(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
Solution: \[\left( c \right)\text{ }\left( A\text{ }\to \text{ }5 \right),\text{ }\left( B\text{ }\to \text{ }1 \right),\text{ }\left( C\text{ }\to \text{ }4 \right),\text{ }\left( D\text{...
Solution: \[\left( d \right)\text{ }\left( A\text{ }\to 4 \right),\text{ }\left( B\to \text{ }1 \right),\text{ }\left( C\text{ }\to 2 \right),\text{ }\left( D\text{ }\to 3 \right)\] Isomerism...
Solution: (i) In the form of conditional statement, expression is If $p$, then $q$ So now, In the statement $p$ and $q$ are $p$: The number is odd. $q$: The square of odd number is odd. As a result,...
Solution: \[\left( a \right)\text{ }\left( A\text{ }\to \text{ }3 \right),\text{ }\left( B\text{ }\to 1 \right),\text{ }\left( C\text{ }\to \text{ }5 \right),\text{ }\left( D\to \text{ }2...
Solution: (i) The given statement is compound statement then components are, $P$: $\mid{x}\mid$ is equal to $x$. $\sim p$: $\mid{x}\mid$ is not equal to $x$. $q$: $\mid{x}\mid$ is equal to $–x$....
Solution: \[\left( b \right)\text{ }\left( A\text{ }\to 5 \right),\text{ }\left( B\text{ }\to \text{ }4 \right),\text{ }\left( C\text{ }\to \text{ }1 \right),\text{ }\left( D~\to 2 \right)\] ...
Solution: (i) The statement given is compound statement whose components are, $P$: 35 is a prime number $\sim p$: 35 is not a prime number. $q$: 35 is a composite number $\sim q$: 35 is not a...
and 
are the roots of Quadratic equation 
. Solution: (i) The sentence given is a compound statement whose components are $p$: $x = 2$ is a root of the Quadratic equation $x^{2}{-}5x + 6 = 0$. $\sim p$: $x = 2$ is not a root of the Quadratic...
Solution: \[\left( b \right)\text{ }\left( A\text{ }\to 4 \right),\text{ }\left( B\text{ }\to \text{ }3 \right),\text{ }\left( C\text{ }\to \text{ }2 \right),\text{ }\left( D\text{ }\to \text{ }1...
Solution: (i) The statement given is compound statement whose components are, $P$: All rational numbers are real. $\sim p$: All rational numbers are not real. $q$: All rational numbers are complex....
Solution: Ambidendate ligands are those having diverse two restricting destinations. Example: Isothiocyanato Thiocyanato and Nitrite-N Nitrito-O The kind of isomerism when ambidentate ligands are...
Solution: (i) The sentence given is a compound statement whose components are $p$: Hindi is the optional paper $q$: English is the optional paper It can now be represented in symbolic function as,...
Solution: In CuSO4.5H2O, there are water particles that go about as ligands. The electrons will invigorate to higher d orbital and show tone. While, in CuSO4, there are no water particles to go...
Solution: (i) The sentence given is a compound statement whose components are $p$: A number is divisible by 2 $q$: A number is divisible by 3 It can now be represented in symbolic function as,...
Solution: \[{{[CO{{F}_{\mathbf{6}}}]}^{\mathbf{3}-}}:\text{ }C{{o}^{\mathbf{3}+}}~({{d}^{\mathbf{6}}})\]
Solution: (i) The sentence given is a compound statement whose components are $p$: 2 is a factor of 12 $q$: 3 is a factor of 12 $r$: 6 is a factor of 12 It can can be represented in symbolic...
Solution: (a, c) NO2 and SCN are ambidentate ligands subsequently, show linkage isomerism.
Solution: hence, option a, b and c are correct
Solution:
Solution: (i) The sentence given here is a compound statement whose components are $p$: Rahul passed in Hindi $q$: Rahul passed in English It can now be represented in symbolic function as, $p\wedge...
Solution: (b, d) In complexes, [Fe(NH3)4Cl2]+ and [CO(NH3)4Cl2], metal is bonded to more than one sort of ligands subsequently, they are heteroleptic.
Solution: (a, c) In complexes [Co(NH3)6]3+ and [Ni(CN)4]2-, both Co and Ni are connected to one sort of ligands just subsequently, they are homoleptic. hence, option a and c are correct
Solution: (b, c) Aqueous pink arrangement of cobalt (II) chloride is because of electronic progress of electron from t2g to eg energy level of [Co(H2O)6]2+ complex. At the point when overabundance...
