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

Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

$\frac{x}{y}$
$\frac{b^{2} x}{a^{2} y}$
$\frac{b x}{a y}$
$\frac{a^{2} y}{b^{2} x}$

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(2)

Ans. $\frac{b^{2} x}{a^{2} y}$

Explanation:

We have

$x = a (t - \frac{1}{t})$

$y = b (t + \frac{1}{t})$

On differentiating both equations with respect to $t$ , we get

$\frac{d x}{d t} = a [\frac{d t}{d t} - \frac{d (t^{- 1})}{d t}]$

$\frac{d y}{d t} = b [\frac{d t}{d t} + \frac{d (t^{- 1})}{d t}]$

$\frac{d x}{d t} = a [1 + \frac{1}{t^{2}}]$ $\dots (i)$

$\frac{d y}{d t} = b [1 - \frac{1}{t^{2}}]$ $\dots (i i)$

On dividing equation (ii) by equation (i), we get

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b [\frac{t^{2} - 1}{t^{2}}]}{a [\frac{t^{2} + 1}{t^{2}}]}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (t^{2} - 1)}{a (t^{2} + 1)} = \frac{b (t - \frac{1}{t})}{a (t + \frac{1}{t})}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (\frac{x}{a})}{a (\frac{y}{b})} = \frac{b^{2} x}{a^{2} y}$

Q 42 :

Which of the following statements are correct?

(A) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere.

(B) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere but not differentiable at $x = 0$ .

(C) Let $f : R - {0} \to R,$ then $f (x) = \frac{1}{x}$ is continuous everywhere.

(D) Let $f : R \to R$ , then $f (x) =$ $| x - 1 |$ $+$ $| x - 2 |$ is continuous everywhere but not differentiable at exactly two points.

(E) If $f : R \to R$ , then $f (x) = \cot x$ is continuous everywhere.

Choose the correct answer from the options given below:

(A) only
(A), (C) only
(A), (B), (C), (D) only
(D), (E) only

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(3)

Ans. (A), (B), (C), (D) only

Explanation:

(A) - A modulus function is continuous everywhere.

(B) - For $f (x) =$ $| x |$ to be differentiable at $x = 0$ ,

$L.H.S. = \lim_{h \to 0} \frac{f (0 - h) - f (0)}{- h}$

$= \lim_{h \to 0} \frac{| 0 - h | - | 0 |}{- h}$

$= \lim_{h \to 0} \frac{h}{- h} = - 1$

$R.H.S. = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$

$= \lim_{h \to 0} \frac{| 0 + h | - | 0 |}{h}$

$= \lim_{h \to 0} \frac{h}{h} = 1$

$L.H.S. \neq R.H.S.$

So, $f (x) =$ $| x |$ is not differentiable at $x = 0$

(C) - $f (x) = \frac{1}{x}$ is continuous everywhere except at $x = 0$

(D) - Yes, and those points are $x = 1$ and $x = 2$

(E) - No, $\cot x$ is not continuous everywhere.

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

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(1)

Ans. 6

Explanation:

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

Then, $\frac{k \cos x}{π - 2 x} = 3$

$\Rightarrow$ $\frac{k \cos (\frac{π}{2})}{π - 2 (\frac{π}{2})} = 3$

Using L' Hospital rule,

$\frac{k (- \sin (\frac{π}{2}))}{- 2} = 3$

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

$\Rightarrow$ $k = 6$

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is:

$\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{2 \sqrt{x}}$
$\sec^{2} (\tan \sqrt{x})$
$\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{x}$
$\sec^{2} (\tan x^{1 / 3})$

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(1)

Ans. $\frac{s e c (t a n \sqrt{x}) t a n (t a n \sqrt{x}) s e c^{2} \sqrt{x}}{2 \sqrt{x}}$

Explanation:

Let $y = \sec (\tan \sqrt{x})$

So, $\frac{d y}{d x} = \frac{d [\sec (\tan \sqrt{x})]}{d (\tan \sqrt{x})} \times \frac{d (\tan \sqrt{x})}{d (\sqrt{x})} \times \frac{d (\sqrt{x})}{d x}$

$= \sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}$

$= \frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{2 \sqrt{x}}$

Q 45 :

