80228
Suppose that \(f(x)\) is a differentiable function such that \(f^{\prime}(x)\) is continuous, \(f^{\prime}(0)=1\) and \(f^{\prime \prime}(0)\) does not exist. Let \(\mathbf{g}(\mathbf{x})=\mathbf{x} \mathbf{f}^{\prime}(\mathbf{x})\), Then,
1 \(g^{\prime}(0)\) does not exist
2 \(\mathrm{g}^{\prime}(0)=0\)
3 \(\mathrm{g}^{\prime}(0)=1\)
4 \(\mathrm{g}^{\prime}(0)=2\)
Explanation:
(C) : Given, \(\mathrm{f}^{\prime}(0)=1\) and \(\mathrm{f}^{\prime \prime}(0)\) does not exist and \(\mathrm{g}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime}(\mathrm{x})\) \(\therefore \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\) Put, \(\mathrm{x}=0\), \(\mathrm{g}^{\prime}(0) =0\left(\mathrm{f}^{\prime \prime}(0)\right)+\mathrm{f}^{\prime}(0)\) \(=0+1=1\) Hence, \(\mathrm{g}^{\prime}(0)=1\)
WB JEE-2014
Limits, Continuity and Differentiability
80229
Let \(\boldsymbol{f}(\mathrm{x})\) be a differentiable function and \(\boldsymbol{f}^{\prime}(4)=\) 5. Then \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) equals
1 0
2 5
3 20
4 -20
Explanation:
(D) : Given, \(\mathrm{f}^{\prime}(4)=5\) \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) According to L-Hospital's rule, we get - \(\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) 2 x}{1}\) \(\frac{-f^{\prime}(4) 2.2}{1}\) \(=-(5) \cdot 4\) \(=-20 \quad\left[f^{\prime}(4)=5\right]\)
WB JEE-2014
Limits, Continuity and Differentiability
80230
The function \(f(x)=\operatorname{asin}|x|+b e^{|x|}\) is differentiable at \(x=0\) when
1 \(3 \mathrm{a}+\mathrm{b}=0\)
2 \(3 \mathrm{a}-\mathrm{b}=0\)
3 \(\mathrm{a}+\mathrm{b}=0\)
4 \(\mathrm{a}-\mathrm{b}=0\)
Explanation:
(C) : It is given that, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin |\mathrm{x}|+\mathrm{be}^{\mathrm{x} \mid}\) We know that, \(\sin |\mathrm{x}|\) and \(\mathrm{e}^{\mathrm{x} \mid}\) is not differentiable at \(\mathrm{x}=0\) Therefore, for \(\mathrm{f}(\mathrm{x})\) to differentiable at \(\mathrm{x}=0\), we must have \(\mathrm{a}+\mathrm{b}=0\)
WB JEE-2014
Limits, Continuity and Differentiability
80231
Let ' \(f\) ' be a twice differentiable function such that \(f^{\prime \prime}(x)=-f(x)\) and \(f^{\prime}(x)=g(x)=g(x)\). If \(h(x)\) \(=\{f(x)\}^{2}+\{g(x)\}^{2}\) and \(h(5)=11\), then \(h(10)=-\)
80232
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(\mathbf{x}) \geq \mathbf{2}\) for \(x \in[\mathbf{1}, \mathbf{6}]\), then
1 \(f(6)=5\)
2 \(\mathrm{f}(6)\lt 6\)
3 \(\mathrm{f}(6)\lt 8\)
4 \(f(6) \geq 8\)
Explanation:
(D) : Given that, \(f(1)=-2\) and \(f^{\prime}(x) \geq 2 \forall 2 \in[1,6]\) We know, Lagrange's mean value theorem states that if \(f(x)\) be continuous on \([a, b]\) and differentiable on \((a, b)\) then there exists some \(c\) between \(a\) and \(b\) such that - \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) In this question Lagrange's mean value theorem can be applied So, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \geq 2\) \(\mathrm{f}(6)-\mathrm{f}(1) \geq 10\) \(\mathrm{f}(6)-(-2) \geq 10\) \(\mathrm{f}(6) \geq 10-2 \Rightarrow \quad \mathrm{f}(6) \geq 8\)
80228
