Let $f(x)= \begin{cases}\frac{\mathrm{a} x^2+2 \mathrm{a} x+3}{4 x^2+4 x-3} & , x \neq-\frac{3}{2}, \frac{1}{2} \\ \mathrm{~b} & , x=-\frac{3}{2}, \frac{1}{2}\end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f \circ f(x)=\frac{7}{5}$, then $x$ is equal to:
If $\lim\limits_{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to :
Let $[\cdot]$ denote the greatest integer function, and let $f(x)=\min \left\{\sqrt{2} x, x^2\right\}$.
Let $\mathrm{S}=\left\{x \in(-2,2)\right.$ : the function $\mathrm{g}(x)=|x|\left[x^2\right]$ is discontinuous at $\left.x\right\}$.
Then $\sum\limits_{x \in \mathrm{~S}} f(x)$ equals
Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$.
Then $\lim\limits _{x \rightarrow 1}\left(\log _e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)$ is equal to :
JEE Main Subjects
Browse all chapters by subject