How many shoot apical meristems are likely to be present
Question: How many shoot apical meristems are likely to be present in a twig of a plant possessing 4 branches and 26 leaves a. 26 b. 1 c. 5 d. 30 e. 4 Solution: Option (c)5 is the answer...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{1-\cos x} d x$ Solution: Let $I=\int \frac{1}{1-\cos x} d x$ On multiplying and dividing $(1+\cos x)$, we can write the integral as $I=\int \frac{1}{1-\cos x}\left(\frac{1+\cos x}{1+\cos x}\right) d x$ $\Rightarrow I=\int \frac{1+\cos x}{(1-\cos x)(1+\cos x)} d x$ $\Rightarrow I=\int \frac{1+\cos x}{1-\cos ^{2} x} d x$ $\Rightarrow I=\int \frac{1+\cos x}{\sin ^{2} x} d x\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $\Rightar...
Read More →Show that
Question: Show that $\lim _{x \rightarrow 0} \frac{1}{|x|}=\infty$. Solution: Let $x=0+h$, when $x$ is tends to $0^{+}$ Since x tends to 0, h will also tend to 0. Right Hand Limit(R.H.L): $\lim _{x \rightarrow 0^{+}} f(x)$ $=\lim _{x \rightarrow 0^{+}} \frac{1}{|x|}$ $=\lim _{x \rightarrow 0^{+}} \frac{1}{(x)}$ $=\lim _{h \rightarrow 0^{+}} \frac{1}{(0+h)}$ $=\frac{1}{0}$ $=\infty$ Let $x=0-h$, when $x$ is tends to 0 - ince $x$ tends to $0, \mathrm{~h}$ will also tend to 0 . Left Hand Limit(L.H....
Read More →In which of the following pairs of parts
Question: In which of the following pairs of parts of a flowering plant is the epidermis absent? a. Root tip and shoot tip b. Shoot bud and floral bud c. Ovule and seed d. Petiole and pedicel Solution: Option (a)Root tip and shoot tip is the answer....
Read More →Phellogen and Phellem respectively denote
Question: Phellogen and Phellem respectively denote a. Cork and cork cambium b. Cork cambium and cork c. Secondary cortex and cork d. Cork and secondary cortex Solution: Option (b)Cork cambium and cork is the answer....
Read More →Interfascicular cambium and cork cambium
Question: Interfascicular cambium and cork cambium is formed due to a. Cell division b. Cell differentiation c. Cell dedifferentiation d. Redifferentiation Solution: Option (c) Cell dedifferentiation is the answer....
Read More →A conjoint and open vascular bundle
Question: A conjoint and open vascular bundle will be observed in the transverse section of a. Monocot root b. Monocot stem c. Dicot root d. Dicot stem Solution: Option (d)Dicot stem is the answer....
Read More →Epiblema of roots is equivalent to
Question: Epiblema of roots is equivalent to a. Pericycle b. Endodermis c. Epidermis d. Stele Solution: Option (c)Epidermisis the answer....
Read More →Cells of this tissue are living and show angular wall thickening.
Question: Cells of this tissue are living and show angular wall thickening. They also provide mechanical support. The tissue is a. Xylem b. Sclerenchyma c. Collenchyma d. Epidermis Solution: Option (c)Collenchymais the answer....
Read More →Show that
Question: Show that $\lim _{x \rightarrow 0} \frac{1}{x}$ does not exist. Solution: Let $x=0+h$ for $x$ tending to $0^{+}$ Since x 0, h also tends to 0 Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)$ $=\lim _{x \rightarrow 0^{+}} \frac{1}{x}$ $=\lim _{h \rightarrow 0^{+}} \frac{1}{0+h}$ $=\lim _{h \rightarrow 0^{+}} \frac{1}{+h}$ $=+\frac{1}{0}$ $=+\infty$ Let $x=0$-h for $x$ tending to $0^{-}$ Since $x \rightarrow 0$, h also tends to 0 . Left Hand Limit(L.H.L.): $=\lim _{x \rightar...
Read More →Identify the simple tissue system from the following
Question: Identify the simple tissue system from the following a. Parenchyma b. Xylem c. Epidermis d. Phloem Solution: Option (a)Parenchymais the answer....
Read More →Match the following and choose the correct
Question: Match the following and choose the correct option from below Options: a. A-iii, B-iv, C-i, D-ii b. A-i, B-ii, C-iii, D-iv c. A-iii, B-ii, C-iv, D-i d. A-iii, B-ii, C-i, D-iv Solution: Option (a)A-iii, B-iv, C-i, D-ii is the answer....
Read More →Match the followings and choose the correct
Question: Match the followings and choose the correct option from below Options: a. A-i, B-iii, C-v, D-ii, E-iv b. A-iii, B-i, C-ii, D-v, E-iv c. A-ii, B-iv, C-v, D-i, E-iii d. A-v, B-iv, C-iii, D-ii, E-i Solution: Option (b)A-iii, B-i, C-ii, D-v, E-iv is the answer....
Read More →Solve this
Question: Let $f(x)=\left\{\begin{array}{l}\cos x, x \geq 0 \\ x+k, x0\end{array}\right.$ Find the value of $k$ for which $\lim _{x \rightarrow 0} f(x)$ exist. Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x+k$ $=0+k$ $=k$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \cos x$ $=\cos (0)$ $=1$ It is given that $\lim _{x \rightarrow 0} f(x)$ exists. Therefore, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \ri...
