M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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25 questions · 2 auto-graded MCQ + 23 self-marked written.

Question 11 Mark

The function f(x) = x − [x], where [⋅] denotes the greatest integer function is:

Continuous everywhere.
Continuous at integer points only.
Continuous at non-integer points only.
Differentiable everywhere.

Answer

Continuous at non-integer points only.

Solution:
f(x) = x - x
Consider n be an integer.
$\text{f(x)}=\text{x}-[\text{x}]=\begin{cases}\text{x}-(\text{n}-1)&\text{n}-1\leq\text{x}<\text{n}\\0&\text{x}=\text{n}\\\text{x}-\text{n}&\text{n}\leq\text{x}<\text{n}+1\end{cases}$
Now,
LHL at x = n
$=\lim\limits_{\text{x}\rightarrow\text{n}^{-}}\text{f(x)}=\text{x}-\text{n}-1=\text{x}-\text{n}+1$
RHL at x = n
$=\lim\limits_{\text{x}\rightarrow\text{n}^{+}}\text{f(x)}=\text{x}-\text{n}=\text{x}-\text{nAs},$
$\text{LHL}\neq\text{RHL}$ at x = n
i.e., given function is not continuous at n.
Now, n is any integer. Therefore, given function is not continuous at integers.
Therefore, given points are continuous at non-integer points only.

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Question 21 Mark

Let f(x) = (x + |x|) |x|. Then, for all x:

f is continuous.
f is differentiable for some x
f' is continuous.
f'' is continuous.

Answer

f is continuous and,

f' is continuous.

Solution :
$\text{f(x)}=(\text{x}+|\text{x}|)|\text{x}|$
$\Rightarrow\text{f(x)}=2\text{x}^2,\text{ x}>0$
$=0,\text{ x}<0$
$\lim\limits_{\text{x}\rightarrow0}2\text{x}^2=0$
Function is continuous at x = 0.
Also, differentiable at x = 0 as it is polynomial function.

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Question 31 Mark

If f(x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f(x) is:

Continuous and differentiable at x = 3
Continuous but not differentiable at x = 3
Differentiable nut not continuous at x = 3
Neither differentiable nor continuous at x = 3

Answer

Neither differentiable nor continuous at x = 3

Solution:
Given function can be writter as
$\text{f(x)}=-\text{x}+9\ \text{ x}<3$
$=\text{x}+4\ \text{ x}>3$
$\lim\limits_{\text{x}\rightarrow3^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}-\text{x}+9=6$
$\lim\limits_{\text{x}\rightarrow3^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}+4=7$
Function is not continuous at x = 3
⇒ Function is not diffentiable at x = 3.

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Question 41 Mark

If $\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$and if f(x) is differentiable at x = 0, then:

$\text{a}=\text{b}=\text{c}=0$
$\text{a}=0,\text{b}=0;\text{c}\in\text{R}$
$\text{b}=\text{c}=0,\text{a}\in\text{R}$
$\text{c}=0,\text{a}=0,\text{b}\in\text{R}$

Answer

$\text{a}=0,\text{b}=0;\text{c}\in\text{R}$

We have,
$\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$
$=\begin{cases}\text{a}\sin\text{x}+\text{bx}^\text{x}+\text{cx}^3 & 0<\text{x}<\frac{\pi}{2}\\-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3 & -\frac{\pi}{2}<\text{x}<0\end{cases}$
Here, f(x) is differentiable at x = 0
Therefore, (LHL at x = 0) = (RHL at x = 0)
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3-\text{b}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}\sin\text{x}+\text{be}^{\text{x}}-\text{cx}^3-\text{b}}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}\sin(0-\text{h})+\text{be}^{-(0-\text{h)}}-\text{c}(0-\text{h})^3-\text{b}}{0-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin(0+\text{h})+\text{be}^{(0+\text{h)}}+\text{c}(0+\text{h})^3-\text{b}}{0+\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{\text{h}} $
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{-1}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{1} $ (By L'Hospital rule)
$\Rightarrow-(\text{a}+\text{b})=\text{a}+\text{b}$
$\Rightarrow-2(\text{a}+\text{b})=0$
$\Rightarrow\text{a}+\text{b}=0$
This is true for all value of c
$\therefore\text{c}\in\text{R}$
In the given option (b) satisfies a + b = 0 and $\text{c}\in\text{R.}$

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Question 51 Mark

The set points where the function f(x) given by $\text{f(x)=}|\text{x}-3|\cos\text{x}$ is diffrentiable, is:

R
R - {3}
$(0,\infty)$
None of these.

