- A0.8952
- B0.9528
- C0.9285
- ✓0.9825
Answer: D.
View full solution →201 questions across 6 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
MCQ
50 Q→02Answer the following questions in short.
12 Q→03Solve the Following Question.(2 Marks)
38 Q→04Solve the Following Question.(3 Marks)
46 Q→05Solve the Following Question.(4 Marks)
25 Q→06Solve the Following Question.(5 Marks)
30 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Answer: D.
View full solution →Answer: D.
View full solution →Answer: A.
View full solution →Answer: D.
View full solution →Answer: C.
View full solution →$f(x)=x-\frac{1}{x}, x \in R, x \neq 0$
$f(x)=2-3 x+3 x^2-x^3, x \in R$.
$3-x^2$ drawn at the point $(1,2)$ and the coordinate axes.
the interval 0 ≤ f ≤ 4 when the instantaneous velocity of the particle is equal to its average velocity in this interval.
point M (1, 2).
$a^2$. Show that the maximum volume of the box is $\frac{a^3}{6 \sqrt{3}}$.
of radius $R$ is $\frac{2 R}{\sqrt{3}}$. Also, find the maximum Volume.
in a sphere of radius $r$ is $\frac{4 r}{3}$.
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