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प्रश्न
The equation of tangent to the curve y = `sin^-1 (2x)/(1 + x^2)` at x = `sqrt(3)` is ______.
पर्याय
`y = -1/2(x - sqrt(3))`
`y - π/3 = -1/2(x - sqrt(3))`
`y + π/3 = -1/2(x - sqrt(3))`
None of these
MCQ
रिकाम्या जागा भरा
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उत्तर
The equation of tangent to the curve y = `sin^-1 (2x)/(1 + x^2)` at x = `sqrt(3)` is `underlinebb(y - π/3 = -1/2(x - sqrt(3)))`.
Explanation:
y = `sin^-1 (2x)/(1 + x^2) = π - 2tan^-1x, "for" x > 1`
or `dy/dx = -2/(1 + x^2)`
or `(dy/dx)_(x = sqrt(3)) = -2/(1 + 3) = -1/2`
Also, when x = `sqrt(3)`, y = `π - 2 xx π/3 = π/3`
Hence, equation of tangent is `y - π/3 = -1/2(x - sqrt(3))`.
shaalaa.com
Application of Derivative in Geometry
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