Sample QuestionsCONTINUITY AND DIFFERENTIABILITY questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Let $\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{V}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ then $\frac{\text{dU}}{\text{dV}}=$
- $\frac{1}{2}$
- $\text{x}$
- $\frac{1-\text{x}^2}{\text{x}^2-4}$
- $1$
View full solution →If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{n}>\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then:
- $\text{m}=1,\text{ n}=0$
- $\text{m}=\frac{\text{n}\pi}{2}+1$
- $\text{n}=\frac{\text{m}\pi}{2}$
- $\text{m}=\text{n}=\frac{\pi}{2}$
View full solution →The function $\text{f(x)}=\tan\text{x}$ is discontinuous on the set:
- $\{\text{n}\pi:\text{n}\in\text{z}\}$
- $\{2\text{n}\pi:\text{n}\in\text{z}\}$
- $\{(2\text{n}+1)\frac{\pi}{2}:\text{n}\in\text{z}\}$
- $\Big\{\frac{\text{n}\pi}{2}:\text{n}\in\text{z}\Big\}$
View full solution →If $\text{f(x)}=\begin{cases}\frac{\sin(\text{a}+1)}{\text{x}},&\text{x}<0\\\text{c},&\text{x}=0\\\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}},&\text{x}>0&\end{cases}$ is continuouse at x = 0, then:
- $\text{a}=-\frac{3}{2},\text{b}=0,\text{c}=\frac{1}{2}$
- $\text{a}=-\frac{3}{2},\text{b}=1,\text{c}=-\frac{1}{2}$
- $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
- $\text{None of these}.$
View full solution →Let $3\sin(\text{xy})+4\cos(\text{xy})=5,$ then $\frac{\text{dy}}{\text{dx}}=$
- $-\frac{\text{y}}{\text{x}}$
- $\frac{3\sin(\text{xy})+4\cos(\text{xy})}{3\cos(\text{xy})-4\sin(\text{xy})}$
- $\frac{3\cos(\text{xy})+4\sin(\text{xy})}{4\cos(\text{xy})-3\sin(\text{xy})}$
- $\text{None of these.}$
View full solution →Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A) $\frac{\text{dx}^{\sin\text{x}}}{\text{dx}}=\text{x}^{\sin\text{x}}[(\cos)\log\text{x}+\frac{\sin\text{x}}{\text{x}}]$
Reason(R) if y = xf(x) then $\frac{\text{dy}}{\text{dx}}=\text{x}^\text{f(x)}[\text{f '(x)}\log\text{x}+\frac{\text{f(x)}}{\text{x}}]$
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false
- A is false but R is true
View full solution →Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A) f(x) = x - 1 + x - 2 is continuous but not differentiable at x = 1, 2.
Reason(R) Every differentiable function is continuous
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false
- A is false but R is true
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion(A): $\text{f(x)}=\sin\text{x}$ is continuous x = 0.
Reason(R): $\sin\text{x}$ is differentiable at x = 0.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A
- A is true but R is false.
- A is false but R is true.
View full solution →Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A) f(x) = [x] greatest integer function is not differentiable at x = 2
Reason(R) The greatest integer function is not continuous at any integer
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false
- A is false but R is true
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion(A): $\text{f(x)}=\tan^2\text{x}$ is continuous at $\text{x}=\frac{\pi}{2}$
Reason(R): 𝑥2 is continuous at $\text{x}=\frac{\pi}{2}$
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A
- A is true but R is false.
- A is false but R is true.
View full solution →Determine the value of ‘k’ for which the following function is continuous at x = 3:
$\text{f(x)} = \begin{cases} \frac{(\text{x}+3)^2-36}{\text{x}-3}\ \ \ \ ,\ \text{x}\neq3\\ \ \ \ \ \ \ \text{k}\ \ \ \ \ \ \ \ \ \ \ ,\ \text{x}=3 \end{cases}$
View full solution →Determine the value of ‘k’ for which the following function is continuous at x = 3:
$ \text{f(x)} = \begin{cases} \frac{(\text{x + 3)}^{2} \text{ - } 36}{\text{x - 3}} & , & \text{x}\neq 3 \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{k}& , & \text{x = 3}\\ \end{cases}$
View full solution →For what value of ‘k’ is the function $\text{f(x)} = \begin{cases} \frac{\sin \text{5x}}{\text{3x}} + \cos \text{x}, & \text{if }& \text{x} \neq 0 \\ \text{ }\text{ }\text{ }\text{ } \text{k}, & \text{if } & \text{x = 0}\\ \end{cases}$ continuous at x = 0?
