Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

The value of b, for which the function $f(x) =\left\{\begin{matrix} 5x-4 & 0< x ≤1\\4x^2 +3bx & 1< x < 2\end{matrix}\right.$ is continuous at every point of its domain is :

Options:

-1

0

1

$\frac{13}{3}$

Correct Answer:

-1

Explanation:

The correct answer is Option (1) → -1

$f(x) =\left\{\begin{matrix} 5x-4 & 0< x ≤1\\4x^2 +3bx & 1< x < 2\end{matrix}\right.$

$f(1)=5(1)-4=1$

$\lim\limits_{x→1^+}f(1)=4(1)^3+3b(1)$

so $4+3b=1$

$b=-1$ for continuity to exist