If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}},|t| \geq 1$, then $\frac{d y}{d x}=$
If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y) =a^2-b^2$, where $a>b>0$, then at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right), \frac{d y}{d x}=$
Consider the quadratic equation $a x^2+b x+c=0$, where $2 a+3 b+6 c=0$ and let $g(x)=\frac{a x^3}{3}+\frac{b x^2}{2}+c x$
Statement I The given quadratic equation $a x^2+b x+c=0$ has atleast one root in $(0,1)$.
Statement II Rolle's theorem is applicable to $g(x){\text {on }}$ [0, 1].
Then
The difference between the absolute maximum and absolute minimum values of the function $f(x)=2 x^3-15 x^2+36 x-30$ on $[-1,4]$ is
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