IIT-JEE 2007 · Paper 1 · Mathematics Q48

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MathematicsSingle correct+3 / -1Question 48 of 66

The tangent to the curve $y=e^x$ drawn at the point $(c,e^c)$ intersects the line joining the points $(c-1,e^{c-1})$ and $(c+1,e^{c+1})$

on the left of

x=c

on the right of

x=c

at no point

at all points

Answer and worked solution

Official answer: A

Lines. Put $t=x-c$ and divide by $e^c$ . The tangent is $Y=1+t$ ; the chord through $(-1,e^{-1})$ and $(1,e)$ has slope $\sinh1$ : $Y=e^{-1}+\sinh1\,(t+1)$ .

Intersection. $1+t=e^{-1}+\sinh1\,(t+1)$ gives
$t+1=\frac{e^{-1}}{1-\sinh1}.$
Since $\sinh1\approx1.175>1$ , $t+1<0$ , so $t<-1<0$ : the point lies to the left of $x=c$ . (Geometrically, the convex curve lies above its tangent, and the chord crosses the tangent outside the chord's own span.)

(A)

Topics:Tangents and normals Equations of a line

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IIT-JEE 2007 · Paper 1, Mathematics question 48