Mathematics · JEE

Limits, continuity and differentiability Mock Test for JEE

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Q1MathsUnit 7: Limit, Continuity and Differentiability
Let f(x)=x3x2+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)},0tx0x13x,1<x2\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 x2x^{2}- 3x+23 x+2 leaves remainder 2x3.2 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)+fx\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
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
Q5MathsUnit 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
Q6MathsUnit 7: Limit, Continuity and Differentiability
If f(x)=cosπ(x+[x]),\boldsymbol{f}(\boldsymbol{x})=\cos \boldsymbol{\pi}(|\boldsymbol{x}|+[\boldsymbol{x}]), then f(x)\boldsymbol{f}(\boldsymbol{x}) is/are (where [.] denotes greatest integer function) This question has multiple correct options
Q7MathsUnit 7: Limit, Continuity and Differentiability
Letf(x+y2)=12[f(x)+f(y)]\operatorname{Let} \boldsymbol{f}\left(\frac{\boldsymbol{x}+\boldsymbol{y}}{\mathbf{2}}\right)=\frac{\mathbf{1}}{\mathbf{2}}[\boldsymbol{f}(\boldsymbol{x})+\boldsymbol{f}(\boldsymbol{y})] for real xx and y.y . If f(0)f^{\prime}(0) exists and equals -1 and f(0)=1f(0)=1 then the value of f(2)f(2) is
Q8MathsUnit 7: Limit, Continuity and Differentiability
Consider the function f(x)=\boldsymbol{f}(\boldsymbol{x})= \left\{\begin{array}{cl}\frac{x+5}{x-2} & \text { if } x \neq 2 \\ 1 & \text { if } x=2\end{array} . \text { Then } f(f(x)) is \right. discontinuous.

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