MCQ

MCQ 11 Mark

If $f ( x )=\frac{x^{50}}{50}+\frac{x^{49}}{49}+\frac{x^{48}}{48}+\ldots+\frac{x^2}{2}+x+1$, then $f ^{\prime}(1)=$

A
48
B
49
✓
50
D
51

Answer

Correct option: C.

(C) 50
Hint:
$
f(x)=\frac{x^{50}}{50}+\frac{x^{49}}{49}+\ldots+\frac{x^2}{2}+x+1
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
f ^{\prime}(x) & =\frac{50 x^{49}}{50}+\frac{49 x^{48}}{49}+\ldots+\frac{2 x}{2}+1+0 \\
& =x^{49}+x^{48}+\ldots+x+1 \\
f^{\prime}(1) & =1+1+\ldots+1+1=50
\end{aligned}
$
49 times

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MCQ 21 Mark

If $f(x)=x^2+\sin x+1$, for $x \leq 0$ $=x^2-2 x+1$, for $x \leq 0$, then

✓
$f$ is continuous at $x=0$, but not differentiable at $x=0$
B
$f$ is neither continuous nor differentiable at $x=0$
C
$f$ is not continuous at $x=0$, but differentiable at $x=0$
D
$f$ is both continuous and differentiable at $x=0$

Answer

Correct option: A.

$f$ is continuous at $x=0$, but not differentiable at $x=0$

(A) $f$ is continuous at $x=0$, but not differentiable at $x=0$
Hint:
$f (x)=x^2+\sin x+1, \quad x \leq 0 $
$\quad=x^2-2 x+1, \quad x>0$
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}}\left(x^2+\sin x+1\right)=0+0+1=1 $
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}}\left(x^2-2 x+1\right)=0-0+1=1$
$f(0)=1$
$\therefore \quad f$ is continuous at $x=0$.
$\operatorname{Lf}^{\prime}(0) $
$ =\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h} $
$ =\lim _{h \rightarrow 0^{-}} \frac{h^2+\sin h+1-1}{h} $
$ =\lim _{h \rightarrow 0^{-}}\left(h+\frac{\sin h}{h}\right)=0+1=1 $
$\operatorname{Rf}^{\prime}(0)$
$ =\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h} $
$ =\lim _{h \rightarrow 0^{+}} \frac{h^2-2 h+1-1}{h} $
$ =\lim _{h \rightarrow 0}(h-2) $
$ =-2$
Here, $\operatorname{Lf}^{\prime}(0) \neq \operatorname{Rf}^{\prime}(0)$
$\therefore \quad f$ is not differentiable at $x=0$.

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MCQ 31 Mark

If $f(x)=2 x+6$, for $0 \leq x \leq 2$
$=a x^2+b x$, for $2is differentiable at $x=2$, then the values of $a$ and $b$ are

A
$a=-\frac{3}{2}, b=3$
B
$a=\frac{3}{2}, b=8$
C
$a=\frac{1}{2}, b=8$
✓
$a=-\frac{3}{2}, b=8$

Answer

Correct option: D.

$a=-\frac{3}{2}, b=8$

(D) $a=-\frac{3}{2}, b=8$
Hint:
$f(x)=2 x+6,0 \leq x \leq 2$
$=a x^2+b x, 2L f^{\prime}(2)=2, R f^{\prime}(2)=4 a+b$
Since $f$ is differentiable at $x=2$,
$L f^{\prime}(2)=\operatorname{Rf}^{\prime}(2)$
$\therefore 2=4 a+b$
$f$ is continuous at $x =2$.
$\therefore \lim _{x \rightarrow 2^{+}} f(x)=f(2)=\lim _{x \rightarrow 2^{-}} f(x) $
$\therefore 4 a+2 b=2(2)+6 $
$\therefore 4 a+2 b=10 $
$\therefore 2 a+b=5 . . . \text { (ii) }$
$Solving (i) and (ii), we get$
$a=-\frac{3}{2}, b=8$

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MCQ 41 Mark

Suppose $f(x)$ is the derivative of $g(x)$ and $g(x)$ is the derivative of $h(x)$.
If $h(x)=a \sin x+b \cos x+c$, then $f(x)+h(x)=$

A
0
✓
$C$
C
$- C$
D
$-2(a \sin x+b \cos x)$

Answer

Correct option: B.

