For $f(x)=\sin x+3x-\dfrac{2}{\pi}(x^{2}+x)$, where $x\in\left[0,\tfrac{\pi}{2}\right]$, consider:
(I) $f$ is increasing in $\left(0,\tfrac{\pi}{2}\right)$.
(II) $f'$ is decreasing in $\left(0,\tfrac{\pi}{2}\right)$.
Let $(2,3)$ be the largest open interval in which the function $f(x)=2\log_e(x-2)-x^2+ax+1$ is strictly increasing and $(b,c)$ be the largest open interval in which the function $g(x)=(x-1)^3(x+2-a)^2$ is strictly decreasing. Then $100(a+b-c)$ is equal to
Let $\lambda^*$ be the largest value of $\lambda$ for which the function $f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_{\lambda^*}(1) + f_{\lambda^*}(-1)$ is equal to:
If the function $f$ given by $f(x)=x^3-3(a-2)x^2+3ax+7$, for some $a\in\mathbb{R}$, is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation $\dfrac{f(x)-14}{(x-1)^2}=0\ (x\ne1)$ is:
Let f be a real valued function, defined on R $-$ {$-$1, 1} and given by f(x) = 3 loge $\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$. Then in which of the following intervals, function f(x) is increasing?
For the function $f(x)=\cos x - x + 1,; x\in\mathbb{R}$, consider the statements
(S1) $f(x)=0$ for only one value of $x$ in $[0,\pi]$.
(S2) $f(x)$ is decreasing in $\left[0,\tfrac{\pi}{2}\right]$ and increasing in $\left[\tfrac{\pi}{2},\pi\right]$.
Which is/are correct?
Let the function $f(x)=\dfrac{x}{3}+\dfrac{3}{x}+3,\ x\ne0$ be strictly increasing in $(-\infty,\alpha_1)\cup(\alpha_2,\infty)$ and strictly decreasing in $(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)$. Then $\displaystyle \sum_{i=1}^{5}\alpha_i^{2}$ is equal to
The function $f(x) = {x^3} - 6{x^2} + ax + b$ is such that $f(2) = f(4) = 0$. Consider two statements : Statement 1 : there exists x1, x2 $\in$(2, 4), x1 < x2, such that f'(x1) = $-$1 and f'(x2) = 0. Statement 2 : there exists x3, x4 $\in$ (2, 4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and $2f'({x_3}) = \sqrt 3 f({x_4})$.Then
If $5f(x)+4f\!\left(\frac{1}{x}\right)=x^{2}-2,\ \forall x\ne 0$ and $y=9x^{2}f(x)$,
then $y$ is strictly increasing in:
$\displaystyle \left(0,\frac{1}{\sqrt5}\right)\cup\left(\frac{1}{\sqrt5},\infty\right)$ $\displaystyle \left(-\frac{1}{\sqrt5},0\right)\cup\left(\frac{1}{\sqrt5},\infty\right)$ $\displaystyle \left(-\frac{1}{\sqrt5},0\right)\cup\left(0,\frac{1}{\sqrt5}\right)$ $\displaystyle \left(-\infty,\frac{1}{\sqrt5}\right)\cup\left(0,\frac{1}{\sqrt5}\right)$ Go to Discussion JEE MAIN JEE Mains PYQJEE MAIN JEE Main 2024 (1 February Morning Shift) PYQ