View full solution →Determine the value of the constant ‘k’ so that the function $\text{f}(x) = \begin{cases} \frac{\text{k}x}{| x|}\text{ }\text{ }, & \text{if } x < 0\\ \text{ }3\text{ }\text{ }\text{ }\text{ }, & \text{if } x\geq 0\\ \end{cases}$ is continuous at x = 0.
View full solution →$\text{If} f(x) = x + 7 \text{and g (x)} = x - 7, x \in \text{R, find (fog) (7)} $
View full solution →Find the value of c in Rolle’s theorem for the function $\text{f(x)} = \text{x}^{3} - \text{3x in} [ -\sqrt{3}, 0].$
View full solution →$\text{Find } \frac{\text{dy}}{\text{dx}} \text{ at t} = \frac{2\pi}{3} \text{ when x 10 (t} -\sin \text{t) and y = 12 } (1 - \cos \text{t}).$
View full solution →$\text{If y}=\text{sin}^{-1}(6\text{x}\sqrt{1-9\text{x}^2}), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}},\text{then find}\frac{\text{dy}}{dx}$
View full solution →Differentiate $\tan^{-1}\Big(\frac{1+\cos\text{x}}{\sin\text{x}}\Big)$ with respect to x.
View full solution →If x = at2, y = 2at, then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
View full solution →If ex+y - x = 0, prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
View full solution →Find the value of k for which the function $\text{f(x)}=\begin{cases}\frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2},&\text{x}\neq2\\\text{k},&\text{x} = {2}\end{cases}$ is continues at x = 2.
View full solution →If $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
View full solution →Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = 2x2 - 3x + 1 on [1, 3]
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem. $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}-\text{x}\text{ on }[0,\pi]$
View full solution →$\text{If} x = \text{a } \sin \text{ 2t}(1 + \cos\text{2t}) \text{ and }y = \text{b}\cos\text{2t}(1- \cos \text{2t}), \text{find }\frac{dy}{dx} \text{ at t} = \frac{\pi}{4}$
View full solution →If ey (x + 1) = 1, then show that $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^{2}.$
View full solution →$\text{ If }\text{ x } =\cos\text{t} (3-2\cos^2\text{t)}\ \text{and}\ \text{y}=\sin\text{t}(3-2\sin\text{ t}),\text{ find the value of}\ \frac{\text{dy}}{\text{dx}} = \text{ at t}=\frac{\pi}{\text{4}}.$
View full solution →$\text{If x = a} (\cos 2\text{t +2t}\sin \text{2t}) \text{and y = a} (\sin \text{2t - 2t}\cos\text{2t}),\text{then find}\frac{\text{d}^{2}{\text{y}}}{\text{dx}^{2}}.$
View full solution →AB is the diameter of a circle and C is any point on the circle. Show that the area of triangle ABC is maximum, when it is an isosceles triangle.
Logarithmic differentiation is a powerful technique to differentiate functions of the form $\text{f}(\text{x})=[\text{u}(\text{x})]^{\text{v}(\text{x})},$ where both u(x) and v(x) are differentiable functions and f and u need to be positive functions. Let function $\text{y}=\text{f}(\text{x})=(\text{u}(\text{x}))^{\text{v}(\text{x})},$ then $\text{y}'=\text{y}\Big[\frac{\text{v}(\text{x})}{\text{u}(\text{x})}\text{u}'(\text{x})+\text{v}'(\text{x})\cdot\log[\text{u}(\text{x})]\Big]$ On the basis of above information, answer the following questions. - Differentiate xx w.r.t. x.
- $\text{x}^\text{x}(1+\log\text{x})$
- $\text{x}^\text{x}(1-\log\text{x})$
- $-\text{x}^\text{x}(1+\log\text{x})$
- $\text{x}^\text{x}\log\text{x}$
- Differentiate xx + ax + xa + aa w.r.t. x.