Answer

R - {3}

Solution:
(LHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{3-\text{h}-3}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{-\text{h}}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3-\text{h}-3|\cos(3-\text{h})-\text{f}(3)}{-\text{h}}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3-\text{h})-0}{-\text{h}}=-\cos3$
(RHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{3+\text{h}-3}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{\text{h}}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3+\text{h}-3|\cos(3+\text{h})-\text{f}(3)}{\text{h}}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3+\text{h})-0}{\text{h}}=\cos3$
So, f(x) is not diffrentiable at x = 3.
Also,f(x) is diffrentiable at all other points because both modulus and cosine function are differentiable and the product of two differentiable function is differentiable.

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Question 61 Mark

If $\text{f(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}, & \text{x}\neq 0\\\frac{1}{2} & \text{x}= 0\end{cases}$ then at x = 0, f(x) is:

Continuous and differentiable.
Differentiable but not continuous.
Continuous but not differentiable.
Neither continuous not differentiale.

Answer

Continuous and differentiable.

Solution:
We have,
$\text{f(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}, & \text{x}\neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$
Continuity at x = 0
(RHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{(\text{h})\sin(\text{h})}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}1-\cos\text{h}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}\sin\text{h}}$
$=1-\cos0.\frac{1}{0\sin0}$
$=0$
Hence, f(x) is continuous at x = 0.
For differentiable at = 0
(LHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\frac{1}{2}}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1-\cos(-\text{h})}{-\text{h}\sin(-\text{h})}-\frac{1}{2}}{-\text{h}}$
$=\frac{1}{\text{h}}\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}-\lim\limits_{\text{h}\rightarrow0}\frac{1}{2}$
$=\frac{1}{2}-0=\frac{1}{2}$
(RHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{h})-\frac{1}{2}}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1-\cos(\text{h})}{-\text{h}\sin(\text{h})}-\frac{1}{2}}{-\text{h}}$
$=-\frac{1}{\text{h}}\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}-\lim\limits_{\text{h}\rightarrow0}\frac{1}{2}$
$=\frac{1}{2}-0=\frac{1}{2}$
(LHL at x = x) $=\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos(-\text{h})}{(-\text{h})\sin(-\text{h})}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}1-\cos\text{h}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}\sin\text{h}}$
$=1-\cos(0).\frac{1}{0\sin0}$

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Question 71 Mark

The function$\text{f(x)}=1+|\cos\text{x}|$ is:

Continuous no where.
Continuous everywhere.
Not differentiable at x = 0
Not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$

Answer

Continuous everywhere.

Solution:
Graph of the function $\text{f(x)}=1+|\cos\text{x}|$ is as show blow:

From the graph, we can see that f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

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Question 81 Mark

If $\text{f(x)}=|\log_\text{e}\text{x}|,$ then:

$\text{f}'(1^+)=1$
$\text{f}'(1^-)=-1$
$\text{f}'(1)=1$
$\text{f}'(1)=-1$

Answer

$\text{f}'(1^+)=1$ and
$\text{f}'(1^-)=-1$

Solution:
$\text{f(x)}=|\log_\text{e}\text{x}|,=\begin{cases}-\log_\text{e}\text{x}, & \text{for}0<\text{x}<1\\\log_\text{e}\text{x}, & \text{for x}\geq1\end{cases}$
Differentiability at x = 1,
We have,
(LHL at x = 1)
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log12}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
(RHL at x = 1)
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log\text{x}-\log1}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{h}}$
$=1$

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Question 91 Mark

Let $\text{f(x)}\begin{cases}\text{ax}^2+1,&\text{x}<1\\\text{x}+\frac{1}{2},&\text{x}\leq1\end{cases}.$ Then, f(x) is derivable at x = 1, if:

a = 2
a = 1
a = 0
$\text{a}=\frac{1}{2}$.

Answer

$\text{a}=\frac{1}{2}.$

Solution:
Given: $\text{f(x)}=\begin{cases}\text{ax}^2+1,&\text{x}<1\\\text{x}+\frac{1}{2},&\text{x}\leq1\end{cases}$
The function is derivable at x = 1, if left hand derivative and right hand derivative of the function are equal at x = 1.
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(1-\text{h}+\frac{1}{2}\Big)-\frac{3}{2}}{-\text{h}}=1$
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1+\text{h})^2+1-\frac{3}{2}}{\text{h}}$
(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h}^2+2\text{h})-\frac{1}{2}}{\text{h}}$
$\therefore\text{LHL}=\text{RHL}$
$\Rightarrow\text{a}-\frac{1}{2}=0$
$\Rightarrow\text{a}-\frac{1}{2}$

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Question 101 Mark

The function $\text{f(x)}=|\cos\text{x}|$ is:

Differentiable at$\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
Continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
Neither differentiable nor continuous at $\text{x}=\text{n}\in\text{Z}$
None of these.

