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, $9h^2=5ab$ Then, $h^2=ab$ $f(x)={\sin }^{-1}[2-4x^2]$ $[.]$ ??.(i) $\left[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right]$ LHS of Eq. (i) is independent of $\left[-\frac{\sqrt{3}}{2},0\right]$ . $\left[-\frac{\sqrt{3}}{2},0\right)\cup \left(0,\frac{\sqrt{3}}{2}\right]$ $\left[-\frac{\sqrt{3}}{2},\infty \right)$ $f(x)=\{\begin{array}{cc}\frac{1-\sqrt{2}\sin x}{\pi -4x},& x\ne \frac{\pi }{4}\\ \frac{4a+1}{4},& x=\frac{\pi }{4}\end{array}$ $x=\frac{\pi }{4}$ From Eq. (i), $f(x)={\cos }^{-1}\left[\frac{1-{({\log }_{e}x)}^2}{1+{({\log }_{e}x)}^2}\right],$ $f^{\prime }(e)$ $\frac{-1}{e}$ $\frac{1}{e}$ $lx+my=1$ $x^2+y^2=a^2$ $(1,m)$ 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 $(q,0)$ . $f(x)=\frac{x^2}{x^2+1}$ $R$ $\left[0,\text{1}\right]$ $\left(0,\text{1}\right]$ $\left[0,\text{1}\right)$ $f(x)=x^3-6x^2+ax+b$