If 0 leq a, b leq 3 and the equation x^2+4+3 | vedclass

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

Question

(C) The given equation is $x^2+4+3 \cos (ax+b)=2x$. Rearranging the terms,we get $(x^2-2x+1) + 3 + 3 \cos (ax+b) = 0$. This simplifies to $(x-1)^2 + 3

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) The given equation is $x^2+4+3 \cos (ax+b)=2x$. Rearranging the terms,we get $(x^2-2x+1) + 3 + 3 \cos (ax+b) = 0$. This simplifies to $(x-1)^2 + 3(1 + \cos (ax+b)) = 0$. Since $(x-1)^2 \geq 0$ and $1 + \cos (ax+b) \geq 0$ for all real $x$,the sum can be zero only if both terms are zero simultaneously. Thus,$(x-1)^2 = 0 \implies x = 1$ and $1 + \cos (ax+b) = 0$. This implies $\cos (ax+b) = -1$,so $ax+b = (2n+1)\pi$ for some integer $n$. Substituting $x=1$,we get $a+b = (2n+1)\pi$. Given $0 \leq a, b \leq 3$,we have $0 \leq a+b \leq 6$. Since $\pi \approx 3.14$,the only possible value for $a+b$ in the range $[0, 6]$ is $\pi$ (where $n=0$).

If $0 \leq a, b \leq 3$ and the equation $x^2+4+3 \cos (ax+b)=2x$ has a real solution,then the value of $(a+b)$ is

Similar Questions

If $\alpha+\beta+\gamma=\pi$,then the expression $\sin^2 \alpha+\sin^2 \beta-\sin^2 \gamma$ is equal to:

In triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C =$

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

Given $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$,then which of the following holds good?

Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta)=4(\sin^{4}(\frac{7\pi}{2}-\theta)+\sin^{4}(11\pi+\theta)) - 2(\sin^{6}(\frac{3\pi}{2}-\theta)+\sin^{6}(9\pi-\theta))$,$\theta \in R$. Then $\alpha+2\beta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module