If $f (x)$ is a continuous function $\forall x \in R$ and the range of $f (x)$ is $(2, \sqrt{26})$ and $g (x) = [\frac{f (x)}{C}]$ is continuous $\forall x \in R$ , then the least positive integral value of $' C'$ is: (where $[.]$ denotes the greatest integer function)

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(3)

$c > \sqrt{26}$

Q 46 :

Match the following Column-I with Column-II

Column-I Column-II

(A) $\lim_{x \to \infty} {(\frac{x^{2} + 2 x - 1}{2 x^{2} - 3 x - 2})}^{\frac{2 x + 1}{2 x - 1}} =$ (P) $\frac{1}{2}$

(B) $\lim_{x \to 0} \frac{\log_{\sec (x / 2)} \cos x}{\log_{\sec x} \cos \frac{x}{2}} =$ (Q) $2$

(C)
$Let f (x) = \max (\cos x, x, 2 x - 1), where x \geq 0, then$

$number of points of non-differentiability of f (x) is$
(R) $5$

(D)
$If f (x) = [2 + 3 \sin x], 0 < x < π, then number of$

$points at which the function is discontinuous, is$
(S) $16$

A-P, B-S, C-Q, D-R
A-S, B-Q, C-P, D-R
A-P, B-Q, C-S, D-R
A-P, B-S, C-R, D-Q

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(1)

$A)$ $\lim_{x \to \infty} {(\frac{x^{2} + 2 x - 1}{2 x^{2} - 3 x - 2})}^{\frac{2 x + 1}{2 x - 1}} = \frac{1}{2}$

$B)$ $\lim_{x \to 0} \frac{\log_{\sec (x / 2)} \cos x}{\log_{\sec x} \cos \frac{x}{2}} = \lim_{x \to 0} {(\log_{\sec \frac{x}{2}} \cos x)}^{2} = \lim_{x \to 0} {(\frac{lncos x}{lnsec \frac{x}{2}})}^{2} = 2$

$C)$

$D)$ $\sin x \neq \frac{1}{3}, \frac{2}{3}, \frac{3}{3}$

Q 47 :

If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

(17)

$F (x) = f (x)$ $x > 1 = \frac{f (x) + g (x)}{2}$ $x = 1$

$= f (x)$ $- 1 < x < 1 = \frac{f (x) + g (x)}{2}$ $x = - 1$

$= g (x)$ $x < - 1$

If $f (x)$ is continuous at $x = 1$ , $F (1^{+}) = F (1) = F (1^{-})$ $b = a + 3$

$F (- 1^{-}) = F (- 1) = F (- 1^{+})$ $a + b = 5$

Q 48 :

$f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3),$ $x \in R$ then which of the following is incorrect?

$f' (1) = 3$
$f' (3) = 10$
$f (0) = - 6$
$f'' (2) = 6$

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(2)

$f' (1) = 3 - 2 a + b = a,$ $f'' (2) = 12 - 2 a = b,$ $f''' (3) = 6$

$∴$ $a = 3,$ $b = 6$

Q 49 :

Statement - I: $f : R \to R$ be a function such that $| f (x) |$ $\leq x^{2},$ $\forall x \in R$ then $f (x)$ is differentiable at $x = 0$ .

Statement - II: $f : R \to R$ be a function such that $| f (x) |$ $\leq x^{p},$ $\forall x \in R$ then $f (x)$ is differentiable at $x = 0$ for any whole number $' p'$

Both statement – I and statement II are true
Both statement- I and statement II are false
Statement I is true and statement II is false
Statement I is false and statement II is true

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Conceptual

Q 50 :

Match the following

Column – I Column - II

1) The number of points of discontinuity of $f (x) = (sgn (x)) (sgn (- x))$ in $R$ is i) 0

2) The number of points of continuity of $f$ $(x)$ $=$ ${\begin{matrix} e^{| x |} - 1, & x is rational \\ \frac{1}{1 + x^{2}}, & x is irrational \end{matrix}$ in $R$ is ii) 1

3) The number of points of discontinuity of $g (x) = \lim_{n \to \infty} \frac{(f {(x))}^{n} - 1}{(f {(x))}^{n} + 1} in (0, \infty)$ where $f (x) = (x - 1) (x - 2) (x - 3) + 1$ is iii) 2

4) The number of points of discontinuity of $f (x) = \max ([x], | x |)$ in $R$ is, where $[.]$ represents the GIF iv) 3

Which of the following match is correct?

1–iii, 2–iv, 3–i, 4–ii
1–ii, 2–iii, 3–iv, 4–i
1–iii, 2–i, 3–iv, 4–ii
1–iv, 2–i, 3–ii, 4–iii

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(2)

conceptual

Q 51 :

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

If $m, n$ are number of points of discontinuity and non-differentiability of $f (x)$ in (0, 2) then ( $[\cdot]$ represents greatest integer function)

$m = 1, n = 2$
$m = 1, n = 1$
$m = 2, n = 2$
$m = 3, n = 2$

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(1)

$By the graph of y = f (x) clearly, f (x) is discontinuous at x = 1$ $and f (x) is non-differentiable at$ $x = \frac{1}{\sqrt{2}} and x = 1$

Q 52 :

Let the function $f : R \to R$ be defined by $f (x) = x^{3} - x^{2} + (x - 1) \sin x$ and let $g : R \to R$ be an arbitrary function.

Let $f g : R \to R$ be the product function defined by $(f g) (x) = f (x) g (x)$ .

Then number of correct statements among the following:

A) If $g$ is discontinuous at $x = 1$ , then $f g$ can never be differentiable at $x = 1$ .

B) If $f g$ is differentiable at $x = 1$ , then $g$ is continuous at $x = 1$ .

C) If $f g$ is differentiable at $x = 1$ , then $g$ must be differentiable at $x = 1$ .

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(4)

$Let f : R \to R is defined by f (x) = (x^{2} + \sin x) (x - 1)$

$Then, f (1^{+}) = f (1^{-}) = f (1) = 0$

$Let (f g) : R \to R is defined by (f g) (x) = f (x) \cdot g (x)$

$Let (f g) = h (x) = f (x) \cdot g (x) then h : R \to R$

$h' (x) = f' (x) g (x) + f (x) g' (x)$

$If g is differentiable at x = 1$

$h' (1) = f' (1) g (1) + 0,$ $[∵ f (1) = 0]$

$If g (x) is differentiable then h (x) is also differentiable (true)$

$If g (x) is differentiable at x = 1, then f g is also differentiable at x = 1$

$If g (x) is continuous at x = 1, then g (1^{+}) = g (1^{-}) = g (1)$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{h}$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h) - 0}{h} = f' (1) g (1)$

$h' (1^{-}) = \lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h) - 0}{- h} = f' (1) g (1)$

$\Rightarrow h (x) = f (x) \cdot g (x) is differentiable at x = 1 (True)$

$So, If g is continuous at x = 1, THEN f g is differentiable at x = 1 .$

$Option (2) (4)$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{- h}$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h)}{h} = f' (1) g (1^{+})$

$h' (1^{-}) = \lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h)}{- h} = f' (1) g (1^{-})$

$\Rightarrow g (1^{+}) = g (1^{-})$

$So, it does not mean that if f g is differentiable at x = 1,$

$then f g is continuous or differentiable at x = 1$

Q 53 :

If $g (x) =$ ${\begin{matrix} \frac{x (3 e^{1 / x} + 4)}{2 - e^{1 / x}}, & x \neq 0 \\ 0, & x = 0 \end{matrix}$ , then $f (x)$ is

Continuous as well as differentiable at $x = 0$
Continuous but not differentiable at $x = 0$
Differentiable but not continuous at $x = 0$
Discontinuous everywhere

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(2)

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

$R f^{1} (0) = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$

$= \lim_{h \to 0} (\frac{h (3 e^{1 / h} + 4)}{2 - e^{1 / h}} - 0) (\frac{1}{h})$

$\lim_{h \to 0} (\frac{- h (3 e^{- 1 / h} + 4)}{2 - e^{- 1 / h}} - 0) (\frac{- 1}{h}) = 2$

$= \lim_{h \to 0} (\frac{3 + 4 e^{- 1 / h}}{2 e^{- 1 / h} - 1}) = - 3$

$Since L f^{1} (0) \neq R f^{1} (0)$

$∴$ $f (x) is differentiable at x = 0 . But f (x) is continuous at x = 0$

Q 54 :

Let $f : R \to R$ be defined as $f (x) = {(2 x - 3 π)}^{3} + \frac{4 x}{3} + \cos x$ and $g = f^{- 1}$ then the value of $7 g' (2 π) + 3 g'' (2 π)$ (where $g'$ and $g''$ represent the $1^{st}$ and $2^{nd}$ order derivatives of $g$ ) is _______

(3)

$g' (2 π) = \frac{3}{7}, g'' (2 π) = 0$

Q 55 :

Let the function

$f (x) =$ ${\begin{matrix} \frac{\log_{e} (1 + 5 x) - \log_{e} (1 + α x)}{x}, & if x \neq 0 \\ 10, & if x = 0 \end{matrix}$

be continuous at $x = 0$ . Then absolute value of $' α'$ is equal to

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5

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(4)

$Given function is$

$\lim_{x \to 0} \frac{(5 x + \dots) - \ln (1 + α x)}{x} = 0$

$f (x) =$ ${\begin{matrix} \frac{\ln (1 + 5 x) - \ln (1 + α x)}{x}, & x \neq 0 \\ 10, & x = 0 \end{matrix}$

$Applying expansion of \ln (1 + x)$

$\lim_{x \to 0} (5 - α) = 10$

$5 - α = 10$

$\Rightarrow α = - 5$

Q 56 :

A function $f$ is defined on $[- 3, 3]$ as $f (x) =$ ${\begin{matrix} \min {| x |, 2 - x^{2}}, & - 2 \leq x \leq 2 \\ [| x |], & 2 < | x | \leq 3 \end{matrix}$ where $[x]$ denotes greatest integer $\leq x$ , number of points where $f$ is not differentiable in $(- 3, 3)$ is

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(3)

It is non diff at 5 point -2, -1, 0, 1, 2

Q 57 :

If $f (x) = \sin^{3} x + \log_{e} (x^{2} + 1)$ , then $f' (x)$ is

$3 \sin^{2} x . \cos x + \frac{2 x}{1 + x^{2}}$
$3 \sin^{2} x . \cos x + \frac{x}{1 + x^{2}}$
$2 \sin^{2} x . \cos x + \frac{2 x}{1 + x^{2}}$
$3 \sin^{2} x . \cos x + \frac{4 x}{1 + x^{2}}$

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(1)

$f' (x) = 3 \sin^{2} x . \cos x + \frac{2 x}{1 + x^{2}}$

Q 58 :

If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2}], a, b > 0,$ $\frac{d y}{d x} |_{x = \frac{π}{2}}$ $=$

$\frac{1}{b}$
$\frac{1}{a}$
$\frac{2}{a}$
$\frac{2}{b}$

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(2)

$y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} (\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2})$

$\frac{d y}{d x} = \frac{2}{\sqrt{a^{2} - b^{2}}} \frac{1}{(1 + \frac{a - b}{a + b} \tan^{2} \frac{x}{2})} \sec^{2} \frac{x}{2} \cdot \frac{1}{2} \sqrt{\frac{a - b}{a + b}}$

$\frac{d y}{d x} |_{x = \frac{π}{2}}$ $= \frac{1}{a}$

Q 59 :

Let $f : (0, π) \to R$ be a function given by $f (x) =$ ${\begin{matrix} \sin^{2} x, & 0 < x < \frac{π}{2} \\ a - 8, & x = \frac{π}{2} \\ {(1 + | \cot x |)}^{\frac{b}{a} | \tan x |}, & \frac{π}{2} < x < π \end{matrix}$

where $a, b \in Z$ . If $f$ is continuous at $x = \frac{π}{2}$ , then $a^{2} + b^{2}$ is equal to _____

(81)

$f$ is continuous at $x = \frac{π}{2} \Rightarrow \lim_{x \to π / 2^{-}} f (x) = \lim_{x \to π / 2^{+}} f (x) = f (\frac{π}{2})$

$\Rightarrow \lim_{x \to π / 2^{-}} {(\frac{8}{7})}^{\frac{\tan 8 x}{\tan 7 x}} = \lim_{x \to π / 2^{+}} {(1 + | \cot x |)}^{\frac{b}{a} | \tan x |} = a - 8$

$\Rightarrow {(\frac{8}{7})}^{0} = e^{\lim_{x \to π / 2^{+}} {(1 + \cot x)}^{- 1} \frac{b}{a} \tan x} = a - 8 \Rightarrow 1 = e^{\lim_{x \to π / 2^{+}} \frac{b}{a}} = a - 8$ $\Rightarrow a - 8 = 1,$ $e^{b / a} = 1$

$\Rightarrow a = 9, b = 0,$ $∴$ $a^{2} + b^{2} = 81 + 0 = 81$

Q 60 :

Let $f (x)$ be a differentiable function at $x = a,$ $f^{1} (a) = 4,$ $f (a) = 2$ .Then $\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$ is equal to

$2 a - 4$
$2 - 4 a$
$4 a - 2$
$4 - 2 a$

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(2)

L. H rule

Q 61 :

If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]

3
–1
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(3)

We have, $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \tan^{- 1} (\frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4} \tan x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \tan^{- 1} (\tan (\frac{3}{4} - x)) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \frac{3}{4} - x + 2 \tan^{- 1} (\tan \frac{t}{2})$ $[Let x = \sin t, then \frac{x}{1 + \sqrt{1 - x^{2}}} = \tan \frac{t}{2}]$

$= \frac{3}{4} - x + \frac{2 t}{2} = \frac{3}{4} - x + \sin^{- 1} x$

$\frac{d y}{d x} = - 1 + \frac{x}{1 + \sqrt{1 - x^{2}}}$

At $x = \frac{\sqrt{3}}{2},$

$\frac{d y}{d x} |_{\frac{\sqrt{3}}{2}} = - 1 + \frac{1}{\sqrt{1 - \frac{3}{4}}} = - 1 + 2 = 1$

Q 62 :

The number of points in the interval [2. 4], at which the function $f (x) = [x^{2} - x - \frac{1}{2}]$ , where [ $\cdot$ ] denotes the greatest integer function, is discontinuous, is ________. [2026]

(10)

We have, $f (x) = [x^{2} - x - \frac{1}{2}]$

At x = 2, $f (x) = [4 - 2 - \frac{1}{2}] = [1.5]$

At x = 4, $f (x) = [16 - 4 - \frac{1}{2}] = [11.5]$

Now, let $g (x) = x^{2} - x - \frac{1}{2}$

$g' (x) = 2 x - 1$

$∴$ g(x) is strictly increasing in [2, 4]

$∴$ 1.5 < k < 11.5

k $\in$ [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

$∴$ No. of discontinuity points = 10.

Q 63 :

The number of points, at which the function $f (x) = \max {6 x, 2 + 3 x^{2}} + | x - 1 | \cos | x^{2} - \frac{1}{4}], x \in (- π, π)$ , is not differentiable, is ________. [2026]

(3)

$f (x) = \max {6 x, 2 + 3 x^{2}} + | x - 1 | \cos | x^{2} - \frac{1}{4}]$

Sharp corner at $3 x^{2} + 2 = 6 x$

$\Rightarrow 3 x^{2} - 6 x + 2 = 0 \Rightarrow x = \frac{6 \pm \sqrt{36 - 24}}{6}$

$\Rightarrow x = \frac{3 \pm \sqrt{3}}{3} \Rightarrow x = 1 \pm \frac{1}{\sqrt{3}} \in (- π, π)$

$∴$ 2 points of non-differentiability

For $| x - 1 | \cos | x^{2} - \frac{1}{4} | : x = 1$ is the point of non-differentiability.

$∴$ Hence, total 3 points of non-differentiability.

Q 64 :

Let $f (x) = {\begin{matrix} e^{x - 1}, & x < 0 \\ x^{2} - 5 x + 6, & x \geq 0 \end{matrix}$ and g(x) = f(|x|) + |f(x)|. If the number of points where g is not continuous and is not differentiable are $α$ and $β$ respectively, then $α + β$ is equal to ________. [2026]

(4)

$f (x) = {\begin{matrix} e^{x - 1}, & x < 0 \\ x^{2} - 5 x + 6, & x \geq 0 \end{matrix}$ ; g(x) = f(|x|) + |f(x)|

Graph of f(x)

(f(x)) is continuous function and it is not differentiable at x = 0

Graph of |f(x)|

|f(x)| is discontinuous at x = 0

$\Rightarrow$ |f(x)| is non-differentiable at x = 0, 2, 3

g(x) = f|x| + |f(x)|

g(x) will be discontinuous at x = 0.

So, $α = 1$

g(x) will be non-differentiable at x = 0, 2, 3.

So, $β = 3$

Hence, $α + β = 4$

Q 65 :

Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]

(3)

$g (f (x)) = {\begin{matrix} (f {(x) - 8)}^{1 / 3}; & f (x) < 0, \\ (f {(x + 4)}^{1 / 2}; & f (x) \geq 0 \end{matrix}$

$g (f (x)) = {\begin{matrix} x & x \in (- \infty, - 2) \\ {(x^{3} + 12)}^{1 / 2} & x \in [- 2, 0) \\ {(x^{2} - 12)}^{1 / 3} & [0, 2) \\ x & x \in [2, \infty) \end{matrix}$

The function is discontinuous at x = –2, 0, 2

Hence, the number of point of discontinuity is 3.

Topic Question Set

Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is:

Q 45 :

If $f (x)$ is a continuous function $\forall x \in R$ and the range of $f (x)$ is $(2, \sqrt{26})$ and $g (x) = [\frac{f (x)}{C}]$ is continuous $\forall x \in R$ , then the least positive integral value of $' C'$ is: (where $[.]$ denotes the greatest integer function)

Q 47 :

If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

Q 48 :

$f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3),$ $x \in R$ then which of the following is incorrect?

Q 51 :

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

If $m, n$ are number of points of discontinuity and non-differentiability of $f (x)$ in (0, 2) then ( $[\cdot]$ represents greatest integer function)

Q 53 :

If $g (x) =$ ${\begin{matrix} \frac{x (3 e^{1 / x} + 4)}{2 - e^{1 / x}}, & x \neq 0 \\ 0, & x = 0 \end{matrix}$ , then $f (x)$ is

Q 54 :

Let $f : R \to R$ be defined as $f (x) = {(2 x - 3 π)}^{3} + \frac{4 x}{3} + \cos x$ and $g = f^{- 1}$ then the value of $7 g' (2 π) + 3 g'' (2 π)$ (where $g'$ and $g''$ represent the $1^{st}$ and $2^{nd}$ order derivatives of $g$ ) is _______

Q 55 :

Let the function

$f (x) =$ ${\begin{matrix} \frac{\log_{e} (1 + 5 x) - \log_{e} (1 + α x)}{x}, & if x \neq 0 \\ 10, & if x = 0 \end{matrix}$

be continuous at $x = 0$ . Then absolute value of $' α'$ is equal to

Q 56 :

A function $f$ is defined on $[- 3, 3]$ as $f (x) =$ ${\begin{matrix} \min {| x |, 2 - x^{2}}, & - 2 \leq x \leq 2 \\ [| x |], & 2 < | x | \leq 3 \end{matrix}$ where $[x]$ denotes greatest integer $\leq x$ , number of points where $f$ is not differentiable in $(- 3, 3)$ is

Q 57 :

If $f (x) = \sin^{3} x + \log_{e} (x^{2} + 1)$ , then $f' (x)$ is

Q 58 :

If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2}], a, b > 0,$ $\frac{d y}{d x} |_{x = \frac{π}{2}}$ $=$

Q 60 :

Let $f (x)$ be a differentiable function at $x = a,$ $f^{1} (a) = 4,$ $f (a) = 2$ .Then $\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$ is equal to

Q 61 :

If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]

Q 62 :

The number of points in the interval [2. 4], at which the function $f (x) = [x^{2} - x - \frac{1}{2}]$ , where [ $\cdot$ ] denotes the greatest integer function, is discontinuous, is ________. [2026]

Q 63 :

The number of points, at which the function $f (x) = \max {6 x, 2 + 3 x^{2}} + | x - 1 | \cos | x^{2} - \frac{1}{4}], x \in (- π, π)$ , is not differentiable, is ________. [2026]

Q 64 :

Let $f (x) = {\begin{matrix} e^{x - 1}, & x < 0 \\ x^{2} - 5 x + 6, & x \geq 0 \end{matrix}$ and g(x) = f(|x|) + |f(x)|. If the number of points where g is not continuous and is not differentiable are $α$ and $β$ respectively, then $α + β$ is equal to ________. [2026]

Q 65 :

Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]

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	Column-I		Column-II
(A)	$\lim_{x \to \infty} {(\frac{x^{2} + 2 x - 1}{2 x^{2} - 3 x - 2})}^{\frac{2 x + 1}{2 x - 1}} =$	(P)	$\frac{1}{2}$
(B)	$\lim_{x \to 0} \frac{\log_{\sec (x / 2)} \cos x}{\log_{\sec x} \cos \frac{x}{2}} =$	(Q)	$2$
(C)	$Let f (x) = \max (\cos x, x, 2 x - 1), where x \geq 0, then$ $number of points of non-differentiability of f (x) is$	(R)	$5$
(D)	$If f (x) = [2 + 3 \sin x], 0 < x < π, then number of$ $points at which the function is discontinuous, is$	(S)	$16$

