Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Let $$\alpha, \beta \epsilon \mathbb R$$ be such that the fonction $$f(x) =\left\{\begin{array}{||}2\alpha(x^{2}-2)+2\beta x & \quad,{x<1}\\(\alpha +3)x+(\alpha -\beta) & \quad ,{x\geq 1}\\\end{array}\right.$$ be differentiable at all $$x\epsilon \beta$$. Then $$34(\alpha +\mathbb R)$$ is equal to
Continuity at (x=1):
$$-2\alpha+2\beta=2\alpha-\beta+3;\Rightarrow;3\beta=4\alpha+3$$
Differentiability at (x=1):
$$4\alpha+2\beta=\alpha+3;\Rightarrow;2\beta=3-3\alpha$$
Solve:
$$\frac{4\alpha+3}{3}=\frac{3-3\alpha}{2}$$
$$;\Rightarrow;\alpha=\frac{3}{17},;\beta=\frac{21}{17}$$
$$\alpha+\beta=\frac{24}{17}$$
$$\quad\Rightarrow\quad34(\alpha+\beta)=48$$
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.