JEE Advanced2007MathematicsApplication of DerivativesActual
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 )
Options
- Aon the left of x=c
- Bon the right of x=c
- Cat no point
- Dat all points
Correct answer
A. on the left of x=c
Step-by-step solution
Slope of the line joining the points (c-1, e^ c-1 ) and (c+1, e^ c+1 ) is equal to e^ c+1 -e^ c-1 2 >e^c Tangent to the curve y=e^x will intersect the given line to the left of the line x=c . ALITER The equation of the tangent to the curve y=e^x at (c, e^c ) is y-e^c=e^c(x-c) Equation of the line joining the given points is y-e^ c-1 = e^c (e-e⁻¹ ) 2 [x-(c-1)] Eliminating y from Eqs. (i) and (ii), we get aligned [x-(c-1)] [2- (e-e⁻¹ ) ] & =2 e⁻¹ x-c & = e+e⁻¹-2 2- (e-e⁻¹ ) < 0 x < c . aligned