	Column – I		Column - II
1)	The number of points of discontinuity of $f (x) = (sgn (x)) (sgn (- x))$ in $R$ is	i)	0
2)	The number of points of continuity of $f$ $(x)$ $=$ ${\begin{matrix} e^{\| x \|} - 1, & x is rational \\ \frac{1}{1 + x^{2}}, & x is irrational \end{matrix}$ in $R$ is	ii)	1
3)	The number of points of discontinuity of $g (x) = \lim_{n \to \infty} \frac{(f {(x))}^{n} - 1}{(f {(x))}^{n} + 1} in (0, \infty)$ where $f (x) = (x - 1) (x - 2) (x - 3) + 1$ is	iii)	2
4)	The number of points of discontinuity of $f (x) = \max ([x], \| x \|)$ in $R$ is, where $[.]$ represents the GIF	iv)	3

Topic Question Set

Q 41 : If x=a(t-1t), y=b(t+1t), then dydx=

Q 43 : If f(x)={kcosxπ-2x,x≠π23,x=π2 is continuous at x=π2, then k is:

Q 44 : The derivative of sec(tanx) with respect to x is:

Q 45 : If f(x) is a continuous function ∀x∈R and the range of f(x) is (2,26) and g(x)=[f(x)C] is continuous ∀x∈R, then the least positive integral value of 'C' is: (where [.] denotes the greatest integer function)

Q 47 : If f(x)=x2+ax+3 and g(x)=x+b, where F(x)=limn→∞f(x)+(x2)ng(x)1+(x2)n. If F(x) is continuous at x=1 and x=-1, then find the value of (a2+b2)

Q 48 : f(x)=x3-x2f'(1)+xf''(2)-f'''(3), x∈R then which of the following is incorrect?

Q 49 : Statement - I: f:R→R be a function such that |f(x)|≤x2, ∀x∈R then f(x) is differentiable at x=0. Statement - II: f:R→R be a function such that |f(x)|≤xp, ∀x∈R then f(x) is differentiable at x=0 for any whole number 'p'

Q 51 : Let f(x)={|1-2x2|,0≤x<1|x2-2x|,1≤x<2 If m,n are number of points of discontinuity and non-differentiability of f(x) in (0, 2) then ([·] represents greatest integer function)

Q 53 : If g(x)={x(3e1/x+4)2-e1/x,x≠00,x=0, then f(x) is

Q 54 : Let f:R→R be defined as f(x)=(2x-3π)3+4x3+cosx and g=f-1 then the value of 7g'(2π)+3g''(2π) (where g' and g'' represent the 1st and 2nd order derivatives of g) is _______

Q 55 : Let the function f(x)={loge(1+5x)-loge(1+αx)x,if x≠010,if x=0 be continuous at x=0. Then absolute value of 'α' is equal to

Q 56 : A function f is defined on [-3,3] as f(x)={min{|x|, 2-x2},-2≤x≤2[|x|],2<|x|≤3 where [x] denotes greatest integer ≤x, number of points where f is not differentiable in (-3,3) is

Q 57 : If f(x)=sin3x+loge(x2+1), then f'(x) is

Q 58 : If y=2a2-b2tan-1[a-ba+btanx2], a,b>0, dydx|x=π2=

Q 59 : Let f:(0,π)→R be a function given by f(x)={sin2x,0<x<π2a-8,x=π2(1+|cotx|)ba|tan x|,π2<x<π where a,b∈Z. If f is continuous at x=π2, then a2+b2 is equal to _____

Q 60 : Let f(x) be a differentiable function at x=a, f1(a)=4, f(a)=2.Then limx→axf(a)-af(x)x-a is equal to

Q 61 : If y=tan–1(3cosx–4sinx4cosx+3sinx)+2tan–1(x1+1–x2), then dydx at x=32 is equal to: [2026]

Q 62 : The number of points in the interval [2. 4], at which the function f(x)=[x2–x–12], where [·] denotes the greatest integer function, is discontinuous, is ________. [2026]

