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

Q 61 :

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]

0%

(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 62 :

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]

0%

(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 63 :

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]

0%

(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 64 :

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]

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(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 65 :

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 66 :

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

2

3

4

(3)

Q 67 :

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]

0%

(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 68 :

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

1

2

3

4

(4)

Q 69 :

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]

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

1

2

3

4