Question

If the 4th term in the expansion of $ \pi $ ; $ \frac{\pi }{2} $ , is $ \frac{13}{56} $ and three normals to the parabola $ \frac{3}{56} $ are drawn through a point (q,0), then

• A. $ \frac{1}{56} $
• B. $ \frac{14}{56} $
• C. $ \frac{91}{15} $
• D. $ (a-b){{x}^{2}}+(c-a)x+(b-c)=0 $

Answer 1

Answer: (B)

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 $