WB JEE 2023 | Differentiation Question 21 | Mathematics

WB JEE 2023

MCQ (More than One Correct Answer)

-0

Let $$f(x) = {x^m}$$, m being a non-negative integer. The value of m so that the equality $$f'(a + b) = f'(a) + f'(b)$$ is valid for all a, b > 0 is

WB JEE 2020

MCQ (More than One Correct Answer)

-0

Let $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$. Then $${{{d^2}y} \over {d{x^2}}}$$ is

$$2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$

$$3\left[ {{2 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 1)}^2}}} - {5 \over {{{(x + 2)}^3}}}} \right]$$

$${6 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 1)}^2}}} + {3 \over {{{(x + 1)}^3}}}$$

$${7 \over {{{(x + 1)}^3}}} - {3 \over {{{(x + 1)}^2}}} + {2 \over {{{(x + 1)}^3}}}$$

WB JEE 2019

MCQ (More than One Correct Answer)

-0

Let f and g be differentiable on the interval I and let a, b $$ \in $$ I, a < b. Then,

If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 is soluble in (a, b)

If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 may not be soluble in (a, b)

If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is soluble in (a, b), k $$ \in $$ R

If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 may not be soluble in (a, b), k $$ \in $$ R

WB JEE 2017

MCQ (More than One Correct Answer)

-0

If f(x) = xⁿ, being a non-negative integer, then the values of n for which f'($$\alpha$$ + $$\beta$$) = f'($$\alpha$$) + f'($$\beta$$) for all $$\alpha$$, $$\beta$$ > 0 is

WB JEE Subjects