Ask & Answer: India's Largest Education Community

Candidates who have passed BE/ BTech examination of VTU or any other recognised university with at least 60% aggregate (50% aggregate for SC/ST and any other category as notified by Government of Karnataka) are eligible to apply for MSc (Engineering) by Research at B.N.M. Institute of Technology.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Options (iii) and (iv) are the correct answers.

As, in the given reaction below-

2KClO₃→2KCl+3O₂

Potassium remains in same oxidation state and oxygen is being oxidized

We have, f (x) = x² e^–x

So, f (x) = $x^2\frac{d}{dx}{e}^{-x}+e^{-x}\frac{d{x}^2}{dx}$

$=x^2{e}^{-x}\frac{d}{dx}(-x)+e^{-x}2\cdot x\cancel{\frac{dx}{dx}}$

= -x² e^-x + e^-x 2x.

= x e^-x ( x + 2).

$=\frac{x(2-x)}{e^{ex}}$

If f (x) = 0.

$\Rightarrow \frac{x(2-x)}{e^x}=0.$

$\Rightarrow$ x = 0, x = 2.

Hence, we get there disjoint interval

$[-\infty ,0),(0,2)(2,\infty )$

When, x $(-\infty ,0),$ we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.

So, f is strictly decreasing.

When x ∈ (0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.

So, f is strictly increasing.

And when x ∈$(2,\infty ),$ f (x) = ( +ve) ( -ve) = ( -ve) < 0.

So, f is strictly decreasing.

Hence, option (D) is correct.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii) is the correct answer.

As, Cl, Br, I all are having -1 to +7 oxidation state. But Oxidation state of F is fixed (-1) as it is the most electronegative element and do not loose electron. Hence, it does not show disproportionate tendency.

Admission is currently open for BSc at MIT School of Vedic Sciences, MIT- ADT University. Students who meet the eligibility criteria can register and begin the admission process.

We have, f (x) = x³- 3x² 3x- 100.

So, f (x) = 3x²- 6x + 3 = 3 (x²- 2x + 1) = 3 (x- 1)²

For $x\in R.$

(x- 10)²$\ge 0$ , 0 for x = 1.

$\Rightarrow$ 3 (x- 1)² $\ge 0$

$\Rightarrow$ f (x) $\ge 0$

∴f (x) is increasing on $?$

Yes, B.N.M. Institute of Technology admissions are currrently open. Aspirants must also keep checking the schedule of the accepted entrance exam for the preferred course. The institute offers admission to candidates based on their entrance exam scores. BNMIT accepts scores in COMEDK, KCET, PGCET, KMAT, etc. for admission to specific programmes.

The deadline for applying to BSc at MIT School of Vedic Sciences, MIT-ADT University is July 31. Students are advised to complete registration before this date. Since eligibility is determined by Class 12 results. The candidates should have documents ready and submit applications early to avoid missing out the last date for MIT BSc admission.

We have, f (x) = log $\left|cosx\right|.$

f (x) = $\frac{1}{cosx}\frac{d}{dx}x=-\frac{sinx}{cosx}=-tanx$

whenx $\left(0,\frac{\pi }{2}\right),$ we get.

tanx> 0 (Ist quadrant).

$\Rightarrow$ tanx< 0

$\Rightarrow$ f (x) < 0.

∴f (x) is decreasing on $\left(0,\frac{\pi }{2}\right).$

When x ∈$\left(\frac{3\pi }{2},2\pi \right)$ we get,

tanx|< |0 (ivth quadrant).

$\Rightarrow$ -tanx|>| 0

$\Rightarrow$ f (x) > 0

∴f (x) is increasing on $\left(\frac{3\pi }{2},2\pi \right).$

Pursuing a PGDM course offered by Muthoot Business School helps students cultivate knowledge of various domains such as HR, Finance, Operations and more. However, today's job market has become highly competitive. Thus, to get a dream job, having only theoretical knowledge might not be enough. Students must develop important employability skills to thrive in the corporate landscape. Listed below are some of the skills demanded in today’s corporate world:

• Digital Marketing
• HR Analytics
• Public Administration
• Foreign Language
• Financial Analytics
• Risk Analysis
• Public Relations Management

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer.

Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-

2NO₂ + 2OH^- →NO₂ ^-+ NO₃^- +H₂O

In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in NO₃⁻?  to +5 in NO₂ ? and decreases from +4 in NO to +3 in NO₂⁻.

Shree Swaminarayan Physiotherapy College offers UG and PG courses in the field of Physiotherapy. Students can apply for BPT or MPT programmes either through State or Mangement Quota. Both the courses are offered under the affiliation of Gujarat University.

We have, f (x) = log sin x

So, f (x) = $\frac{1}{sinx}\frac{dsinx}{dx}=\frac{cosx}{sinx}=cotx$

When $x\in \left(0,\frac{\pi }{2}\right)$

f (x) = cot x> 0 (Ist quadrat )

So, f (x) is strictly increasing on $\left(0,\frac{\pi }{2}\right)$

When x ∈$\left(\frac{\pi }{2},\pi \right)$

f (x) = cot x< 0. (IIndquadrant ).

So, f (x) is strictly decreasing on $\left(\frac{\pi }{2},\pi \right)$

AAA College of Engineering and Technology offers BTech in Cyber Security for 4 years with an annual intake of 60 students. The Institute offers admission to the respective course based on the minimum aggregate of 45% in Class 12 and valid scores in TNEA. Final admission is granted based on the cutoff criteria met by the candidates and after qualifying the counselling round.

We have, f (x) = $\stackrel{?}{x}+\frac{1}{x}$

So, f (x) = $1-\frac{1}{x^2}=\frac{x^2-1}{x^2}=\frac{(x-1)(x+1)}{x^2}$

So, for every x∈ I, where I is disjoint from [-1,1]$$

f (x) = $\frac{(+ve)(+ve)}{(+ve)}=(+ve)>0whenx>1.$

and f (x) = $\frac{(-ve)(-ve)}{(+ve)}$ = ( + ve) > 0 when x< -1.

∴f (x) is strictly increasing on I .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer

For 3d¹ 4s² can exhibit the highest oxidation state as 2+1 = +3,

For 3d³4s²can exhibit the highest oxidation state as 2+3 = +5

For 3d⁵4s¹can exhibit the highest oxidation state as 1+5 = +6

For 3d⁵4s²can exhibit the highest oxidation state as 2+5 = +7

B.N.M Institute of Technology considers KCET scores for admission to BE programme. The closing cutoff ranks for KCET in 2024 was between 13679 and 41037. Computer Science Engineering was the most competitive specialisation with a closing rank of 13679 for General All India Category. The cutoff of CSE saw a declining trend from 8660 in 2022, 9008 in 2023 to 13679 in 2024.