Question : If $\small a^{4}+1=\left [\frac{a^{2}}{b^{2}}\right ]\left (4b^{2}-b^{4}-1\right),$ then what is the value of $a^{4}+b^{4}?$
Option 1: 2
Option 2: 16
Option 3: 32
Option 4: 64
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 2
Solution : Given: $a^{4}+1=\frac{a^{2}}{b^{2}}( 4b^{2}-b^{4}-1)$ ⇒ $\frac{(a^{4}+1)}{a^2}=\frac{1}{b^{2}}( 4b^{2}-b^{4}-1)$ ⇒ $a^2+\frac{1}{a^2}=4 -b^2-\frac{1}{b^2}$ ⇒ $a^2+\frac{1}{a^2}+2+b^2+\frac{1}{b^2} +2 =0$ ⇒ $(a-\frac{1}{a})^2+(b-\frac{1}{b})^2=0$ So, $(a-\frac{1}{a})^2=0 \;$and$\; (b-\frac{1}{b})^2=0$ ⇒ $a-\frac{1}{a}=0 \; $ and$\; b-\frac{1}{b}=0$ ⇒ $a=\frac{1}{a} \; $ and$\; b=\frac{1}{b}$ ⇒ $a^2=1 \;$ and $\; b^2 = 1$ $\therefore a^{4}+b^{4}=2$ Hence, the correct answer is 2.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\frac{1}{x^2+a^2}=x^2-a^2$, then the value of $x$ is:
Question : If $x_{1}x_{2}x_{3}=4(4+x_{1}+x_{2}+x_{3})$, then what is the value of $\left [ \frac{1}{(2+x_{1})} \right ]+\left [ \frac{1}{(2+x_{2})} \right ]+\left [ \frac{1}{(2+x_{3})} \right ]?$
Question : What is the value of
Question : If $M =\left ( \frac{3}{7} \right ) ÷ \left ( \frac{6}{5} \right ) ×\left ( \frac{2}{3} \right ) + \left ( \frac{1}{5} \right ) ×\left ( \frac{3}{2} \right )$ and
Question : The value of $15 \div 8-\frac{5}{4}$ of $\left(\frac{8}{3} \times \frac{9}{16}\right)+\left(\frac{9}{8} \times \frac{3}{4}\right)-\left(\frac{5}{32} \div \frac{5}{7}\right)+\frac{3}{8}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile