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

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$, then $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2$ $+(5 a-8 b+3 c)^2$ is equal to

Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6 x^5+15 x^4-20 x^3+15 x^2-6 x+1=0$ Then $\frac{a^2+b^2}{a+b+1}$ is

If ${x^2} + 2ax + 10 - 3a > 0$ for all $x \in R$, then