The value of constant $ c $ that makes the function $ f $ defined by $f(x) = \ x^2 - c^2$, $\&\ \text{if } x<4$$cx + 20, \&\ \text{if } x \geq 4$ continuous for all real numbers is:

Question

Let $[t]$ denote the greatest integer less than or equal to $t$.
If the function
\[ f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} \] is continuous at $x=0$, then $a^2+b^2$ is equal to

Answer 1

Step 1: Continuity condition at $ x = 4 $. For $ f(x) $ to be continuous at $ x = 4 $, the left-hand limit ($ LHL $), the right-hand limit ($ RHL $), and the function value $ f(4) $ must be equal. \[LHL = \lim_{x \to 4^-} (x^2 - c^2) = 4^2 - c^2 = 16 - c^2\]\[RHL = \lim_{x \to 4^+} (cx + 20) = 4c + 20\]Equating $ LHL $ and $ RHL $:\[16 - c^2 = 4c + 20\]Step 2: Solving the equation. Rearranging the equation:\[c^2 + 4c + 4 = 0 \quad \Rightarrow \quad (c + 2)^2 = 0 \quad \Rightarrow \quad c = -2\]Conclusion: Therefore, the determined value for $ c $ is $ -2 $, aligning with option $ \mathbf{(A)} $.

The value of constant \( c \) that makes the function \( f \) defined by \(f(x) = \ x^2 - c^2\), \(\&\ \text{if } x<4\)\(cx + 20, \&\ \text{if } x \geq 4\) continuous for all real numbers is:

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Continuity

Questions Asked in CBSE Class XII exam