If the equation frac{{{x^2} + 5}}{2} = x - 2c | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$-\cos (a x+b)=\frac{x^{2}-2 x+5}{4}=\frac{(x-1)^{2}}{4}+1$

$\mathrm{Now},-\cos (\mathrm{ax}+\mathrm{b}) \in[-1,1]$

And $\frac{(x-1)^{2}}{4}+1 \

Vedclass · Accepted Answer

$\pi $

Vedclass · Answer

$-\cos (a x+b)=\frac{x^{2}-2 x+5}{4}=\frac{(x-1)^{2}}{4}+1$

$\mathrm{Now},-\cos (\mathrm{ax}+\mathrm{b}) \in[-1,1]$

And $\frac{(x-1)^{2}}{4}+1 \in[1, \infty)$

$\Rightarrow-\cos (a x+b)=1$ And $\frac{(x-1)^{2}}{4}+1=1$

$\Rightarrow x=1$ And $ a x+b=(2 k+1) \pi, k \in I$

$	herefore a+b=\pi$

If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to

Similar Questions

Let $\alpha$ and $\beta$ be the roots of the equation $5 x^{2}+6 x-2=0 .$ If $S_{n}=\alpha^{n}+\beta^{n}, n=1,2,3 \ldots$ then :

If $x,\;y,\;z$ are real and distinct, then $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - zxy$ is always

If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then

If $x$ is real, then the maximum and minimum values of the expression $\frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$ will be

Let $\alpha$ and $\beta$ be the two disinct roots of the equation $x^3 + 3x^2 -1 = 0.$ The equation which has $(\alpha \beta )$ as its root is equal to