Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks.
If $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^{2}}{2!}+\frac{\text{x}^{3}}{3!}+....$ then ____________.

Answer

If $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^{2}}{2!}+\frac{\text{x}^{3}}{3!}+....$then $\frac{\text{dy}}{\text{dx}}=\text{y}.$Solution:
Given that $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^{2}}{2!}+\frac{\text{x}^{3}}{3!}+....$ $\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1!}+\frac{2 \text{x}}{2!}+\frac{3\text{x}^{2}}{3!}+......$ $=1+\frac{\text{x}}{1!}+\frac{\text{x}^{2}}{2!}+\frac{\text{x}^{3}}{3!}+....=\text{y}$ Hence, the value of the filler is y.

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Question 21 Mark

Fill in the blanks.
$\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ then ___________.

Answer

$\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ then $\text{m}=\frac{2\sqrt{3}}{3}.$ Solution:
Given $\lim\limits_{\pi \rightarrow 0}\Big(\sin\text{mx}\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ $=\lim\limits_{\pi \rightarrow 0}\frac{\sin\text{mx}}{\text{mx}}\times\text{mx}\lim\limits_{\text{x} \rightarrow 0}\Big(\cot\frac{\text{x}}{\sqrt{3}}\Big)=2$ $=1\times\text{mx}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{\tan\frac{\text{x}}{\sqrt{3}}}=2$ $=1\times\text{mx}\times\frac{\frac{\text{x}}{\sqrt{3}}}{\frac{\text{x}}{\sqrt{3}.\tan\frac{\text{x}}{\sqrt{3}}}}=2$ $=\frac{\text{mx}}{\frac{\text{x}}{\sqrt{3}}}(1)=2$ $=\sqrt{2}\text{m}=2$ $\text{m}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$ Hence, the value of the filler is $\frac{2\sqrt{3}}{3}.$

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Question 31 Mark

Fill in the blanks.
$\lim\limits_{\text{x} \rightarrow 3^{+}}\frac{\text{x}}{[\text{x}]}=$ ___________.

Answer

$\lim\limits_{\text{x} \rightarrow 3^{+}}\frac{\text{x}}{[\text{x}]}=1$Solution:
Given $\lim\limits_{\text{x} \rightarrow 3^{+}}\frac{\text{x}}{[\text{x}]}$ $\lim\limits_{\text{x} \rightarrow 3^{+}}\frac{\text{x}}{[\text{x}]}=1$ Hence, the value of the filler is 1.

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Question 41 Mark

Fill in the blanks.
If $\text{f}(\text{x})=\frac{\tan\text{x}}{\text{x}-\pi}$ then $\lim\limits_{\text{x} \rightarrow \pi}\text{f}(\text{x})=$ ____________.

Answer

If $\text{f}(\text{x})=\frac{\tan\text{x}}{\text{x}-\pi}$ then $\lim\limits_{\text{x} \rightarrow \pi}\text{f}(\text{x})=\lim\limits_{\pi \rightarrow 0}\frac{-\tan(\pi-\text{x})}{-(\pi-\text{x})}$Solution:
Given $\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow \pi}\text{f}(\text{x})=\lim\limits_{\pi \rightarrow 0}\frac{-\tan(\pi-\text{x})}{-(\pi-\text{x})}$ $=1$ Hence, the value of the filler is 1.

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Maths Fill In The Blanks[1 Marks ]

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