$C$

(B) $C$
Hint:
$h(x)=a \sin x+b \cos x+c$
Differentiating w.r.t. $x$, we get
$h^{\prime}(x)=a \cos x-b \sin x=g(x) \ldots .$. [given]
Differentiating w.r.t. $x$, we get
$g^{\prime}(x)=-a \sin x-b \cos x=f(x) \ldots . . .[\text { given] }$
$\therefore f(x)+h(x)=-a \sin x-b \cos x+a \sin x+b \cos x+c $
$\therefore f(x)+h(x)=c$

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MCQ 51 Mark

If $y =\frac{5 \sin x-2}{4 \sin x+3}$, then $\frac{d y}{d x}=$

A
$\frac{7 \cos x}{(4 \sin x+3)^2}$
✓
$\frac{23 \cos x}{(4 \sin x+3)^2}$
C
$-\frac{7 \cos x}{(4 \sin x+3)^2}$
D
$-\frac{15 \cos x}{(4 \sin x+3)^2}$

Answer

Correct option: B.

$\frac{23 \cos x}{(4 \sin x+3)^2}$

(B) $\frac{23 \cos x}{(4 \sin x+3)^2}$
Hint:
$
y=\frac{5 \sin x-2}{4 \sin x+3}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{(4 \sin x+3) \frac{d}{d x}(5 \sin x-2)-(5 \sin x-2) \frac{d}{d x}(4 \sin x+3)}{(4 \sin x+3)^2} \\
& =\frac{(4 \sin x+3)(5 \cos x)-(5 \sin x-2)(4 \cos x)}{(4 \sin x+3)^2} \\
& =\frac{20 \sin x \cos x+15 \cos x-20 \sin x \cos x+8 \cos x}{(4 \sin x+3)^2} \\
& =\frac{23 \cos x}{(4 \sin x+3)^2}
\end{aligned}
$

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MCQ 61 Mark

If $y =\frac{3 x+5}{4 x+5}$, then $\frac{d y}{d x}=$

A
$-\frac{15}{(3 x+5)^2}$
B
$-\frac{15}{(4 x+5)^2}$
✓
$-\frac{5}{(4 x+5)^2}$
D
$-\frac{13}{(4 x+5)^2}$

Answer

Correct option: C.

$-\frac{5}{(4 x+5)^2}$

(C) $-\frac{5}{(4 x+5)^2}$
Hint:
$
y=\frac{3 x+5}{4 x+5}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{(4 x+5) \frac{d}{d x}(3 x+5)-(3 x+5) \frac{d}{d x}(4 x+5)}{(4 x+5)^2} \\
& =\frac{3(4 x+5)-4(3 x+5)}{(4 x+5)^2} \\
& =-\frac{5}{(4 x+5)^2}
\end{aligned}
$

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MCQ 71 Mark

If $y =\frac{a x+b}{c x+d}$, then $\frac{d y}{d x}=$

A
$\frac{a b-c d}{(c x+d)^2}$
B
$\frac{a x-c}{(c x+d)^2}$
C
$\frac{a c-b d}{(c x+d)^2}$
✓
$\frac{a d-b c}{(c x+d)^2}$

Answer

Correct option: D.

$\frac{a d-b c}{(c x+d)^2}$

(D) $\frac{a d-b c}{(c x+d)^2}$
Hint:
$
y=\frac{a x+b}{c x+d}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{(c x+d) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(c x+d)}{(c x+d)^2} \\
& =\frac{a(c x+d)-c(a x+b)}{(c x+d)^2}=\frac{a d-b c}{(c x+d)^2}
\end{aligned}
$

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MCQ 81 Mark

If $y =\frac{x-4}{\sqrt{x+2}}$, then $\frac{d y}{d x}$ is

A
$\frac{1}{x+4}$
B
$\frac{\sqrt{x}}{\left(\sqrt{x+2)^2}\right.}$
✓
$\frac{1}{2 \sqrt{x}}$
D
$\frac{x}{(\sqrt{x}+2)^2}$

Answer

Correct option: C.

$\frac{1}{2 \sqrt{x}}$

(C) $\frac{1}{2 \sqrt{x}}$
Hint:
$
\begin{aligned}
y & =\frac{x-4}{\sqrt{x}+2}=\frac{(\sqrt{x})^2-(2)^2}{\sqrt{x}+2} \\
& =\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+2)}
\end{aligned}
$
$y=\sqrt{x}-2$
Differentiating w.r.t. $x$, we get
$
\frac{ d y}{ d x}=\frac{1}{2 \sqrt{x}}
$

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