JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

Let

f:[0,2 \pi] \rightarrow[-3,3]

be a given function defined as

f(x)=3 \cos \frac{x}{2}

. The slope of the tangent to the curve

y=f^{-1}(x)

at the point where the curve crosses the

y

-axis is :

View answer and explanation

Correct option: B

-\frac{2}{3}

Option B (-

\frac{2}{3}

) is correct because finding the inverse function and its derivative, then evaluating at the y-intercept, gives the slope of the tangent as -2/3.

More Questions from TANGENTS & NORMALS

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z