Solution: hence, option a and c are correct for [Fe(CN)6]3- complex
Solution: so, option a and c are the correct answer
Solution: hence , a and c are the correct answer
Solution: (c) [Pt(NH3)2Cl(NO2)] is diamminechloridonitrito-N-platinum (II)
Solution: (b) The given mixtures have diverse number of water atoms inside and outside the organize circle.
Solution: (b) Ligand should give a pair of electrons or inexactly held electron pair to metal and shape a M – L bond,
Solution: (a) Thiosulphate or S2O3–isn't a chelating specialist since it is a monodentate ligand.
Solution: (d) [Co(SO4)(NH3)5]Br and [CO(SO4)(NH3)5]Cl address no isomerism since they are different compounds.
Solution: (a) The ligands having two distinctive holding locales are known as ambident ligands e.g., NCS, NO2, and so forth Here, NCS has two restricting locales at N and S. Subsequently, NCS...
Solution:
Solution:
Solution: (c) arrangement of cycle by linkage between metal particle and ligand balances out the coordination compound. The ligand which chelates the metal particle are known as chelating ligand....
Solution: (a) [Pt(NH3)2Cl2] is diamminedichloridoplatinum (II) .
Solution: (d) 3 mol of AgCl implies 3Cl are given in the arrangement henceforth, the equation of the complex will be [Cr(H2O)6]Cl3.
Solution: (b) One mole of AgNO3 accelerates one mole of chloride particle. In the above response, when 0.1 mole COCl3(NH3)5 is treated with abundance of AgNO3, 0.2 mole of AgCl are obtained hence,...
Solution:
Solution: (b) The greater the value of log K, the more prominent will be strength of complicated compound shaped. For response, For this response, log K has most elevated worth among the...
, 
, and 
terms of an A.P. and G.P. are both a, b and c respectively, show that 
Solution: Let m be the first term of the AP and d be the common difference Let I be the GP’s first term and s be the common ration The Ap’s nth term is given as $t_{n}=a+{(n-1)}d$ in which the first...
Solution: An AP’s sum of n terms is given by ${S_{n}}={\frac{n}{2}}(2a+(n-1)d)$ Where the first term is 'a' and the common difference is 'd'. It is given that $S_{p}=q$ and $S_{q}=p$ $\Rightarrow...
are in A.P., whose common difference is d, show that 
Solution: It is given that $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in the form of A.P., and $d$ is the common difference, We now need to prove that $\operatorname{Sec}...
are in A.P., where 
for all i, show that 
Solution: It is given that $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in the form of AP in which $a_{i}\lt{0}$ for all i. To prove that: $\begin{array}{l}...
(i) "Not p" is the negation of the assertion p. The negation of p is represented by $\sim p$. The truth value of $\sim p$ is the opposite of the truth value of p. The negation of the statement is...
(i) "Not p" is the negation of the assertion p. The negation of p is represented by $\sim p$. The truth value of $\sim p$ is the opposite of the truth value of p. The negation of the statement is...
(i) A compound statement is made up of two or more statements (Components). As a result, the elements of the given sentence "All living things have two eyes and two legs" are as follows: p: All...
(i) A compound statement is made up of two or more statements (Components). As a result, the elements of the provided statement "57 is divisible by 2 or 3" are as follows: p: 57 is divisible by 2....
A compound statement is made up of two or more statements (Components). As a result, the elements of the given statement "A rectangle is a quadrilateral or a 5-sided polygon" are as follows: p: A...
So, \[7\text{ }=\text{ }2k\] \[k\text{ }=\text{ }7/2\text{ }=\text{ }3.5\] SO VALUE OF K IS \[3.5\]
to (i) 
terms (ii) 10 termsSolution: As per the question $\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots$ Let's assume the series be...
and 
terms of a G.P. are 
and 
respectively, show that its 
term is 
Solution: The $n^{\text {th }}$ term of Geometric Progression is given by $t_{n}=a r^{n-1}$ in which the first term is $a$ and the common difference is $r$ $\mathrm{p}^{\text {th }}$ term is given...
(i) A compound statement is made up of two or more statements (Components). As a result, the parts of the supplied statement "Plants use sunshine, water, and carbon dioxide for photosynthesis" are...
