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

Q 11 :

Consider the function $f : (0, \infty) \to R$ defined by $f (x) = e^{- | \log_{e} x |}$ . If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is [2024]

1
0
2
3

1

2

3

4

(1)

Given, $f (x) = e^{- | \log_{e} x |}$

$\Rightarrow$ $f (x) =$ ${\begin{matrix} 1 / e^{- \log_{e} x}, & 0 < x < 1 \\ 1 / e^{\log_{e} x}, & x \geq 1 \end{matrix} = {\begin{matrix} x, & 0 < x < 1 \\ 1 / x, & x \geq 1 \end{matrix}$

$∴$ $f (x)$ is continuous everywhere for $x > 0$ but not differentiable at $x = 1 .$

Thus, $m = 0, n = 1$

Hence, $m + n = 1$

Q 12 :

Let $f (x) = 2^{x} - x^{2}, x \in R .$ If $m$ and $n$ are respectively the number of points at which the curves $y = f (x)$ and $y = f' (x)$ intersect the $x -$ axis, then the value of $m + n$ is _________. [2024]

(5)

We have, $f (x) = 2^{x} - x^{2}$

By graph, since $f (x)$ intersects the $x$ -axis at 3 points. So, number of solutions of $f (x) = 3$

$∴$ $m = 3$

Also, $f' (x) = 2^{x} \log 2 - 2 x$

By graph, since $f' (x)$ intersects $x$ -axis at 2 points. So, number of solutions of $f' (x) = 2$

$∴$ $n = 2$

Thus, $m + n = 3 + 2 = 5$

Q 13 :

Let $f : R \to R$ be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y $\in$ R.

If $f' (0) = \frac{1}{2}$ , then the value of $24 f'' (\frac{5 π}{3})$ is : [2025]

–3
–2
3
2

1

2

3

4

(1)

We have,

(sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y))

$\Rightarrow$ f(2x + 2y) sin (x – y) = f(2x – 2y) sin (x + y)

$\Rightarrow \frac{f (2 x + 2 y)}{\sin (x + y)} = \frac{f (2 x - 2 y)}{\sin (x - y)}$

Put 2x + 2y = m and 2x – 2y = n, we get

$\frac{f (m)}{\sin (\frac{m}{2})} = \frac{f (n)}{\sin (\frac{n}{2})} = K$

$\Rightarrow f (m) = K \sin (\frac{m}{2}) and f (n) = K \sin (\frac{n}{2})$

$∴$ $f (x) = K \sin (\frac{x}{2}) \Rightarrow f' (x) = \frac{K}{2} \cos (\frac{x}{2})$

Now, $f' (0) = \frac{1}{2} \Rightarrow \frac{1}{2} = \frac{K}{2} \Rightarrow K = 1$

$∴ f' (x) = \frac{1}{2} \cos \frac{x}{2} \Rightarrow f'' (x) = - \frac{1}{4} \sin \frac{x}{2}$

$∴ 24 f'' (\frac{5 π}{3}) = 24 (- \frac{1}{4} \sin (\frac{5 π}{6})) = \frac{- 24}{8} = - 3$ .

Q 14 :

Let $f (x)$ $= {\begin{matrix} {(1 + a x)}^{1 / x}, & x < 0 \\ 1 + b, & x = 0 \\ \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} & x > 0 \end{matrix} \begin{matrix} \end{matrix}$ be continuous at x = 0. Then $e^{a} b c$ is equal to : [2025]

48
72
36
64

1

2

3

4

(1)

L.H.L. = $f (0^{-}) = e^{\lim_{x \to 0} \frac{1}{x} \log (1 + a x)} = e^{\lim_{x \to 0} \frac{a}{1 + a x}} = e^{a}$

R.H.L. = $f (0^{+}) = \lim_{x \to 0} \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} = \frac{0}{c^{1 / 3} - 2}$

Since, f(x) is continuous at x = 0

$∴$ Right hand limit exists

$\Rightarrow c^{1 / 3} - 2 = 0 \Rightarrow c^{1 / 3} = 2 \Rightarrow c = 8$ ... (i)

Now, $f (0^{+}) = \frac{2 - 2}{2 - 2}$ $(\frac{0}{0} form)$

$= \frac{\frac{1}{2 \sqrt{0 + 4}}}{\frac{1}{3} {(0 + c)}^{- 2 / 3}}$ [Using L'Hospital's Rule]

$= \frac{3}{4} c^{2 / 3} = \frac{3}{4} {(8)}^{2 / 3} = 3$ [From (i)]

Now, $f (0) = f (0^{-}) = f (0^{+})$

$\Rightarrow 1 + b = e^{a} = 3 \Rightarrow b = 2 and e^{a} = 3$

$∴ e^{a} b c = 3 \times 2 \times 8 = 48$ .

Q 15 :

If $y (x)$ $=$ $| \begin{matrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{matrix} |$ $, x \in ℝ$ , then $\frac{d^{2} y}{d x^{2}} + y$ is equal to [2025]

–1
27
1
28

1

2

3

4

(1)

We have,

$y (x)$ $=$ $| \begin{matrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{matrix} |$

$\Rightarrow$ y(x) = sin x (28 – 27) – cos x (27 – 27) + (sin x + cos x + 1)(27 – 28)

y(x) = – cos x – 1

On differentiate w.r.t. x, we get

$\frac{d y}{d x} = \sin x \Rightarrow \frac{d^{2} y}{d x^{2}} = \cos x$

$∴ \frac{d^{2} y}{d x^{2}} + y = \cos x - 1 - \cos x = - 1$ .

Q 16 :

Let $f : R \to R$ be a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x for all x $\in$ R. If $\lim_{n \to \infty} {f (x) - f (\frac{x}{2^{n}})} = G (x)$ , then $\sum_{r = 1}^{10} G (r^{2})$ is equal to [2025]

540
420
385
215

1

2

3

4

(3)

We have, f(2x) – f(x) = x

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

$\Rightarrow f (\frac{x}{2}) - f (\frac{x}{4}) = \frac{x}{4}$

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

On adding all the above statements, we get

$f (2 x) - f (\frac{x}{2^{n}}) = x + \frac{x}{2} + \frac{x}{4} + . . . + \frac{x}{2^{n}}$

$= x {\frac{1 - {(\frac{1}{2})}^{n + 1}}{1 - \frac{1}{2}}} = 2 x [1 - {(\frac{1}{2})}^{n + 1}]$

$\Rightarrow f (x) + x - f (\frac{x}{2^{n}})$

$= 2 x [1 - {(\frac{1}{2})}^{n + 1}]$

$\Rightarrow \lim_{n \to \infty} [f (x) - f (\frac{x}{2^{n}})]$

$= \lim_{n \to \infty} [2 x (1 - {(\frac{1}{2})}^{n + 1}) - x]$

$\Rightarrow G (x) = x \Rightarrow \sum_{r = 1}^{10} r^{2} = 385$ .

Q 17 :

Let f(x) be a real differentiable function such that f(0) = 1 and $f (x + y) = f (x) f' (y) + f' (x) f (y)$ for all x, y $\in$ R. Then $\sum_{n = 1}^{100} \log_{e} f (n)$ is equal to : [2025]

2406
5220
2525
2384

1

2

3

4

(3)

$f (x + y) = f (x) f' (y) + f' (x) f (y)$

When x = 0, y = 0, we have

$f (0) = f (0) f' (0) + f' (0) f (0)$

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

When y = 0, $f (x) = f (x) f' (0) + f (0) f' (x)$

$\Rightarrow \frac{1}{2} f (x) = f' (x)$ [ $∵$ f(0) = 1]

Integrating both sides, we get $f (x) = e^{x / 2} \cdot C$

Now, $f (0) = 1 \Rightarrow f (x) = e^{x / 2} \Rightarrow \log_{e} f (x) = \frac{x}{2}$

$∴ \sum_{n = 1}^{100} \log_{e} f (n)$

$= \sum_{n = 1}^{100} \frac{n}{2} = \frac{1}{2} \frac{(100) (101)}{2} = 2525$ .

Q 18 :

If the function $f (x)$ $=$ ${\begin{matrix} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}), & x > 0 \end{matrix}$ is continuous at x = 0, then $k_{1}^{2} + k_{2}^{2}$ is equal to [2025]

20
5
10
8

1

2

3

4

(3)

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x} = 4$

$\Rightarrow 2 (k_{1} + 1) + 2 (k_{2} - 1) = 4$

$\Rightarrow 2 [k_{1} + 1 + k_{2} - 1] = 4$

$\Rightarrow k_{1} + k_{2} = 2$ ... (i)

$\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}) = 4$

$\lim_{x \to 0^{+}} \frac{1}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}) = 2$ ;

$\lim_{x \to 0^{+}} \frac{1}{x} \log_{e} (1 + \frac{(k_{1} - k_{2}) x}{2 + k_{2} x}) = 2$

$\Rightarrow \frac{k_{1} - k_{2}}{2} = 2 \Rightarrow k_{1} - k_{2} = 4$ ... (ii)

Adding (i) and (ii), we get

$k_{1} = 3 \Rightarrow k_{2} = - 1$

$∴ k_{1}^{2} + k_{2}^{2} = {(3)}^{2} + {(- 1)}^{2} = 9 + 1 = 10$ .

Q 19 :

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f (x) = [x] + | x - 2 |, - 2 < x < 3$ , is not continuous and not differentiable Then m + n is equal to : [2025]

7
8
6
9

1

2

3

4

(2)

Given : $f (x) = [x] + | x - 2 |, - 2 < x < 3$

The function can also be written as follows:

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

Here, function f(x) is not continuous at x = –1, 0, 1 and 2.

Hence, function f(x) is not differentiable at x = –1, 0, 1 and 2.

So, we have m = n = 4.

$∴$ m + n = 4 + 4 = 8.

Q 20 :

Let the function $f (x) = (x^{2} - 1) | x^{2} - a x + 2 | + \cos | x |$ be not differentiable at the two points $x = α = 2$ and $x = β$ . Then the distance of the point $(α, β)$ from the line 12x + 5y +10 = 0 is equal to : [2025]

4
3
2
5

1

2

3

4

(*)

We have,

$f (x) = (x^{2} - 1) | x^{2} - a x + 2 | + \cos | x |$

Now, cos |x| is always differentiable

So, we will check for $| x^{2} - a x + 2 |$ and it is not differentiable at its roots.

It is given that $x = α = 2$

$\Rightarrow {(2)}^{2} - a (2) + 2 = 0 \Rightarrow 4 - 2 a + 2 = 0 \Rightarrow a = 3$

The other root of $x^{2} - 3 x + 2 is x = 1$ .

Note: There is error in question, f(x) is differentiable at x = 1.

Q 21 :

Let m and n be number of points at which the function $f (x) = \max {x, x^{3}, x^{5}, . . ., x^{21}}, x \in R$ , is not differentiable and not continuous, respectively, Then m + n is equal to __________. [2025]

(3)

Here, $f (x)$ $= {\begin{matrix} x, & x < - 1 \\ x^{21}, & - 1 \leq x < 0 \\ x, & 0 \leq x < 1 \\ x^{21}, & x \geq 1 \end{matrix}$ is continuous everywhere.

Then, n = 0

$\Rightarrow f' (x)$ $= {\begin{matrix} 1, & x < - 1 \\ 21 x^{20}, & - 1 \leq x < 0 \\ 1, & 0 \leq x < 1 \\ 21 x^{20}, & x \geq 1 \end{matrix}$ is not differentiable at

$x = - 1, 0, 1 \Rightarrow m = 3$

So, m + n = 3.

Q 22 :

The number of points of discontinuity of the function $f (x)$ $= [\frac{x^{2}}{2}] - [\sqrt{x}], x \in [0, 4]$ , where $[\cdot]$ denotes the greatest integer function, is __________. [2025]

(8)

Values of x, where $[\frac{x^{2}}{2}]$ may be discontinuous on $x \in [0, 4]$ are $\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, 2 \sqrt{3}, \sqrt{14}, 4$

And for $[\sqrt{x}]$ , values of x are = 1, 4

On checking for continuity at these points, we get f(x) is discontinuous at $x = 1, \sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, 2 \sqrt{3}, \sqrt{14}$ and continuous at x = 4.

Hence, f(x) is discontinuous for 8 values of $x \in [0, 4]$ .

Q 23 :

If the function $f (x) = \frac{\tan (\tan x) - \sin (\sin x)}{\tan x - \sin x}$ is continuous at x = 0, then f(0) is equal to ________. [2025]

(2)

$f (0) = \lim_{x \to 0} \frac{\tan (\tan x) - \sin (\sin x) - \tan x + \tan x - \sin x + \sin x}{\tan x - \sin x}$

$= \lim_{x \to 0} \frac{\frac{\tan (\tan x) - \tan x}{\tan^{3} x} \times \frac{\tan^{3} x}{x^{3}} + \frac{\tan x - \sin x}{x^{3}} + \frac{\sin x - \sin (\sin x)}{\sin^{3} x} \times \frac{\sin^{3} x}{x^{3}}}{\frac{\tan x - \sin x}{x^{3}}}$

$= 1 + \frac{(\frac{1}{3} + \frac{1}{6})}{(\frac{1}{3} + \frac{1}{6})} = 2$ . [From L'Hospital's Rule]

Q 24 :

Let $f (x)$ $= {\begin{matrix} 3 x, & x < 0 \\ \min {1 + x + [x], x + 2 [x]}, & 0 \leq x \leq 2 \\ 5, & x > 2 \end{matrix}$ , where $[\cdot]$ denotes greatest integer function. If $α$ and $β$ are the number of points, where f is not continuous and is not differentiable, respectively, then $α + β$ equals __________. [2025]

(5)

$f (x)$ $= {\begin{matrix} 3 x, & x < 0 \\ \min {1 + x + [x], x + 2 [x]}, & 0 \leq x \leq 2 \\ 5, & x > 2 \end{matrix}$

$\Rightarrow$ $f (x)$ $= {\begin{matrix} 3 x, & x < 0 \\ x, & 0 \leq x < 1 \\ x + 2, & 1 \leq x < 2 \\ 5, & x > 2 \end{matrix}$

$∴$ By graph, we have f(x) is not continuous at $x \in {1, 2} \Rightarrow α = 2$

f(x) is not differentiable at $x \in {0, 1, 2} \Rightarrow β = 3$

$∴ α + β = 5$ .

Q 25 :

$If 2 x^{y} + 3 y^{x} = 20, then \frac{d y}{d x} at (2, 2) is equal to$ [2023]

$- (\frac{3 + \log_{e} 8}{2 + \log_{e} 4})$
$- (\frac{3 + \log_{e} 16}{4 + \log_{e} 8})$
$- (\frac{3 + \log_{e} 4}{2 + \log_{e} 8})$
$- (\frac{2 + \log_{e} 8}{3 + \log_{e} 4})$

1

2

3

4

(4)

Q 26 :

$Let f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}, x \in [0, π] - {\frac{π}{4}} .$ Then $f (\frac{7 π}{12}) f'' (\frac{7 π}{12})$ is equal to [2023]

$\frac{- 1}{3 \sqrt{3}}$
$\frac{2}{9}$
$\frac{2}{3 \sqrt{3}}$
$\frac{- 2}{3}$

1

2

3

4

(2)

$Given f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$

$\Rightarrow$ $f (x) = \frac{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x - 1}{\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x}$

$= \frac{\sin (\frac{π}{4}) \cdot \sin x + \cos (\frac{π}{4}) \cdot \cos x - 1}{\cos \frac{π}{4} \cdot \sin x - \sin \frac{π}{4} \cdot \cos x}$

$\Rightarrow f (x) = \frac{\cos (x - \frac{π}{4}) - 1}{\sin (x - \frac{π}{4})} = \frac{- 2 \sin^{2} (\frac{x}{2} - \frac{π}{8})}{2 \sin (\frac{x}{2} - \frac{π}{8}) \cdot \cos (\frac{x}{2} - \frac{π}{8})}$

$\Rightarrow f (x) = - \tan (\frac{x}{2} - \frac{π}{8}) \Rightarrow f' (x) = - \frac{1}{2} \sec^{2} (\frac{x}{2} - \frac{π}{8})$

$and f'' (x) = - \frac{1}{2} \times 2 \sec (\frac{x}{2} - \frac{π}{8}) \cdot \sec (\frac{x}{2} - \frac{π}{8}) \cdot \tan (\frac{x}{2} - \frac{π}{8}) \times \frac{1}{2}$

$\Rightarrow f'' (x) = - \frac{1}{2} \sec^{2} (\frac{x}{2} - \frac{π}{8}) \cdot \tan (\frac{x}{2} - \frac{π}{8})$

$∴$ $f'' (\frac{7 π}{12}) = - \frac{1}{2} \sec^{2} (\frac{7 π}{24} - \frac{π}{8}) \cdot \tan (\frac{7 π}{24} - \frac{π}{8})$

$= - \frac{1}{2} \sec^{2} (\frac{π}{6}) \cdot \tan (\frac{π}{6}) = - \frac{1}{2} \times \frac{4}{3} \cdot \frac{1}{\sqrt{3}} = - \frac{2}{3 \sqrt{3}}$

$Also, f (\frac{7 π}{12}) = - \tan (\frac{π}{6}) = - \frac{1}{\sqrt{3}}$

$∴$ $f (\frac{7 π}{12}) \cdot f'' (\frac{7 π}{12}) = \frac{- 2}{3 \sqrt{3}} \times \frac{- 1}{\sqrt{3}} = \frac{2}{9}$

Q 27 :

Let $f (x) = [x^{2} - x] +$ $| - x + [x] |$ , where $x \in ℝ$ and $[t]$ denotes the greatest integer less than or equal to t. Then, $f$ is: [2023]

continuous at $x = 0$ , but not continuous at $x = 1$
continuous at $x = 0$ and $x = 1$
not continuous at $x = 0$ and $x = 1$
continuous at $x = 1$ , but not continuous at $x = 0$

1

2

3

4

(4)

We have, $f (x) = [x^{2} - x] +$ $| - x + [x] |$

$= [x (x - 1)] +$ $| - x + x - {x} |$

$\Rightarrow f (x) = [x (x - 1)] + {x}$

$f (0^{+}) = - 1 + 0 = - 1$	$R.H.L. = f (1^{+}) = 0 + 0 = 0$
$f (0) = 0$	$f (1) = 0$
	$L.H.L. = f (1^{-}) = - 1 + 1 = 0$

$∴$ $f (x) is continuous at x = 1, and discontinuous at x = 0 .$

Q 28 :

$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

continuous everywhere but not differentiable exactly at one point
not continuous at $x = - 1$
continuous everywhere but not differentiable at $x = 1$
differentiable everywhere

