power indices and surds Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (b)

$4^x .2^y = 128 and 3^{3x}.3^{2y} – 9^{xy} = 0$

$2^x. 2^y = 128 =(2)^7 3^{3x + 2y} = 3^{2xy} [∵ 9 = 3^2]$

⇒2x + y = 7⇒3x + 2y = 2xy ...(2)

⇒y = 7 – 2x ...(1)

Substitute this value of y in (2) we get

3x + 2(7 – 2x) = 2x (7 – 2x)

3x + 14 – 4x = 14x – $4x^2$

$4x^2$ – 15x + 14 = 0

(4x – 7) (x – 2) = 0

either 4x – 7 = 0 or x – 2 = 0

⇒ x = $7/4$ = or x = 2

⇒x ≠ $7/4$⇒x = 2

y = 7 - 2(2) = 3

⇒x + y = 2 + 3 = 5

∴ Option (b) is correct.

Answer: (d)

$x^{√x} = ^n√{x^x}$

⇒$(x^x)^{1/2} = x^{x/n}$

Take log on both the sides,

⇒$log(x^{√x}) = log(x)^{x/n}$

⇒$√x log x = x/n$.log x

⇒$√x log x - x/n$ log x = 0

⇒log x ($√x - x/n$) = 0

⇒(log x)$(√x)(1 - {√x}/n)$ = 0

⇒log x = 0⇒x = 1,

$√x = 0⇒x = 0 or √x = n ⇒x = n^2$

Thus, x = 0, 1, $n^2$

x = 0 is not admissible

Since, logx is not defined

∴ x = 1, $n^2$

Answer: (d)

Given, x = $(a + √{a^2 + b^3})^{1/3} + (a - √{a^2 + b^3})^{1/3}$

On cubing both sides, we get

$x^3 = (a + √{a^2 + b^3}) + (a - √{a^2 + b^3})$

$+3(a + √{a^2 + b^3})^{1/3}(a - √{a^2 + b^3})^{1/3}$

$[(a + √{a^2 + b^3})^{1/3} + (a - √{a^2 + b^3})^{1/3}]$

⇒$x^3$ = 2a - 3b(x)

⇒$x^3$ + 3bx - 2a = 0

Answer:(d)

LCM of 2, 3 and 6 = 12

Now, $√2 = 2^{1/2 × {12}/{12}} = ^12√{2^6} = ^12√{64}$

$^3$$√3 = 3^{1/3 × {12}/{12}} = ^12√{3^4} = ^12√{81}$

$^6$$√6 = 6^{1/6 × {12}/{12}} = ^12√{6^2} = ^12√{36}$

So, $√2$ is not smallest and $^6$$√6$ is not greatest. So neither I nor II correct.

Answer: (d)

$√{7.84} + √{0.0784} + √{0.000784} + √{0.00000784}$

= $√{{784}/{100}} + √{{784}/{10000}} + √{{784}/{1000000}} + √{{784}/{100000000}}$

= ${28}/{10} + {28}/{100} + {28}/{1000} + {28}/{10000}$

= 2.8 + 0.28 + 0.028 + 0.0028 = 3.1108