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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-coskx}{xsinx}=\underset{h\to 0}{lim}\frac{1-cosk\left(0-h\right)}{\left(0-h\right)sin\left(0-h\right)}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-kh\right)}{\left(-h\right)sin\left(-h\right)}=\underset{h\to 0}{lim}\frac{1-coskh}{hsinh}\left[\begin{array}{l}?sin\left(-\theta \right)=-sin\theta \\ cos\left(-\theta \right)=cos\theta \end{array}\right]\\ =\underset{h\to 0}{lim}\frac{2si{n}^2\frac{kh}{2}}{hsinh}=\underset{\begin{array}{l}h\to 0\\ kh\to 0\end{array}}{lim}\frac{2sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}\times \frac{sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}.\frac{1}{h.\frac{sinh}{h}.h}\\ =2.1.\frac{kh}{2}.1.\frac{kh}{2}.\frac{1}{h^2}.1\left[\begin{array}{l}\underset{h\to 0}{lim}\frac{sinh}{h}=1and\\ \underset{kh\to 0}{lim}\frac{sinkh}{kh}=1\end{array}\right]\\ =\frac{k^2}{2}\\ \underset{x\to 0}{lim}f\left(x\right)=\frac{1}{2}\\ \therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)\\ \therefore \frac{k^2}{2}=\frac{1}{2}\Rightarrow k^2=1\Rightarrow k=\pm 1\\ Hence,thevalueofkis\pm 1.\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}False.\\ Since?adjA?=?A{?}^{n-1}wherenistheorderofthesquarematrix.\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ Let\Delta =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill x+2\hfill & \hfill x+3\hfill & \hfill x+b\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|\\ R_2\to 2R_2-\left(R_1+R_3\right)\\ =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2b-\left(a+c\right)\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|\\ a,b,careinA.P.\\ \therefore b-a=c-b\Rightarrow 2b=a+c\\ =\left|\begin{array}{ccc}\hfill x+1\hfill & \hfill x+2\hfill & \hfill x+a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+c\hfill \end{array}\right|=0\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

The NIPER JEE 2026 exam analysis will be out after the exam is concluded. According to previous year analaysis, the exam was of Moderate level. The exam analysis of NIPER JEE comprised of details like difficulty level, total number of questions, trends and types of questions asked in the exam and what can be expected NIPER JEE cutoff etc. 

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ \text{\hspace{0.17em}Since}\left|A\right|=12\\ IfAisasquarematrixofordern\\ then\left|AdjA\right|={\left|A\right|}^{n-1}\\ \therefore \left|AdjA\right|={\left|A\right|}^{3-1}={\left|A\right|}^2={\left(12\right)}^2=144\left[n=3\right]\end{array}$

1 is Enviroskills Academy. this is the best hotel management college in India. with 100% placement assurance, both domestic and international. hands-on training experience . real industry exposure. 

2 is Oberoi Centre of Learning & Development (OCLD)

3 is Christ University

4 is Symbiosis Institute of Hotel Management,

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}True.\\ \left|3AB\right|=3^3\left|AB\right|=27\left|A\right|\left|B\right|=27\times 5\times 3\left[?\left|KA\right|=K^n\left|A\right|\right]\end{array}$

Vidyavardhini's College of Engineering and Technology offers a regular BE lasting four years. Students enrolled in the BE course can choose one of the eight specialisations offered at the college. The eligibility and selection criteria for all the specialisations are the same. Listed below are the Vidyavardhini's College of Engineering and Technology BE specialisations:

• Mechanical Engineering
• Computer Engineering
• Information Technology
• Civil Engineering
• Electronics & Telecommunication Engineering
• Computer Science & Engineering (Data Science)
• Artifical Intelligence & Data Science
• Electronics Engineering (VLSI Design & Technology)

