79983
If the function \(f(x)=\left\{\frac{\log _{e}\left(1-x+x^{2}\right)+\log _{e}\left(1+x+x^{2}\right)}{\sec x-\cos x}\right.\), \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\}\) is continuous at \(x=0\), then \(k\) is equal to :
79984
If \(f(x)=\left\{\begin{aligned} x+a, x \leq 0 \\ |x-4|, x>0\end{aligned}\right.\) and \(g(x)=\left\{\begin{array}{r}x+1, x\lt 0 \\ (x-4)^{2}+b, x \geq 0\end{array}\right.\) are continuous on \(R\), then (got) (2) + (fog) (-2) is equal to :
79985
Lef \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a continuous function such that \(f(3 x)-f(x)=x\). If \(f(8)=7\), then \(f(14)\) is equal to :
1 4
2 10
3 11
4 16
Explanation:
(B) : Given, \(\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=\mathrm{x} \tag{i}\) \(\mathrm{f}(\mathrm{x})=7\) We know that, Lagrangels mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) When function is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable \((a, b)\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})}{3 \mathrm{x}-\mathrm{x}}\) \(f^{\prime}(x)=\frac{x}{2 x} \Rightarrow f^{\prime}(x)=\frac{1}{2}\) Integrating both sides, \(\int f^{\prime}(x) d x=\int \frac{d x}{2}\) \(f(x)=\frac{x}{2}+C\) Now, \(\quad \mathrm{x}=8 \Rightarrow \mathrm{f}(8)=7\) \(7=\frac{8}{2}+C \Rightarrow C=3\) \(\therefore \quad \mathrm{f}(14)=\frac{14}{2}+3 \Rightarrow f(14)=10\)
JEE Main-2022-26.07.2022
Limits, Continuity and Differentiability
79986
Let \(f(x)=\left[x^{2}-x\right]+|-x+[x]|\), where \(x \in \mathbb{R}\) and [t] denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
1 continuous at \(x=0\), but not continuous at \(x=1\)
2 continuous at \(x=0\) and \(x=1\)
3 not continuous at \(x=0\) and \(x=1\)
4 continuous at \(x=1\), but not continuous at \(x=0\)
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\left[\mathrm{x}^{2}-\mathrm{x}\right]+|-\mathrm{x}+[\mathrm{x}]|\) We know that, \(f(x)=[x(x-1)]+\{x\}\) Now, at \(\mathrm{x}=0\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}(\mathrm{x}+1) ; & -0.5\lt \mathrm{x}\lt 0 \\ 0 ; & \mathrm{x}=0 \\ -1+\mathrm{x} ; & 0\lt \mathrm{x}\lt 1 \\ 0 ; & \mathrm{x}=1 \\ \mathrm{x}-1 ; & 1\lt \mathrm{x}\lt 1.5\end{array}\right.\) Left hand limit (LHL) \(=\lim _{x \rightarrow 0^{-}} f(x)=1\) Right hand limit (RHL) \(=\lim _{x \rightarrow 0^{+}} f(x)=-1\) and, \(\quad \mathrm{f}(0)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=0\) At, \(\quad \mathrm{x}=1\) Left hand limit (LHL) \(=\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0\) Right hand \(\operatorname{limit}(\mathrm{RHL})=\lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=1-1=0\) And, \(\mathrm{f}(1)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) but not continuous at \(\mathrm{x}=0\)
79983
If the function \(f(x)=\left\{\frac{\log _{e}\left(1-x+x^{2}\right)+\log _{e}\left(1+x+x^{2}\right)}{\sec x-\cos x}\right.\), \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\}\) is continuous at \(x=0\), then \(k\) is equal to :
79984
If \(f(x)=\left\{\begin{aligned} x+a, x \leq 0 \\ |x-4|, x>0\end{aligned}\right.\) and \(g(x)=\left\{\begin{array}{r}x+1, x\lt 0 \\ (x-4)^{2}+b, x \geq 0\end{array}\right.\) are continuous on \(R\), then (got) (2) + (fog) (-2) is equal to :
79985
Lef \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a continuous function such that \(f(3 x)-f(x)=x\). If \(f(8)=7\), then \(f(14)\) is equal to :
1 4
2 10
3 11
4 16
Explanation:
(B) : Given, \(\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=\mathrm{x} \tag{i}\) \(\mathrm{f}(\mathrm{x})=7\) We know that, Lagrangels mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) When function is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable \((a, b)\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})}{3 \mathrm{x}-\mathrm{x}}\) \(f^{\prime}(x)=\frac{x}{2 x} \Rightarrow f^{\prime}(x)=\frac{1}{2}\) Integrating both sides, \(\int f^{\prime}(x) d x=\int \frac{d x}{2}\) \(f(x)=\frac{x}{2}+C\) Now, \(\quad \mathrm{x}=8 \Rightarrow \mathrm{f}(8)=7\) \(7=\frac{8}{2}+C \Rightarrow C=3\) \(\therefore \quad \mathrm{f}(14)=\frac{14}{2}+3 \Rightarrow f(14)=10\)
JEE Main-2022-26.07.2022
Limits, Continuity and Differentiability
79986
Let \(f(x)=\left[x^{2}-x\right]+|-x+[x]|\), where \(x \in \mathbb{R}\) and [t] denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
1 continuous at \(x=0\), but not continuous at \(x=1\)
2 continuous at \(x=0\) and \(x=1\)
3 not continuous at \(x=0\) and \(x=1\)
4 continuous at \(x=1\), but not continuous at \(x=0\)
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\left[\mathrm{x}^{2}-\mathrm{x}\right]+|-\mathrm{x}+[\mathrm{x}]|\) We know that, \(f(x)=[x(x-1)]+\{x\}\) Now, at \(\mathrm{x}=0\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}(\mathrm{x}+1) ; & -0.5\lt \mathrm{x}\lt 0 \\ 0 ; & \mathrm{x}=0 \\ -1+\mathrm{x} ; & 0\lt \mathrm{x}\lt 1 \\ 0 ; & \mathrm{x}=1 \\ \mathrm{x}-1 ; & 1\lt \mathrm{x}\lt 1.5\end{array}\right.\) Left hand limit (LHL) \(=\lim _{x \rightarrow 0^{-}} f(x)=1\) Right hand limit (RHL) \(=\lim _{x \rightarrow 0^{+}} f(x)=-1\) and, \(\quad \mathrm{f}(0)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=0\) At, \(\quad \mathrm{x}=1\) Left hand limit (LHL) \(=\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0\) Right hand \(\operatorname{limit}(\mathrm{RHL})=\lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=1-1=0\) And, \(\mathrm{f}(1)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) but not continuous at \(\mathrm{x}=0\)
79983
If the function \(f(x)=\left\{\frac{\log _{e}\left(1-x+x^{2}\right)+\log _{e}\left(1+x+x^{2}\right)}{\sec x-\cos x}\right.\), \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\}\) is continuous at \(x=0\), then \(k\) is equal to :
79984
If \(f(x)=\left\{\begin{aligned} x+a, x \leq 0 \\ |x-4|, x>0\end{aligned}\right.\) and \(g(x)=\left\{\begin{array}{r}x+1, x\lt 0 \\ (x-4)^{2}+b, x \geq 0\end{array}\right.\) are continuous on \(R\), then (got) (2) + (fog) (-2) is equal to :
79985
Lef \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a continuous function such that \(f(3 x)-f(x)=x\). If \(f(8)=7\), then \(f(14)\) is equal to :
1 4
2 10
3 11
4 16
Explanation:
(B) : Given, \(\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=\mathrm{x} \tag{i}\) \(\mathrm{f}(\mathrm{x})=7\) We know that, Lagrangels mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) When function is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable \((a, b)\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})}{3 \mathrm{x}-\mathrm{x}}\) \(f^{\prime}(x)=\frac{x}{2 x} \Rightarrow f^{\prime}(x)=\frac{1}{2}\) Integrating both sides, \(\int f^{\prime}(x) d x=\int \frac{d x}{2}\) \(f(x)=\frac{x}{2}+C\) Now, \(\quad \mathrm{x}=8 \Rightarrow \mathrm{f}(8)=7\) \(7=\frac{8}{2}+C \Rightarrow C=3\) \(\therefore \quad \mathrm{f}(14)=\frac{14}{2}+3 \Rightarrow f(14)=10\)
JEE Main-2022-26.07.2022
Limits, Continuity and Differentiability
79986
Let \(f(x)=\left[x^{2}-x\right]+|-x+[x]|\), where \(x \in \mathbb{R}\) and [t] denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
1 continuous at \(x=0\), but not continuous at \(x=1\)
2 continuous at \(x=0\) and \(x=1\)
3 not continuous at \(x=0\) and \(x=1\)
4 continuous at \(x=1\), but not continuous at \(x=0\)
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\left[\mathrm{x}^{2}-\mathrm{x}\right]+|-\mathrm{x}+[\mathrm{x}]|\) We know that, \(f(x)=[x(x-1)]+\{x\}\) Now, at \(\mathrm{x}=0\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}(\mathrm{x}+1) ; & -0.5\lt \mathrm{x}\lt 0 \\ 0 ; & \mathrm{x}=0 \\ -1+\mathrm{x} ; & 0\lt \mathrm{x}\lt 1 \\ 0 ; & \mathrm{x}=1 \\ \mathrm{x}-1 ; & 1\lt \mathrm{x}\lt 1.5\end{array}\right.\) Left hand limit (LHL) \(=\lim _{x \rightarrow 0^{-}} f(x)=1\) Right hand limit (RHL) \(=\lim _{x \rightarrow 0^{+}} f(x)=-1\) and, \(\quad \mathrm{f}(0)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=0\) At, \(\quad \mathrm{x}=1\) Left hand limit (LHL) \(=\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0\) Right hand \(\operatorname{limit}(\mathrm{RHL})=\lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=1-1=0\) And, \(\mathrm{f}(1)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) but not continuous at \(\mathrm{x}=0\)
79983
If the function \(f(x)=\left\{\frac{\log _{e}\left(1-x+x^{2}\right)+\log _{e}\left(1+x+x^{2}\right)}{\sec x-\cos x}\right.\), \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\}\) is continuous at \(x=0\), then \(k\) is equal to :
79984
If \(f(x)=\left\{\begin{aligned} x+a, x \leq 0 \\ |x-4|, x>0\end{aligned}\right.\) and \(g(x)=\left\{\begin{array}{r}x+1, x\lt 0 \\ (x-4)^{2}+b, x \geq 0\end{array}\right.\) are continuous on \(R\), then (got) (2) + (fog) (-2) is equal to :
79985
Lef \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a continuous function such that \(f(3 x)-f(x)=x\). If \(f(8)=7\), then \(f(14)\) is equal to :
1 4
2 10
3 11
4 16
Explanation:
(B) : Given, \(\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})=\mathrm{x} \tag{i}\) \(\mathrm{f}(\mathrm{x})=7\) We know that, Lagrangels mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) When function is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable \((a, b)\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(3 \mathrm{x})-\mathrm{f}(\mathrm{x})}{3 \mathrm{x}-\mathrm{x}}\) \(f^{\prime}(x)=\frac{x}{2 x} \Rightarrow f^{\prime}(x)=\frac{1}{2}\) Integrating both sides, \(\int f^{\prime}(x) d x=\int \frac{d x}{2}\) \(f(x)=\frac{x}{2}+C\) Now, \(\quad \mathrm{x}=8 \Rightarrow \mathrm{f}(8)=7\) \(7=\frac{8}{2}+C \Rightarrow C=3\) \(\therefore \quad \mathrm{f}(14)=\frac{14}{2}+3 \Rightarrow f(14)=10\)
JEE Main-2022-26.07.2022
Limits, Continuity and Differentiability
79986
Let \(f(x)=\left[x^{2}-x\right]+|-x+[x]|\), where \(x \in \mathbb{R}\) and [t] denotes the greatest integer less than or equal to \(t\). Then, \(f\) is
1 continuous at \(x=0\), but not continuous at \(x=1\)
2 continuous at \(x=0\) and \(x=1\)
3 not continuous at \(x=0\) and \(x=1\)
4 continuous at \(x=1\), but not continuous at \(x=0\)
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\left[\mathrm{x}^{2}-\mathrm{x}\right]+|-\mathrm{x}+[\mathrm{x}]|\) We know that, \(f(x)=[x(x-1)]+\{x\}\) Now, at \(\mathrm{x}=0\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}(\mathrm{x}+1) ; & -0.5\lt \mathrm{x}\lt 0 \\ 0 ; & \mathrm{x}=0 \\ -1+\mathrm{x} ; & 0\lt \mathrm{x}\lt 1 \\ 0 ; & \mathrm{x}=1 \\ \mathrm{x}-1 ; & 1\lt \mathrm{x}\lt 1.5\end{array}\right.\) Left hand limit (LHL) \(=\lim _{x \rightarrow 0^{-}} f(x)=1\) Right hand limit (RHL) \(=\lim _{x \rightarrow 0^{+}} f(x)=-1\) and, \(\quad \mathrm{f}(0)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=0\) At, \(\quad \mathrm{x}=1\) Left hand limit (LHL) \(=\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0\) Right hand \(\operatorname{limit}(\mathrm{RHL})=\lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=1-1=0\) And, \(\mathrm{f}(1)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) but not continuous at \(\mathrm{x}=0\)