1

2

3

4

(1)

We have,

$f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$

$f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ x - 1, & x \geq 1 \\ 1 - x, & 0 \leq x < 1 \end{matrix}$ $∴$ $(g o f) (x) =$ ${\begin{matrix} x + 2, & x < - 1 \\ 1, & x \geq - 1 \end{matrix}$

$Hence, (g o f) (x) is not differentiable at x = - 1, but it is continuous everywhere.$

Q 29 :

For the differentiable function $f : ℝ - {0} \to ℝ, let 3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10, then$ $| f (3) + f' (\frac{1}{4}) |$ is equal to [2023]

13
7
$\frac{33}{5}$
$\frac{29}{5}$

1

2

3

4

(1)

Given, $3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10 ...(i)$

Put $x = \frac{1}{x}$ in (i), we get $3 f (\frac{1}{x}) + 2 f (x) = x - 10 ...(ii)$

Multiply (i) by 3 and (ii) by 2, then (i) − (ii):

$5 f (x) = \frac{3}{x} - 2 x - 10 \Rightarrow f (x) = \frac{1}{5} (\frac{3}{x} - 2 x - 10)$

Differentiate it w.r.t. $x,$ $f' (x) = \frac{1}{5} (\frac{- 3}{x^{2}} - 2)$

$∴$ $| f (3) + f' (\frac{1}{4}) |$ $=$ $| \frac{1}{5} (\frac{3}{3} - 2 (3) - 10) + \frac{1}{5} (- 48 - 2) |$

$=$ $| \frac{1}{5} (1 - 6 - 10 - 50) |$ $=$ $| \frac{1}{5} (- 65) |$ $= 13$

Q 30 :

Let $[x]$ denote the greatest integer function and $f (x) = \max {1 + x + [x], 2 + x, x + 2 [x]}, 0 \leq x \leq 2$ . Let $m$ be the number of points in $[0, 2]$ where $f$ is not continuous and $n$ be the number of points in (0, 2) where $f$ is not differentiable. Then ${(m + n)}^{2} + 2$ is equal to [2023]

2
3
6
11

1

2

3

4

(2)

In [0,1]

$f (x) = \max {1 + x, 2 + x, x} = 2 + x$

In (1, 2)

$f (x) = \max {1 + x + 1, 2 + x, x + 2} = 2 + x$

At $x$ = 2

$f (x) = \max {1 + x + 2, 2 + x, x + (2 \times 2)}$

$= \max {x + 3, x + 2, x + 4}$

$= x + 4$

In [0, 2], $f (x)$ is not continuous at $x$ = 2

In (0, 2), $f (x)$ is not differentiable function.

$∴$ $m = 1, n = 0$

$So, {(m + n)}^{2} + 2 = {(1 + 0)}^{2} + 2 = 1 + 2 = 3$

Q 31 :

$If y (x) = x^{x}, x > 0, then y'' (2) - 2 y' (2) is equal to$ [2023]

$4 {(\log_{e} 2)}^{2} - 2$
$8 \log_{e} 2 - 2$
$4 {(\log_{e} 2)}^{2} + 2$
$4 \log_{e} 2 + 2$

1

2

3

4

(1)

Given, $y = x^{x}$

Taking log on both sides

$\log y = x \log_{e} x,$ $y' = x^{x} (1 + \log_{e} x)$

$y'' = x^{x} {(1 + \log_{e} x)}^{2} + x^{x} \cdot \frac{1}{x}$

Now, find $y'' (2)$ and $y' (2)$

$y'' (2) = 4 {(1 + \log_{e} 2)}^{2} + 2$

$y' (2) = 4 (1 + \log_{e} 2)$

Now, $y'' (2) - 2 y' (2) = 4 {(1 + \log_{e} 2)}^{2} + 2 - 2 [4 (1 + \log_{e} 2)]$

$= 4 {(1 + \log_{e} 2)}^{2} + 2 - 8 (1 + \log_{e} 2)$

$= 4 (1 + \log_{e} 2) [1 + \log_{e} 2 - 2] + 2$

$= 4 {(\log_{e} 2)}^{2} - 1 + 2 = 4 {(\log_{e} 2)}^{2} - 2$

Q 32 :

Let $f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$ Then at $x$ = 0 [2023]

$f$ is continuous but not differentiable
$f$ is continuous but $f'$ is not continuous
$f'$ is continuous but not differentiable
$f$ and $f'$ both are continuous

1

2

3

4

(2)

$f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$

LHD = $\lim_{h \to 0} \frac{f (0 - h) - (0)}{- h} = \lim_{h \to 0^{-}} - \frac{h^{2} \sin (1 / h)}{- h} = 0$

RHD = $\lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0^{+}} \frac{h^{2} \sin (1 / h)}{h} = 0$

$\Rightarrow f (x) is continuous and differentiable at x = 0 .$

Now, $f' (x) = {\begin{matrix} 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$

$\lim_{x \to 0} f' (x) = \lim_{x \to 0} [2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})] = 0 - [- 1, 1] \neq 0$

$f' (0) = 0$

$\Rightarrow f' (x) is discontinuous at x = 0 .$

Q 33 :

If $f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3), x \in R,$ then [2023]

$2 f (0) - f (1) + f (3) = f (2)$
$f (3) - f (2) = f (1)$
$3 f (1) + f (2) = f (3)$
$f (1) + f (2) + f (3) = f (0)$

1

2

3

4

(1)

Given, $f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3)$

$f' (x) = 3 x^{2} - 2 x f' (1) + f'' (2)$ ...(i)

$f'' (x) = 6 x - 2 f' (1)$ ...(ii)

$f''' (x) = 6 \Rightarrow f''' (3) = 6$ ...(iii)

From (iii),

$f'' (2) = 12 - 2 f' (1)$ ...(iv)

From (ii),

$f' (1) = 3 {(1)}^{2} - 2 f' (1) + f'' (2)$

$\Rightarrow f'' (2) = 3 f' (1) - 3$ ...(v)

$\Rightarrow$ $12 - 2 f' (1) = 3 f' (1) - 3$

$\Rightarrow f' (1) = 3;$ $f'' (2) = 12 - 6 = 6$ and $f' (1) = 3$

$f (x) = x^{3} - 3 x^{2} + 6 x - 6$ $\Rightarrow f (0) = - 6$

$f (1) = - 2, f (2) = 2, f (3) = 12$

$∴$ $2 f (0) - f (1) + f (3) = 2 = f (2)$

Q 34 :

Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y' - y''$ at $x = - 1$ is equal to [2023]

976
944
496
464

1

2

3

4

(3)

Given, $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$

$= \frac{1 - x^{32}}{1 - x} = 1 + x + x^{2} + x^{3} + \dots + x^{31}$

$y' (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots + 31 x^{30}$

$y' (- 1) = 1 - 2 + 3 - 4 + \dots - 30 + 31$

$= - 1 - 1 - 1 \dots (upto 15 times) + 31 = 16$

$y'' (x) = 2 + 6 x + 12 x^{2} + 20 x^{3} + \dots + 31 \times 30 x^{29}$

$y'' (- 1) = 2 - 6 + 12 - 20 + \dots - 31 \times 30$

$= - 4 - 8 - 12 - \dots (15 terms) = - 4 \times \frac{15 \times 16}{2} = - 480$

$y' - y'' = 16 + 480 = 496$

Q 35 :

If the function

$f (x) =$ ${\begin{matrix} (1 + | \cos x |) \frac{λ}{| \cos x |}, & 0 < x < \frac{π}{2} \\ μ, & x = \frac{π}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{π}{2} < x < π \end{matrix}$

is continuous at $x = \frac{π}{2}$ , then $9 λ + 6 \log_{e} μ + μ^{6} - e^{6 λ}$ is equal to [2023]

$2 e^{4} + 8$
11
8
10

1

2

3

4

(4)

Given question is incorrect. It should be $e^{(\frac{\cot 6 x}{\cot 4 x})}$ in place of $\frac{\cot 6 x}{e^{\cot 4 x}} .$

$f (x) =$ ${\begin{matrix} \frac{(1 + | \cos x |) λ}{| \cos x |}, & 0 < x < \frac{π}{2} \\ μ, & x = \frac{π}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{π}{2} < x < π \end{matrix}$

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

$R.H.L. at x = \frac{π}{2}$

$\lim_{x \to π / 2^{+}} e^{(\frac{\cot 6 x}{\cot 4 x})} = \lim_{x \to π / 2^{+}} e^{(\frac{\cos 6 x}{\sin 6 x} \times \frac{\sin 4 x}{\cos 4 x})} = e^{2 / 3}$

$L.H.L. = \lim_{x \to π / 2} (1 + | \cos x |) \frac{λ}{| \cos x |} = e^{λ}$

Value of the function, $f (π / 2) = μ$ (given)

By the definition of continuity, $e^{2 / 3} = e^{λ} = μ$ $[∵ R.H.L = L.H.L = value of the function]$

$\Rightarrow λ = \frac{2}{3}, μ = e^{2 / 3}$

$∴$ $9 λ + 6 \log_{e} μ + μ^{6} - e^{6 λ}$

$= 9 \times \frac{2}{3} + 6 \times \frac{2}{3} + {(e^{2 / 3})}^{6} - e^{6 \times \frac{2}{3}} = 6 + 4 + e^{4} - e^{4} = 10$

Q 36 :

Let $a \in ℤ$ and $[t]$ be the greatest integer $\leq t$ . Then the number of points, where the function $f (x) = [a + 13 \sin x], x \in (0, π)$ is not differentiable, is _______. [2023]

(25)

Let $f (x) = [a + 13 \sin x]$

$∵$ $a \in ℤ$ and $x \in (0, π)$ and $p$ is prime.

Total number of points of non-differentiability of $[a + p \sin x] = 2 p - 1$

Here $p = 13$

Thus, total number of non-differentiability points are $2 \times 13 - 1 = 25 .$

Q 37 :

Let $f$ and $g$ be twice differentiable functions on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x,$

$f' (1) = 4 g' (1) - 3 = 9,$

$f (2) = 3 g (2) = 12 .$

Then which of the following is NOT true? [2023]

$| f' (x) - g' (x) |$ $< 6 \Rightarrow - 1 < x < 1$
If $- 1 < x < 2$ , then $| f (x) - g (x) |$ $< 8$
There exists $x_{0} \in (1, 3 / 2)$ such that $f (x_{0}) = g (x_{0})$
$g (- 2) - f (- 2) = 20$

1

2

3

4

(2)

Suppose $f$ and $g$ be twice differentiable on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x$ ...(1)

$f' (1) = 4 g' (1) - 3 = 9$ ...(2)

$f (2) = 3 g (2) = 12$ ...(3)

Firstly, we integrate equation (1).

So, $f' (x) = g' (x) + 6 \cdot \frac{x^{2}}{2} + C$ At $x$ = 1, we have

$f' (1) = g' (1) + 3 + C,$ where C is the constant of integration.

$\Rightarrow 9 = 3 + 3 + C \Rightarrow C = 3 (Using equation (2))$

$∴$ $f' (x) = g' (x) + 3 x^{2} + 3$

Again, by integrating, $f (x) = g (x) + \frac{3 x^{3}}{3} + 3 x + D,$ D is another constant of integration.

At $x$ = 2, we have $f (2) = g (2) + 8 + 6 + D$

$\Rightarrow 12 = 4 + 8 + 6 + D \Rightarrow D = - 6 (from (3))$

So, $f (x) = g (x) + x^{3} + 3 x - 6$

At $x = - 2$ , $\Rightarrow g (- 2) - f (- 2) = 20$

Hence, option (4) is true.

Now, for $- 1 < x < 2$

Let $b (x) = f (x) - g (x) = x^{3} + 3 x - 6 \Rightarrow b' (x) = 3 x^{2} + 3$

So, $b (- 1) < b (x) < b (2) \Rightarrow - 10 < b (x) < 8 \Rightarrow | b (x) | < 10$

So, option (2) is not true.

Now, $b' (x) = f' (x) - g' (x) = 3 x^{2} + 3$

If $| b' (x) | < 6 \Rightarrow$ $| 3 x^{2} + 3 |$ $< 6$

$\Rightarrow 3 x^{2} + 3 < 6 \Rightarrow x^{2} < 1 \Rightarrow - 1 < x < 1$

So, option (1) is also true.

For $x \in (- 1, 1)$ , we have $| f' (x) - g' (x) |$ $< 6$

We have to solve the equation $f (x) - g (x) = 0$

$\Rightarrow x^{3} + 3 x - 6 = 0 \Rightarrow b (x) = x^{3} + 3 x - 6 = 0$

$b (x) = x^{3} + 3 x - 6$

Here we have $b (1) = - ve$ and $b (\frac{3}{2}) = + ve$

$So, there exists x_{0} \in (1, \frac{3}{2}) such that f (x_{0}) = g (x_{0})$

Hence, option (3) is also true.

Q 38 :

If $a_{α}$ is the greatest term in the sequence $a_{n} = \frac{n^{3}}{n^{4} + 147}, n = 1, 2, 3, \dots,$ then $α$ is equal to ________. [2023]

(5)

We have, $a_{n} = \frac{n^{3}}{n^{4} + 147}$

Differentiate both sides w.r.t. $n$ , we get

$a_{n}' = \frac{(n^{4} + 147) 3 n^{2} - n^{3} \cdot 4 n^{3}}{{(n^{4} + 147)}^{2}}$

Put $a_{n}^{'} = 0$

$\Rightarrow \frac{n^{2} {3 (n^{4} + 147) - 4 n^{4}}}{{(n^{4} + 147)}^{2}} = 0$

$\Rightarrow 3 n^{4} + 441 - 4 n^{4} = 0 \Rightarrow n^{4} = 441 \Rightarrow n^{2} = 21 \Rightarrow n = \sqrt{21}$

$\Rightarrow 4 < \sqrt{21} < 5 \Rightarrow a_{5} > a_{4}$

So, $α = 5$

Q 39 :

Let $k$ and $m$ be positive real numbers such that the function $f (x) =$ ${\begin{matrix} 3 x^{2} + k \sqrt{x + 1}, & 0 < x < 1 \\ m x^{2} + k^{2}, & x \geq 1 \end{matrix}$ is differentiable for all $x > 0$ . Then $\frac{8 f' (8)}{f' (\frac{1}{8})}$ is equal to ______ . [2023]

(309)

Q 40 :

Let $f : (- 2, 2) \to R$ be defined by $f (x) =$ ${\begin{matrix} x [x], & - 2 < x < 0 \\ (x - 1) [x], & 0 \leq x < 2 \end{matrix}$ where $[x]$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in (- 2, 2) at which $y =$ $| f (x) |$ is not continuous and not differentiable, then $m + n$ is equal to _____. [2023]

(4)

$f (x) or$ $| f (x) |$ $=$ ${\begin{matrix} - 2 x, & - 2 < x < - 1 \\ - x, & - 1 \leq x < 0 \\ 0, & 0 \leq x \leq 1 \\ x - 1, & 1 \leq x < 2 \end{matrix}$

$Point of discontinuity: m = - 1 i.e., one in number$ $Point of non-differentiability: n = - 1, 0, 1 i.e., 3 in number.$

$∴$ $m + n = 1 + 3 = 4$

Topic Question Set

Q 11 :

Consider the function $f : (0, \infty) \to R$ defined by $f (x) = e^{- | \log_{e} x |}$ . If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is [2024]

Q 12 :

Let $f (x) = 2^{x} - x^{2}, x \in R .$ If $m$ and $n$ are respectively the number of points at which the curves $y = f (x)$ and $y = f' (x)$ intersect the $x -$ axis, then the value of $m + n$ is _________. [2024]

Q 13 :

Let $f : R \to R$ be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y $\in$ R.

If $f' (0) = \frac{1}{2}$ , then the value of $24 f'' (\frac{5 π}{3})$ is : [2025]

Q 14 :

Let $f (x)$ $= {\begin{matrix} {(1 + a x)}^{1 / x}, & x < 0 \\ 1 + b, & x = 0 \\ \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} & x > 0 \end{matrix} \begin{matrix} \end{matrix}$ be continuous at x = 0. Then $e^{a} b c$ is equal to : [2025]

Q 15 :

If $y (x)$ $=$ $| \begin{matrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{matrix} |$ $, x \in ℝ$ , then $\frac{d^{2} y}{d x^{2}} + y$ is equal to [2025]

Q 16 :

Let $f : R \to R$ be a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x for all x $\in$ R. If $\lim_{n \to \infty} {f (x) - f (\frac{x}{2^{n}})} = G (x)$ , then $\sum_{r = 1}^{10} G (r^{2})$ is equal to [2025]

Q 17 :

Let f(x) be a real differentiable function such that f(0) = 1 and $f (x + y) = f (x) f' (y) + f' (x) f (y)$ for all x, y $\in$ R. Then $\sum_{n = 1}^{100} \log_{e} f (n)$ is equal to : [2025]

Q 18 :

If the function $f (x)$ $=$ ${\begin{matrix} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}), & x > 0 \end{matrix}$ is continuous at x = 0, then $k_{1}^{2} + k_{2}^{2}$ is equal to [2025]

Q 19 :

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f (x) = [x] + | x - 2 |, - 2 < x < 3$ , is not continuous and not differentiable Then m + n is equal to : [2025]

Q 20 :

Let the function $f (x) = (x^{2} - 1) | x^{2} - a x + 2 | + \cos | x |$ be not differentiable at the two points $x = α = 2$ and $x = β$ . Then the distance of the point $(α, β)$ from the line 12x + 5y +10 = 0 is equal to : [2025]

Q 21 :

Let m and n be number of points at which the function $f (x) = \max {x, x^{3}, x^{5}, . . ., x^{21}}, x \in R$ , is not differentiable and not continuous, respectively, Then m + n is equal to __________. [2025]

Q 22 :

The number of points of discontinuity of the function $f (x)$ $= [\frac{x^{2}}{2}] - [\sqrt{x}], x \in [0, 4]$ , where $[\cdot]$ denotes the greatest integer function, is __________. [2025]

Q 23 :

If the function $f (x) = \frac{\tan (\tan x) - \sin (\sin x)}{\tan x - \sin x}$ is continuous at x = 0, then f(0) is equal to ________. [2025]

Q 25 :

$If 2 x^{y} + 3 y^{x} = 20, then \frac{d y}{d x} at (2, 2) is equal to$ [2023]

(4)

Q 26 :

$Let f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}, x \in [0, π] - {\frac{π}{4}} .$ Then $f (\frac{7 π}{12}) f'' (\frac{7 π}{12})$ is equal to [2023]

Q 27 :

Let $f (x) = [x^{2} - x] +$ $| - x + [x] |$ , where $x \in ℝ$ and $[t]$ denotes the greatest integer less than or equal to t. Then, $f$ is: [2023]

