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We have, f (x) = x² + x + 1

So, f (x) = 2x + a

If f (x) is strictly increasing onx $(1,2),f^{\prime }(x)>0.$

So, 1

$\to 2\times 1<2\times x<2\times 2$

$\Rightarrow 2<2x<4$

$\Rightarrow 2+a<2x+a<4+a.$

$\Rightarrow 2+a<f^{\prime }(x)<4+a.$

The minimum value of f (x) is 2 + a and that of men value is 4 + a.

∴ 2 +a> 0and 4 + a> 0

$\Rightarrow$ a> -2 and a> -4.

$\Rightarrow$ a> -2.

∴The best value of a is -2.

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Yes, admissions are open at Shree Swaminarayan Physiotherapy College. Eligible students can apply for direct admission either through State or Management quota.

MIT School of Vedic Sciences, MIT-ADT University BSc students get internships, research projects, and skill building courses. It improves academic knowledge and practical experience, preparing students for careers in psychology, mental health services, etc.

We have, f (x) = x 100 + sin x - 1.

So, f (x) = 100x⁹⁹ + cosx.

(A) When x (0,1). We get.

x>0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100 x⁹⁹> 0.

Now, 0 radian = 0 degree

and 1 radian = $\frac{180^{°}\times 1}{\pi }=\frac{180^{°}}{\left(\raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\right)}=57^{°}.$

So, cosx> 0 for x∈ (0,1) radian = (0,57)

∴f (x) > 0 for x∈ (0,1).

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

So, x> 1

$\Rightarrow$ x⁹⁹> 1

$\Rightarrow$ 100x⁹⁹> 100. $\left\{\begin{array}{cc}\hfill ?\left(\frac{\pi }{2},\pi \right)=(1.57,3.14)\hfill & \hfill which is great than 1\hfill \end{array}\right\}$

And cosx is negative between -1 and 0.

So, f (x) = 100x⁹⁹ + cosx> 100 - 1 = 99 > 0.

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

(c) When x ∈$\left(0,\frac{\pi }{2}\right),$ we get,

$\Rightarrow$ x> 0

$\Rightarrow$ x⁹⁹> 0

$\Rightarrow$ 100x⁹⁹> 0

and cosx> 0. (firstquadrant).

I e, f(x) > 0.

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

Hence, option (D) is correct.

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The application process for Shree Swaminarayan Physiotherapy College is conducted online. Students can apply through the official ACPMEC website or opt for the Management Quota by applying directly via the college's website. Additionally, the college offers the option to fill out the admission form through its official website.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii) is the correct answer.

The oxidation states are given below-

In NH₂OH oxidation of N is -1.

NH₄NO₃ exists as NH₄⁺.NO₃^-,    

Thus, the oxidation state of N in NH₄⁺ is -3 while in NO₃^? is +5.

In N₂H₄ oxidation of N is -2

In N₃H oxidation of N is $\frac{-1}{3}$

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B.N.M. Institute of Technology offers seats to candidates based on their entrance exam scores. Aspirants are first required to register and appear for any of the accepted entrance exams. The institute considers KCET/ COMEDK for BE, PGCET for ME and KMAT/ Karnataka PGCET for MBA programmes. Candidates have to further participate in counselling by respective exam-conducting bodies. Aspirants who are allotted seat in the institute need to pay a course fee to confirm their admission.

Bharati Vidyapeeth's Institute of Management Studies and Research provides great internships to its students. The placement rate for summer internships in the last three years is given in the table below:

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) is the correct answer

In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.

We have, f (x) = x²-x + 1.

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

Atf (x) = 0.

$\Rightarrow$ 2x - 1 = 0

I e, x = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ divides the real line into two disjoint interval the interval ( -1, 1) into

Two disjoint interval

$\left(-1,\frac{1}{2}\right)\left(\frac{1}{2},1\right)$

And f (x) is strictly increasing in $\left(\frac{1}{2},1\right)$ and strictly decreasing in $\left(-1,\frac{1}{2}\right)$

Hence, f (x) is with a increasing or decreasing on ( -1,1).

We have, f (x) = log x.

So, f (x) = $\frac{d}{dx}(logx)=\frac{1}{x}.$

i e, f (x) $=\frac{1}{x}$ > 0. When $x\ne 0x\in (0,\infty )$

Hence, the logarithmic fx is strictly increasing on $(0,\infty ).$

MIT School of Vedic Sciences, MIT-ADT University BSc (H) is a four-year UG course. It offers specialisations in Psychology with a curriculum designed to integrate academic learning and skill-based modules for comprehensive student development.

The highest package recorded during the placement drive in the recent years is given in the table below:

We have, y = $\frac{4sin\theta }{(2+cos\theta )}-\theta$

Differentiating w rt.Ø we get,

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left[\frac{4sin\theta }{2+cos\theta }-\theta \right]$

$=\frac{(2+cor\theta )\frac{d}{d\theta }(4sin\theta )-4sin\theta \frac{d}{d\theta }(2+cor\theta )}{{(2+cos\theta )}^2}-\frac{d\theta }{d\theta }.$

$=\frac{(2+cos\theta )(4cos\theta )-4sin\theta (-sin\theta )}{{(2+cos\theta )}^2}-1$

$=\frac{8cos\theta +4co{s}^2\theta +4si{n}^2\theta \cdot }{{(2+cos\theta )}^2.}-1$

$=\frac{8cos\theta +4\left(co{s}^2\theta +si{n}^2\theta \right)-{(2+cos\theta )}^2}{{(2+cos\theta )}^2}$

$=\frac{8cor\theta +4-\left[4+co{s}^2\theta +4cos\theta \right]}{{(2+cor\theta )}^2}.$

$=\frac{8cos\theta +4-4-co{s}^2\theta -4cos\theta }{{(2+cos\theta )}^2}$

$=\frac{4cos\theta -co{s}^2\theta }{{(2+cos\theta )}^2}=\frac{cos\theta (4-cos\theta )}{{(2+cos\theta )}^2}$

When $\theta \in \left[0,\frac{\pi }{2}\right]$ we know that, $0\le cos\theta \le 1.$

So, $4-cos\theta >0$

And also, (2 + cos$\theta$)²> 0.

$\therefore \frac{dy}{d\theta }\ge 0\forall \theta \in \left[0,\frac{\pi }{2}\right]$

Hence, y is an increasing fxⁿ of $\theta$ in $\left[0,\frac{\pi }{2}\right]$