Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x$ Solution: The denominator is factorized, so let separate the fraction through partial fraction, hence let $\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A}{(1+\sin x)}+\frac{B}{2+\sin x} \ldots \ldots$ (i) $\Rightarrow \frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A(2+\sin x)+B(1+\sin x)}{(1+\sin x)(2+\sin x)}$ $\Rightarrow \sin 2 x=A(2+\sin x)+B(1+\sin x)=2 A+A \sin x+B+B \sin x$ $\Rightarrow 2 \sin x \cos x...
Read More →Which of the following reactions of glucose
Question: Which of the following reactions of glucose can be explained only by its cyclic structure? (i) Glucose forms pentaacetate. (ii) Glucose reacts with hydroxylamine to form an oxime. (iii) Pentaacetate of glucose does not react with hydroxylamine. (iv) Glucose is oxidised by nitric acid to gluconic acid. Solution: Option (iii)Pentaacetate of glucose does not react with hydroxylamine is the answer....
Read More →Three cyclic structures of monosaccharides
Question: Three cyclic structures of monosaccharides are given below which of these are anomers. (i) I and II (ii) II and III (iii) I and III (iv) III is anomer of I and II Solution: Option (i)I and II is the answer....
Read More →Which of the following bases is not
Question: Which of the following bases is not present in DNA? (i) Adenine (ii) Thymine (iii) Cytosine (iv) Uracil Solution: Option (iv)Uracilis the answer....
Read More →Define * on Z by a * b = a – b + ab
Question: Define $*$ on $Z$ by $a * b=a-b+a b$. Show that $*$ is a binary operation on $Z$ which is neither commutative nor associative. Solution: * is an operation as $a * b=a-b+a b$ where $a, b \in Z$. Let $a=\frac{1}{2}$ and $b=2$ two integers. $\mathrm{a} * \mathrm{~b}=\frac{1}{2} * 2=\frac{1}{2}-2+\frac{1}{2} \cdot 2 \Rightarrow \frac{1-4}{2}+1=\frac{-3+2}{2} \Rightarrow \frac{-1}{2} \in \mathrm{Z}$ So, $*$ is a binary operation from $Z \times Z \rightarrow Z$. For commutative, $\mathrm{b}^...
Read More →Which of the following B group vitamins
Question: Which of the following B group vitamins can be stored in our body? (i) Vitamin B1 (ii) Vitamin B2 (iii) Vitamin B6 (iv) Vitamin B12 Solution: Option (iv) Vitamin B12is the answer....
Read More →DNA and RNA contain four bases each.
Question: DNA and RNA contain four bases each. Which of the following bases is not present in RNA? (i) Adenine (ii) Uracil (iii) Thymine (iv) Cytosine Solution: Option (iii)Thymine is the answer....
Read More →Each polypeptide is a protein has amino acids
Question: Each polypeptide is a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be ____________. (i) primary structure of proteins. (ii) secondary structure of proteins. (iii) the tertiary structure of proteins. (iv) quaternary structure of proteins. Solution: Option (i) primary structure of proteins. is the answer....
Read More →Which of the following statements is not
Question: Which of the following statements is not true about glucose? (i) It is an aldohexose. (ii) On heating with HI, it forms n-hexane. (iii) It is present in furanose form. (iv) It does not give 2,4-DNP test. Solution: Option (iii)It is present in furanose form is the answer....
Read More →Define * on N by m * n = 1 cm (m, n).