- $(1+\log\text{x})+(\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1})$
- $\text{x}^\text{x}(1+\log\text{x})+\log\text{a}+\text{ax}^{\text{a}-1}$
- $\text{x}^\text{x}(1+\log\text{x})+\text{x}^\text{a}\log\text{x}+\text{ax}^{\text{a}-1}$
- $\text{x}^\text{x}(1+\log\text{x})+\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$
- If $\text{x}=\text{e}^\frac{\text{x}}{\text{y}},$ then find $\frac{\text{dy}}{\text{dx}}.$
- $-\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
- $-\frac{(\text{x}-\text{y})}{\text{x}\log\text{x}}$
- $\frac{(\text{x}+\text{y})}{\text{x}\log\text{x}}$
- $\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
- If y = (2 - x)3(3 + 2x)5, then find $\frac{\text{dy}}{\text{dx}}.$
- $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}-\frac{8}{2-\text{x}}\Big]$
- $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{15}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
- $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$
- $(2-\text{x})^3(3+2\text{x})^5\cdot\Big[\frac{10}{3+2\text{x}}+\frac{3}{2-\text{x}}\Big]$
- If $\text{y}=\text{x}^\text{x}\cdot\text{e}^{(2\text{x}+5)},$ then find $\frac{\text{dy}}{\text{dx}}.$
- $\text{x}^\text{x}\text{e}^{2\text{x}+5}$
- $\text{x}^\text{x}\text{e}^{2\text{x}+5}(3-\log\text{x})$
- $\text{x}^\text{x}\text{e}^{2\text{x}+5}(1-\log\text{x})$
-
$\text{x}^\text{x}\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$
View full solution →If a relation between x and y is such that y cannot be expressed in terms of x, then y is called an implicit function of x. When a given relation expresses y as an implicit function of x and we want to find $\frac{\text{dy}}{\text{dx}},$ then we differentiate every term of the given relation w.r.t. x, remembering that a tenn in y is first differentiated w.r.t. y and then multiplied by $\frac{\text{dy}}{\text{dx}}.$
Based on the ab:ve information, find the value of $\frac{\text{dy}}{\text{dx}}$ in each of the following questions.
- x3 + x2y + xy2 + y3 = 81
- $\frac{(3\text{x}^2+2\text{xy}+\text{y}^2)}{\text{x}^2+2\text{xy}+3\text{y}^2}$
- $\frac{-(3\text{x}^2+2\text{xy}+\text{y}^2)}{\text{x}^2+2\text{xy}+3\text{y}^2}$
- $\frac{(3\text{x}^2+2\text{xy}-\text{y}^2)}{\text{x}^2-2\text{xy}+3\text{y}^2}$
- $\frac{3\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2+\text{xy}+3\text{y}^2}$
- xy = ex-y
- $\frac{\text{x}-\text{y}}{(1+\log\text{x})}$
- $\frac{\text{x}+\text{y}}{(1+\log\text{x})}$
- $\frac{\text{x}-\text{y}}{\text{x}(1+\log\text{x})}$
- $\frac{\text{x}+\text{y}}{\text{x}(1+\log\text{x})}$
- $\text{e}^{\sin\text{y}}=\text{xy}$
- $\frac{-\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$
- $\frac{\text{y}}{\text{y}\cos\text{y}-1}$
- $\frac{\text{y}}{\text{y}\cos\text{y}+1}$
- $\frac{\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$
- $\sin^2\text{x}+\cos^2\text{y}=1$
- $\frac{\sin2\text{y}}{\sin2\text{x}}$
- $-\frac{\sin2\text{x}}{\sin2\text{y}}$
- $-\frac{\sin2\text{y}}{\sin2\text{x}}$
- $\frac{\sin2\text{x}}{\sin2\text{y}}$
- $\text{y}=(\sqrt{\text{x}})^{\sqrt{\text{x}}^\sqrt{\text{x}}...\infty}$
- $\frac{-\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$
- $\frac{\text{y}^2}{2+\text{y}\log\text{x}}$
- $\frac{\text{y}^2}{\text{x}(2+\text{y}\log\text{x})}$
- $\frac{\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$
View full solution →Derivative of y = f(x) w.r.t. x (if exists) is denoted by $\frac{\text{dy}}{\text{dx}}$ or f'(x) and is called the first order derivative of y. If we take derivative of $\frac{\text{dy}}{\text{dx}}$ again, then we get $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}^2\text{y}}{\text{dx}^2}$ or f''(x) and is called the second order derivative of y. Similarly, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$ is denoted and defined as $\frac{\text{d}^3\text{y}}{\text{dx}^3}$ or f'''(x) and is known as third order derivative of y and so on.
Based on the above information, answer the following questions.
