Let $a = \min \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Then the value of $\sum_{r=0}^n a^r b^{n-r}$ is

  • A
    $\frac{2^{n+1} + 1}{3 \cdot 2^n}$
  • B
    $\frac{2^{n+1} - 1}{3 \cdot 2^n}$
  • C
    $\frac{2^n - 1}{3 \cdot 2^n}$
  • D
    $\frac{4^{n+1} - 1}{3 \cdot 2^n}$

Explore More

Similar Questions

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

Let $A$ be a $3 \times 3$ matrix such that $A^2 - 5A + 7I = 0$.
Statement-$I$: ${A^{-1}} = \frac{1}{7}(5I - A)$.
Statement-$II$: The polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5(A - 4I)$.

If $\alpha , \beta \neq 0$ and $f(n) = \alpha^n + \beta^n$ and $\begin{vmatrix} 3 & 1 + f(1) & 1 + f(2) \\ 1 + f(1) & 1 + f(2) & 1 + f(3) \\ 1 + f(2) & 1 + f(3) & 1 + f(4) \end{vmatrix} = K(1 - \alpha)^2 (1 - \beta)^2 (\alpha - \beta)^2$,then $K = \dots$

If $A+B=\left[\begin{array}{cr}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ where $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix,then the matrix $\left(A^{-1} B+A B^{-1}\right)$ at $\theta=\frac{\pi}{6}$ is given by

Let $S = \{ m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6} \}$,where $A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. Then $n(S)$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo