MCQ

MCQ 11 Mark

The derivative of $2 x^4+x$ is

✓
$8 x^3+1$
B
$8 x^3$
C
$8 x^3-1$
D
$8 x^2+1$

Answer

Correct option: A.

$8 x^3+1$

(A) $8 x^3+1$
Explanation : Let $\quad y=2 x^4+x$
$\frac{d y}{d x}=\frac{d}{d x}\left(2 x^4+x\right)$
$\begin{array}{l}=\frac{d}{d x} 2 x^4+\frac{d}{d x} x \\ =2 \frac{d}{d x} x^4+1 \\ =2 \cdot 4 \cdot x^3+1=8 x^3+1 .\end{array}$

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MCQ 21 Mark

If $f(x)=1+x+\frac{x^2}{2}+\ldots+\frac{x^{100}}{100}$, then $f^{\prime}(1)$ is equal to :

A
$\frac{1}{100}$
✓
$1 0 0$
C
$0$
D
Does not exist

Answer

Correct option: B.

$1 0 0$

(B) $1 0 0$
Explanation : Given,
$\begin{array}{l}f(x)=1+x+\frac{x^2}{2}+....+\frac{x^{100}}{100} \\ f^{\prime}(x)=0+1+\frac{2 x}{2}+....+\frac{100 x^{99}}{100}\end{array}$
$\left[f^{\prime}(x)\right]_{ at x=1}=1+\frac{2.1}{2}+....+\frac{100.1}{100}$
$\begin{array}{l}=1+1+....+1 \text { (100 times) } \\ =100\end{array}$

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MCQ 31 Mark

If $x=t^2$ and $y=t^3$ then $\frac{d^2 y}{d x^2}$ is

A
$\frac{3}{2}$
✓
$\frac{3}{4 t}$
C
$\frac{3}{2 t}$
D
$\frac{3}{4}$

Answer

Correct option: B.

$\frac{3}{4 t}$

(B) $\frac{3}{4 t}$
Explanation : Given that,
$x=t^2$ and $y=t^3$
Then, $\frac{d x}{d t}=2 t$ and $\frac{d y}{d t}=3 t^2$
Thus, $\quad \frac{d y}{d x}=\frac{3 t^2}{2 t}=\frac{3 t}{2}$
$\frac{d^2 y}{d x^2}=\frac{3}{2} \frac{d t}{d x}$
$=\frac{3}{2} \cdot \frac{1}{2 t}=\frac{3}{4 t}$.

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MCQ 41 Mark

If $f(x)=\frac{x-4}{2 \sqrt{x}}$, then $f^{\prime}(1)$ is equal to :

✓
$\frac{5}{4}$
B
$\frac{4}{5}$
C
1
D
$0$

Answer

Correct option: A.

$\frac{5}{4}$

(A) $\frac{5}{4}$
Explanation : Given, $f(x)=\frac{x-4}{2 \sqrt{x}}$
$f^{\prime}(x)=\frac{1}{2}\left[\frac{\sqrt{x} \cdot 1-(x-4) \cdot \frac{1}{2 \sqrt{x}}}{x}\right]$
$\begin{array}{l}=\frac{1}{2}\left[\frac{2 x-x+4}{2 \sqrt{x} \cdot x}\right] \\ =\frac{1}{2}\left[\frac{x+4}{2 x^{3 / 2}}\right]\end{array}$
Therefore, $\left(f^{\prime}(x)\right)_{\text {at } x=1}=\frac{1}{2}\left[\frac{1+4}{2 \times 1}\right]=\frac{5}{4}$.

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MCQ 51 Mark

If $y=\sqrt{x}+\frac{1}{\sqrt{x}}$, then $\frac{d y}{d x}$ at $x=1$ is equal to:

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

Answer

Correct option: D.

$0$

(D) $0$
Explanation : Given, $y=\sqrt{x}+\frac{1}{\sqrt{x}}$
$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3 / 2}}$
$\left(\frac{d y}{d x}\right)_{ at x=1}=\frac{1}{2}-\frac{1}{2}=0$

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MCQ 61 Mark

If $f(x)=2 x$ and $g(x)=\frac{x^2}{2}+1$ then which of the following can be a discontinuous function at $x=0$ ?

A
$f(x)+g(x)$
B
$f(x)-g(x)$
C
$f(x) \cdot g(x)$
✓
$\frac{g(x)}{f(x)}$

Answer

Correct option: D.

$\frac{g(x)}{f(x)}$

(D) $\frac{g(x)}{f(x)}$
Explanation : Since $f(x)=2 x$ and $g(x)=\frac{x^2}{2}+1$ are continuous functions, then by using the algebra of continuous function, the functions $f(x)+g(x)$, $f(x)-g(x), f(x) \cdot g(x)$ are also continuous functions but $\frac{g(x)}{f(x)}$ is discontinuous function at $x=0$.

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MCQ 71 Mark

The function $f(x)=\frac{4-x^2}{4 x-x^3}$

A
discontinuous at only one point
B
discontinuous at exactly two points
✓
discontinuous at exactly three points
D
none of these

Answer

Correct option: C.

discontinuous at exactly three points

(C) discontinuous at exactly three points
Explanation : Given that,
$f(x)=\frac{4-x^2}{4 x-x^3}$, then it is discontinuous if
$\Rightarrow \quad 4 x-x^3=0$
$\Rightarrow \quad x\left(4-x^2\right)=0$
$\Rightarrow \quad x(2+x)(2-x)=0$
$\Rightarrow \quad x=0,-2,2$
Thus, the given function is discontinuous at exactly three points.

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MCQ 81 Mark

$\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}$ is equal to :

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

Answer

Correct option: B.

$\frac{-1}{10}$

(B) $\frac{-1}{10}$
Explanation : Given,
$\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}$
$
\begin{array}{l}
=\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{(2 x+3)(x-1)} \\
=\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(\sqrt{x}+1)(2 x-3)}{(2 x+3)(x-1)(\sqrt{x}+1)} \\
=\lim _{x \rightarrow 1} \frac{(x-1)(2 x-3)}{(2 x+3)(x-1)(\sqrt{x}+1)} \\
=\lim _{x \rightarrow 1} \frac{(2 x-3)}{(2 x+3)(\sqrt{x}+1)} \\
=\frac{(2.1-3)}{(2.1+3)(\sqrt{1}+1)}=-\frac{1}{10}
\end{array}
$

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MCQ 91 Mark

$\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}$ is equal to :

✓
$n$
B
1
C
$-n$
D
$0$

Answer

Correct option: A.

$n$

(A) $n$
Explanation : Given,
$
\begin{array}{l}
\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} \\
=\lim _{x \rightarrow 0} \frac{(1+x)^n-(1)^n}{(1+x)-(1)}
\end{array}
$
Let $(1+x=y)$$
=\lim _{x \rightarrow 1} \frac{y^n-1}{y-1}
$
$\begin{array}{l}=n(1)^{n-1}=n \\ \quad\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]\end{array}$

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MCQ 101 Mark

$\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ is equal to :

A
1
B
2
✓
-1
D
-2

Answer

Correct option: C.

-1

(C) -1
Explanation : Given,
$\begin{array}{l}
\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi} \\
\quad=\lim _{\pi-x \rightarrow 0} \frac{\sin (\pi-x)}{-(\pi-x)}=-1 \\
{\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \pi-x \rightarrow 0 \Rightarrow x \rightarrow \pi\right]}
\end{array}$

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