Answer

Continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

Solution:
$\text{f(x)}=|\cos\text{x}|$
Given function is trigonometric function.
⇒ Hence, it is continuous.
Function is not differentiable at odd multiples of $\frac{\pi}{2}$
⇒ f(x) is not differentiable at $\text{x}=(2+\text{n}+1)\frac{\pi}{2}.$

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Question 111 Mark

Let $\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4,$ where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if:

a = 0
b = 0
c = 0
None of these.

Answer

b = 0

Solution:
We have,
$\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4$
$\text{f(x)}=\begin{cases}\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}\geq0\\\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}<0\end{cases}$
Here, f(x) is differentiable at x = 0
$\therefore$ (LHL at x = 0) = (RHL at x = 0)
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{a}-\text{bx}+\text{cx}^4-\text{a}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}+\text{bx}^4-\text{a}}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}-\text{b}(0-\text{h})+\text{c}(0-\text{h})^4-\text{a}}{0-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}(0+\text{h})+\text{c}(0+\text{h})^4-\text{a}}{0+\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}(-\text{b}-\text{bh}^3)=\lim\limits_{\text{h}\rightarrow0}(\text{b}+\text{ch}^3)$
$\Rightarrow-\text{b}=\text{b}$
$\Rightarrow2\text{b}=0$
$\Rightarrow\text{b}=0$

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MCQ 121 Mark

Let $f(x) = |x|$ and $g(x) = |x^3|,$ then:

✓
$f(x)$ and $g(x)$ both are continuous at $x = 0$
B
$f(x)$ and $g(x)$ both are differentiable at $x = 0$
C
$f(x)$ is differentiable but $g(x)$ is not differentiable at $x = 0$
D
$f(x)$ and $g(x)$ both are not differentiable at $x = 0$

Answer

Correct option: A.

$f(x)$ and $g(x)$ both are continuous at $x = 0$

Absolute value function is continuous on $R.$

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Question 131 Mark

If $\text{f(x)}=\begin{cases}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$ then f(x) is:

Continuous as well as differentiable at x = 0
Continuous but not differentiable at x = 0
Differentiable but not continuous at x = 0
None of these.

Answer

None of these.

Solution:
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits _{\text{x}\rightarrow0}\frac{1}{1+\text{e}^{\frac{-1}{\text{x}}}}=1\ \Big(\because\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}=0\Big)$
$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\text{f}(0)$
Function is not continuous,
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\text{f}(0)}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{e}^{\frac{1}{\text{h}}}}-0}{-\text{h}}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{-\Big(1+\text{e}^{\frac{1}{\text{h}}}\Big)\text{h}}=-\infty$
Similarly,
$\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\infty$

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Question 141 Mark

Let $\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$ then, f is:

Continuous at x = -1
Differentible at x = -1
Everywhere continuous.
Everywhere diffrentiable.

Answer

Differentible at x = -1

Solution:
$\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$
$\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\text{x}+1}{\text{x}+1}=0$
Similarly,
$\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{x}+1}{\text{x}+1}=0$
Function is diffrentiable at x = -1.

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Question 151 Mark

The function $\text{f(x)}=\sin^{-1}(\cos\text{x})$ is:

Discontinuous at x = 0
Continuous at x = 0
Differentiable at x = 0
None of these.

Answer

Continuous at x = 0

Solution:
$\text{f(x)}=\sin^{-1}(\cos\text{x})$
$\text{f(x)}=\sin^{-1}\Big[\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$\text{f(x)}=\frac{\pi}{2}-\text{x}$
Function is continuous at x = 0.

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Question 161 Mark

If $\text{f(x)}=|\log_\text{e}|\text{x}||,$ then:

f(x) is continuous and differentiable for all x in its domain.
f(x) is continuous for all for all × in its domain but not differentiable at $\text{x}=\pm1$
f(x) is neither continuous nor differentiable at $\text{x}=\pm1$
None of these.

