Question

Let f(x) = $\left\{ \begin{matrix} x^{2}e^{2(x - 1)},x \leq 1 \\ a\cos(2x - 2) + bx^{2}x > 1 \end{matrix} \right.\$f(x) will be

differentiable at x = 1, if

• A. a = – 1, b = 2
• B. a = 1, b = – 2
• C. a = 1, b = 2
• D. None of these

Answer 1

Answer: (A)

a = – 1, b = 2

Answer 2

Function is diff. at x = 1 it means function is continuous at x = 1, diff. at x = 1

R.H.L. at x = 1 = L.H.L. at x = 1, R.H.D. = L.H.D.

a cos (0) + b = 1

a + b = 1 ........(1)

(–2 a sin (2x – 2) + 2bx)_{x = 1}

= (2x² e^{2(x – 1)} + 2xe^{2(x – 1)})_{x = 1}

0 + 2b = 2 + 2

2b = 4

b = 2

So a = 1 – b = 1 – 2 = – 1