Question: Define $*$ on $\mathrm{N}$ by $\mathrm{m} * \mathrm{n}=1 \mathrm{~cm}(\mathrm{~m}, \mathrm{n})$. Show that $*$ is a binary operation which is commutative as well as associative. Solution: $*$ is an operation as $m * n=\operatorname{LCM}(m, n)$ where $m, n \in N$. Let $m=2$ and $b=3$ two natural numbers. m*n = 2*3 $=\operatorname{LCM}(2,3)$ $=6 \in \mathrm{N}$ So, $*$ is a binary operation from $\mathrm{N} \times \mathrm{N} \rightarrow \mathrm{N}$. For commutative, $\mathrm{n} * \mathrm...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{3}}{(x-1)(x-2)(x-3)} d x$ Solution: First we simplify numerator, we will rewrite denominator as shown below $\frac{x^{3}}{(x-1)(x-2)(x-3)}=\frac{x^{3}}{x^{3}-6 x^{2}+11 x-6}$ Add and subtract numerator with $\left(-6 x^{2}+11 x-6\right)$, we get $\frac{x^{3}-6 x^{2}+11 x-6+\left(6 x^{2}-11 x+6\right)}{x^{3}-6 x^{2}+11 x-6}$ $\Rightarrow=1+\frac{6 x^{2}-11 x+6}{x^{3}-6 x^{2}+11 x-6}$ $\Rightarrow=1+\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}$ T...
Read More →Nucleic acids are the polymers
Question: Nucleic acids are the polymers of ______________. (i) Nucleosides (ii) Nucleotides (iii) Bases (iv) Sugars Solution: Option (ii)Nucleotides is the answer....
Read More →Dinucleotide is obtained by joining two nucleotides
Question: Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are Are these linkages present? (i) 5 and 3 (ii) 1 and 5 (iii) 5 and 5 (iv) 3 and 3' Solution: Option (i)5 and 3 is the answer....
Read More →Which of the following acids is a vitamin?
Question: Which of the following acids is a vitamin? (i) Aspartic acid (ii) Ascorbic acid (iii) Adipic acid (iv) Saccharic acid Solution: Option (ii)Ascorbic acidis the answer....
Read More →Proteins are found to have two different types
Question: Proteins are found to have two different types of secondary structures viz. -helix and -pleated sheet structure. -helix structure of the protein is stabilised by : (i) Peptide bonds (ii) van der Waals forces (iii) Hydrogen bonds (iv) Dipole-dipole interactions Solution: Option (iii)Hydrogen bonds is the answer....
Read More →A binary operation * on the set
Question: A binary operation $*$ on the set $(0,1,2,3,4,5)$ is defined as $a * b= \begin{cases}a+b ; \text { if } a+b6 \\ a+b-6 ; \text { if } a+b \geq 6\end{cases}$ Show that 0 is the identity for this operation and each element a has an inverse (6 - a) Solution: To find: identity and inverse element For a binary operation if $a * e=a$, then e $s$ called the right identity If $\mathrm{e}^{*} \mathrm{a}=\mathrm{a}$ then $\mathrm{e}$ is called the left identity For the given binary operation, $e^...
Read More →Which of the following pairs
Question: Which of the following pairs represents anomers? Solution: Option (3) is the answer....
Read More →Let X be a nonempty set and * be a binary operation on
Question: Let $X$ be a nonempty set and $*$ be a binary operation on $P(X)$, the power set of $X$, defined by $A * B=$ $A \cap B$ for all $A, B \in P(X)$. (i) Find the identity element in P(X). (ii) Show that X is the only invertible element in P(X). Solution: $\mathrm{e}$ is the identity of $*$ if $\mathrm{e}^{*} \mathrm{a}=\mathrm{a}$ From the above Venn diagram, $A^{*} X=A \cap X=A$ $X^{*} A=X \cap A=A$ $\Rightarrow X$ is the identity element for binary operation $*$ Let B be the invertible e...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)} d x$ Solution: $\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}$ The denominator is factorized, so let separate the fraction through partial fraction, hence let $\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A}{(x-1)}+\frac{B}{x+1}+\frac{C}{2 x+3} \ldots \ldots$ (i) $\Rightarrow \frac{2 x-3}{(x-1)(x+1)(2 x+3)}$ $=\frac{A(x+1)(2 x+3)+B(x-1)(2 x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2 x+3)}$ $\Rightar...
Read More →Show that
Question: For $a l l a, b \in N$, we define $a * b=a^{3}+b^{3}$. Show that $*$ is commutative but not associative. Solution: let $a=1, b=2 \in N$ $a^{*} b=1^{3}+2^{3}=9$ And $b * a=2^{3}+1^{3}=9$ Hence * is commutative. Let c = 3 $\left(a^{*} b\right)^{*} c=9^{*} c=9^{3}+3^{3}$ $a^{*}\left(b^{*} c\right)=a^{*}\left(2^{3}+3^{3}\right)=1 * 35=1^{3}+35^{3}$ $\left(a^{*} b\right)^{*} c \neq a^{*}\left(b^{*} c\right)$ Hence * is not associative....
Read More →Solve this
Question: For all $a, b \in R$, we define $a^{*} b=|a-b|$. Show that $*$ is commutative but not associative. Solution: a*b = a - b if ab = - (a - b) if ba b*a = a - b if ab = - (a - b) if ba So a*b = b*a So * is commutative To show that * is associative we need to show $(a * b) * c=a *(b * c)$ Or ||$a-b|-c|=|a-| b-c||$ Let us consider cab Eg a = 1,b = - 1,c = 5 LHS: $|a-b|=|1+1|=2$ ||$a-b|-c|=|2-5|=3$ RHS $|b-c|=|-1-5|=6$ $|a-| b-c||=|1-6|=|-5|=5$ As LHS is not equal to RHS * is not associative...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x$ Solution: $\frac{x^{2}+1}{x\left(x^{2}-1\right)}=\frac{x^{2}+1}{x(x-1)(x+1)}$ The denominator is factorized, so let separate the fraction through partial fraction, hence let $\frac{\mathrm{x}^{2}+1}{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}-1}+\frac{\mathrm{C}}{\mathrm{x}+1} \ldots \ldots$ (i) $\Rightarrow \frac{\mathrm{x}^{2}+1}{\mathrm{x}(\mathrm{...
Read More →Show that * on R –{ - 1}, defined by
Question: Show that $*$ on $R-\{-1\}$, defined by $(a * b)=\frac{a}{(b+1)}$ is neither commutative nor associative. Solution: let $a=1, b=0 \in R-\{-1\}$ $a^{*} b=\frac{1}{0+1}=1$ And $b^{*} a=\frac{0}{1+1}=0$ Hence * is not commutative. Let c = 3. $(a * b) * c=1 * c=\frac{1}{3+1}=\frac{1}{4}$ $a *(b * c)=a * \frac{0}{3+1}=1 * 0=\frac{1}{0+1}=1$ Hence * is not associative....
Read More →Show that the set
Question: Show that the set $A=\{-1,0,1)$ is not closed for addition. Solution: For a set to be closed for addition, For any 2 elements of the set, say a and $b, a+b$ must also be a member of the given set, where a and $\mathrm{b}$ may be same or distinct In the given problem let $a=1$ and $b=1$ $a+b=2$ which is not in the given in set So the set is not closed for addition. Hence proved....
Read More →Let Q be the set of all positive rational numbers.
Question: Let Q be the set of all positive rational numbers. (i) Show that the operation $*$ on $\mathrm{Q}^{+}$defined by $\mathrm{a} * \mathrm{~b}=\frac{1}{2}(\mathrm{a}+\mathrm{b})$ is a binary operation. (ii) Show that $*$ is commutative. (iii) Show that $*$ is not associative. Solution: (i) Let $a=1, b=2 \in Q+$ $a * b=\frac{1}{2}(1+2)=1.5 \in Q+$ * is closed and is thus a binary operation on Q + (ii) $a * b=\frac{1}{2}(1+2)=1.5$ And $b^{*} a=\frac{1}{2}(2+1)=1.5$ Hence $*$ is commutative. ...
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