MCQ
If $\left(a+\frac{1}{a}+2\right)^2=4$, then $a^2+\frac{1}{a^2}=$
- A12
- B13
- ✓14
- D-14
✓
Answer
Correct option: C.
14
(c)
We have,
$\left(a+\frac{1}{a}+2\right)^2=4$
$\Rightarrow \quad a+\frac{1}{a}+2= \pm 2 \Rightarrow a+\frac{1}{a}=0$ or, $a+\frac{1}{a}=-4$
Now, $\quad a+\frac{1}{a}=0 \Rightarrow a^2+1=0$, which is impossible. Therefore, $a+\frac{1}{a} \neq 0$
$\therefore \quad a+\frac{1}{a}=-4 \Rightarrow\left(a+\frac{1}{a}\right)^2=16 \Rightarrow a^2+\frac{1}{a^2}+2=16 \Rightarrow a^2+\frac{1}{a^2}=14$
We have,
$\left(a+\frac{1}{a}+2\right)^2=4$
$\Rightarrow \quad a+\frac{1}{a}+2= \pm 2 \Rightarrow a+\frac{1}{a}=0$ or, $a+\frac{1}{a}=-4$
Now, $\quad a+\frac{1}{a}=0 \Rightarrow a^2+1=0$, which is impossible. Therefore, $a+\frac{1}{a} \neq 0$
$\therefore \quad a+\frac{1}{a}=-4 \Rightarrow\left(a+\frac{1}{a}\right)^2=16 \Rightarrow a^2+\frac{1}{a^2}+2=16 \Rightarrow a^2+\frac{1}{a^2}=14$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the area of an equilateral triansle is $81 \sqrt{3} cm^2$, then its semi - perimeter is
→$\left(x^2-4 x-21\right)=?$
→Axioms are assumed
If the perimeter of an isosceles triangle is 32 cm and the ratio of the equal side to its base is $3: 2$, then area of the triangle is
→In Fig., AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If $\angle A P R=25^{\circ}, \angle R Q C=30^{\circ}$ and $\angle C Q F=65^{\circ}$, then
→In Fig, $PQ \| RS$, $\angle\text{AEF}=95^\circ,\angle\text{BHS}=110^\circ,$ and $\angle\text{ABC}=\text{x}^\circ.$ Then the value of $x$ is:
→Write the correct answer in the following: A rational number between $\sqrt{2}$ and $\sqrt{3}$ is
→In the adjoining figure, $AB = AC$ and $AD$ is median of $\triangle\text{ABC},$ then $\triangle\text{ADC}$ is equal to:
→The simplest form of $0.12 \overline{3}$ is
→The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is: