LFM Pure

jee-main 2021 Q84 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \frac { a x - \left( e ^ { 4 x } - 1 \right) } { a x \left( e ^ { 4 x } - 1 \right) }$ exists and is equal to $b$, then the value of $a - 2 b$ is $\underline{\hspace{1cm}}$.

jee-main 2021 Q84 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to $\_\_\_\_$.

jee-main 2021 Q86 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \left[ \frac { \alpha x e ^ { x } - \beta \log _ { e } ( 1 + x ) + \gamma x ^ { 2 } e ^ { - x } } { x \sin ^ { 2 } x } \right] = 10 , \alpha , \beta , \gamma \in R$, then the value of $\alpha + \beta + \gamma$ is $\underline{\hspace{1cm}}$.

jee-main 2022 Q3 Chain Rule with Composition of Explicit Functions View

If $t = \sqrt { x } + 4$, then $\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) _ { t = 4 }$ is:
(1) 4
(2) Zero
(3) 8
(4) 16

jee-main 2022 Q66 Limit Involving Derivative Definition of Composed Functions View

Let $f(x)$ be a polynomial function such that $f(x) + f ^ { \prime } (x) + f ^ { \prime \prime } (x) = x ^ { 5 } + 64$. Then, the value of $\lim _ { x \rightarrow 1 } \frac { f(x) } { x - 1 }$ is equal to
(1) $- 15$
(2) $15$
(3) $- 60$
(4) $60$

jee-main 2022 Q66 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow 0 } \frac { \cos ( \sin x ) - \cos x } { x ^ { 4 } }$ is equal to
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 12 }$

jee-main 2022 Q69 Limit Evaluation Involving Composition or Substitution View

$\lim_{x \to \frac{\pi}{2}} \tan^2 x \left[(2\sin^2 x + 3\sin x + 4)^{\frac{1}{2}} - (\sin^2 x + 6\sin x + 2)^{\frac{1}{2}}\right]$ is equal to
(1) $\frac{1}{12}$
(2) $-\frac{1}{18}$
(3) $-\frac{1}{12}$
(4) $\frac{1}{6}$

jee-main 2022 Q69 Limit Evaluation Involving Composition or Substitution View

$\lim_{x \rightarrow \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}$ is equal to
(1) 14
(2) 7
(3) $14\sqrt{2}$
(4) $7\sqrt{2}$

jee-main 2022 Q72 Piecewise Function Differentiability Analysis View

$f , g : R \rightarrow R$ be two real valued function defined as $f ( x ) = \left\{ \begin{array} { c l } - | x + 3 | & , x < 0 \\ e ^ { x } & , x \geq 0 \end{array} \right.$ and $g ( x ) = \left\{ \begin{array} { l l } x ^ { 2 } + k _ { 1 } x , & x < 0 \\ 4 x + k _ { 2 } , & x \geq 0 \end{array} \right.$, where $k _ { 1 }$ and $k _ { 2 }$ are real constants. If $gof$ is differentiable at $x = 0$, then $gof ( - 4 ) + gof ( 4 )$ is equal to
(1) $4 \left( e ^ { 4 } + 1 \right)$
(2) $2 \left( 2 e ^ { 4 } + 1 \right)$
(3) $4 e ^ { 4 }$
(4) $2 \left( 2 e ^ { 4 } - 1 \right)$

jee-main 2023 Q72 Limit Evaluation Involving Composition or Substitution View

Let $x = 2$ be a root of the equation $x ^ { 2 } + p x + q = 0$ and $f ( x ) = \left\{ \begin{array} { c l } \frac { 1 - \cos \left( x ^ { 2 } - 4 p x + q ^ { 2 } + 8 q + 16 \right) } { ( x - 2 p ) ^ { 4 } } , & x \neq 2 p \\ 0 , & x = 2 p \end{array} \right.$. Then $\lim _ { x \rightarrow 2 p ^ { + } } [ f ( x ) ]$ where $[ \cdot ]$ denotes greatest integer function, is
(1) 2
(2) 1
(3) 0
(4) $- 1$

jee-main 2023 Q72 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow 0 } \left( \left( \frac { 1 - \cos ^ { 2 } ( 3 x ) } { \cos ^ { 3 } ( 4 x ) } \right) \left( \frac { \sin ^ { 3 } ( 4 x ) } { \left( \log _ { e } ( 2 x + 1 ) \right) ^ { 5 } } \right) \right)$ is equal to
(1) 15
(2) 9
(3) 18
(4) 24

jee-main 2023 Q73 Chain Rule with Composition of Explicit Functions View

Let $y = f(x) = \sin^3\left(\frac{\pi}{3}\cos\left(\frac{\pi}{3\sqrt{2}}\left(-4x^3 + 5x^2 + 1\right)^{3/2}\right)\right)$. Then, at $x = 1$,
(1) $2y' + \sqrt{3}\pi^2 y = 0$
(2) $2y' + 3\pi^2 y = 0$
(3) $\sqrt{2}y' - 3\pi^2 y = 0$
(4) $y' + 3\pi^2 y = 0$

jee-main 2023 Q79 Chain Rule Combined with Fundamental Theorem of Calculus View

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f\left(\frac{\pi}{2}\right) = 0$ and $f'\left(\frac{\pi}{2}\right) = 1$ and let $g(x) = \int_x^{\pi/4} (f'(t)\sec t + \tan t \cdot f(t)\sec t)\,dt$. Then $\lim_{x \to \pi/2} \frac{g(x)}{(x - \pi/2)^2}$ is equal to $\_\_\_\_$.

jee-main 2023 Q79 Higher-Order Derivatives of Products/Compositions View

Let $y ( x ) = ( 1 + x ) \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 4 } \right) \left( 1 + x ^ { 8 } \right) \left( 1 + x ^ { 16 } \right)$. Then $y ^ { \prime } - y ^ { \prime \prime }$ at $x = - 1$ is equal to
(1) 976
(2) 464
(3) 496
(4) 944

jee-main 2023 Q85 Higher-Order Derivatives of Products/Compositions View

If $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$, then the value of $f(4) - g(4)$ is equal to $\_\_\_\_$.

jee-main 2024 Q67 Limit Evaluation Involving Composition or Substitution View

$\lim _ { x \rightarrow 0 } \frac { e ^ { 2 \sin x } - 2 \sin x - 1 } { x ^ { 2 } }$
(1) is equal to $-1$
(2) does not exist
(3) is equal to 1
(4) is equal to 2

jee-main 2024 Q68 Limit Involving Derivative Definition of Composed Functions View

Let $f : ( - \infty , \infty ) - \{ 0 \} \rightarrow \mathbb { R }$ be a differentiable function such that $f ^ { \prime } ( 1 ) = \lim _ { a \rightarrow \infty } a ^ { 2 } f \left( \frac { 1 } { a } \right)$. Then $\lim _ { a \rightarrow \infty } \frac { a ( a + 1 ) } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { a } \right) + a ^ { 2 } - 2 \log _ { e } a$ is equal to
(1) $\frac { 3 } { 2 } + \frac { \pi } { 4 }$
(2) $\frac { 3 } { 4 } + \frac { \pi } { 8 }$
(3) $\frac { 3 } { 8 } + \frac { \pi } { 4 }$
(4) $\frac { 5 } { 2 } + \frac { \pi } { 8 }$

jee-main 2024 Q68 Limit Evaluation Involving Composition or Substitution View

If $\lim _ { x \rightarrow 0 } \frac { 3 + \alpha \sin x + \beta \cos x + \log _ { e } ( 1 - x ) } { 3 \tan ^ { 2 } x } = \frac { 1 } { 3 }$, then $2 \alpha - \beta$ is equal to :
(1) 2
(2) 7
(3) 5
(4) 1

jee-main 2024 Q69 Limit Evaluation Involving Composition or Substitution View

If $a = \lim _ { x \rightarrow 0 } \frac { \sqrt { 1 + \sqrt { 1 + x ^ { 4 } } } - \sqrt { 2 } } { x ^ { 4 } }$ and $b = \lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { \sqrt { 2 } - \sqrt { 1 + \cos x } }$, then the value of $a b ^ { 3 }$ is:
(1) 36
(2) 32
(3) 25
(4) 30

jee-main 2024 Q72 Chain Rule with Composition of Explicit Functions View

Suppose for a differentiable function $h , h ( 0 ) = 0 , h ( 1 ) = 1$ and $h ^ { \prime } ( 0 ) = h ^ { \prime } ( 1 ) = 2$. If $\mathrm { g } ( x ) = h \left( \mathrm { e } ^ { x } \right) \mathrm { e } ^ { h ( x ) }$, then $g ^ { \prime } ( 0 )$ is equal to:
(1) 5
(2) 4
(3) 8
(4) 3

jee-main 2024 Q73 Continuity Conditions via Composition View

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \frac { 1 + x } { 2 + x } , & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is
(1) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { e^{1/3} }$
(2) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { 9 } + 1$
(3) $\log _ { e } \frac { 4 } { 9 } - 1$
(4) $\log _ { e } \frac { 4 } { 9 e ^ { 1/3 } }$

jee-main 2024 Q73 Chain Rule with Composition of Explicit Functions View

Suppose $f ( x ) = \frac { \left( 2 ^ { x } + 2 ^ { - x } \right) \tan x \sqrt { \tan ^ { - 1 } \left( x ^ { 2 } - x + 1 \right) } } { \left( 7 x ^ { 2 } + 3 x + 1 \right) ^ { 3 } }$. Then the value of $f ^ { \prime } ( 0 )$ is equal to
(1) $\pi$
(2) 0
(3) $\sqrt { \pi }$
(4) $\frac { \pi } { 2 }$

jee-main 2024 Q85 Higher-Order Derivatives of Products/Compositions View

Let $f ( x ) = x ^ { 3 } + x ^ { 2 } f ^ { \prime } ( 1 ) + x f ^ { \prime \prime } ( 2 ) + f ^ { \prime \prime \prime } ( 3 ) , x \in R$. Then $f ^ { \prime } ( 10 )$ is equal to

jee-main 2024 Q85 Limit Evaluation Involving Composition or Substitution View

Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$

jee-main 2024 Q85 Limit Evaluation Involving Composition or Substitution View

The value of $\lim _ { x \rightarrow 0 } 2 \left( \frac { 1 - \cos x \sqrt { \cos 2 x } \sqrt [ 3 ] { \cos 3 x } \ldots \ldots \sqrt [ 10 ] { \cos 10 x } } { x ^ { 2 } } \right)$ is $\_\_\_\_$

1
2
3
4
5

Load questions...

LFM Pure

Straight Lines & Coordinate Geometry 247

Circles 443

Simultaneous equations 79

Proof 711

Proof by induction 25

Sine and Cosine Rules 195

Radians, Arc Length and Sector Area 35

Trig Graphs & Exact Values 75

Standard trigonometric equations 106

Trig Proofs 53

Quadratic trigonometric equations 36

Trigonometric equations in context 18

Modulus function 49

Matrices 1553

Linear transformations 26

Invariant lines and eigenvalues and vectors 188

Reciprocal Trig & Identities 44

Addition & Double Angle Formulae 98

Harmonic Form 12

Small angle approximation 20

Chain Rule 113

Product & Quotient Rules 18

Differentiating Transcendental Functions 183

Connected Rates of Change 56

Standard Integrals and Reverse Chain Rule 43

Integration by Substitution 116

Integration by Parts 74

Implicit equations and differentiation 43

Differential equations 352

3x3 Matrices 144