1
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

If $$f(x)=\left\{\begin{array}{cc}\frac{(1-\cos 4 x)}{x^2}, & \text { if } x < 0 \\ a, & \text { if } x=0, \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}, & \text { if } x > 0\end{array}\right.$$ then $$f(x)$$ is continuous at $$x=0$$, for $$a$$

A
4
B
$$\sqrt{32}$$
C
8
D
16
2
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

The value of $$\int\limits_0^{\pi / 2} \frac{d x}{1+\tan x}$$ is

A
$$\frac{\pi}{2}$$
B
0
C
$$\frac{\pi}{4}$$
D
$$\frac{\pi}{8}$$
3
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

Evaluate $$\int \frac{3 x-2}{(x+3)(x+1)^2} d x$$.

A
$$\frac{11}{4} \log [|x+1||x+3|]+\frac{5}{2(x+1)}+C$$
B
$$\frac{11}{4} \log \left|\frac{x+3}{x+1}\right|+\frac{1}{x+1}+C$$
C
$$\frac{11}{4} \log |x+2|+\frac{5}{2}(x+3)+\frac{1}{x+1}+C$$
D
$$\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$$
4
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

The general solution of the linear differential equation $$\frac{d y}{d x}+\sec x \cdot y=\tan x\left(0 \leq x \leq \frac{\pi}{2}\right)$$ is

A
$$y=-x(\sec x+\tan x)^{-1}+\frac{c}{\sec x+\tan x}+1$$
B
$$y=x+\frac{C}{\sec x+\tan x}+\frac{1}{\tan x}$$
C
$$y=\frac{x+1}{\sec x+\tan x}+C$$
D
$$y=x+\sec x+\tan x+C$$
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