Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 21 :

Let $[x]$ be the greatest integer $\leq x$ . Then the number of points in the interval $(- 2, 1)$ , where the function $f (x) =$ $| [x] |$ $+ \sqrt{x - [x]}$ is discontinuous, is _______ . [2023]

(2)

$We have, f (x) =$ $| [x] |$ $+ \sqrt{x - [x]}$

Where $[x]$ is G.I.F. discontinuous at $x \in I$ only. Then,

at $x = - 1, f (- 1^{+}) = 1 + 0 = 1$ and $f (- 1^{-}) = 2 + 1 = 3$

at $x = 0, f (0^{+}) = 0 + 0 = 0$ and $f (0^{-}) = 1 + 1 = 2$

Hence, $f (x)$ is discontinuous at two points.

Q 22 :

Let $f (x) = \sum_{k = 1}^{10} k x^{k}, x \in ℝ .$ If $2 f (2) + f' (2) = 119 {(2)}^{n} + 1,$ then $n$ is equal to _________ . [2023]

(10)

$f (x) = \sum_{k = 1}^{10} k \cdot x^{k}, x \in ℝ$

$f (x) = x + 2 x^{2} + 3 x^{3} + \dots + 10 x^{10}$

$x \cdot f (x) = x^{2} + 2 x^{3} + 3 x^{4} + \dots + 9 x^{10} + 10 x^{11}$

$f (x) (1 - x) = x + x^{2} + x^{3} + \dots + x^{10} - 10 x^{11}$

$f (x) = \frac{x (1 - x^{10})}{{(1 - x)}^{2}} - \frac{10 x^{11}}{1 - x}$

$f (x) = \frac{x - x^{11} - 10 x^{11} (1 - x)}{{(1 - x)}^{2}}$

$f (x) = \frac{x - x^{11} - 10 x^{11} + 10 x^{12}}{{(1 - x)}^{2}} = \frac{10 x^{12} - 11 x^{11} + x}{{(1 - x)}^{2}}$

$f' (x) = \frac{{(1 - x)}^{2} \times [120 x^{11} - 121 x^{10} + 1] + 2 (1 - x) [10 x^{12} - 11 x^{11} + x]}{{(1 - x)}^{4}}$

$Hence, 2 f (2) + f' (2) = 119 \cdot 2^{10} + 1$

$∴$ $n = 10$

Q 23 :

If $f (x) = x^{2} + g' (1) x + g'' (2)$ and $g (x) = f (1) x^{2} + x f' (x) + f'' (x)$ , then the value of $f (4) - g (4)$ is equal to [2023]

(14)

$Let g' (1) = A$

$g'' (2) = B, f (x) = x^{2} + A x + B, f (1) = A + B + 1$

$f' (x) = 2 x + A, f'' (x) = 2$ ,

$g (x) = (A + B + 1) x^{2} + x (2 x + A) + 2$

$\Rightarrow g (x) = x^{2} (A + B + 3) + A x + 2$

$g' (x) = 2 x (A + B + 3) + A, g' (1) = A$

$\Rightarrow 2 (A + B + 3) + A = A$

$A + B = - 3 ...(i)$

$g'' (x) = 2 (A + B + 3), g'' (2) = B$

$\Rightarrow 2 (A + B + 3) = B$

$\Rightarrow 2 A + B = - 6 ...(ii)$

$From (i) and (ii): A = - 3, B = 0$

$f (x) = x^{2} - 3 x, f (4) = 16 - 12 = 4$

$g (x) - 3 x + 2, g (4) = - 12 + 2 = - 10$

$f (4) - g (4) = 4 - (- 10) = 14$

Q 24 :

Let $f : ℝ \to ℝ$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in ℝ .$ If $f' (0) = 2$ , then $| f (- 2) |$ is equal to _______ . [2023]

(3)

$Given f (x + y) = f (x) + f (y) - 1$

$Putting x = y = 0 \Rightarrow f (0) = 1$

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$f' (0) = \lim_{h \to 0} \frac{f (h) - f (0)}{h} \Rightarrow f' (0) = 2$

$f' (x) = 2 \Rightarrow f (x) = 2 x + C \Rightarrow y = 2 x + C$

$Now, f (0) = 1$

$∴$ $1 = 2 (0) + C \Rightarrow C = 1$

So, $f (x) = 2 x + 1$

$| f (- 2) |$ $=$ $| 2 (- 2) + 1 |$ $=$ $| - 3 |$ $= 3$

Q 25 :

Let $f : ℝ \to ℝ$ be a twice differentiable function such that $f'' (x) \sin (\frac{x}{2}) + f' (2 x - 2 y) = (\cos x) \sin (y + 2 x) + f (2 x - 2 y)$ for all $x, y \in R$ . If $f (0) = 1$ , then the value of $24 f^{(4)} (\frac{5 π}{3})$ is: [2025]

2
-3
1
3

1

2

3

4

(2)

Q 26 :

Let $f (x)$ $=$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{4 x^{2} + 4 x - 3}, & x \neq - \frac{3}{2}, \frac{1}{2} \\ b, & x = - \frac{3}{2}, \frac{1}{2} \end{matrix}$

be continuous at $x = - \frac{3}{2}$ . If $f o f (x) = \frac{7}{5}$ , then $x$ is equal to: [2026]

1.4
0
1
2

1

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3

4

(3)

$f (x) =$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$

$For continuity at x = \frac{- 3}{2}$

$LHL = RHL$

$\Rightarrow \lim_{x \to \frac{- 3}{2}} \frac{(a x^{2} + 2 a x + 3)}{(2 x - 1) (2 x + 3)}$

$At x = \frac{- 3}{2} \Rightarrow Numerator = 0$

$a {(\frac{- 3}{2})}^{2} + 2 a (\frac{- 3}{2}) + 3 = 0$

$\frac{9}{4} a - 3 a + 3 = 0$

$\frac{3 a}{4} = 3 \Rightarrow a = 4$

$∴$ $f (x) =$ ${\begin{matrix} \frac{4 x^{2} + 8 x + 3}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$

$f (x) =$ ${\begin{matrix} \frac{(2 x + 1) (2 x + 3)}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$
$f o f (x)$ $= f (\frac{2 x + 1}{2 x - 1})$ $= \frac{2 (\frac{2 x + 1}{2 x - 1}) + 1}{2 (\frac{2 x + 1}{2 x - 1}) - 1}$

$= \frac{\frac{4 x + 2}{2 x - 1} + 1}{\frac{4 x + 2}{2 x - 1} - 1}$ $= \frac{\frac{6 x + 1}{2 x - 1}}{\frac{2 x + 3}{2 x - 1}}$

$= \frac{6 x + 1}{2 x + 3}$ $= \frac{7}{5}$

$\Rightarrow 5 (6 x + 1) = 7 (2 x + 3)$

$30 x + 5 = 14 x + 21$

$16 x = 16$

$\Rightarrow x = 1$

$Ans. x = 1$

Q 27 :

Let $f : R \to R$ be a twice differentiable function such that the quadratic equation $f (x) m^{2} - 2 f' (x) m + f'' (x) = 0$ in $m$ , has two equal roots for every $x \in R$ . If $f (0) = 1, f' (0) = 2$ and $(α, β)$ is the largest interval in which the function $f (\log_{e} x - x)$ is increasing, then $α + β$ is equal to [2026]

(1)

Given quadratic equation has equal roots, thus

$D = 0 \Rightarrow (f' {(x))}^{2} = f'' (x) \cdot f (x)$

$\frac{f' (x)}{f (x)} = \frac{f'' (x)}{f' (x)}$

Integrate,

$l n (f (x)) = l n (f' (x)) + \ln C \Rightarrow f (x) = c f' (x)$

Put $x = 0$ ,

$1 = c \cdot 2 \Rightarrow c = \frac{1}{2}$

Now, $2 f (x) = f' (x)$

$\Rightarrow \frac{f' (x)}{f (x)} = 2$

Integrate,

$l n (f (x)) = 2 x + d$

$\Rightarrow d = 0$

$\Rightarrow l n (f (x)) = 2 x \Rightarrow f (x) = e^{2 x}$

Now let $g (x) = f (l n x - x) = e^{2 (l n x - x)}$

$∴$ $g' (x) = 2 e^{2 (l n x - x)} (\frac{1}{x} - 1) \geq 3$

$\Rightarrow \frac{1 - x}{x} \geq 0$

$\Rightarrow x \in (0, 1]$

$\Rightarrow α = 0, β = 1$

$α + β = 1 .$

Q 28 :

Let $α, β \in ℝ$ be such that the function $f (x)$ $=$ ${\begin{matrix} 2 α (x^{2} - 2) + 2 β x, & x < 1 \\ (α + 3) x + (α - β), & x \geq 1 \end{matrix}$

be differentiable at all $x \in ℝ$ . Then $34 (α + β)$ is equal to [2026]

84
24
36
48

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2

3

4

(4)

$f (x)$ $=$ ${\begin{matrix} 2 α x^{2} + 2 β x - 4 α, & x < 1 \\ (α + 3) x + α - β, & x \geq 1 \end{matrix}$

$f (1^{+}) = 2 α - β + 3, f (1^{-}) = - 2 α + 2 β$

$2 α - β + 3 = 2 β - 2 α \Rightarrow 4 α - 3 β + 3 = 0 . . . (1)$

$f' (1^{+}) = 4 α + 2 β, f' (1^{-}) = α + 3$

$4 α + 2 β = α + 3 \Rightarrow 3 α + 2 β - 3 = 0 . . . (2)$

$Solving (1) & (2)$

We get $α = \frac{3}{17}, β = \frac{21}{17}$

$\Rightarrow 34 (α + β) = 34 \times \frac{27}{17} = 48$

Q 29 :

If the function $f (x)$ $= \frac{e^{x} (e^{\tan x - x} - 1) + \log_{e} (\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$ , then the value of $f (0)$ is equal to [2026]

$\frac{2}{3}$
$\frac{3}{2}$
$2$
$\frac{1}{2}$

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(2)

$f (0) = \lim_{x \to 0} \frac{e^{\tan x} - e^{x} + l n (\sec x + \tan x) - x}{\tan x - x}$

$Applying L'Hospital rule$

$\Rightarrow f (0) = \lim_{x \to 0} \frac{e^{\tan x} . \sec^{2} x - e^{x} + \sec x - 1}{\sec^{2} x - 1}$

$\Rightarrow f (0) = \lim_{x \to 0} \frac{e^{\tan x} (\sec^{2} x - 1) + (e^{\tan x} - e^{x}) + \sec x - 1}{\tan^{2} x}$

$\Rightarrow f (0) = \lim_{x \to 0} (e^{\tan x} + \frac{e^{x} (e^{\tan x - x} - 1)}{\tan^{2} x} + \frac{1}{\sec x + 1})$

$\Rightarrow f (0) = 1 + 0 + \frac{1}{2} = \frac{3}{2}$

Q 30 :

Consider the following three statements for the function $f : (0, \infty) \to ℝ$ defined by
$f (x) =$ $| \log_{e} x |$ $-$ $| x - 1 |$ $:$

(I) $f$ is differentiable at all $x > 0$ .

(II) $f$ is increasing in (0, 1).

(III) $f$ is decreasing in $(1, \infty)$ .

Then. [2026]

Only (I) and (III) are TRUE.
Only (II) and (III) are TRUE.
All (I), (II) and (III) are TRUE.
Only (I) is TRUE.

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(1)

$f (x) =$ $| l n x |$ $-$ $| x - 1 |$

$=$ ${\begin{matrix} l n x - (x - 1) & x \geq 1 \\ - l n x + (x - 1) & 0 < x < 1 \end{matrix}$

$=$ ${\begin{matrix} l n x - x + 1 & x \geq 1 \\ - l n x + x - 1 & 0 < x < 1 \end{matrix}$

$f' (x)$ $=$ $(\begin{matrix} \frac{1}{x} - 1 & x \geq 1 \\ - \frac{1}{x} + 1 & 0 < x < 1 \end{matrix}$

$f' (1^{+}) = f' (1^{-}) = 0 \Rightarrow f (x) is differentiable \forall x > 0$

$f' (x) < 0 \forall x > 1$

$f' (x) < 0 \forall 0 < x < 1$

$\Rightarrow f (x) is decreasing \forall x \in (0, \infty)$

$Option (1)$

Q 31 :

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $\lim_{t \to x} (\frac{t^{2} y (x) - x^{2} y (t)}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to [2026]

23
27
18
12

1

2

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4

(1)

$\lim_{t \to x} \frac{2 t f (x) - x^{2} f' (t)}{- 1} = 3$

$x^{2} f' (x) - 2 x f (x) = 3$

$\frac{d y}{d x} - \frac{2 y}{x} = \frac{3}{x^{2}}$

$I.F. = e^{\int \frac{- 2}{x} d x} = e^{- 2 \log_{e} x} = \frac{1}{x^{2}}$

$y \cdot \frac{1}{x^{2}} = \int \frac{3}{x^{4}} d x$

$\frac{y}{x^{2}} = - \frac{1}{x^{3}} + c \Rightarrow y = c x^{2} - \frac{1}{x} = f (x)$

$f (1) = 2 = c - 1 \Rightarrow c = 3$

$f (x) = 3 x^{2} - \frac{1}{x}$

$f (2) = 12 - \frac{1}{2} \Rightarrow 2 f (2) = 23$

Q 32 :

Let $[t]$ denote the greatest integer less than or equal to $t$ . If the function

$f (x) =$ ${\begin{matrix} b^{2} \sin (\frac{π}{2} [\frac{π}{2} (\cos x + \sin x) \cos x]), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2 x}{x^{3}}, & x > 0 \\ a, & x = 0 \end{matrix}$

is continuous at $x = 0$ , then $a^{2} + b^{2}$ is equal to [2026]

$\frac{9}{16}$
$\frac{1}{2}$
$\frac{5}{8}$
$\frac{3}{4}$

1

2

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4

(4)

$f (0) = a$

$RHL = \lim_{x \to 0^{+}} \frac{\sin x (1 - \cos x)}{x^{3}} = \frac{1}{2}$

$LHL = \lim_{x \to 0^{-}} (b^{2} \sin \frac{π}{2} [\frac{π}{2} (\sin x + \cos x) \cos x]) = b^{2}$

$∴$ $a = \frac{1}{2}, b^{2} = \frac{1}{2}$

$so (a^{2} + b^{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

Q 33 :

If $f (x)$ $=$ ${\begin{matrix} \frac{a | x | + x^{2} - 2 (\sin | x |) (\cos | x |)}{x} & , x \neq 0 \\ b & , x = 0 \end{matrix}$

is continuous at $x = 0$ , then $a + b$ is equal to [2026]

0
4
2
1

1

2

3

4

(3)

$f (x)$ $=$ ${\begin{matrix} \frac{a | x | + x^{2} - 2 \sin | x | \cos | x |}{x}, & x \neq 0 \\ b, & x = 0 \end{matrix}$

$For continuity:$

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0)$

$\lim_{x \to 0^{-}} \frac{a h + h^{2} - 2 (\sin h) \cos h}{- h}$

$= \lim_{x \to 0^{+}} \frac{a h + h^{2} - 2 (\sin h) \cos h}{h}$

$or - a + 2 = a - 2 = b$

$2 a = 4$

$a = 2, b = 0$

$∴$ $a + b = 2$

Q 34 :

Let $[\cdot]$ denote the greatest integer function, and let $f (x) = \min {\sqrt{2 x}, x^{2}} .$

Let $S = {x \in (- 2, 2) : the function g (x) = | x | [x^{2}] is discontinuous at x} .$ Then $\sum_{x \in S} f (x)$ equals: [2026]

$\sqrt{6} - 2 \sqrt{2}$
$1 - \sqrt{2}$
$2 - \sqrt{2}$
$2 \sqrt{6} - 3 \sqrt{2}$

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(2)

$g (x) =$ $| x |$ $[x^{2}]$

$Points of discontinuity of g (x) in (- 2, 2) are (\pm 1, \pm \sqrt{2}, \pm \sqrt{3})$

$∴$ $S = {- 1, 1, - \sqrt{2}, \sqrt{2}, - \sqrt{3}, \sqrt{3}}$

$∵$ $f (x) = \min {\sqrt{2} x, x^{2}}$

$∴$ $\sum_{x \in S} f (x) = - \sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6}$

$= 1 - \sqrt{2}$

Q 35 :

Check whether the function $f (x) = 3 x - 5$ is continuous at $x = 0$ , $x = - 5$ and at $x = 3$ .

$f (x)$ is continuous at $x = 3$
$f (x)$ is not continuous at $x = 0$
$f (x)$ is not continuous at $x = - 5$
$f (x)$ is not continuous at $x = 3$

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(1)

Ans. $f (x)$ is continuous at $x = 3$

Explanation:

$f (x) = 3 x - 5$

At $x = 0$ ,

$f (0) = 3 \times 0 - 5 = - 5$

$\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} (3 x - 5) = 3 \times 0 - 5$

$= - 5$

$∴$ $\lim_{x \to 0^{+}} f (x) = f (0)$

Therefore, $f$ is continuous at $x = 0$ .

At $x = - 5$ ,

$f (- 5) = 3 \times (- 5) - 5 = - 20$

$\lim_{x \to 5^{+}} f (x) = \lim_{x \to 5} (3 x - 5) = 3 \times (- 5) - 5$

$= - 20$

$∴$ $\lim_{x \to - 5^{+}} f (x) = f (- 5)$

Therefore, $f$ is continuous at $x = - 5$ .

At $x = 3$ , $f (3) = 3 \times (3) - 5 = 4$

$\lim_{x \to 3^{+}} f (x) = \lim_{x \to 3} (3 x - 5) = 3 \times (3) - 5 = 4$

$∴$ $\lim_{x \to 3^{+}} f (x) = f (3)$

Therefore, $f$ is continuous at $x = 3$ .

Q 36 :

Find the continuity of $f (x) = x$ at $x = k$ , where $k$ is any positive value:

$f (x)$ is continuous at $x = k$
$f (x)$ is not continuous at $x = k$
$f (x)$ is continuous at $x = 0$
None of the above

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(1)

Ans. $f (x)$ is continuous at $x = k$

Explanation:

The given function is,

$f (x) = x$

At $x = k$ ,

$f (k) = k$

$\lim_{x \to k} f (x) = \lim_{x \to k} (x) = k$

$∴$ $\lim_{x \to k} f (x) = f (k)$

$∴$ $f (x) is continuous at x = k$

Q 37 :

Find the continuity of $f (x) = \frac{x^{2} - 16}{x + 4}, x \neq - 4$ at $x = k$ , where $k$ is any positive value:

$f (x)$ is not continuous at $x = k$
$f (x)$ is continuous at $x = k$
$f (x)$ is continuous at $x = - 4$
$f (x)$ is not continuous at $x = - 4$

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(2)

Ans. $f (x)$ is continuous at $x = k$

Explanation:

The given function is, $f (x) = \frac{x^{2} - 16}{x + 4}$

At $x = k$ ,

$f (k) = \frac{k^{2} - 16}{k + 4} = \frac{(k + 4) (k - 4)}{k + 4} = k - 4$

$\lim_{x \to k} f (x) = \lim_{x \to k} (\frac{x^{2} - 16}{x + 4}) = \frac{(k + 4) (k - 4)}{k + 4} = k - 4$

$∴$ $\lim_{x \to k} f (x) = f (k)$

$∴$ $f (x) is continuous at x = k$

Q 38 :

Find all points of discontinuous of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

$f$ is discontinuous at $x = 3$
$f$ is continuous at $x = 3$
$f$ is continuous at $x = - 3$
$f$ is discontinuous at $x = - 3$

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(1)

Ans. $f$ is discontinuous at $x = 3$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

$(i) c < 3 (ii) c = 3 (iii) c > 3$

where $c$ be a point on the real line. Then these three cases are valid.

Case (i): $c < 3$

Then, $f (x) = x + 3$

$f (c) = c + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x + 3) = c + 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x < 3$

Case (ii): $c = 3$

Then, $f (x) = x + 3$

$\lim_{x \to c^{-}} f (x) = \lim_{x \to 3^{-}} (x + 3) = 3 + 3 = 6$

$\lim_{x \to c^{+}} f (x) = \lim_{x \to 3^{+}} (x - 3) = 3 - 3 = 0$

The value of the function $f$ at $c = 3$ is $f (3) = 3 + 3 = 6$

Since $\lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x),$ the limit does not exist.

Hence, $f$ is discontinuous at $x = 3$ .

Case (iii): $c > 3$

Then, $f (x) = x - 3$

$f (c) = c - 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x - 3) = c - 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x > 3$ .

The given function $f$ is discontinuous at $x = 3$ .

Q 39 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

$f$ is discontinuous at $x = - 3$
$f$ is discontinuous at $x = 3$
$f$ is continuous at $x = - 3$
$f$ is continuous at $x = 3$

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(2)

Ans. $f$ is discontinuous at $x = 3$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

$(i) c < 3 (ii) c = 3 (iii) c > 3$

where $c$ be a point on the real line. Then these three cases are valid.

Case (i): $c < 3$

Then, $f (x) = x^{2} + 3$

$f (c) = c^{2} + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 3) = c^{2} + 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x < 3$ .

Case (ii): $c = 3$

Then, $f (x) = x^{2} + 3$

$\lim_{x \to c^{-}} f (x) = \lim_{x \to 3^{-}} (x^{2} + 3) = 3^{2} + 3 = 12$

The value of the function at $c = 3$ is $f (3) = 3^{2} + 3 = 12 .$

$\lim_{x \to c^{+}} f (x) = \lim_{x \to 3^{+}} (x^{2} - 3) = 3^{2} - 3 = 6$

Since $\lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x),$ the limit does not exist.

Hence, $f$ is discontinuous at $c = 3$ .

Case (iii): $c > 3$

Then, $f (x) = x^{2} - 3$

$f (c) = c^{2} - 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} - 3) = c^{2} - 3$

$∴$ $\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x > 3$ .

$∴$ The given function $f$ is discontinuous at point $c = 3$ .

Q 40 :

The function $f (x) = 2 - 3 x$ is:

Increasing
Decreasing
Neither increasing nor decreasing
None of the above

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(2)

Ans. Decreasing

Explanation:

Given function,

$f (x) = 2 - 3 x$

Its derivative is,

$f' (x) = - 3 < 0$

Hence, the function is strictly decreasing.

Q 41 :

The function $f (x) =$ ${\begin{matrix} 1, & if x \neq 0 \\ 2, & if x = 0 \end{matrix}$ is not continuous at:

$x = 0$
$x = 1$
$x = - 1$
None of these

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(1)

Ans. $x = 0$

Explanation:

$\lim_{x \to 0^{-}} f (x) = 1 = \lim_{x \to 0^{+}} f (x)$

and $f (0) = 2$

Here, $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \neq f (0)$

$∴$ $f (x)$ is not continuous at $x = 0$ .

Q 42 :

The point of discontinuity of the function

$f (x) =$ ${\begin{matrix} 2 x + 3, & if x \leq 2 \\ 2 x - 3, & if x > 2 \end{matrix}$ is:

$x = 0$
$x = 1$
$x = 2$
None of these

1

2

3

4

(3)

Ans. $x = 2$

Explanation:

$\lim_{x \to 2^{-}} f (x) = 2 (2) + 3 = 7$ and $\lim_{x \to 2^{+}} f (x) = 2 (2) - 3 = 1$

$∴$ $\lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x)$

Hence, the function is discontinuous at $x = 2$

Q 43 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

$f$ is discontinuous at all points of $c$
$f$ is continuous at all points of $c$
$f$ is discontinuous at $x = 0$
$f$ is continuous at $x = 0$

1

2

3

4

(1)

Ans. $f$ is discontinuous at all points of $c$

Explanation:

The given function $f$ is

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

$(i) c < 0 (ii) c > 1 (iii) 0 \leq c \leq 1$

where $c$ be a point on the real line. Then these three cases are valid.

Case (i): $c < 0$

then $f (x) = x^{2} + x + 3$

$f (c) = c^{2} + c + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + x + 3) = c^{2} + c + 3$

$\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $x < 0$ .

Case (ii): $c > 1$

then $f (x) = x^{2} - x + 3$

$\lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} - x + 3) = c^{2} - c + 3$

Therefore, $f$ is continuous at all points $x$ , such that $x > 1$ .

Case (iii): $0 \leq c \leq 1$

then $f (x) = 0$

$f (c) = 0$

$\lim_{x \to c} f (x) = \lim_{x \to c} 0 = 0$

$\lim_{x \to c} f (x) = f (c)$

Therefore, $f$ is continuous at all points $x$ , such that $0 \leq x \leq 1$ .

But, the given function $f$ is discontinuous at all points of $c$ .

Q 44 :

If $f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix},$ then which one of the following is correct?

$f (x)$ is continuous at $x = 0$ for any value of $λ$
$f (x)$ is discontinuous at $x = 0$ for any value of $λ$
$f (x)$ is discontinuous at $x = 1$ for any value of $λ$
None of the above

1

2

3

4

(2)

Ans. $f (x)$ is discontinuous at $x = 0$ for any value of $λ$

Explanation:

$\lim_{x \to 0^{-}} f (x) = 0$ and $\lim_{x \to 0^{+}} f (x) = 1$

Q 45 :

The function $f (x) = c o t x$ is discontinuous on the set:

${x = n π : n \in Z}$
${x = 2 n π : n \in Z}$
${x = (2 n + 1) \frac{π}{2} : n \in Z}$
${x = \frac{n π}{2} : n \in Z}$

1

2

3

4

(1)

Ans. ${x = n π : n \in Z}$

Explanation:

We know that, $f (x) = \cot x$ is continuous in $R - {n π : n \in Z}$

Since, $f (x) = \cot x = \frac{\cos x}{\sin x}$ $[Since, \sin x = 0 at n π, n \in Z]$

Hence, $f (x) = \cot x$ is discontinuous on the set ${x = n π : n \in Z}$

Q 46 :

The function defined by $g (x) = x - [x]$ is discontinuous at:

all rational points
all irrational points
all integral points
None of the above

1

2

3

4

(3)

Ans. all integral points

Explanation:

The function $f (x) = [x]$ is discontinuous at every integral point. Hence, the function $g (x) = x - [x]$ is also discontinuous at every integral point.

Q 47 :

The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals:

- 6
6
5
- 5

1

2

3

4

(2)

Ans. 6

Explanation:

We have, $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$

$f (x)$ is continuous at $x = \frac{π}{2}$

$∴$ $\lim_{x \to \frac{π^{-}}{2}} \frac{k \cos x}{π - 2 x} = 3$

$\Rightarrow \lim_{h \to 0} \frac{k \cos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} = 3$

$\Rightarrow \lim_{h \to 0} \frac{k \sin h}{2 h} = 3$

$∴$ $\frac{k}{2} = 3 \Rightarrow k = 6$

Q 48 :

The number of points at which the function $f (x) = \frac{1}{x - [x]}$ where $[.]$ denotes the greatest integer function, is not continuous, is:

1
2
3
None of these

1

2

3

4

(4)

Ans. None of these

Explanation:

$x - [x] = 0$ when $x$ is an integer, so that $f (x)$ is discontinuous for all $x \in I,$ i.e., $f (x)$ is discontinuous at infinite number of points.

Q 49 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & for - 1 \leq x < 0 \\ 2 x^{2} + 3 x - 2 & for 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then $k$ is equal to:

- 4
- 3
- 2
- 1

1

2

3

4

(3)

Ans. - 2

Explanation:

$LHL = \lim_{x \to 0^{-}} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}$

$= \lim_{x \to 0^{-}} \frac{2 k x}{x (\sqrt{1 + k x + \sqrt{1 - k x}})} = k$

$RHL = \lim_{x \to 0^{+}} (2 x^{2} + 3 x - 2) = - 2$

$f (0) = - 2$

$∵$ It is given that $f (x)$ is continuous at $x = 0$

$∴$ $LHL = RHL = f (0) \Rightarrow k = - 2$

Q 50 :

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

$f (x) + g (x)$
$f (x) - g (x)$
$f (x) \cdot g (x)$
$\frac{g (x)}{f (x)}$

1

2

3

4

(4)

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

Explanation:

We know that, if $f$ and $g$ are continuous functions, then

(1) $f + g$ is continuous.

(2) $f - g$ is continuous.

(3) $f g$ is continuous.

(4) $\frac{f}{g}$ is continuous at those points where $g (x) \neq 0$

Here, $\frac{g (x)}{f (x)} = \frac{\frac{x^{2}}{2} + 1}{2 x} = \frac{x^{2} + 2}{4 x},$ which is discontinuous at $x = 0$ .

Topic Question Set

Q 21 :

Let $[x]$ be the greatest integer $\leq x$ . Then the number of points in the interval $(- 2, 1)$ , where the function $f (x) =$ $| [x] |$ $+ \sqrt{x - [x]}$ is discontinuous, is _______ . [2023]

Q 22 :

Let $f (x) = \sum_{k = 1}^{10} k x^{k}, x \in ℝ .$ If $2 f (2) + f' (2) = 119 {(2)}^{n} + 1,$ then $n$ is equal to _________ . [2023]

Q 23 :

If $f (x) = x^{2} + g' (1) x + g'' (2)$ and $g (x) = f (1) x^{2} + x f' (x) + f'' (x)$ , then the value of $f (4) - g (4)$ is equal to [2023]

Q 24 :

Let $f : ℝ \to ℝ$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in ℝ .$ If $f' (0) = 2$ , then $| f (- 2) |$ is equal to _______ . [2023]

Q 25 :

Let $f : ℝ \to ℝ$ be a twice differentiable function such that $f'' (x) \sin (\frac{x}{2}) + f' (2 x - 2 y) = (\cos x) \sin (y + 2 x) + f (2 x - 2 y)$ for all $x, y \in R$ . If $f (0) = 1$ , then the value of $24 f^{(4)} (\frac{5 π}{3})$ is: [2025]

(2)

Q 26 :

Let $f (x)$ $=$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{4 x^{2} + 4 x - 3}, & x \neq - \frac{3}{2}, \frac{1}{2} \\ b, & x = - \frac{3}{2}, \frac{1}{2} \end{matrix}$

be continuous at $x = - \frac{3}{2}$ . If $f o f (x) = \frac{7}{5}$ , then $x$ is equal to: [2026]

Q 28 :

Let $α, β \in ℝ$ be such that the function $f (x)$ $=$ ${\begin{matrix} 2 α (x^{2} - 2) + 2 β x, & x < 1 \\ (α + 3) x + (α - β), & x \geq 1 \end{matrix}$

be differentiable at all $x \in ℝ$ . Then $34 (α + β)$ is equal to [2026]

Q 29 :

If the function $f (x)$ $= \frac{e^{x} (e^{\tan x - x} - 1) + \log_{e} (\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$ , then the value of $f (0)$ is equal to [2026]

Q 30 :

Consider the following three statements for the function $f : (0, \infty) \to ℝ$ defined by
$f (x) =$ $| \log_{e} x |$ $-$ $| x - 1 |$ $:$

(I) $f$ is differentiable at all $x > 0$ .

(II) $f$ is increasing in (0, 1).

(III) $f$ is decreasing in $(1, \infty)$ .

Then. [2026]

Q 31 :

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $\lim_{t \to x} (\frac{t^{2} y (x) - x^{2} y (t)}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to [2026]

Q 33 :

If $f (x)$ $=$ ${\begin{matrix} \frac{a | x | + x^{2} - 2 (\sin | x |) (\cos | x |)}{x} & , x \neq 0 \\ b & , x = 0 \end{matrix}$

is continuous at $x = 0$ , then $a + b$ is equal to [2026]

Q 34 :

Let $[\cdot]$ denote the greatest integer function, and let $f (x) = \min {\sqrt{2 x}, x^{2}} .$

Let $S = {x \in (- 2, 2) : the function g (x) = | x | [x^{2}] is discontinuous at x} .$ Then $\sum_{x \in S} f (x)$ equals: [2026]

Q 35 :

Check whether the function $f (x) = 3 x - 5$ is continuous at $x = 0$ , $x = - 5$ and at $x = 3$ .

Q 36 :

Find the continuity of $f (x) = x$ at $x = k$ , where $k$ is any positive value:

Q 37 :

Find the continuity of $f (x) = \frac{x^{2} - 16}{x + 4}, x \neq - 4$ at $x = k$ , where $k$ is any positive value:

Q 38 :

Find all points of discontinuous of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

Q 39 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

Q 40 :

The function $f (x) = 2 - 3 x$ is:

Q 41 :

The function $f (x) =$ ${\begin{matrix} 1, & if x \neq 0 \\ 2, & if x = 0 \end{matrix}$ is not continuous at:

Q 42 :

The point of discontinuity of the function

$f (x) =$ ${\begin{matrix} 2 x + 3, & if x \leq 2 \\ 2 x - 3, & if x > 2 \end{matrix}$ is:

Q 43 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

Q 44 :

If $f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix},$ then which one of the following is correct?

Q 45 :

The function $f (x) = c o t x$ is discontinuous on the set:

Q 46 :

The function defined by $g (x) = x - [x]$ is discontinuous at:

Q 47 :

The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals:

Q 48 :

The number of points at which the function $f (x) = \frac{1}{x - [x]}$ where $[.]$ denotes the greatest integer function, is not continuous, is:

Q 49 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & for - 1 \leq x < 0 \\ 2 x^{2} + 3 x - 2 & for 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then $k$ is equal to:

Q 50 :

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

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Topic Question Set

Q 21 : Let [x] be the greatest integer ≤x. Then the number of points in the interval (-2,1), where the function f(x)=|[x]|+x-[x] is discontinuous, is _______ . [2023]

Q 22 : Let f(x)=∑k=110kxk, x∈ℝ. If 2f(2)+f'(2)=119(2)n+1, then n is equal to _________ . [2023]

Q 23 : If f(x)=x2+g'(1)x+g''(2) and g(x)=f(1)x2+xf'(x)+f''(x), then the value of f(4)-g(4) is equal to [2023]

Q 24 : Let f:ℝ→ℝ be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1,∀x,y∈ℝ. If f'(0)=2, then |f(-2)| is equal to _______ . [2023]

Q 25 : Let f:ℝ→ℝ be a twice differentiable function such that f''(x)sin(x2)+f'(2x-2y)=(cosx)sin(y+2x)+f(2x-2y) for all x,y∈R. If f(0)=1, then the value of 24 f(4)(5π3) is: [2025]

(2)

Q 26 : Let f(x)={ax2+2ax+34x2+4x-3,x≠-32,12b ,x=-32,12 be continuous at x=-32. If fof(x)=75, then x is equal to: [2026]

Q 27 : Let f:R→R be a twice differentiable function such that the quadratic equation f(x)m2-2f'(x)m+f''(x)=0 in m, has two equal roots for every x∈R. If f(0)=1,f'(0)=2 and (α, β) is the largest interval in which the function f(logex-x) is increasing, then α+β is equal to [2026]

Q 28 : Let α,β∈ℝ be such that the function f(x)={2α(x2-2)+2βx,x<1(α+3)x+(α-β),x≥1 be differentiable at all x∈ℝ. Then 34(α+β) is equal to [2026]

Q 29 : If the function f(x)=ex(etanx-x-1)+loge(secx+tanx)-xtanx-x is continuous at x=0, then the value of f(0) is equal to [2026]

Q 30 : Consider the following three statements for the function f:(0,∞)→ℝ defined by f(x)=|logex|-|x-1|: (I) f is differentiable at all x>0. (II) f is increasing in (0, 1). (III) f is decreasing in (1,∞). Then. [2026]

Q 31 : Let y=y(x) be a differentiable function in the interval (0,∞) such that y(1)=2, and limt→x(t2y(x)-x2y(t)x-t)=3 for each x>0. Then 2y(2) is equal to [2026]

Q 32 : Let [t] denote the greatest integer less than or equal to t. If the function f(x)={b2sin(π2[π2(cosx+sinx)cosx]),x<0sinx-12sin2xx3,x>0a,x=0 is continuous at x=0, then a2+b2 is equal to [2026]

Q 33 : If f(x)={a|x|+x2-2(sin|x|)(cos|x|)x, x≠0b, x=0 is continuous at x=0, then a+b is equal to [2026]

Q 34 : Let [·] denote the greatest integer function, and let f(x)=min{2x, x2}. Let S={x∈(-2,2): the function g(x)=|x|[x2] is discontinuous at x}. Then ∑x∈Sf(x) equals: [2026]

Q 35 : Check whether the function f(x)=3x−5 is continuous at x=0, x=−5 and at x=3.

Q 36 : Find the continuity of f(x)=x at x=k, where k is any positive value:

Q 37 : Find the continuity of f(x)=x2-16x+4, x≠-4 at x=k, where k is any positive value:

Q 38 : Find all points of discontinuous of f, where f is defined by f(x)={x+3,if x≤3x-3,if x>3

Q 39 : Find all points of discontinuity of f, where f is defined by f(x)={x2+3,if x≤3x2-3,if x>3

Q 40 : The function f(x)=2−3x is:

Q 41 : The function f(x)={1,if x≠02,if x=0 is not continuous at:

Q 42 : The point of discontinuity of the function f(x)={2x+3,if x≤22x-3,if x>2 is:

Q 43 : Find all points of discontinuity of f, where f is defined by f(x)={x2+x+3,if x<30,if 0≤x≤1x2-x+3,if x>1

Q 44 : If f(x)={λ(x2-2x),if x≤04x+1,if x>0, then which one of the following is correct?

Q 45 : The function f(x)=cot x is discontinuous on the set:

Q 46 : The function defined by g(x)=x−[x] is discontinuous at:

Q 47 : The function f(x)={kcosxπ-2x,if x≠π23,if x=π2 is continuous at x=π2, when k equals:

Q 48 : The number of points at which the function f(x)=1x-[x] where [.] denotes the greatest integer function, is not continuous, is:

Q 49 : If f(x)={1+kx-1-kxx,for -1≤x<02x2+3x-2for 0≤x≤1 is continuous at x=0, then k is equal to:

Q 50 : If f(x)=2x and g(x)=x22+1, then which of the following can be discontinuous functions?