Read More →A transverse section of the stem is stained
Question: A transverse section of the stem is stained first with safranin and then with fast green following the usual schedule of double staining for the preparation of a permanent slide. What would be the colour of the stained xylem and phloem? a. Red and green b. Green and red c. Orange and yellow d. Purple and orange Solution: Option (a)Red and greenis the answer....
Read More →Find the value
Question: Let $f(x)= \begin{cases}\frac{3 x}{|x|+2 x} , x \neq 0 \\ 0, x=0\end{cases}$ Show that $\lim _{x \rightarrow 0} f(x)$ does not exist. Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{3 x}{|x|+2 x}$ $=\lim _{x \rightarrow 0^{-}} \frac{3 x}{(-x)+2 x}$ $=\lim _{x \rightarrow 0^{-}} \frac{3 x}{x}$ $=\lim _{x \rightarrow 0^{-}} 3$ $=3$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{3 x}{|x|+...
Read More →Find the value
Question: Let $f(x)= \begin{cases}\frac{3 x}{|x|+2 x} , x \neq 0 \\ 0, x=0\end{cases}$ Show that $\lim _{x \rightarrow 0} f(x)$ does not exist. Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{3 x}{|x|+2 x}$ $=\lim _{x \rightarrow 0^{-}} \frac{3 x}{(-x)+2 x}$ $=\lim _{x \rightarrow 0^{-}} \frac{3 x}{x}$ $=\lim _{x \rightarrow 0^{-}} 3$ $=3$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{3 x}{|x|+...
Read More →Prove that
Question: Let $f(x)=\left\{\begin{array}{l}4 x-5, x \leq 2 \\ x-a, x2\end{array}\right.$ If $\lim _{x \rightarrow 2} f(x)$ exists then find the value of $a$. Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 4 x-5$ $=4(2)-5$ $=8-5$ $=3$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} x-a$ $=2-\mathrm{a}$ Since $\lim _{x \rightarrow 2} f(x)$ it exists, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+...
Read More →Seeds of some plants germinate immediately after shedding
Question: Seeds of some plants germinate immediately after shedding from the plants while in other plants they require a period of rest before germination. The latter phenomenon is called dormancy. Give the reasons for seed dormancy and some methods to break it. Solution: Reasons for seed dormancy - Impermeable and hard seed coat Presence of chemical inhibitors like abscisic acid Immature embryo Methods to break seed dormancy are Washing away of inhibitors due to rain Maturation of embryo Inacti...
Read More →How do you distinguish between hypogeal germination
Question: How do you distinguish between hypogeal germination and epigeal germination? What is the role of the cotyledon (s) and the endosperm in the germination of seeds? Solution: Role of Cotyledons and Endosperm in the germination of seeds They contain reserved food materials. When seed imbibes water, enzymes get activated, hydrolyze reserve food material and make it available for the germinating seed. Hypogeal germination is that epicotyl grows first and the only plumule is pushed out of the...
Read More →Find the value
Question: Let $f(x)=\left\{\begin{array}{l}5 x-4, \quad 0x \leq 1 \\ 4 x^{3}-3 x, 1x2\end{array}\right.$ Find $\lim _{x \rightarrow 1} f(x)$ Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 5 x-4$ $=5(1)-4$ $=5-4$ $=1$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 4 x^{3}-3 x$ $=4(1)^{3}-3(1)$ $=4-3$ $=1$ $\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$ Thus, $\lim _{x \rightar...
Read More →Sunflower is not a flower.
Question: Sunflower is not a flower. Explain. Solution: Sunflower is not a flower but a type of inflorescence called capitulum in which the receptacle is flattened. The whole cluster of florets gets surrounded by bracts, known as an involucre. Two kinds of florets are recognised in sunflower: (i) Ray Florets Arranged on the rim of a receptacle having distinct yellow and strap-shaped petals. These florets are female, sterile and are always zygomorphic and may be arranged in one or more whorls. (i...
Read More →Find the value
Question: Let $f(x)= \begin{cases}\frac{x-|x|}{x}, x \neq 0 \\ 2, x=0\end{cases}$ Show that $\lim _{x \rightarrow 0} f(x)$ does not exist Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x-|x|}{x}$ $=\lim _{x \rightarrow 0^{-}} \frac{x-(-x)}{x}$ $=\lim _{x \rightarrow 0^{-}} \frac{x+x}{x}$ $=\lim _{x \rightarrow 0^{-}} \frac{2 x}{x}$ $=\lim _{x \rightarrow 0^{-}} 2$ $=2$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \ri...
Read More →Stolon, offset and rhizome are different forms of stem modifications.
Question: Stolon, offset and rhizome are different forms of stem modifications. How can these modified forms of stem be distinguished from each other? Solution: Stolon is underground stems which spread to new niches and when old plants die, new are formed. It is a creeper. Offset is a lateral branch with short internodes and each node bearing a rosette of leaves and tuft of roots is found in aquatic plants. The rhizome is an underground stem growing parallel to soil surface which is differentiat...
Read More →Find the value
Question: Let $f(x)=\left\{\begin{array}{l}1+x^{2}, 0 \leq x \leq 1 \\ 2-x, x1\end{array}\right.$ Show that $\lim _{x \rightarrow 1} f(x)$ does not exist. Solution: Left Hand Limit(L.H.L.): $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1+x^{2}$ $=1+(1)^{2}$ $=1+1$ $=2$ Right Hand Limit(R.H.L.): $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2-x$ $=2-(1)$ $=2-1$ $=1$ $\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$ Thus, $\lim _{x \rightarrow...
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