$ or $t=3*log 2$

i.e $log x=log 4$ or $log x=log 8$

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. $\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$

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

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

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

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

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 $(1 – x)^{-4}$

$\Rightarrow$ $\dfrac{(-4)*(-4-1)*(-4-2)* … *(-4-11)}{12!}$

$\Rightarrow$ $\dfrac{15!}{12!*3!}$

$\Rightarrow$ $\dfrac{15*14*13}{3*2*1}$

$\Rightarrow$ $455$

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

$\Rightarrow$ $\dfrac{(-4)*(-4-1)*(-4-2)* … *(-4-5)}{6!}$

$\Rightarrow$ $\dfrac{9!}{6!*3!}$

$\Rightarrow$ $\dfrac{7*8*9}{3*2*1}$

$\Rightarrow$ $84$

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)

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

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

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

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 $\log_{10}{a}+\log_{10}{b}=\log_{10}{ab}$

$\log_{10}{3*(4x+1)}=\log_{10}{(x+1)*10}$

$12x+3=10x+10$

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

18) Answer (B)

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

19) Answer (C)

Let, $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-…..\infty}}}}$ = $x$
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$\neq(-6)^2$
x=2 => 5$\neq(-3)^2$
x=3 => 4=$(9-7)^2$
x=4 => 3$\neq(9)^2$
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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