Q 28 :

$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

Q 29 :

For the differentiable function $f : ℝ - {0} \to ℝ, let 3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10, then$ $| f (3) + f' (\frac{1}{4}) |$ is equal to [2023]

Q 31 :

$If y (x) = x^{x}, x > 0, then y'' (2) - 2 y' (2) is equal to$ [2023]

Q 32 :

Let $f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$ Then at $x$ = 0 [2023]

Q 33 :

If $f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3), x \in R,$ then [2023]

Q 34 :

Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y' - y''$ at $x = - 1$ is equal to [2023]

Q 36 :

Let $a \in ℤ$ and $[t]$ be the greatest integer $\leq t$ . Then the number of points, where the function $f (x) = [a + 13 \sin x], x \in (0, π)$ is not differentiable, is _______. [2023]

Q 37 :

Let $f$ and $g$ be twice differentiable functions on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x,$

$f' (1) = 4 g' (1) - 3 = 9,$

$f (2) = 3 g (2) = 12 .$

Then which of the following is NOT true? [2023]

Q 38 :

If $a_{α}$ is the greatest term in the sequence $a_{n} = \frac{n^{3}}{n^{4} + 147}, n = 1, 2, 3, \dots,$ then $α$ is equal to ________. [2023]

Q 39 :

Let $k$ and $m$ be positive real numbers such that the function $f (x) =$ ${\begin{matrix} 3 x^{2} + k \sqrt{x + 1}, & 0 < x < 1 \\ m x^{2} + k^{2}, & x \geq 1 \end{matrix}$ is differentiable for all $x > 0$ . Then $\frac{8 f' (8)}{f' (\frac{1}{8})}$ is equal to ______ . [2023]

(309)

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : Consider the function f:(0,∞)→R defined by f(x)=e-|logex|. If m and n be respectively the number of points at which f is not continuous and f is not differentiable, then m+n is [2024]

Q 12 : Let f(x)=2x-x2, x∈R. If m and n are respectively the number of points at which the curves y=f(x) and y=f'(x) intersect the x-axis, then the value of m+n is _________. [2024]

Q 13 : Let f:R→R be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y ∈ R. If f'(0)=12, then the value of 24f''(5π3) is : [2025]

Q 14 : Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/2–2(x+c)1/3–2x>0 be continuous at x = 0. Then eabc is equal to : [2025]

Q 15 : If y(x)=|sinxcosxsinx+cosx+1272827111|, x∈ℝ, then d2ydx2+y is equal to [2025]

Q 16 : Let f:R→R be a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x for all x∈ R. If limn→∞{f(x)–f(x2n)}=G(x), then ∑r=110G(r2) is equal to [2025]

Q 17 : Let f(x) be a real differentiable function such that f(0) = 1 and f(x+y)=f(x)f'(y)+f'(x)f(y) for all x, y ∈ R. Then ∑n=1100loge f(n) is equal to : [2025]

Q 18 : If the function f(x)={2x{sin(k1+1)x+sin(k2–1)x},x<04, x=02xloge(2+k1x2+k2x),x>0 is continuous at x = 0, then k12+k22 is equal to [2025]

Q 19 : Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x–2|,–2<x<3, is not continuous and not differentiable Then m + n is equal to : [2025]

Q 20 : Let the function f(x)=(x2–1)|x2–ax+2|+cos|x| be not differentiable at the two points x=α=2 and x=β. Then the distance of the point (α, β) from the line 12x + 5y +10 = 0 is equal to : [2025]

Q 21 : Let m and n be number of points at which the function f(x)=max {x,x3,x5,...,x21}, x∈R, is not differentiable and not continuous, respectively, Then m + n is equal to __________. [2025]

Q 22 : The number of points of discontinuity of the function f(x)=[x22]–[x], x∈[0,4], where [·] denotes the greatest integer function, is __________. [2025]

Q 23 : If the function f(x)=tan(tanx)–sin(sinx)tanx–sinx is continuous at x = 0, then f(0) is equal to ________. [2025]

Q 24 : Let f(x)={3x,x<0min {1+x+[x], x+2[x]},0≤x≤25,x>2, where [·] denotes greatest integer function. If α and β are the number of points, where f is not continuous and is not differentiable, respectively, then α+β equals __________. [2025]

Q 25 : If 2xy+3yx=20, then dydx at (2,2) is equal to [2023]

(4)

Q 26 : Let f(x)=sinx+cosx-2sinx-cosx, x∈[0,π]-{π4}. Then f(7π12)f''(7π12) is equal to [2023]

Q 27 : Let f(x)=[x2-x]+|-x+[x]|, where x∈ℝ and [t] denotes the greatest integer less than or equal to t. Then, f is: [2023]

Q 28 : Let f and g be two functions defined by f(x)={x+1,x<0|x-1|, x≥0 and g(x)={x+1,x<01,x≥0 Then (gof)(x) is [2023]

Q 29 : For the differentiable function f:ℝ-{0}→ℝ, let 3f(x)+2f(1x)=1x-10, then |f(3)+f'(14)| is equal to [2023]

Q 30 : Let [x] denote the greatest integer function and f(x)=max{1+x+[x], 2+x, x+2[x]}, 0≤x≤2. Let m be the number of points in [0,2] where f is not continuous and n be the number of points in (0, 2) where f is not differentiable. Then (m+n)2+2 is equal to [2023]

Q 31 : If y(x)=xx, x>0, then y''(2)-2y'(2) is equal to [2023]

Q 32 : Let f(x)={x2sin(1x),x≠00,x=0 Then at x = 0 [2023]

Q 33 : If f(x)=x3-x2f'(1)+xf''(2)-f'''(3), x∈R, then [2023]

Q 34 : Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y'-y'' at x=-1 is equal to [2023]

Q 35 : If the function f(x)={(1+|cosx|)λ|cosx|,0<x<π2μ,x=π2cot6xecot4x,π2<x<π is continuous at x=π2, then 9λ+6logeμ+μ6-e6λ is equal to [2023]

Q 36 : Let a∈ℤ and [t] be the greatest integer ≤t. Then the number of points, where the function f(x)=[a+13sinx], x∈(0,π) is not differentiable, is _______. [2023]

Q 37 : Let f and g be twice differentiable functions on ℝ such that f''(x)=g''(x)+6x, f'(1)=4g'(1)-3=9, f(2)=3g(2)=12. Then which of the following is NOT true? [2023]

Q 38 : If aα is the greatest term in the sequence an=n3n4+147, n=1,2,3,…, then α is equal to ________. [2023]

Q 39 : Let k and m be positive real numbers such that the function f(x)={3x2+kx+1,0<x<1mx2+k2,x≥1 is differentiable for all x>0. Then 8f'(8)f'(18) is equal to ______ . [2023]

(309)

Q 40 : Let f:(-2,2)→R be defined by f(x)={x[x],-2<x<0(x-1)[x],0≤x<2 where [x] denotes the greatest integer function. If m and n respectively are the number of points in (- 2, 2) at which y=|f(x)| is not continuous and not differentiable, then m+n is equal to _____. [2023]

Want Us to Email you About Special Offers & Updates?

