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प्रश्न
The equation of tangent to the curve y=`y=x^2+4x+1` at
(-1,-2) is...............
(a) 2x -y = 0 (b) 2x+y-5 = 0
(c) 2x-y-1=0 (d) x+y-1=0
उत्तर
(a)
`y=x^2+4x+1`
Differentiating w.r.t 'x', we get
`dy/dx=2x+4`
`dy/dx|_(x->-1) =2(-1)+4=2`
Hence, slope of tangent at (-1, -2) is 2.
So equation of tangent line is
y -(-2)= 2( x-(- 1))
2x - y = 0
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