M.C.Q (1 Marks) - Page 4 - Maths STD 11 Science Questions - Vidyadip

Questions · Page 4 of 4

M.C.Q (1 Marks)

MCQ 1511 Mark

Find the derivative of ${e^x}^2$:

A
${e^x}^2$
B
$2x$
C
${2e^x}^2$
✓
${2xe^x}^2$

Answer

Correct option: D.

${2xe^x}^2$

We apply chain rule. First we differentiate $x^2$.
$\frac{\text{d}}{\text{dx}} (\text{x}^2) = 2\text{x}$
Now, we know that $\frac{\text{d}}{\text{dx}} (\text{e}^x) = \text{e}^\text{x}$
We differentiate ${e^x}^2$ in the same manner and then multiply with the derivative of $x^2$
$\frac{\text{d}}{\text{dx}} (\text{e}^\text{x}) = 2\text{x}\text{e}^\text{x}$

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MCQ 1521 Mark

What is the value of $\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\tan\text{x}}{\cot\text{x}} ?$

✓
$0$
B
$1$
C
$2$
D
$\frac{1}{2}$

Answer

Correct option: A.

$0$

$\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\tan\text{x}}{\cot\text{x}}=\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\sin\text{x}}}$
$\lim_\limits{\text{x} \rightarrow 0}\text{x}= 0$

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MCQ 1531 Mark

$\lim\limits_{\text{x}\rightarrow\infty}\sin\text{x}$ equals:

A
$1$
B
$0$
C
$\infty$
✓
does not exist

Answer

Correct option: D.

does not exist

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MCQ 1541 Mark

Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:

A
$2$
B
$0$
✓
$1$
D
$-1$

Answer

Correct option: C.

$1$

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$
$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$
Taking limit, we get
$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]$
$=\frac{1}{2}\times1\times2$
$=1$

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MCQ 1551 Mark

Choose the correct answer. $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$ is:

A
$1$
B
$2$
✓
$-1$
D
$-2$

Answer

Correct option: C.

$-1$

Given, $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{\sin(\pi)-\text{x}}{-(\pi-\text{x})}$
$=-1$

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MCQ 1561 Mark

$\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$

A
$-\frac{5}{3}$
✓
$\frac{5}{3}$
C
$-\frac{7}{3}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{5}{3}$

$\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$
$=\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{5\text{x}}}\times\frac{\text{3x}}{\tan\text{x}}\times\frac{5}{3}$
we know that $ =\displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=1$
$=\mathop {\lim }\limits_{\text{x} \to 0}\times\frac{\text{3x}}{\tan\text{x}}=1$
$= \text{L}=1\times 1\times \dfrac {5}{3}$
$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=\frac{5}{3}$

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MCQ 1571 Mark

What is the value of $ \lim_\limits{\text{y} \rightarrow \infty}\frac{2}{\text{y}}?$

✓
$0$
B
$1$
C
$2$
D
Infinity

Answer

Correct option: A.

$0$

Any number divided by infinity gives us $0.$
Here, since the number $2$ is divided by $y,$ as $y$ approaches infinity, we get $0.$

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MCQ 1581 Mark

Evaluate the following limit $\lim_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\text{x}^2}:$

A
$0$
B
$1$
✓
$2$
D
None of these

Answer

Correct option: C.

$2$

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MCQ 1591 Mark

Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ is:

A
$\frac{1}{10}$
✓
$-\frac{1}{10}$
C
$1$
D
None of these.

Answer

Correct option: B.

$-\frac{1}{10}$

Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$
Taking limit, we get
$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}$
$=\frac{-1}{2\times5}$
$=\frac{-1}{10}$

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MCQ 1601 Mark

If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is

A
$1$
B
$\frac{1}{2}$
C
$\frac{1}{\sqrt{\text{2}}}$
✓
$0$

Answer

Correct option: D.

$0$

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x$, we get
$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$
Putting $x = 1,$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is $0.$

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MCQ 1611 Mark

$\lim_{x\rightarrow 0} x \sin \frac{1}{\text{x}}$ is equal to:

✓
$0$
B
$1$
C
$ \frac{1}{2}$
D
does not exist

Answer

Correct option: A.

$0$

We know that,
$= \lim_\limits{\text{x}\rightarrow 0} \text{x} = 0$
And
$= -1 \leq \sin \frac{1}{\text{x}} \leq 1$
By Sandwich theorem,
$=\lim_\limits{\text{x}\rightarrow 0} \text{x} \sin \frac{1}{\text{x}}$
$=0$

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MCQ 1621 Mark

What is the value of $\lim_\limits{\text{x} \rightarrow 3}\frac{\text{x}^2-9}{\text{x}-3}?$

A
$0$
B
$3$
C
Infinity
✓
$6$

Answer

Correct option: D.

$6$

When $x$ tends to $3$, both the numerator and,
the denominator become $0$ and it becomes of the form, $0.$
Therefore, we use L’Hospital’s rule,
which states the we differentiate the numerator and the denominator,
until a definite answer is reached.

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MCQ 1631 Mark

Find the derivative of $ e^{x^2} $:

A
$ e^{x^2} $
B
$2^x$
C
$ 2 e^{x^2} $
✓
$2xe^{x^2}$

Answer

Correct option: D.

$2xe^{x^2}$

We apply chain rule.
First we differentiate $x^2$.
$\frac{\text{d}}{\text{dx}} (x^2 ) = 2\text{x}$
Now, we know that $\frac{\text{d}}{\text{dx}}(e^x ) = e^x$
We differentiate $ex^2$ in the same manner and then,
multiply with the derivative of $\frac{{\text{x}}^2 \text{d}}{\text{dx} (\text{e}^{x^2})} = \text{2xe}^{x^2}.$

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MCQ 1641 Mark

Choose the correct answer.
If $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ then the quadeatic equation whose roots are $\lim\limits_{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$ and $\lim\limits_{\text{x} \rightarrow 2^{+}}\text{f}(\text{x})$ is:

A
$\text{x}^{2}-6\text{x}+9=0$
B
$\text{x}^{2}-7\text{x}+8=0$
C
$\text{x}^{2}+14\text{x}+49=0$
✓
$\text{x}^{2}-10\text{x}+21=0$

Answer

Correct option: D.

$\text{x}^{2}-10\text{x}+21=0$

Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$
$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]$
$=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$
$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$
$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]$
$=7$
Therefore, the quadratic equation whose roots are $3$ and $7$ is $\text{x}^{2}-10\text{x}+21=0$

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MCQ 1651 Mark

Choose the correct answer. $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}$ is equal to:

✓
$2$
B
$\frac{3}{2}$
C
$-\frac{3}{2}$
D
$1$

Answer

Correct option: A.

$2$

Given $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}^{2}\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\text{x}^{2}}{4}\times4\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\text{x}}{2}\Big)\cdot2\cos\text{x}}{\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\bigg)\cdot2\cos\text{x}$
$=2\cos\text{x}$
$=2\times1$
$=2$

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MCQ 1661 Mark

If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is:

✓
$-2$
B
$0$
C
$\frac{1}{2}$
D
does not exist

Answer

Correct option: A.

$-2$

$\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$=\frac{(\sin\text{x}-\cos\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-(\cos^2\text{x}+\sin^2\text{x}-2\cos\text{x}\sin\text{x})-(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-1+2\cos\text{x}\sin\text{x}-1-2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^2}$
Putting $x = 0$ is $-2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=0}$
$=\frac{-2}{(\sin0-\cos0)}$
$=\frac{-2}{(0-1)^2}$
$=-2$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $-2$

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MCQ 1671 Mark

If $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots,$ then $\frac{\text{dy}}{\text{dx}}=$

A
$y + 1$
B
$y - 1$
✓
$y$
D
$\text{y}^2$

Answer

Correct option: C.

$y$

$\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^4}{4!}\Big)+\dots$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{1}{1!}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2!}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{1}{3 !}\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{1}{4!}\frac{\text{d}}{\text{dx}}(\text{x}^4)+\dots$
$=0+\frac{1}{1!}\times1+\frac{1}{2!}\times2\text{x}+\frac{1}{3!}\times3\text{x}^2+\frac{1}{4!}\times4\text{x}^3+\dots$
$=1+\frac{1}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\ \Big[\frac{\text{n}}{\text{n}!}=\frac{1}{(\text{n}-1)!}\Big]$
$=\text{y}$

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MCQ 1681 Mark

$\lim_\limits{\text{x} \rightarrow\infty}\frac{\sqrt{\text{x}^2+1}-^3\sqrt{\text{x}^2+1}}{^4\sqrt{\text{x}^4+1}-^5\sqrt{\text{x}^4+1}}$ is equal to:

✓
$1$
B
$-1$
C
$0$
D
none of these

Answer

Correct option: A.

$1$

$\lim_\limits{\text{x} \rightarrow\infty}\frac{\sqrt{\text{x}^2+1}-^3\sqrt{\text{x}^2+1}}{^4\sqrt{\text{x}^4+1}-^5\sqrt{\text{x}^4+1}}$
$=\displaystyle \lim_{\text{x}\rightarrow \infty}\frac {\sqrt {1+1/\text{x}^2}-\sqrt [3]{1/\text{x}+1/\text{x}^3}}{\sqrt [4]{1+1/\text{x}^4}-\sqrt [5]{1/\text{x}-1/\text{x}^5}}$
$ =\frac{1-0}{1-0}$
$=1$

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M.C.Q (1 Marks) - Page 4 - Maths STD 11 Science Questions - Vidyadip