Answer

f(x) is continuous for all for all × in its domain but not differentiable at $\text{x}=\pm1$

Solution:
We have,
$\text{f(x)}=|\log_\text{e}|\text{x}||$
We know that log function is defined for posirive value.
Here, |x| is positive for all non zero x.
Therefore, domian of function is R - {0}
And we know that logarithmic function continuous in its domain.
Therefore, $|\log_\text{e}|\text{x}||$ is continuous in its domain.
We will check the differentiability at its critical points.
$|\log_\text{e}|\text{x}||=\begin{cases}\log_\text{e}(-\text{x}) & -\infty<\text{x<-1}\\-\log_\text{e}(-\text{x}) &-1<\text{x}<0\\-\log_\text{e}(\text{x})&0<\text{x}<1\\\log_\text{e}(\text{x})&1<\text{x}<\infty\end{cases}$
(LHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\log_\text{e}(-\text{x})-0}{\text{x}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[-(-1-\text{h})]}{-1-\text{h}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{-\text{h}}$
$=-1$
(RHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$
$=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{-\log_\text{e}(-\text{x})-0}{\text{x}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[-(-1+\text{h})]}{-1+\text{h}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}(1-\text{h})}{\text{h}}$
$=-\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$
$=-1\times-1=1$
Here, $\text{LHL}\neq\text{RHL}$
Therefore, the given function is not differentiable at x = -1.
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{-\log_\text{e}(\text{x})-0}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[(1-\text{h})]}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$
$=-1$
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-(1)}$
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log_\text{e}(\text{x})-0}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[(1+\text{h})]}{1+\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{\text{h}}$
$=1$
Here, $\text{LHL}\neq\text{RHL}$
Therefore, the given function is not differentiable at x =1.
Therefore, given function is continuous for all x in its domain but not differentiable at $\text{x}=\pm1.$

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MCQ 171 Mark

The set of points where the function $f(x) = x |x|$ is differentiable is:

✓
$(-\infty,\infty)-\infty,\infty$
B
$(-\infty,0)\cup(0,\infty)-\infty,0\cup0,\infty$
C
$(0,\infty)0,\infty$
D
$[0,\infty]0,\infty$

Answer

Correct option: A.

$(-\infty,\infty)-\infty,\infty$

We have,$\text{f(x)}=\text{x}|\text{x}|$
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$
When, $x < 0,$ we have
$f(x) = -x^2$ which being a polynomial function is continuous and differentable in $(-\infty,0)$
When, $x > 0,$ we have
$f(x) = -x^2$ which being a polynomial function is continuous and differentable in $(0,\infty,)$
Thus possible point of non$-$differentiability of $f(x)$ is $x = 0$
Now, $\text{LHL} (at x = 0) =\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
And $\text{RHL} (at x = 0) =\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
$\therefore \text{LHL} ($at $x = 0) = \text{RHL} ($at $x = 0)$
So, $f(x)$ is also differentiable at $x = 0$
i.e. $f(x)$ is differentiable in $(-\infty,\infty).$

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Question 181 Mark

Let $\text{f(x)}=|\cos\text{x}|.$ Then,

f(x) is everywhere differentiable.
f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
None of these.

Answer

f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

Solution:
$\text{f}(\text{x)} = |\cos\text{x}|$
Given function is trigonometric function.
⇒ Hence, it is continuous.
Function is not differentiable at odd multiples of $\frac{\pi}{2}.$
⇒ f(x) is not differentiable at $\text{x} = (2\text{n} + 1) \frac{\pi}{2}$

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Question 191 Mark

The function $\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2},$ where[.] denotes the greatest integer function, is:

Continuous as well as differentiable for all $\text{x}\in\text{R}$
Continuous for all x but differentiable at some x
Differentiable for all x but not continuous at some x
None of these.

Answer

Continuous as well as differentiable for all $\text{x}\in\text{R}$

Solution:
Here,
$\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2}$
Since, we know that $\pi(\text{x}-\pi)=\text{n}\pi$ and $\sin\text{n}\pi=0.$
$\because4+\text{x}[\text{x}]^2\neq0$
$\therefore\text{f(x)}=0$ for all x
Thus, f(x) is a constant function and it is continuous and differentible everywhere.

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Question 201 Mark

The function $\text{f(x)}=|\cos\text{x}|$ is:

Everywhere continuous and differentiable.
Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
Neither continuous nor differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
None of these.

