JEE Main 2021 (Online) 4th August Morning ShiftQuestion 1
The equation of the tangent to the curve $y=\int_{x^{2}}^{x^{3}}\frac{dt}{\sqrt{1+t^{2}}}$ at the point whose abscissa is $1$, is
Differentiation and its Application
Solution
Concept. At $x=1$ both limits are $1$, so $y=0$. Differentiate with the Leibniz rule for the slope. Formulas used. (1) $\dfrac{d}{dx}\int_{u(x)}^{v(x)}h(t)\,dt=h(v)\,v'-h(u)\,u'$Step 1 - Point. At $x=1$: $y=\int_1^1(\dots)=0$. Step 2 - Slope. By (1), $$y'=\frac{3x^2}{\sqrt{1+x^6}}-\frac{2x}{\sqrt{1+x^4}},\qquad y'(1)=\frac{3}{\sqrt2}-\frac{2}{\sqrt2}=\frac{1}{\sqrt2}.$$ Step 3 - Tangent. $y=\frac{1}{\sqrt2}(x-1)$, i.e. $x-\sqrt2\,y-1=0$. Answer: (D)