Let $$y = f(x)$$ be a parabola, having its axis parallel to y-axis, which is touched by the line $$y =x$$ at $$x = 1$$, then
The general equation of a parabola having its axis parallel to y-axis is
$$y = ax^2 + bx + c$$ ........ (i)
The line $$y = x$$ touches the required parabola. $$y = ax^2 + bx + c$$ at $$(1,1)$$.
Hence, the slope of the parabola at $$x=1$$ is $$1$$.
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx}\right )_{(1, 1)} = 1$$
$$\Rightarrow 2a + b = 1$$ .....(ii)
Also, (1, 1) lies on the parabola.
$$\Rightarrow a+ b + c = 1$$ ...(iii)
From the equations (ii) and (iii)
$$ a - c = 0$$
$$\Rightarrow a = c$$
Using $$a=c$$ in (iii), we get $$2c + b = 1$$.
$$\Rightarrow 2f(0) + f' (0) = 1$$ $$[\because f(0) = c \ and \ f'(0) = b]$$
or, $$ 2f(0) = 1- f' (0)$$
The other options are not satisfied.
Hence, option A and C are correct.
A helicopter is flying along the curve given by $$y - x^{3/2} =7, (x \ge 0)$$. A solider positioned at the point $$\left(\dfrac{1}{2}, 7\right)$$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:
A tangent to the curve, $$y=f(x)$$ at $$P(x, y)$$ meets $$x-$$axis at $$A$$ and $$y-$$axis at $$B$$. If $$AP:BP=1:3$$ and $$f(1)=1$$, then the curve also passes through the point.
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Let $$f(x)=\dfrac{\sin \pi X}{x^2},x>0$$
Let $$x_1
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