Answer

Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

Solution:
As cos x is even function it is continuous everywhere but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
$\cos\Big[(2\text{n}+1)\frac{\pi}{2}=\cos\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\sin\text{n}\pi$
For n as an integer $\Rightarrow\sin\text{n}\pi=0$
For n as rational $\Rightarrow\sin\text{n}\pi=-1$

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Question 211 Mark

Let $\text{f(x)}=|\sin\text{x}|.$ then,

f(x) is everywhere differentiable.
f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
None of these.

Answer

f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Solution:
$\text{f(x)}=|\sin\text{x}|$
Given function is continuous and differentiable on $(2\text{n}\pi,(2\text{n}+1)\pi)$
But not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$
As $\sin\text{n}\pi=0$ for $\text{n}\in\text{Z}.$

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Question 221 Mark

Let $\text{f(x)}=\begin{cases}\frac{1}{|\text{x}|} & \text{for |x|}\geq1\\\text{ax}^2+\text{b} & \text{for |x|}<1\end{cases}$ if f(x) is continuous and differentiable at any point, then:

$\text{a}=\frac{1}{2},\text{b}=-\frac{3}{2}$
$\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$
$\text{a}=1,\text{b}=-1$
None of these.

Answer

$\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$

Solution:
Given function is continuous at x = 1.
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{1}{\text{x}}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{ax}^2+\text{b}$
$\Rightarrow1=\text{a}+\text{b}\ \dots(1)$
Function is derivable at x = 1.
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\text{f}(1+\text{h}-\text{f}(1))}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{f}(0+\text{h}-\text{f}(1))}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\frac{1}{1+\text{h}}+1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{a}(1+\text{h})^2-\text{a}}{\text{h}}$
$\Rightarrow-1=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{h}(2\text{a}+\text{h})}{\text{h}}$
$\Rightarrow2\text{a}=-1$
$\Rightarrow\text{a}=\frac{-1}{2}$
$\text{a}+\text{b}=1\ (\text{From(1)})$
$\frac{-1}{2}+\text{a}=1$
$\Rightarrow\text{b}=\frac{3}{2}$

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Question 231 Mark

If $\text{f(x)}=\begin{cases}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}, & \text{x}\neq-2\\2, & \text{x}=-2\end{cases},$ then f(x) is:

Continuous at x = -2
Not continuous at x = -2
Diffrentiable at x = -2
Continuous but nit derivable at x = -2

Answer

Not continuous at x = -2

Solution:
$\lim\limits_{\text{x}\rightarrow-2}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}$
Let, x = -2 + h
x → -2 ⇒ h → 0
$\lim\limits_{\text{h}\rightarrow0}\frac{|-2+\text{h}+2|}{\tan^{-1}(-2+\text{h}+2)}$
$\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\tan^{-1}\text{h}}=1$
$\lim\limits_{\text{h}\rightarrow0}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}\neq\text{f}(-2)$
Function is not continuous at x = -2.

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Question 241 Mark

If $\text{f(x)}=\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)}+....+\frac{\text{x}^2}{(1+\text{x}^2)}+....,$ then at x = 0, f(x):

Has not limit.
Is discontinuous.
Is continuous but not differentiable.
Is differentiable.

Answer

Is discontinuous.

Solution:
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\Big(1+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\bigg(\frac{1}{1-\frac{1}{1+\text{x}^2}}\bigg)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(1+\text{x}^2)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}=1$
But, f(0)=0
$\text{f}(0)\neq\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$
Function is discontinuous.

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Question 251 Mark

If $\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}},$ then f(x) is:

Continuous on [-1, 1] and differentiable on (-1, 1)
Continuous on [-1, 1] and differentiable on $(-1,0)\cup(0,1)$
Continuous and differentiable on [-1, 1]
None of these.

Answer

Continuous on [-1, 1] and differentiable on $(-1,0)\cup(0,1)$

Solution:
We have,
$\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$
Here, function will be defined for those values of x for which
$1-\text{x}^2\geq0$
$\Rightarrow1\geq\text{x}^2$
$\Rightarrow\text{x}^2\leq1$
$\Rightarrow|\text{x}\leq1|$
$\Rightarrow-1\leq\text{x}\leq1$
Therefore, function is continuous in -1, 1
Now, we need to check the differentiability of $\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$ in the interval -1, 1
Now,
We will check the differentiable at x = 0
LHL at x = 0
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0-\text{h})^2}}}{0-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{-\text{h}}$
$=-\infty$
RHL at x = 0
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0+\text{h})^2}}}{0+\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{\text{h}}$
$=\infty$
So, the function is not differentiable at x = 0.

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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip