ISI2015-MMA-13

Question

The number of real roots of the equation

$$2 \cos \left( \frac{x^2+x}{6} \right) = 2^x +2^{-x} \text{ is }$$

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Comments

Same question asked in ISI 2016: https://gateoverflow.in/242727/isi2016-mma-3

https://math.stackexchange.com/questions/763481/no-of-real-solutions-of-the-equation-2-cos-fracx2-x6-2x-2-x

Answer 1

Answer: $A$

Minimum value of $2^x + 2^{-x} = 2$

$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) = 2^x + 2^{-x}$$

$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) \ge 2$$

Now, $\because \cos 0 = 1$

$$\implies \frac{x^2+x}{6} =  0 \implies x(x+1) = 0 \implies x = 0, x = -1$$

$$

So, $x = 0$ is the only solution as the equation is not satisfying at $x = -1$

Comments

@ankitgupta.1729

Can you please check-out this solution?

---

@`JEET  this is rightly done brother, but in LHS its given that 2cos( x^2 + x^6). The cosine is also going to take some anglular value. Lets say it takes from 0 to pi/4 or any range of angles then only one value will satisfy the inequality . Hence the answer should be 1. option B should be correct according to me . 

Answer 2

as cosx=(e^ix+e^-ix)/2 so I got (B).

Comments

I think answer will be D

---

How please explain @ankitgupta

---

@ankitgupta.1729

no, it is 1 and i.e. $0$

https://math.stackexchange.com/questions/3286446/finding-root-of-an-equation?noredirect=1#comment6759409_3286446