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
Arrange the following limits in the ascending order: (1) limx(1+x2+x)x+2\lim _{x \rightarrow \infty}\left(\frac{1+x}{2+x}\right)^{x+2} (2) limx0(1+2x)3/x\lim _{x \rightarrow 0}(1+2 x)^{3 / x} (3) limθ0sinθ2θ\lim _{\boldsymbol{\theta} \rightarrow \mathbf{0}} \frac{\sin \boldsymbol{\theta}}{\mathbf{2} \boldsymbol{\theta}} (4) limx0loge(1+x)x\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{x}
Q5MathsUnit 7: Limit, Continuity and Differentiability
Assertion limx01cos2xx\lim _{\boldsymbol{x} \rightarrow \mathbf{0}} \frac{\sqrt{1-\cos 2 x}}{\boldsymbol{x}} does not exist. Reason sinx={sinx;0<x<π2sinx;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)= {1,2x0x1,0<x2 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
Q9MathsUnit 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
Q10MathsUnit 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
Q11MathsUnit 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.
Q12MathsUnit 7: Limit, Continuity and Differentiability
The function f(x)=sin1(cosx)f(x)=\sin ^{-1}(\cos x) is :
Q13MathsUnit 7: Limit, Continuity and Differentiability
The set of all values of 'a' for which limxa[x]\lim _{x \rightarrow a}[x] does not exist is ([x]([x] denotes greatest integer function)
Q14MathsUnit 7: Limit, Continuity and Differentiability
g(x)={1x212x2<x<4 .then x,x4\boldsymbol{g}(\boldsymbol{x})=\left\{\begin{array}{ll}\mathbf{1} & \boldsymbol{x} \leq-\boldsymbol{2} \\ \frac{\mathbf{1}}{\mathbf{2}} \boldsymbol{x} & -\boldsymbol{2}<\boldsymbol{x}<\mathbf{4} \text { .then } \\ \sqrt{\boldsymbol{x}} & , \boldsymbol{x} \geq \mathbf{4}\end{array}\right.
Q15MathsUnit 7: Limit, Continuity and Differentiability
Which one of the following is correct in respect of the function f(x)=x2x\boldsymbol{f}(\boldsymbol{x})=\frac{\boldsymbol{x}^{2}}{|\boldsymbol{x}|} for x0\boldsymbol{x} \neq \mathbf{0} and f(0)=0?\boldsymbol{f}(\mathbf{0})=\mathbf{0} ?

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