- If $\text{y}=\tan^{-1}\Big(\frac{\log(\frac{\text{e}}{\text{x}^2})}{\log(\text{ex}^2)}\Big)+\tan^{-1}\Big(\frac{3+2\log\text{x}}{1-6\log\text{x}}\Big),$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equal to:
- 2
- 1
- 0
- -1
- If u = x2 + y2 and x = s + 3t, y = 2s - t, then $\frac{\text{d}^2\text{u}}{\text{ds}^2}$ is equal to:
- 12
- 32
- 36
- 10
- If $\text{f}(\text{x})=2\log\sin\text{x},$ then f''(x) is equal to:
- $2\text{cosec}^3\text{x}$
- $2\cot^2\text{x}-4\text{x}^2\text{cosec}^2\text{x}^2$
- $2\text{x}\cot\text{x}^2$
- $-2\text{cosec}^2\text{x}$
- If $\text{f}(\text{x})=\text{e}^\text{x}\sin\text{x},$ then f'''(x) =
- $2\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$
- $2\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
- $2\text{e}^\text{x}(\sin\text{x}-\cos\text{x})$
- $2\text{e}^\text{x}\cos\text{x}$
- If $\text{y}^2=\text{ax}^2+\text{bx}+\text{c},$ then $\frac{\text{d}}{\text{dx}}(\text{y}^3\text{y}_2)=$
- 1
- -1
- $\frac{4\text{ac}-\text{b}^2}{\text{a}^2}$
- 0
View full solution →The function f(x) will be discontinuous at x = a if f(x) has
- Discontinuity of first kind : $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ both exist but are not equal. If is also known as irremovable discontinuity.
- Discontinuity of second kind : If none of the limits $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ exist.
- Removable discontinuity : $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}-\text{h})$ and $\lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})$ both exist and equal but not equal to f(a).
Based on the above information, answer the following questions.
- If $\text{f}(\text{x})=\begin{cases}\frac{\text{x}^2-9}{\text{x}-3},&\text{for x}\neq3\\4,&\text{for x}=3\end{cases},$ then at x = 3
- f has removable discontinuity.
- f is continuous.
- f has irremovable discontinuity.
- None of these.
- Let $\text{f}(\text{x})=\begin{cases}\text{x}+2,&\text{if x}\leq4\\\text{x}+4,&\text{if x}\geq4\end{cases}$ then at x = 4
- f is continuous.
- f has removable discontinuit.
- f has irremovable discontinuit.
- None of thesee.
- Consider the function f(x) defined as $\text{f}(\text{x})=\begin{cases}\frac{\text{x}^2-4}{\text{x}-2},&\text{for x}\neq2\\5,&\text{for x}=2\end{cases},$ then at x = 2
- f has removable discontinuity.
- f has irremovable discontinuity.
- f is continuous.
- f is continuous if f(2) = 3
- If $\text{f}(\text{x})=\begin{cases}\frac{\text{x}-|\text{x}|}{\text{x}},&\text{x}\neq0\\2,&\text{x}=0\end{cases},$ then at x = 0
- f is continuous.
- f has removable discontinuity.
- f has irremovable discontinuity.
- None of these.
- If $\text{f}(\text{x})=\begin{cases}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})},&\text{if x}\neq0\\7,&\text{if x}=0\end{cases},$ then at x = 0
- fis continuous if f(0) = 2
- f is continuous
- f has irremovable discontinuity.
- f has removable discontinuity.
View full solution →Let f(x) be a real valued function, then its
- Left Hand Derivative (L.H.D.) : $\text{Lf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}-\text{h})-\text{f}(\text{a})}{-\text{h}}$
- Right Hand Derivative (R.H.D.) : $\text{Rf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f}(\text{a})}{\text{h}}$
Also, a function f(x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and are equal.
For the function $\text{f}(\text{x})=\begin{cases}|\text{x}-3|,\text{x}\geq1\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},\text{x}<1\end{cases},$ answer the following questions.
- R.H.D. of f(x) at x = 1 is:
- 1
- -1
- 0
- 2
- L.H.D. of f(x) at x = 1 is:
- 1
- -1
- 0
- 2
- f(x) is non-differentiable at:
- x = 1
- x = 2
- x = 3
- x = 4
- Find the value of f'(2).
- 1
- 2
- 3
- -1
- The value of f'(-1) is:
- 2
- 1
- -2
- -1
View full solution →