Suppose that \(f(x)\) is a differentiable function such that \(f^{\prime}(x)\) is continuous, \(f^{\prime}(0)=1\) and \(f^{\prime \prime}(0)\) does not exist. Let \(\mathbf{g}(\mathbf{x})=\mathbf{x} \mathbf{f}^{\prime}(\mathbf{x})\), Then,
1 \(g^{\prime}(0)\) does not exist
2 \(\mathrm{g}^{\prime}(0)=0\)
3 \(\mathrm{g}^{\prime}(0)=1\)
4 \(\mathrm{g}^{\prime}(0)=2\)
Explanation:
(C) : Given, \(\mathrm{f}^{\prime}(0)=1\) and \(\mathrm{f}^{\prime \prime}(0)\) does not exist and \(\mathrm{g}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime}(\mathrm{x})\) \(\therefore \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\) Put, \(\mathrm{x}=0\), \(\mathrm{g}^{\prime}(0) =0\left(\mathrm{f}^{\prime \prime}(0)\right)+\mathrm{f}^{\prime}(0)\) \(=0+1=1\) Hence, \(\mathrm{g}^{\prime}(0)=1\)
WB JEE-2014
Limits, Continuity and Differentiability
80229
Let \(\boldsymbol{f}(\mathrm{x})\) be a differentiable function and \(\boldsymbol{f}^{\prime}(4)=\) 5. Then \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) equals
1 0
2 5
3 20
4 -20
Explanation:
(D) : Given, \(\mathrm{f}^{\prime}(4)=5\) \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) According to L-Hospital's rule, we get - \(\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) 2 x}{1}\) \(\frac{-f^{\prime}(4) 2.2}{1}\) \(=-(5) \cdot 4\) \(=-20 \quad\left[f^{\prime}(4)=5\right]\)
WB JEE-2014
Limits, Continuity and Differentiability
80230
The function \(f(x)=\operatorname{asin}|x|+b e^{|x|}\) is differentiable at \(x=0\) when
1 \(3 \mathrm{a}+\mathrm{b}=0\)
2 \(3 \mathrm{a}-\mathrm{b}=0\)
3 \(\mathrm{a}+\mathrm{b}=0\)
4 \(\mathrm{a}-\mathrm{b}=0\)
Explanation:
(C) : It is given that, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin |\mathrm{x}|+\mathrm{be}^{\mathrm{x} \mid}\) We know that, \(\sin |\mathrm{x}|\) and \(\mathrm{e}^{\mathrm{x} \mid}\) is not differentiable at \(\mathrm{x}=0\) Therefore, for \(\mathrm{f}(\mathrm{x})\) to differentiable at \(\mathrm{x}=0\), we must have \(\mathrm{a}+\mathrm{b}=0\)
WB JEE-2014
Limits, Continuity and Differentiability
80231
Let ' \(f\) ' be a twice differentiable function such that \(f^{\prime \prime}(x)=-f(x)\) and \(f^{\prime}(x)=g(x)=g(x)\). If \(h(x)\) \(=\{f(x)\}^{2}+\{g(x)\}^{2}\) and \(h(5)=11\), then \(h(10)=-\)
80232
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(\mathbf{x}) \geq \mathbf{2}\) for \(x \in[\mathbf{1}, \mathbf{6}]\), then
1 \(f(6)=5\)
2 \(\mathrm{f}(6)\lt 6\)
3 \(\mathrm{f}(6)\lt 8\)
4 \(f(6) \geq 8\)
Explanation:
(D) : Given that, \(f(1)=-2\) and \(f^{\prime}(x) \geq 2 \forall 2 \in[1,6]\) We know, Lagrange's mean value theorem states that if \(f(x)\) be continuous on \([a, b]\) and differentiable on \((a, b)\) then there exists some \(c\) between \(a\) and \(b\) such that - \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) In this question Lagrange's mean value theorem can be applied So, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \geq 2\) \(\mathrm{f}(6)-\mathrm{f}(1) \geq 10\) \(\mathrm{f}(6)-(-2) \geq 10\) \(\mathrm{f}(6) \geq 10-2 \Rightarrow \quad \mathrm{f}(6) \geq 8\)
80228