Q 11 :

Consider the function $f : (0, \infty) \to R$ defined by $f (x) = e^{- | \log_{e} x |}$ . If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is [2024]

Q 12 :

Let $f (x) = 2^{x} - x^{2}, x \in R .$ If $m$ and $n$ are respectively the number of points at which the curves $y = f (x)$ and $y = f' (x)$ intersect the $x -$ axis, then the value of $m + n$ is _________. [2024]

Q 13 :

Let $f : R \to R$ be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y $\in$ R.

If $f' (0) = \frac{1}{2}$ , then the value of $24 f'' (\frac{5 π}{3})$ is : [2025]

Q 14 :

Let $f (x)$ $= {\begin{matrix} {(1 + a x)}^{1 / x}, & x < 0 \\ 1 + b, & x = 0 \\ \frac{{(x + 4)}^{1 / 2} - 2}{{(x + c)}^{1 / 3} - 2} & x > 0 \end{matrix} \begin{matrix} \end{matrix}$ be continuous at x = 0. Then $e^{a} b c$ is equal to : [2025]

Q 15 :

If $y (x)$ $=$ $| \begin{matrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{matrix} |$ $, x \in ℝ$ , then $\frac{d^{2} y}{d x^{2}} + y$ is equal to [2025]

Q 16 :

Let $f : R \to R$ be a continuous function satisfying f(0) = 1 and f(2x) – f(x) = x for all x $\in$ R. If $\lim_{n \to \infty} {f (x) - f (\frac{x}{2^{n}})} = G (x)$ , then $\sum_{r = 1}^{10} G (r^{2})$ is equal to [2025]

Q 17 :

Let f(x) be a real differentiable function such that f(0) = 1 and $f (x + y) = f (x) f' (y) + f' (x) f (y)$ for all x, y $\in$ R. Then $\sum_{n = 1}^{100} \log_{e} f (n)$ is equal to : [2025]

Q 18 :

If the function $f (x)$ $=$ ${\begin{matrix} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}), & x > 0 \end{matrix}$ is continuous at x = 0, then $k_{1}^{2} + k_{2}^{2}$ is equal to [2025]

Q 19 :

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f (x) = [x] + | x - 2 |, - 2 < x < 3$ , is not continuous and not differentiable Then m + n is equal to : [2025]

Q 20 :

Let the function $f (x) = (x^{2} - 1) | x^{2} - a x + 2 | + \cos | x |$ be not differentiable at the two points $x = α = 2$ and $x = β$ . Then the distance of the point $(α, β)$ from the line 12x + 5y +10 = 0 is equal to : [2025]

Q 21 :

Let m and n be number of points at which the function $f (x) = \max {x, x^{3}, x^{5}, . . ., x^{21}}, x \in R$ , is not differentiable and not continuous, respectively, Then m + n is equal to __________. [2025]

Q 22 :

The number of points of discontinuity of the function $f (x)$ $= [\frac{x^{2}}{2}] - [\sqrt{x}], x \in [0, 4]$ , where $[\cdot]$ denotes the greatest integer function, is __________. [2025]

Q 23 :

If the function $f (x) = \frac{\tan (\tan x) - \sin (\sin x)}{\tan x - \sin x}$ is continuous at x = 0, then f(0) is equal to ________. [2025]

Q 25 :

$If 2 x^{y} + 3 y^{x} = 20, then \frac{d y}{d x} at (2, 2) is equal to$ [2023]

Q 26 :

$Let f (x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}, x \in [0, π] - {\frac{π}{4}} .$ Then $f (\frac{7 π}{12}) f'' (\frac{7 π}{12})$ is equal to [2023]

Q 27 :

Let $f (x) = [x^{2} - x] +$ $| - x + [x] |$ , where $x \in ℝ$ and $[t]$ denotes the greatest integer less than or equal to t. Then, $f$ is: [2023]

Q 28 :

$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

Q 29 :

For the differentiable function $f : ℝ - {0} \to ℝ, let 3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10, then$ $| f (3) + f' (\frac{1}{4}) |$ is equal to [2023]

Q 31 :

$If y (x) = x^{x}, x > 0, then y'' (2) - 2 y' (2) is equal to$ [2023]

Q 32 :

Let $f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$ Then at $x$ = 0 [2023]

Q 33 :

If $f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3), x \in R,$ then [2023]

Q 34 :

Let $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ . Then $y' - y''$ at $x = - 1$ is equal to [2023]

Q 36 :

Let $a \in ℤ$ and $[t]$ be the greatest integer $\leq t$ . Then the number of points, where the function $f (x) = [a + 13 \sin x], x \in (0, π)$ is not differentiable, is _______. [2023]

Q 37 :

Let $f$ and $g$ be twice differentiable functions on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x,$

$f' (1) = 4 g' (1) - 3 = 9,$

$f (2) = 3 g (2) = 12 .$

Then which of the following is NOT true? [2023]

Q 38 :

If $a_{α}$ is the greatest term in the sequence $a_{n} = \frac{n^{3}}{n^{4} + 147}, n = 1, 2, 3, \dots,$ then $α$ is equal to ________. [2023]

Q 39 :

Let $k$ and $m$ be positive real numbers such that the function $f (x) =$ ${\begin{matrix} 3 x^{2} + k \sqrt{x + 1}, & 0 < x < 1 \\ m x^{2} + k^{2}, & x \geq 1 \end{matrix}$ is differentiable for all $x > 0$ . Then $\frac{8 f' (8)}{f' (\frac{1}{8})}$ is equal to ______ . [2023]