Q 63 : The number of points, at which the function f(x)=max{6x,2+3x2}+|x–1|cos|x2–14],x∈(–π,π), is not differentiable, is ________. [2026]

Q 64 : Let f(x)={ex–1,x<0x2–5x+6,x≥0 and g(x) = f(|x|) + |f(x)|. If the number of points where g is not continuous and is not differentiable are α and β respectively, then α+β is equal to ________. [2026]

Q 65 : Let f(x)={x2+8;x<0,x2–4;x≥0 and g(x)={(x–8)1/3;x<0,(x+4)1/2;x≥0. Then the number of points, where the function gof is discontinuous, is ________. [2026]

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Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is:

Q 45 :

If $f (x)$ is a continuous function $\forall x \in R$ and the range of $f (x)$ is $(2, \sqrt{26})$ and $g (x) = [\frac{f (x)}{C}]$ is continuous $\forall x \in R$ , then the least positive integral value of $' C'$ is: (where $[.]$ denotes the greatest integer function)

Q 47 :

If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

Q 48 :

$f (x) = x^{3} - x^{2} f' (1) + x f'' (2) - f''' (3),$ $x \in R$ then which of the following is incorrect?

Q 51 :

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

If $m, n$ are number of points of discontinuity and non-differentiability of $f (x)$ in (0, 2) then ( $[\cdot]$ represents greatest integer function)

Q 53 :

If $g (x) =$ ${\begin{matrix} \frac{x (3 e^{1 / x} + 4)}{2 - e^{1 / x}}, & x \neq 0 \\ 0, & x = 0 \end{matrix}$ , then $f (x)$ is

Q 54 :

Let $f : R \to R$ be defined as $f (x) = {(2 x - 3 π)}^{3} + \frac{4 x}{3} + \cos x$ and $g = f^{- 1}$ then the value of $7 g' (2 π) + 3 g'' (2 π)$ (where $g'$ and $g''$ represent the $1^{st}$ and $2^{nd}$ order derivatives of $g$ ) is _______

Q 55 :

Let the function

$f (x) =$ ${\begin{matrix} \frac{\log_{e} (1 + 5 x) - \log_{e} (1 + α x)}{x}, & if x \neq 0 \\ 10, & if x = 0 \end{matrix}$

be continuous at $x = 0$ . Then absolute value of $' α'$ is equal to

Q 56 :

A function $f$ is defined on $[- 3, 3]$ as $f (x) =$ ${\begin{matrix} \min {| x |, 2 - x^{2}}, & - 2 \leq x \leq 2 \\ [| x |], & 2 < | x | \leq 3 \end{matrix}$ where $[x]$ denotes greatest integer $\leq x$ , number of points where $f$ is not differentiable in $(- 3, 3)$ is

Q 57 :

If $f (x) = \sin^{3} x + \log_{e} (x^{2} + 1)$ , then $f' (x)$ is

Q 58 :

If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2}], a, b > 0,$ $\frac{d y}{d x} |_{x = \frac{π}{2}}$ $=$

Q 60 :

Let $f (x)$ be a differentiable function at $x = a,$ $f^{1} (a) = 4,$ $f (a) = 2$ .Then $\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$ is equal to

Q 61 :

If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]

Q 62 :

The number of points in the interval [2. 4], at which the function $f (x) = [x^{2} - x - \frac{1}{2}]$ , where [ $\cdot$ ] denotes the greatest integer function, is discontinuous, is ________. [2026]

Q 63 :

The number of points, at which the function $f (x) = \max {6 x, 2 + 3 x^{2}} + | x - 1 | \cos | x^{2} - \frac{1}{4}], x \in (- π, π)$ , is not differentiable, is ________. [2026]

Q 64 :

Let $f (x) = {\begin{matrix} e^{x - 1}, & x < 0 \\ x^{2} - 5 x + 6, & x \geq 0 \end{matrix}$ and g(x) = f(|x|) + |f(x)|. If the number of points where g is not continuous and is not differentiable are $α$ and $β$ respectively, then $α + β$ is equal to ________. [2026]

Q 65 :

Let $f (x) = {\begin{matrix} x^{2} + 8; & x < 0, \\ x^{2} - 4; & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} {(x - 8)}^{1 / 3}; & x < 0, \\ {(x + 4)}^{1 / 2}; & x \geq 0 \end{matrix}$ . Then the number of points, where the function gof is discontinuous, is ________. [2026]