JEE MAIN Increasing Decreasing Function Previous Year Questions (PYQs) – Page 1 of 2
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$f:R \to R$ be defined as$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.$ Let A = {x $ \in $ R : f is increasing}. Then A is equal to :
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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 $\left(0,\tfrac{\pi}{2}\right)$.
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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
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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:
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The function $f(x) = x e^{\,x(1-x)}, \, x \in \mathbb{R}$ is :
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The interval in which the function $f(x)=x^{x}$, $x>0$, is strictly increasing is:
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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:
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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?
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Solution
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?
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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
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Solution
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