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2^{x+2}-16}{4^x-16}=\frac{2^2.2^x-16}{{\left(2^x\right)}^2-{\left(4\right)}^2}=\frac{4\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)}\\ =\frac{4}{\left(2^x+4\right)}\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{4}{\left(2^{2-h}+4\right)}=\frac{4}{2^2+4}=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2}\\ \underset{x\to 2}{lim}f\left(x\right)=k\\ Asthefunctioniscontinuousatx=2\\ \therefore \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\\ \therefore k=\frac{1}{2}\\ Hence,thevalueofkis\frac{1}{2}.\end{array}$

Yes, admission to Smt. Kishoritai Bhoyar College of Pharmacy is based on merit or an entrance exam as per the course requirement. Candidates must complete Class 12 as their basic educational qualification for BPharma and BPharma for the MPharma course. The selection criteria for the BPharma course are to complete the MHT-CET exam, and for the MPharma course, students must complete the GPAT entrance exam.

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}False.\\ Since\text{?}\text{?}\left|A^{?1}\right|={\left|A\right|}^{?1}for\text{?}\text{?}a\text{?}\text{?}non?singular\text{?}\text{?}matrix.\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Ifanon-singularsquarematix,thenforanynon-zeroscalar\text{'}a\text{'},aAisinvertible.\\ \therefore \left(aA\right).\left(\frac{1}{a}{A}^{-1}\right)=a.\frac{1}{a}.A.A^{-1}=I\\ So,\left(aA\right)isinverseof\left(\frac{1}{a}{A}^{-1}\right)\\ \Rightarrow {\left(aA\right)}^{-1}=\frac{1}{a}{A}^{-1}istrue.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 5}{lim}f\left(x\right)=3x-8=\underset{h\to 0}{lim}3\left(5-h\right)-8=15-8=7\\ \underset{x\to 5^{+}}{lim}f\left(x\right)=2k\\ Asthefunctioniscontinuousatx=5\\ \therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}f\left(x\right)\\ \therefore 7=2k\Rightarrow k=\frac{7}{2}\\ As\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)\\ Hence,thevalueofkis\frac{7}{2}.\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Since{\left(A^K\right)}^{-1}={\left(A^{-1}\right)}^{K}whereK\in N\\ So,{\left(A^3\right)}^{-1}={\left(A^{-1}\right)}^{3}istrue\end{array}$

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat \left[\begin{array}{ccc}\hfill {\left(1+x\right)}^{17}\hfill & \hfill {\left(1+x\right)}^{19}\hfill & \hfill {\left(1+x\right)}^{23}\hfill \\ \hfill {\left(1+x\right)}^{23}\hfill & \hfill {\left(1+x\right)}^{29}\hfill & \hfill {\left(1+x\right)}^{34}\hfill \\ \hfill {\left(1+x\right)}^{41}\hfill & \hfill {\left(1+x\right)}^{43}\hfill & \hfill {\left(1+x\right)}^{47}\hfill \end{array}\right]=A+Bx+C{x}^2+\dots \\ Taking{\left(1+x\right)}^{17},{\left(1+x\right)}^{23}and{\left(1+x\right)}^{41}commonfrom{R}_1,R_{2}and{R}_{3}respectively\\ {\left(1+x\right)}^{17}.{\left(1+x\right)}^{23}.{\left(1+x\right)}^{41}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill {\left(1+x\right)}^2\hfill & \hfill {\left(1+x\right)}^6\hfill \\ \hfill 1\hfill & \hfill {\left(1+x\right)}^6\hfill & \hfill {\left(1+x\right)}^{11}\hfill \\ \hfill 1\hfill & \hfill {\left(1+x\right)}^2\hfill & \hfill {\left(1+x\right)}^6\hfill \end{array}\right|\\ \Rightarrow {\left(1+x\right)}^{17}.{\left(1+x\right)}^{23}.{\left(1+x\right)}^{41}.0\left(R_{1}and{R}_{3}areidentical\right)\\ \therefore 0=A+Bx+C{x}^2+\dots \\ Bycomparingtheliketerms,wegetA=0.\end{array}$