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Q 21 :

Let $[x]$ be the greatest integer $\leq x$ . Then the number of points in the interval $(- 2, 1)$ , where the function $f (x) =$ $| [x] |$ $+ \sqrt{x - [x]}$ is discontinuous, is _______ . [2023]

Q 22 :

Let $f (x) = \sum_{k = 1}^{10} k x^{k}, x \in ℝ .$ If $2 f (2) + f' (2) = 119 {(2)}^{n} + 1,$ then $n$ is equal to _________ . [2023]

Q 23 :

If $f (x) = x^{2} + g' (1) x + g'' (2)$ and $g (x) = f (1) x^{2} + x f' (x) + f'' (x)$ , then the value of $f (4) - g (4)$ is equal to [2023]

Q 24 :

Let $f : ℝ \to ℝ$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in ℝ .$ If $f' (0) = 2$ , then $| f (- 2) |$ is equal to _______ . [2023]

Q 25 :

Let $f : ℝ \to ℝ$ be a twice differentiable function such that $f'' (x) \sin (\frac{x}{2}) + f' (2 x - 2 y) = (\cos x) \sin (y + 2 x) + f (2 x - 2 y)$ for all $x, y \in R$ . If $f (0) = 1$ , then the value of $24 f^{(4)} (\frac{5 π}{3})$ is: [2025]

Q 26 :

Let $f (x)$ $=$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{4 x^{2} + 4 x - 3}, & x \neq - \frac{3}{2}, \frac{1}{2} \\ b, & x = - \frac{3}{2}, \frac{1}{2} \end{matrix}$

be continuous at $x = - \frac{3}{2}$ . If $f o f (x) = \frac{7}{5}$ , then $x$ is equal to: [2026]

Q 28 :

Let $α, β \in ℝ$ be such that the function $f (x)$ $=$ ${\begin{matrix} 2 α (x^{2} - 2) + 2 β x, & x < 1 \\ (α + 3) x + (α - β), & x \geq 1 \end{matrix}$

be differentiable at all $x \in ℝ$ . Then $34 (α + β)$ is equal to [2026]

Q 29 :

If the function $f (x)$ $= \frac{e^{x} (e^{\tan x - x} - 1) + \log_{e} (\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$ , then the value of $f (0)$ is equal to [2026]

Q 30 :

Consider the following three statements for the function $f : (0, \infty) \to ℝ$ defined by
$f (x) =$ $| \log_{e} x |$ $-$ $| x - 1 |$ $:$

(I) $f$ is differentiable at all $x > 0$ .

(II) $f$ is increasing in (0, 1).

(III) $f$ is decreasing in $(1, \infty)$ .

Then. [2026]

Q 31 :

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $\lim_{t \to x} (\frac{t^{2} y (x) - x^{2} y (t)}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to [2026]

Q 33 :

If $f (x)$ $=$ ${\begin{matrix} \frac{a | x | + x^{2} - 2 (\sin | x |) (\cos | x |)}{x} & , x \neq 0 \\ b & , x = 0 \end{matrix}$

is continuous at $x = 0$ , then $a + b$ is equal to [2026]

Q 34 :

Let $[\cdot]$ denote the greatest integer function, and let $f (x) = \min {\sqrt{2 x}, x^{2}} .$

Let $S = {x \in (- 2, 2) : the function g (x) = | x | [x^{2}] is discontinuous at x} .$ Then $\sum_{x \in S} f (x)$ equals: [2026]

Q 35 :

Check whether the function $f (x) = 3 x - 5$ is continuous at $x = 0$ , $x = - 5$ and at $x = 3$ .

Q 36 :

Find the continuity of $f (x) = x$ at $x = k$ , where $k$ is any positive value:

Q 37 :

Find the continuity of $f (x) = \frac{x^{2} - 16}{x + 4}, x \neq - 4$ at $x = k$ , where $k$ is any positive value:

Q 38 :

Find all points of discontinuous of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x + 3, & if x \leq 3 \\ x - 3, & if x > 3 \end{matrix}$

Q 39 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + 3, & if x \leq 3 \\ x^{2} - 3, & if x > 3 \end{matrix}$

Q 40 :

The function $f (x) = 2 - 3 x$ is:

Q 41 :

The function $f (x) =$ ${\begin{matrix} 1, & if x \neq 0 \\ 2, & if x = 0 \end{matrix}$ is not continuous at:

Q 42 :

The point of discontinuity of the function

$f (x) =$ ${\begin{matrix} 2 x + 3, & if x \leq 2 \\ 2 x - 3, & if x > 2 \end{matrix}$ is:

Q 43 :

Find all points of discontinuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} x^{2} + x + 3, & if x < 3 \\ 0, & if 0 \leq x \leq 1 \\ x^{2} - x + 3, & if x > 1 \end{matrix}$

Q 44 :

If $f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix},$ then which one of the following is correct?

Q 45 :

The function $f (x) = c o t x$ is discontinuous on the set:

Q 46 :

The function defined by $g (x) = x - [x]$ is discontinuous at:

Q 47 :

The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals:

Q 48 :

The number of points at which the function $f (x) = \frac{1}{x - [x]}$ where $[.]$ denotes the greatest integer function, is not continuous, is:

Q 49 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & for - 1 \leq x < 0 \\ 2 x^{2} + 3 x - 2 & for 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then $k$ is equal to:

Q 50 :

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