Question

Let \(f(x) =   \begin{cases}     \tan x,       & if\ 0\leq x\leq \frac{\pi}{4} \\      ax+b,       & if\ \frac{\pi}{4}\lt x\lt \frac{\pi}{2}   \end{cases}\), If f(x) is differentiable at \(x=\frac{\pi}{4}\), then the values of a and b are respectively

• A. \(2,\frac{2-\pi}{2}\)
• B. \(2,\frac{4-\pi}{4}\)
• C. \(1,\frac{-\pi}{4}\)
• D. \(2,\frac{-\pi}{4}\)
• E. 2,1-π

Answer 1

The Correct Option is A

For the function to be differentiable at $x = \frac{\pi}{4}$, it must be continuous and have matching derivatives at $x = \frac{\pi}{4}$ from both sides.
Step 1: Continuity at } $x = \frac{\pi}{4}$
The function must be continuous at $x = \frac{\pi}{4}$.
Thus, the left-hand limit at $x = \frac{\pi}{4}$ should be equal to the right-hand limit at $x = \frac{\pi}{4}$.
For $x \leq \frac{\pi}{4}$, $f(x) = \tan(x)$, so: \[ f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \] For $\frac{\pi}{4} < x < \frac{\pi}{2}$, $f(x) = ax + b$. Substituting $x = \frac{\pi}{4}$: \[ f\left(\frac{\pi}{4}\right) = a\left(\frac{\pi}{4}\right) + b \] For continuity, these must be equal: \[ a\left(\frac{\pi}{4}\right) + b = 1 \] Thus, we get the equation: \[ \frac{a\pi}{4} + b = 1 \quad \text{(Equation 1)} \] \textbf{Step 2: Differentiability at } $x = \frac{\pi}{4}$ For the function to be differentiable at $x = \frac{\pi}{4}$, the left-hand derivative should equal the right-hand derivative. For $x \leq \frac{\pi}{4}$, $f(x) = \tan(x)$, so: \[ f'(x) = \sec^2(x) \] At $x = \frac{\pi}{4}$: \[ f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = 2 \] For $\frac{\pi}{4} < x < \frac{\pi}{2}$, $f(x) = ax + b$, so: \[ f'(x) = a \] For differentiability, we require: \[ a = 2 \] So, $a = 2$.
Step 3: Solving for } $b$ Substitute $a = 2$ into Equation 1: \[ \frac{2\pi}{4} + b = 1 \quad \Rightarrow \quad \frac{\pi}{2} + b = 1 \quad \Rightarrow \quad b = 1 - \frac{\pi}{2} \] Thus, $b = \frac{2 - \pi}{2}$.

The correct option is (A) : \(2,\frac{2-\pi}{2}\)

Answer 2

• We are given: \[f(x) = \begin{cases} \tan x, & 0 \leq x \leq \frac{\pi}{4} \\ ax + b, & \frac{\pi}{4} < x < \frac{\pi}{2} \end{cases}\]
• For \(f(x)\) to be differentiable at \(x=\frac{\pi}{4}\), it must first be continuous at \(x=\frac{\pi}{4}\). This means the left-hand limit and the right-hand limit must be equal at \(x=\frac{\pi}{4}\).
• Continuity condition: \[\tan(\frac{\pi}{4}) = a\frac{\pi}{4} + b\] \[1 = a\frac{\pi}{4} + b\]
• Now, for differentiability, the left-hand derivative and the right-hand derivative must be equal at \(x=\frac{\pi}{4}\).
• Left-hand derivative: \[\frac{d}{dx}(\tan x) = \sec^2 x\] At \(x = \frac{\pi}{4}\), the left-hand derivative is \(\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2\).
• Right-hand derivative: \[\frac{d}{dx}(ax + b) = a\] So, the right-hand derivative is simply \(a\).
• Differentiability condition: \[a = 2\]
• Substitute \(a = 2\) into the continuity equation: \[1 = 2\cdot\frac{\pi}{4} + b\] \[1 = \frac{\pi}{2} + b\] \[b = 1 - \frac{\pi}{2} = \frac{2-\pi}{2}\]
• Therefore, \(a = 2\) and \(b = \frac{2-\pi}{2}\).