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Limits, continuity and differentiability Previous Year Questions for JEE

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Part of Limit, Continuity and Differentiability in JEE Main Mathematics.

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Q1MathsUnit 7: Limit, Continuity and Differentiability
Let f(x)=x3−x2+x+1\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^{3}-\boldsymbol{x}^{2}+\boldsymbol{x}+\mathbf{1} and g(x)=\boldsymbol{g}(\boldsymbol{x})= {max⁡{f(t)},0≤t≤x0≤x≤13−x,1<x≤2\left\{\begin{array}{l}\max \{f(t)\}, \quad 0 \leq t \leq x \quad 0 \leq x \leq 1 \\ 3-x, \quad 1<x \leq 2\end{array}\right. Then in the interval [0,2],g(x)[0,2], g(x) is This question has multiple correct options
Q2MathsUnit 7: Limit, Continuity and Differentiability
A polynomial p(x)p(x) when divided by x2−x^{2}- 3x+23 x+2 leaves remainder 2x−3.2 x-3 . Then
Q3MathsUnit 7: Limit, Continuity and Differentiability
\operatorname{Let} f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0<x \leq 2\end{array} and \right. g(x)=∣f(x)∣+f∣x∣\boldsymbol{g}(\boldsymbol{x})=|\boldsymbol{f}(\boldsymbol{x})|+\boldsymbol{f}|\boldsymbol{x}| then the number of points which g(x)g(x) is non differentiable, is
Q4MathsUnit 7: Limit, Continuity and Differentiability
Arrange the following limits in the ascending order: (1) lim⁡x→∞(1+x2+x)x+2\lim _{x \rightarrow \infty}\left(\frac{1+x}{2+x}\right)^{x+2} (2) lim⁡x→0(1+2x)3/x\lim _{x \rightarrow 0}(1+2 x)^{3 / x} (3) lim⁡θ→0sin⁡θ2θ\lim _{\boldsymbol{\theta} \rightarrow \mathbf{0}} \frac{\sin \boldsymbol{\theta}}{\mathbf{2} \boldsymbol{\theta}} (4) lim⁡x→0log⁡e(1+x)x\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{x}
Q5MathsUnit 7: Limit, Continuity and Differentiability
Assertion lim⁡x→01−cos⁡2xx\lim _{\boldsymbol{x} \rightarrow \mathbf{0}} \frac{\sqrt{1-\cos 2 x}}{\boldsymbol{x}} does not exist. Reason ∣sin⁡x∣={sin⁡x;0<x<π2−sin⁡x;−π2<x<0|\sin x|=\left\{\begin{array}{cc}\sin x ; & 0<x<\frac{\pi}{2} \\ -\sin x ; & -\frac{\pi}{2}<x<0\end{array}\right.
Q6MathsUnit 7: Limit, Continuity and Differentiability
Let f(x)\boldsymbol{f}(\boldsymbol{x}) be defined in the interval [-2,2] such that f(x)=f(x)= {−1,−2≤x≤0x−1,0<x≤2 and g(x)=\left\{\begin{array}{ll}-1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2\end{array} \text { and } g(x)=\right. f(∣x∣)+∣f(x)∣\boldsymbol{f}(|\boldsymbol{x}|)+|\boldsymbol{f}(\boldsymbol{x})| Test the differentiablity of g(x)g(x) in (-2,2)
Q7MathsUnit 7: Limit, Continuity and Differentiability
The function f(x)=f(x)= \left\{\begin{array}{l}\frac{\cos 3 x-\cos 4 x}{x^{2}}, \text { for } x \neq 0 \\ \frac{7}{2}, \text { for } x=0\end{array} at \right. x=0\boldsymbol{x}=\mathbf{0} is
Q8MathsUnit 7: Limit, Continuity and Differentiability
Assertion If f(x)=0f(x)=0 has two distinct positive real roots then number of non- differentiable points of y=∣f(−∣x∣)∣\boldsymbol{y}=|\boldsymbol{f}(-|\boldsymbol{x}|)| is 1\mathbf{1} Reason Graph of y=f(∣x∣)\boldsymbol{y}=\boldsymbol{f}(|\boldsymbol{x}|) is symmetrical about y-axis

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