Question

The point M (x, y) of the graph of the function y = e^–|x| so that area bounded by the tangent at M and the coordinate axes is greatest is –

• A. (1, e^–1)
• B. (2, e^–2)c
• C. (–2, e²)
• D. (0, 1)

Answer 1

Answer: (A)

(1, e^–1)

Answer 2

For x ³ 0, y = e^–x. The equation of tangent is Y – y = – e^–x

(X – x). This will intersect coordinate axes at (x + ye^x, 0) and (0, y + xe^–x). Hence the area of the required triangle A is $\frac{1}{2}$ (y + xe^–x) (x + ye^x)

= $\frac{1}{2}$ (1 + x)² e^–x [Q y = e^–x]

Now $\frac{dA}{dx}$= $\frac{1}{2}$ [– (1 + x)² e^–x + 2(1 + x)e^–x]

= $\frac{1}{2}$ (1 + x) e^–x (1 – x)

Note that $\frac{dA}{dx}$ = 0 Ž x = 1, –1

Also, $\frac{dA}{dx}$ > 0, if 0 £ x < 1 and $\frac{dA}{dx}$ < 0 if x < 1. Hence A is maximum when x = 1 so

y = e^–1. Since y is even function other possibility of M is

(–1, e^–1).