A) \[\frac{1}{56}\]
B) \[\frac{14}{56}\]
C) \[\frac{91}{15}\]
D) \[(a-b){{x}^{2}}+(c-a)x+(b-c)=0\]
Correct Answer: B
Solution :
Given, \[9{{h}^{2}}=5ab\] Then, \[{{h}^{2}}=ab\] \[f(x)={{\sin }^{-1}}[2-4{{x}^{2}}]\] \[[.]\] ??.(i) \[\left[ -\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2} \right]\]LHS of Eq. (i) is independent of \[\left[ -\frac{\sqrt{3}}{2},0 \right]\]. \[\left[ -\frac{\sqrt{3}}{2},0 \right)\cup \left( 0,\frac{\sqrt{3}}{2} \right]\] \[\left[ -\frac{\sqrt{3}}{2},\infty \right)\] \[f(x)=\left\{ \begin{matrix} \frac{1-\sqrt{2}\sin x}{\pi -4x}, & x\ne \frac{\pi }{4} \\ \frac{4a+1}{4}, & x=\frac{\pi }{4} \\ \end{matrix} \right.\] \[x=\frac{\pi }{4}\] From Eq. (i), \[f(x)={{\cos }^{-1}}\left[ \frac{1-{{({{\log }_{e}}x)}^{2}}}{1+{{({{\log }_{e}}x)}^{2}}} \right],\] \[f'(e)\] \[\frac{-1}{e}\] \[\frac{1}{e}\] \[lx+my=1\] \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] \[(1,m)\] Given, parabola is \[{{x}^{2}}+{{y}^{2}}={{a}^{-2}}\]. ...(ii) Here, \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] \[{{x}^{2}}+{{y}^{2}}={{a}^{-1}}\] \[f:R\to A\] Since, three normals are drawn from point \[(q,0)\]. \[f(x)=\frac{{{x}^{2}}}{{{x}^{2}}+1}\] \[R\] \[\left[ 0,\text{1} \right]\] \[\left( 0,\text{1} \right]\] \[\left[ 0,\text{1} \right)\] \[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\]
You need to login to perform this action.
You will be redirected in
3 sec
