If the function $f(x)=\left\{\begin{array}{lc}k_{1}(x-\pi)^{2}-1 & x \leq \pi \\ k_{2} \cos x & x>\pi\end{array}\right.$ is twice differentiable, then the ordered pair $(k_{1}, k_{2})$ is equal to:

(D) $\left(\frac{1}{2}, 1\right)$

(C) $\left(\frac{1}{2},-1\right)$

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Continuity and Differentiability

If the function

f (x) = {k_{1} (x - π)^{2} - 1 k_{2} cos x x \leq π x > π

is twice differentiable, then the ordered pair

(k_{1}, k_{2})

is equal to:

Answer: (D)

Match the two pieces and their derivatives at

x = π

. Continuity requires

k_{1} (π - π)^{2} - 1 = k_{2} cos π

, that is

- 1 = - k_{2}

, so

k_{2} = 1

. The first derivatives are

2 k_{1} (x - π)

and

- k_{2} sin x

, both zero at

π

, so differentiability is automatic. The second derivatives are

2 k_{1}

and

- k_{2} cos x

; equating them at

π

gives

2 k_{1} = - k_{2} cos π = k_{2} = 1

, so

k_{1} = \frac{1}{2}

. The ordered pair is

(\frac{1}{2}, 1)

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option