simplification Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (b)

Given, a = ${1 + x}/{2 - x}$

∴ $1/{a + 1} + {2a + 1}/{a^2 - 1} = {3a}/{a^2 - 1} = {3({1 + x}/{2 - x})}/{({1 + x}/{2 - x})^2- 1}$

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

= ${3(1 + x)(2 - x)}/{6x - 3} = {(1 + x) (2 - x)}/{(2x - 1)}$

Answer: (c)

For x = 1,

$(1)^{29} – (1)^{24} + (1)^{13}$ – 1 = 1 – 1 + 1 – 1 = 0

So, (x – 1) is a factor of $x^{29} – x^{24} + x^{13}$ – 1.

For x = –1,

$(–1)^{29} – 1(–1)^{24} + (–1)^{13}$ – 1 = –1 – 1 – 1 – 1 = –4

So, (x + 1) is not a factor of $x^{29} – x^{24} + x^{13}$ – 1.

Answer: (d)

Approx value = 9999 ÷ 99

= 101 ÷ 9 = 11.2 ≈ 11

Answer: (a)

The expression $3x^3 – kx^2$ + 4x + 16 is divisible by

x - $k/2$.

Then, x = $k/2$ satisfy the equation

⇒$3(k/2)^3 - k (k/2)^2 + 4 (k/2)$ + 16 = 0

⇒${3k^3 - 2k^3 + 16k + 128}/8$ = 0

⇒$k^3$ + 16k + 128 = 0

⇒(k + 4) ($k^2$ – 4k + 32) = 0

⇒k + 4 = 0

⇒k = – 4

Answer: (c)

$x^4 + 4y^4 = x^4 + 4y^4 + 4x^2 y^2 – 4x^2 y^2$

= $(x^2 + 2y^2)^2 – (2xy)^2$

= $(x^2 + 2y^2 – 2xy) (x^2 + 2y^2$ + 2xy)

From above it is clear that $x^4 + 4y^4$ is divisible by $x^2 + 2y^2$ + 2xy.