1708No ketan , actually answer is
f_(max) = 2 and f_(min) = 0
\hspace{-16}$Find Max. and Min. value of $\mathbf{f(x)=\frac{2\cos x+2}{\sin x+\cos x+2}}$
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1057my logic is a bit vague but still...
first of f(x) cant be negative for any value of x since 2cosx≤-2 and sinx+cosx≤-√2
so to attain min we have to see if 0 is possible...
for f(x) to be zero 2cos x shud be equal to (-2)
so cos x shud be -1 which is clearly possible so min(f(x))=0
for max i can just say that we have to make de denominator as small as possible at the same time not affecting de numerator much....
so we can keep sin x = -1
so cos x will be 0
so f(x) will be 2/1=2
1708I have seems you are right
\hspace{-16}$ Let $\mathbf{y=\frac{2\cos x+2}{\sin x+\cos x+2}}$\\\\\\ $\mathbf{y\sin x+(y-2)\cos x = (2-2y)}$\\\\ Now Using Cauchy-Schwartz Inequality\\\\ $\mathbf{\left\{y^2+(y-2)^2\right\}.\left\{\sin^2 x+\cos ^2 x\right\}\geq \left\{y\sin x+(y-2)\cos x\right\}^2}$\\\\ $\mathbf{y^2+(y-2)^2\geq (2-2y)^2}$\\\\ $\mathbf{2y^2-4y\leq 0}$\\\\ $\mathbf{y.(y-2)\leq 0}$\\\\ So $\mathbf{0\leq y \leq 2}$\\\\ So $\mathbf{\text{Min}(y) = 0}$ and $\mathbf{\text{Max}(y) = 2}$
71@ketan. Better to know it. But the problem is that one could not figure out that it can be used is many so serious cases. It requires some practice which I too fail for.
1708yes vivek you are saying right
so it is better to know the given Inequalities.
1708Question from bansal Test
\hspace{-16}$ If $a^2+b^2=1$ and $u$ is a Minimum value of $\frac{b+1}{a+b-2}$.\\\\ Then Find the value of $u^2$