(i) A compound statement is made up of two or more statements (Components). As a result, the components of the provided statement "The number 100 is divisible by 3, 11, and 5" are as follows. p: 100...
terms is zero, show that the sum of its next q terms is 
. [Hint: Required sum ![\left.=\mathrm{S}_{\mathrm{p}+\mathrm{q}}-\mathrm{S}_{\mathrm{p}}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-43a61029dbbcc74f00d776544ddd71be_l3.png "Rendered by QuickLaTeX.com")
Solution: It is given that the first term is ' $a$ ' and the sum of first '$p$' terms is $S_{p}=0$ We now need to find the sum of the next 'q' terms So, the total terms are $p+q$ As a result, Sum of...
(i) If a statement is true or false but not both, it is a declarative sentence. "The sum of opposite angles of a cyclic quadrilateral is 180°," according to quadrilateral characteristics. As a...
(i) If a statement is true or false but not both, it is a declarative sentence. "Where is your bag?" is a question in this context. As a result, it is not a statement. (ii) Every square is a...
(i) If a statement is true or false but not both, it is a declarative sentence. As a result, the expression "15 + 8 > 23" is incorrect. Because the L.H.S result will always equal the R.H.S...
(i) If a statement is true or false but not both, it is a declarative sentence. The sentence "sky is red" is incorrect. As a result, the statement is false. (ii) If a statement is true or false but...
Solution: They vary in the quantity of combined and unpaired electrons. A solid field ligand will cause blending of electrons while a feeble field ligand won't cause matching. Blending or not...
Solution: The expanding request of gem field energy is [Cr(Cl)6]3–<[Cr(NH3)6]3+ <[Cr(CN)6]3– This is additionally the request for field strength of the ligands as indicated by the...
Solution: For [Fe(H2O)6]3+, H2O is a powerless field ligand will not cause matching of electrons. Along these lines, the quantity of unpaired electrons will be 5. [Fe(CN)6]3–, Fe3+ has six unpaired...
Solution: For tetrahedral edifices, the gem field parting energy is excessively low. It is lower than blending energy in this way, the matching of electrons isn't supported and consequently the...
Solution: The electronic design will be t42g e2g. It has 4 unpaired electron and paramagnetic. With weal ligand Δ0 < p. The design with solid field ligand will be t62g e0g. the Δ0 > p and...
The curves are y = √x and line x = 2y + 3 Solving y = √x and x = 2y + 3, we get
Solution: An attractive snapshot of 5.92 BM implies there are 5 unpaired electrons in light of the fact that \[Attractive\text{ }Moment\text{ }=\text{ }\surd \text{ }n\left( n+2 \right)\] The...
The curve y = –x2 or x2 = –y and the line x + y + 2 = 0 Solving the two equation, we get \[x\text{ }-\text{ }{{x}^{2}}~+\text{ }2\text{ }=\text{ }0\] \[{{x}^{2}}~\text{ }-x\text{ }\text{ }-2\text{...
Solution: The design must be cis-octahedral. Model for a particularly mind boggling is [Co(en)2Cl2]+ which is optically dynamic.
Solution: Assuming it structures silver chloride, there is without one chlorine iota outside the coordination circle. The primary recipe must be [Cr(H2O)4Cl2]Cl. The name of this complex is...
Solution: The expanding request of conductivity is as per the following: [Co(NH3)3Cl3]<[Co(NH3)4Cl2]Cl< [Cr(NH3)5Cl]Cl2<[Co(NH3)6]Cl3
The equations of curve y = √x and line y = x Solving the equations y = √x ⇒ y2 = x and y = x, we get \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }x \\ {{x}^{2}}~-\text{ }x\text{ }=\text{ }0 \\...
Solution: Viable atomic charge increments as we move along the period from left to right. Because of the explanation, the size likewise lessens. Subsequently the electrons will be held all the more...
Solution: Electrons are filled by the n+l rule. Assuming an orbital has lower n+l esteem, the electron will enter that orbital. \[\begin{array}{*{35}{l}} For\text{ }3d,\text{ }n+l=\text{...
\[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( {{a}^{2}}~-~{{x}^{2}} \right)\text{ }\Rightarrow ~{{y}^{2}}~=~{{a}^{2}}~-~{{x}^{2}} \\ {{x}^{2~}}+~{{y}^{2}}~=~{{a}^{2}} \\ \end{array}\] which is...
Solution: Halides become more covalent with expanding oxidation state. As the oxidation state expands, the charge on the iota increments and the size of the particle of progress component...
Solution: The decreased type of Cu2+ is more steady than the oxidized type of Cu. In this manner, the worth of E° is positive for Cu. Eliminating two electrons gives a steady setup [Ar]3d10 with...
Solution: In acidic medium, permanganate changes to manganous particle which is lackluster. \[MnO4-+8H+\text{ }+\text{ }5e-\to \text{ }Mn2+\text{ }+\text{ }4H2O\] (drab) In basic...
The curve is \[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( x~-\text{ }1 \right) \\ \Rightarrow \text{ }{{y}^{2}}~=\text{ }x\text{ }-\text{ }1 \\ \end{array}\] Plotting the curve and finding...
Solution: At the point when Cr2O72–is treated with an antacid: \[\left( orange \right)\text{ }Cr2O72+\text{ }OH-\to \text{ }2CrO42-\left( yellow \right)\] At the point when the yellow...
Solution: This is a redox titration. The profound purple shade of KMnO4 vanishes because of the development of MnSO4. \[\begin{array}{*{35}{l}} ~ \\ 5H2C2O4\text{ }+\text{ }2KMnO4\text{...
Given, {(x, 0) : \[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( 4\text{ }~-{{x}^{2}} \right)\ \\ So,\text{ }{{y}^{2}}~=\text{ }4\text{ }-\text{ }{{x}^{2}} \\ {{x}^{2}}~+\text{...
Solution: At the point when earthy colored co pound of manganese (A) is treated with HCl it gives chlorine gas. \[MnO2\text{ }+\text{ }4HCl\text{ }\to \text{ }MnCl2\text{ }+\text{ }Cl2\text{...
The equation of line x = 2 and parabola y2 = 8x Putting value of x in the other equation, we have \[\begin{array}{*{35}{l}} {{y}^{2}}~=\text{ }8\left( 2 \right) \\ {{y}^{2}}~=\text{ }16 \\...
Solution: \[~2Cu2+\text{ }+\text{ }4I-\to \text{ }Cu2I2\text{ }+\text{ }I2\] Cu2+ gets decreased to Cu+, and I–gets oxidized to I2.
The curves are y2 = 9x and y = x Solving the above equations, we have \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }9x\text{ }\Rightarrow \text{ }{{x}^{2}}~\text{ }-9x\text{ }=\text{ }0 \\ x\left(...
Solution: (b, c) Cerium shows +4 oxidation state likewise on the grounds that it tends to accomplish respectable gas setup and achieve f° design. Ce – 4f15d'6s2 (Ce4+–4f°)
Solution: (b, c) An animal types can go about as an oxidizing specialist just when metal is available in high oxidation state however lower oxidation state show strength. As higher oxidations...
The curves are y2 = 4x … (i) and x2 = 4y … (ii) On solving the equations, we get From (ii), y = x2/4 Putting value of y in (i), we have \[\begin{array}{*{35}{l}} {{\left( {{x}^{2}}/4...
Solution: (a, b) Cr and Co structure MF3 kind of mixtures. The capacity of fluorine to balance out the most elevated oxidation state is because of higher grid energy in CoF3 and higher bond enthalpy...
Solution: (b, c) Mn2+ (3d5) and Fe2+ (3d6) have 5 and 4 unpaired electrons thus higher upsides of twist just attractive second when contrasted with Ti3+ (3d1) and Co2+ (3d7).
The curves are y = x3, y = x + 6 and x = 0 On solving y = x3 and y = x + 6, we have \[\begin{array}{*{35}{l}} {{x}^{3}}~=\text{ }x\text{ }+\text{ }6 \\ {{x}^{3}}~\text{ }-x\text{ }-\text{ }6\text{...
Given curves are y2 = 9x and y = 3x solving the two equations we have \[\begin{array}{*{35}{l}} {{\left( 3x \right)}^{2}}~=\text{ }9x \\ 9{{x}^{2}}~=\text{ }9x \\ 9{{x}^{2}}~\text{ }-9x\text{...
Solution: (b, c)
Solution:
Solution: (b, d) Np and Pu show +7 oxidation state.
Solution:
Solution: (b, c) In d-block components, for heavier components, the higher oxidation states are more steady. Thus, Mo(VI) and W(VI) are more steady than Cr (VI). That is the reason, Cr (VI) as...
