1
WB JEE 2024
+1
-0.25

If $$\alpha, \beta$$ are the roots of the equation $$a x^2+b x+c=0$$ then $$\lim _\limits{x \rightarrow \beta} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\beta)^2}$$ is

A
$$(\alpha-\beta)^2$$
B
$$\frac{1}{2}(\alpha-\beta)^2$$
C
$$\frac{a^2}{4}(\alpha-\beta)^2$$
D
$$\frac{\mathrm{a}^2}{2}(\alpha-\beta)^2$$
2
WB JEE 2023
+1
-0.25

$$\mathop {\lim }\limits_{x \to \infty } \left\{ {x - \root n \of {(x - {a_1})(x - {a_2})\,...\,(x - {a_n})} } \right\}$$ where $${a_1},{a_2},\,...,\,{a_n}$$ are positive rational numbers. The limit

A
does not exist
B
is $${{{a_1} + {a_2}\, + \,...\,{a_n}} \over n}$$
C
is $$\root n \of {{a_1}{a_2}\,...\,{a_n}}$$
D
is $${n \over {{a_1} + {a_2}\, + \,...\,{a_n}}}$$
3
WB JEE 2023
+1
-0.25

Let $$f:[1,3] \to R$$ be continuous and be derivable in (1, 3) and $$f'(x) = {[f(x)]^2} + 4\forall x \in (1,3)$$. Then

A
$$f(3) - f(1) = 5$$ holds
B
$$f(3) - f(1) = 5$$ does not hold
C
$$f(3) - f(1) = 3$$ holds
D
$$f(3) - f(1) = 4$$ holds
4
WB JEE 2023
+1
-0.25

f(x) is a differentiable function and given $$f'(2) = 6$$ and $$f'(1) = 4$$, then $$L = \mathop {\lim }\limits_{h \to 0} {{f(2 + 2h + {h^2}) - f(2)} \over {f(1 + h - {h^2}) - f(1)}}$$

A
does not exist
B
equal to $$-3$$
C
equal to 3
D
equal to 3/2
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