Suppose that \(f(x)\) is a differentiable function such that \(f^{\prime}(x)\) is continuous, \(f^{\prime}(0)=1\) and \(f^{\prime \prime}(0)\) does not exist. Let \(\mathbf{g}(\mathbf{x})=\mathbf{x} \mathbf{f}^{\prime}(\mathbf{x})\), Then,
1 \(g^{\prime}(0)\) does not exist
2 \(\mathrm{g}^{\prime}(0)=0\)
3 \(\mathrm{g}^{\prime}(0)=1\)
4 \(\mathrm{g}^{\prime}(0)=2\)
Explanation:
(C) : Given, \(\mathrm{f}^{\prime}(0)=1\) and \(\mathrm{f}^{\prime \prime}(0)\) does not exist and \(\mathrm{g}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime}(\mathrm{x})\) \(\therefore \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\) Put, \(\mathrm{x}=0\), \(\mathrm{g}^{\prime}(0) =0\left(\mathrm{f}^{\prime \prime}(0)\right)+\mathrm{f}^{\prime}(0)\) \(=0+1=1\) Hence, \(\mathrm{g}^{\prime}(0)=1\)
WB JEE-2014
Limits, Continuity and Differentiability
80229
Let \(\boldsymbol{f}(\mathrm{x})\) be a differentiable function and \(\boldsymbol{f}^{\prime}(4)=\) 5. Then \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) equals
1 0
2 5
3 20
4 -20
Explanation:
(D) : Given, \(\mathrm{f}^{\prime}(4)=5\) \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) According to L-Hospital's rule, we get - \(\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) 2 x}{1}\) \(\frac{-f^{\prime}(4) 2.2}{1}\) \(=-(5) \cdot 4\) \(=-20 \quad\left[f^{\prime}(4)=5\right]\)
WB JEE-2014
Limits, Continuity and Differentiability
80230
The function \(f(x)=\operatorname{asin}|x|+b e^{|x|}\) is differentiable at \(x=0\) when
1 \(3 \mathrm{a}+\mathrm{b}=0\)
2 \(3 \mathrm{a}-\mathrm{b}=0\)
3 \(\mathrm{a}+\mathrm{b}=0\)
4 \(\mathrm{a}-\mathrm{b}=0\)
Explanation:
(C) : It is given that, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin |\mathrm{x}|+\mathrm{be}^{\mathrm{x} \mid}\) We know that, \(\sin |\mathrm{x}|\) and \(\mathrm{e}^{\mathrm{x} \mid}\) is not differentiable at \(\mathrm{x}=0\) Therefore, for \(\mathrm{f}(\mathrm{x})\) to differentiable at \(\mathrm{x}=0\), we must have \(\mathrm{a}+\mathrm{b}=0\)
WB JEE-2014
Limits, Continuity and Differentiability
80231
Let ' \(f\) ' be a twice differentiable function such that \(f^{\prime \prime}(x)=-f(x)\) and \(f^{\prime}(x)=g(x)=g(x)\). If \(h(x)\) \(=\{f(x)\}^{2}+\{g(x)\}^{2}\) and \(h(5)=11\), then \(h(10)=-\)
80232
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(\mathbf{x}) \geq \mathbf{2}\) for \(x \in[\mathbf{1}, \mathbf{6}]\), then
1 \(f(6)=5\)
2 \(\mathrm{f}(6)\lt 6\)
3 \(\mathrm{f}(6)\lt 8\)
4 \(f(6) \geq 8\)
Explanation:
(D) : Given that, \(f(1)=-2\) and \(f^{\prime}(x) \geq 2 \forall 2 \in[1,6]\) We know, Lagrange's mean value theorem states that if \(f(x)\) be continuous on \([a, b]\) and differentiable on \((a, b)\) then there exists some \(c\) between \(a\) and \(b\) such that - \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) In this question Lagrange's mean value theorem can be applied So, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \geq 2\) \(\mathrm{f}(6)-\mathrm{f}(1) \geq 10\) \(\mathrm{f}(6)-(-2) \geq 10\) \(\mathrm{f}(6) \geq 10-2 \Rightarrow \quad \mathrm{f}(6) \geq 8\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
80228
Suppose that \(f(x)\) is a differentiable function such that \(f^{\prime}(x)\) is continuous, \(f^{\prime}(0)=1\) and \(f^{\prime \prime}(0)\) does not exist. Let \(\mathbf{g}(\mathbf{x})=\mathbf{x} \mathbf{f}^{\prime}(\mathbf{x})\), Then,
1 \(g^{\prime}(0)\) does not exist
2 \(\mathrm{g}^{\prime}(0)=0\)
3 \(\mathrm{g}^{\prime}(0)=1\)
4 \(\mathrm{g}^{\prime}(0)=2\)
Explanation:
(C) : Given, \(\mathrm{f}^{\prime}(0)=1\) and \(\mathrm{f}^{\prime \prime}(0)\) does not exist and \(\mathrm{g}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime}(\mathrm{x})\) \(\therefore \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\) Put, \(\mathrm{x}=0\), \(\mathrm{g}^{\prime}(0) =0\left(\mathrm{f}^{\prime \prime}(0)\right)+\mathrm{f}^{\prime}(0)\) \(=0+1=1\) Hence, \(\mathrm{g}^{\prime}(0)=1\)
WB JEE-2014
Limits, Continuity and Differentiability
80229
Let \(\boldsymbol{f}(\mathrm{x})\) be a differentiable function and \(\boldsymbol{f}^{\prime}(4)=\) 5. Then \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) equals
1 0
2 5
3 20
4 -20
Explanation:
(D) : Given, \(\mathrm{f}^{\prime}(4)=5\) \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) According to L-Hospital's rule, we get - \(\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) 2 x}{1}\) \(\frac{-f^{\prime}(4) 2.2}{1}\) \(=-(5) \cdot 4\) \(=-20 \quad\left[f^{\prime}(4)=5\right]\)
WB JEE-2014
Limits, Continuity and Differentiability
80230
The function \(f(x)=\operatorname{asin}|x|+b e^{|x|}\) is differentiable at \(x=0\) when
1 \(3 \mathrm{a}+\mathrm{b}=0\)
2 \(3 \mathrm{a}-\mathrm{b}=0\)
3 \(\mathrm{a}+\mathrm{b}=0\)
4 \(\mathrm{a}-\mathrm{b}=0\)
Explanation:
(C) : It is given that, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin |\mathrm{x}|+\mathrm{be}^{\mathrm{x} \mid}\) We know that, \(\sin |\mathrm{x}|\) and \(\mathrm{e}^{\mathrm{x} \mid}\) is not differentiable at \(\mathrm{x}=0\) Therefore, for \(\mathrm{f}(\mathrm{x})\) to differentiable at \(\mathrm{x}=0\), we must have \(\mathrm{a}+\mathrm{b}=0\)
WB JEE-2014
Limits, Continuity and Differentiability
80231
Let ' \(f\) ' be a twice differentiable function such that \(f^{\prime \prime}(x)=-f(x)\) and \(f^{\prime}(x)=g(x)=g(x)\). If \(h(x)\) \(=\{f(x)\}^{2}+\{g(x)\}^{2}\) and \(h(5)=11\), then \(h(10)=-\)
80232
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(\mathbf{x}) \geq \mathbf{2}\) for \(x \in[\mathbf{1}, \mathbf{6}]\), then
1 \(f(6)=5\)
2 \(\mathrm{f}(6)\lt 6\)
3 \(\mathrm{f}(6)\lt 8\)
4 \(f(6) \geq 8\)
Explanation:
(D) : Given that, \(f(1)=-2\) and \(f^{\prime}(x) \geq 2 \forall 2 \in[1,6]\) We know, Lagrange's mean value theorem states that if \(f(x)\) be continuous on \([a, b]\) and differentiable on \((a, b)\) then there exists some \(c\) between \(a\) and \(b\) such that - \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) In this question Lagrange's mean value theorem can be applied So, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \geq 2\) \(\mathrm{f}(6)-\mathrm{f}(1) \geq 10\) \(\mathrm{f}(6)-(-2) \geq 10\) \(\mathrm{f}(6) \geq 10-2 \Rightarrow \quad \mathrm{f}(6) \geq 8\)
80228
Suppose that \(f(x)\) is a differentiable function such that \(f^{\prime}(x)\) is continuous, \(f^{\prime}(0)=1\) and \(f^{\prime \prime}(0)\) does not exist. Let \(\mathbf{g}(\mathbf{x})=\mathbf{x} \mathbf{f}^{\prime}(\mathbf{x})\), Then,
1 \(g^{\prime}(0)\) does not exist
2 \(\mathrm{g}^{\prime}(0)=0\)
3 \(\mathrm{g}^{\prime}(0)=1\)
4 \(\mathrm{g}^{\prime}(0)=2\)
Explanation:
(C) : Given, \(\mathrm{f}^{\prime}(0)=1\) and \(\mathrm{f}^{\prime \prime}(0)\) does not exist and \(\mathrm{g}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime}(\mathrm{x})\) \(\therefore \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xf} \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\) Put, \(\mathrm{x}=0\), \(\mathrm{g}^{\prime}(0) =0\left(\mathrm{f}^{\prime \prime}(0)\right)+\mathrm{f}^{\prime}(0)\) \(=0+1=1\) Hence, \(\mathrm{g}^{\prime}(0)=1\)
WB JEE-2014
Limits, Continuity and Differentiability
80229
Let \(\boldsymbol{f}(\mathrm{x})\) be a differentiable function and \(\boldsymbol{f}^{\prime}(4)=\) 5. Then \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) equals
1 0
2 5
3 20
4 -20
Explanation:
(D) : Given, \(\mathrm{f}^{\prime}(4)=5\) \(\lim _{\mathrm{x} \rightarrow 2} \frac{f(4)-f\left(\mathrm{x}^{2}\right)}{\mathrm{x}-2}\) According to L-Hospital's rule, we get - \(\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) 2 x}{1}\) \(\frac{-f^{\prime}(4) 2.2}{1}\) \(=-(5) \cdot 4\) \(=-20 \quad\left[f^{\prime}(4)=5\right]\)
WB JEE-2014
Limits, Continuity and Differentiability
80230
The function \(f(x)=\operatorname{asin}|x|+b e^{|x|}\) is differentiable at \(x=0\) when
1 \(3 \mathrm{a}+\mathrm{b}=0\)
2 \(3 \mathrm{a}-\mathrm{b}=0\)
3 \(\mathrm{a}+\mathrm{b}=0\)
4 \(\mathrm{a}-\mathrm{b}=0\)
Explanation:
(C) : It is given that, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin |\mathrm{x}|+\mathrm{be}^{\mathrm{x} \mid}\) We know that, \(\sin |\mathrm{x}|\) and \(\mathrm{e}^{\mathrm{x} \mid}\) is not differentiable at \(\mathrm{x}=0\) Therefore, for \(\mathrm{f}(\mathrm{x})\) to differentiable at \(\mathrm{x}=0\), we must have \(\mathrm{a}+\mathrm{b}=0\)
WB JEE-2014
Limits, Continuity and Differentiability
80231
Let ' \(f\) ' be a twice differentiable function such that \(f^{\prime \prime}(x)=-f(x)\) and \(f^{\prime}(x)=g(x)=g(x)\). If \(h(x)\) \(=\{f(x)\}^{2}+\{g(x)\}^{2}\) and \(h(5)=11\), then \(h(10)=-\)
80232
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(\mathbf{x}) \geq \mathbf{2}\) for \(x \in[\mathbf{1}, \mathbf{6}]\), then
1 \(f(6)=5\)
2 \(\mathrm{f}(6)\lt 6\)
3 \(\mathrm{f}(6)\lt 8\)
4 \(f(6) \geq 8\)
Explanation:
(D) : Given that, \(f(1)=-2\) and \(f^{\prime}(x) \geq 2 \forall 2 \in[1,6]\) We know, Lagrange's mean value theorem states that if \(f(x)\) be continuous on \([a, b]\) and differentiable on \((a, b)\) then there exists some \(c\) between \(a\) and \(b\) such that - \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) In this question Lagrange's mean value theorem can be applied So, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(6)-\mathrm{f}(1)}{6-1} \geq 2\) \(\mathrm{f}(6)-\mathrm{f}(1) \geq 10\) \(\mathrm{f}(6)-(-2) \geq 10\) \(\mathrm{f}(6) \geq 10-2 \Rightarrow \quad \mathrm{f}(6) \geq 8\)