Solution: (a, d) Co2+ (3d7) and Cr3+ (3d3) have 3 unpaired electrons. Thus they have practically same twist just attractive second.
Solution:
Solution:
Solution: (a, b) KMnO4 and Ce(S04)2 are shaded because of charge move
Solution: (b) HCl isn't utilized to make the medium acidic in oxidation responses of KMnO4 in acidic medium. The explanation is that in case HCl is utilized, the oxygen delivered from KMnO4 + HCl is...
Solution:
Solution: (c) Zirconium and hafnium have comparable nuclear range thus they show comparable physical and synthetic properties.
Solution:
audio cassettes and 
videocassettes cost 
, while 
audio cassettes and 
videocassettes cost 
. Find the cost of audio cassettes and a video cassette.Let’s assume the cost of an audio cassette and that of a video cassette be ₹a and ₹b, respectively. Then forming equations according to the question, we have $7a+3b=1110$…(a) $5a+4b=1350$… (b) On...
Solution: (d) The most noteworthy oxidation condition of manganese in fluoride is +4 (MnF4) yet in oxides it is +7 (Mn2O7) in light of the fact that in covalent mixtures fluorine can frame single...
Solution: (c) When fermented K2Cr2O7 arrangement is added to Sn2+ salt, Sn2+ changes to Sn4+. The response is given underneath:
Solution: (a) Copper doesn't free hydrogen from acids. \[\begin{array}{*{35}{l}} Cu\text{ }+\text{ }2H2S04\text{ }\text{ }>\text{ }CuSO4\text{ }+\text{ }S02\text{ }+\text{ }2H2O \\ ~ \\...
Solution: \[\left( c \right)\text{ }2KMnO4\text{ }+\text{ }KI\text{ }+\text{ }H2O\text{ }\text{ }>\text{ }2K0H\text{ }+\text{ }2MnO2\text{ }+\text{ }KI03\]
Solution: (b) The attractive second is related with its twist precise force and orbital rakish energy.
Solution: (d) Interstitial mixtures are synthetically idle.
Solution: (a) 64Gd: [Xe] 4f7 5d1 6s2
Solution: (a) V2O5 and Cr2O3 are amphoteric oxides on the grounds that both respond with alkalies just as acids. Keep in mind: In lower oxides, the fundamental person is transcendent while in higher...
Solution:
Solution: (c) Tm (Thulium) is a lanthanoid.
Solution: (d) When KMnO4 arrangement is added to oxalic corrosive arrangement, the decolourisation is delayed first and foremost yet becomes immediate after some time in light of the fact that Mn2+...
Solution:
Solution: (b) +3 oxidation state is generally normal for all lanthanoids.
Solution: (b) The more noteworthy the quantity of unpaired electron, the higher will be its worth of attractive second. Since, 3d5 has 5 unpaired electrons subsequently most noteworthy attractive...
Solution: \[\left( a \right)\text{ }2KMnO4\text{ }+\text{ }2H2SO4\left( Conc\text{ } \right)\text{ }\text{ }>\text{ }Mn2O7\text{ }+\text{ }2KHSO4\text{ }+\text{ }H2O\]
Solution: (b) Cu2+ has 1 unpaired electron in CuF2, consequently, it is hued in strong state.
Solution: (d) On moving passed on to directly along period, metallic span diminishes while mass increments. Diminishes in metallic span combined with expansion in nuclear mass outcomes in...
Solution: (a) Cu(II) is more steady because of more noteworthy compelling atomic charge of Cu(II).
Solution:
, and coefficients of 
and 
terms in the binomial expansion of 
are equal, then
(A) $n=2 r$ Explanation: Given expression is $(1+x)^{2 n}$ $T_{3 r}=T_{(3 r-1)+1}={ }^{2 n} C_{3 r-1} x^{3 r-1}$ $T_{r+2}=T_{(r+1)+1}={ }^{2 n} \mathrm{C}_{r+1} x^{r+1}$ ${ }^{2 n} C_{3 r-1}={ }^{2...
in the expansion of 
Given expression is, $\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Considering $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Using standard formula, we get...
occurs in the expansion of 
. Prove that its coefficient is 
Given expression is $\left(x^{2}+\frac{1}{x}\right)^{2 n}$ Using the standard formula, we get, $T_{r+1}={ }^{2 n} C_{r}\left(x^{2}\right)^{2 n-r}\left(\frac{1}{x}\right)^{r}={ }^{2 n} C_{r} x^{4 n-3...
if the sum of odd terms is denoted by 
and the sum of even term by 
. Then prove that
(i) We know, $(x+a)^{n}=$ ${ }^{n}C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots$ Sum of odd terms will be, $O={ }^{n} C_{0} x^{n}+{ }^{n}...
(B) 
(C) 
(D) None of theseSolution: Option (C) 45 is correct. Explanation: It is given that the third term of G.P, ${{T}_{3}}~=\text{ }4$ We need to find the product of first 5 terms It is known that, ${{T}_{n}}~=\text{...
is 
Given, expression is $\left(x-\frac{1}{x}\right)^{2 n}$. Since the index is given as $2 n$, which is even. So, there is only one middle term, i.e., $\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th...
is a real number and if the middle term in the expansion of 
Is 1120 , find .Given expansion is $\left(\frac{p}{2}+2\right)^{8}$ Since the index is given as $n=8$, there is only one middle term, i.e., $\left(\frac{8}{2}+1\right)$ th $=5^{\text {th }}$ term $T_{5}=T_{4+1}={...
in the expansion of 
Given expression is $\left(1+x+x^{2}+x^{3}\right)^{11}$ $=\left[(1+x)+x^{2}(1+x)\right]^{11}=\left[(1+x)\left(1+x^{2}\right)\right]^{11}=(1+x)^{11} \cdot\left(1+x^{2}\right)^{11}$ $=\left({ }^{11}...
, if the coefficients of 
and 
terms in the expansion of 
are equal.Given function is $(1+x)^{18}$ Now, $(2 r+4)^{\text {th }}$ term will be equal to $T_{(2 r+3)+1}$ $T_{(2 r+3)+1}={ }^{18} C_{2 r+3}(x)^{2 r+3}$ And $(r-2)^{\text {th }}$ term, that is $T_{(r-3)...
in the expansion of 
Given function is $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ We may write the provided expression as $T_{r+1}$ using the conventional formula, $T_{r+1}={ }^{15}...
Solution: Let the sum got by the lead position group be a Rs and d be distinction in sum As the thing that matters is same subsequently the runner up will get a – d and the third spot a – 2d, etc...
Solution: Given at start he needs to run 24m to get the main potato then 28 m as the following potato is 4m away from first, and so on Thus the succession of his running will be 24, 28, 32 … There...
in the expansion of 
Given function is $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$ Expansion of the function is, $=\left(1-3 x+7 x^{2}\right)\left({ }^{16} C_{0}-{ }^{16} C_{1} x^{1}+{ }^{16} C_{2} x^{2}+\ldots+{ }^{16}...
Solution: Say ABC is a triangle with $AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }20$ cm Let’s say that D, E and F are respectively the midpoints of AC, CB and AB that are joined to...
Solution: It is given that the sum of interior angles of a polygon having ‘n’ sides is denoted by $(n-2)\times {{180}^{\circ }}$ The sum of angles having 3 sides i.e n $=\text{ }3\text{ }is\text{...
, in the expansion of 
Given function is $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$ We know, the standard formula of $T_{r+1}$ will be, $T_{r+1}={ }^{15} C_{r}\left(\frac{3...
Solution: It is given that on the day first he made five frames then 2 frames more than the previous i.e. 7 and then 9 and so on Therefore the making of frames each day forms a sequence of 5, 7, 9…...
Solution: It is given to us that in first month the man’s salary is Rs.5200 and then it increases by Rs.320 every month Therefore, 5200, 5200 + 320, 5200 + 640… will be the sequence so formed of his...
Solution: It is given that the total amount saved is 66000 in 20 years, therefore ${{S}_{20}}~=\text{ }66000$ Let ‘a’ be the amount saved in first year. Then every year if he increases Rs. 200 every...
Solution: Option (B) 12 is correct. Explanation: It is known that, ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ $=\frac{n !}{r !(n-r) !}$ Let total time of handshakes $= {}^n{C}_2 = 66$...
Solution: Option (C) 7200 is correct. Explanation: It is known that, ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ $=\frac{n !}{r !(n-r) !}$ 4 be the total number of given vowel 5 be the total number...
Solution: Option (A) 216 is correct Explanation: Using digits $0,1,2,4,5$ the 5-digit numbers that can be formed $$\begin{tabular}{|l|l|l|l|l|} \hline 4 & 4 & 3 & 2 & 1 \\ \hline \end{tabular}$$ $4...
Solution: Option (C) 24 is correct. Explanation: It is known that, $\begin{array}{l} { }^{n} P_{r}= \\ \frac{n !}{(n-r) !} \end{array}$ ${ }^{4} \mathrm{P}_{4}=4 !=24$ is the four digit numbers that...
Solution: Option (B) 64 is correct. Explanation: The coin is tossed 6 times. 2 is the number of possible outcomes When a coin is tossed 6 times the number of possible outcomes is $2^ {6} = 64$ As a...
, then is equal to A. 20 B. 12 C. 6 D. 30Solution: Option (A)20 is correct. Explanation: As per the question, ${ }^{\mathrm{n}} \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_{8}$ It is known that, $\begin{array}{l} { }^{\mathrm{n}}...
Solution: It is known that ${ }^{n} C_{r}$ $=\frac{n !}{r !(n-r) !}$ (i) At least three girls The total no. of ways in which the team can have at least three girls $={ }^{4} \mathrm{C}_{3}{ }^{7}...
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Solution: It is known that ${ }^{n} C_{r}$ $=\frac{n !}{r !(n-r) !}$ (i) No girls The total no. of ways the team can have no girls $={ }^{4} \mathrm{C}_{0}{ }^{7} \mathrm{C}_{5}=21$ (ii) at least...
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Solution: It is known that, $=\frac{n !}{r^{n}} C_{r}$ In the following two ways a team of 11 students can be constituted (i) Five students from class $\mathrm{XI}$ and six students from...
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Solution: It is known that, $=\frac{{ }^{n} C_{r}}{r !(n-r) !}$ As per the question, Out of 16, 11 players can be selected $={ }^{16} \mathrm{C}_{11}$ (i) include 2 particular players $={ }^{14}...
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Solution: We know that, $\begin{array}{l} { }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \\ =\frac{n !}{r !(n-r) !} \end{array}$ As per the question, No. of white marbles is 6 No. of red marbles is 5...
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(a) The distance of any point on the parabola from its focus and its directrix is same. Given that, directrix, x = 0 and focus = (6, 0) If a parabola has a vertical axis, the standard form of the...
Solution: We know that, $\begin{array}{l} { }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \\ =\frac{n !}{r !(n-r) !} \end{array}$ As per the question, No. of white marbles is 6 No. of red marbles is 5...
the equation of a circle having centre (h, k), having radius as r units, is (x – h)2 + (y – k)2 = r2Centre lies on the line i.e., y = x – 1, Co – Ordinates are (h, k) = (h, h – 1)...
Given \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }-2x\text{ }+\text{ }{{y}^{2}}~\text{ }-4y\text{ }-\text{ }20\text{ }=\text{ }0 \\ {{x}^{2}}~\text{ }-2x\text{ }+\text{ }1\text{ }+{{y}^{2}}~\text{...
Using Pythagoras Theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 = (3)2 + (√29)2 = 29 + 9 = √38 Hypotenuse = √38 units (radius) Since, the radius bisects the chord into two equal halves, Since,...
the equation of a circle having centre (h, k), having radius as r units, is \[{{\left( x\text{ }-\text{ }h \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }k \right)}^{2}}~=\text{ }{{r}^{2}}\ldots...
Since, diameters of a circle intersect at the centre of a circle, \[\begin{array}{*{35}{l}} 2x\text{ }-\text{ }3y\text{ }=\text{ }5\text{ }\ldots \ldots \ldots 1 \\ 3x\text{ }-\text{ }4y\text{...
equations are, y = mx + 1 & y2 = 4x By solving given equations we get (mx + 1)2 = 4x Expanding the above equation we get \[{{m}^{2}}{{x}^{2}}~+\text{ }2mx\text{ }+\text{ }1\text{ }=\text{ }4x\]...
Given, \[L\text{ }=\text{ }200\left( 10\text{ }\text{ }t \right)2\] where L addresses the quantity of liters of water in the pool. On separating both the sides w.r.t, t, we get \[dL/dt\text{...
equation of an ellipse is y2 = 4ax, Also we have length of latus rectum = 4a Now by comparing the above two equations, 4a = 8 Therefore \[\begin{array}{*{35}{l}} a\text{ }=\text{ }2 \\...