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Binomial Theorem — JEE MCQs with answers

NCERT Class 11 · Mathematics · Ch 710 free questions508 in the full bank

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Question 1 · 7.2 Binomial Theorem for Positive Integral Indices · Hard
Using the expansion of $(x-y)^n$ printed on p.130, the coefficient of $x^{3}y^{7}$ in $(x-y)^{10}$ is
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Answer: A — Option A is correct: the $y^7$ term is $(-1)^7\,{}^{10}C_{7}x^{3}y^{7}$, and ${}^{10}C_{7} = 120$, so the coefficient is $-120$. Option B is wrong: this drops the binomial coefficient altogether and keeps only the sign, but the printed expansion attaches ${}^{n}C_{k}$ to every term. Option C is wrong: this makes both slips at once: coefficient ${}^{10}C_{6}=210$ and no sign factor. Option D is wrong: this uses ${}^{10}C_{2}=45$ and also drops the minus sign that an odd power of $y$ carries.

NCERT Class 11 Mathematics · Ch 7 · §7.2.2 · p.130 ("Thus (x-y)n = nC0xn - nC1xn - 1 y + nC2xn - 2 y2 + ... + (-1)n nCn yn")

Question 2 · 7.1 Introduction · Medium
The portrait printed beside Section 7.1 on the opening page of Chapter 7 carries which caption?
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Answer: D — Option D is correct. The caption reads 'Blaise Pascal (1623-1662)'. Option B gives the birth year as 1632, transposing the digits of 1623. Option C gives the death year as 1666, changing the printed 1662. Option A keeps the printed years but puts C.P. Steinmetz's name on them; Steinmetz is the author of the epigraph, not the subject of the portrait.

Blaise Pascal (1623-1662)

Question 3 · 7.2 Binomial Theorem for Positive Integral Indices · Medium
The sum of all the coefficients in the expansion of $(2x-3)^6$, that is its value at $x=1$, is
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Answer: A — Option A is correct: putting $x=1$ gives $(2-3)^6=(-1)^6=1$. Option D is wrong: it is what you get if you use the odd index $5$, evaluating $(2-3)^5$. Option B is wrong: it is what you get if you add the first and last terms $2^6+3^6$ and stop there. Option C is wrong: it is what you get if you evaluate $-(2+3)^6$, keeping an outside minus instead of the one inside the bracket. Adding the coefficients $64-576+2160-4320+4860-2916+729$ gives $1$.

(2x - 3)6

Question 4 · 7.1 Introduction · Hard
A student reads Section 7.1 and concludes: 'the binomial theorem cannot even be stated when $n = 3/4$.' Judged against what Section 7.1 says, is the student right?
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Answer: A — Option A is correct. Section 7.1 describes the theorem as giving an easier way to expand $(a+b)^n$ 'where $n$ is an integer or a rational number', and $3/4$ is a rational number, so the theorem as described does cover it; what is restricted is the Chapter, not the theorem. Option C and option D both answer yes and put the restriction on the theorem itself, which reverses the two sentences. Option B answers no for the wrong reason: $3/4$ is not a positive integral index, so it is not inside the Chapter's scope at all.

It gives an easier way to expand (a + b)n, where n is an integer or a rational number.

Question 5 · 7.2 Binomial Theorem for Positive Integral Indices · Medium
According to Observation 4 on p.130, what is the index of $b$ in the first term of the expansion of $(a+b)^n$?
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Answer: A — Option A is correct because Observation 4 says the index of $b$ increases by unity, 'starting with zero in the first term'; Observation 1 makes the same point by writing that term as $^nC_0a^nb^0$ with $b^0 = 1$. Option B is wrong because $2$ is the index of $b$ in the third term $^nC_2a^{n-2}b^2$; the count starts one step earlier, at zero. Option C is wrong because $1$ is the index of $b$ in the second term; Observation 4 says the index of $b$ is $1$ 'in the second'. Option D is wrong because $n$ is the index of $b$ in the last term, where Observation 4 says the increase ends.

NCERT Class 11 Mathematics · Ch 7 · §7.2 · p.130 ("At the same time the index of b increases by unity, starting with zero in the first term, 1 in the second and so on ending with n in the last term.")

Question 6 · 7.1 Introduction · Hard
In the sentence 'In this Chapter, we study binomial theorem for positive integral indices only', what is the word 'only' doing?
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Answer: C — Option C is correct. The word restricts the Chapter's own coverage; the preceding sentence has already allowed the theorem itself an integer or rational index. Option D puts the restriction on the theorem, which contradicts the preceding sentence. Option A makes a claim about other chapters of the book that this sentence does not make. Option B claims $(a+b)^n$ is undefined otherwise, which is a claim about the expression, not about what the Chapter chooses to study.

In this Chapter, we study binomial theorem for positive integral indices only.

Question 7 · 7.2 Binomial Theorem for Positive Integral Indices · Easy
In the expansion of $(x-2y)^5$ printed on p.130, the coefficient of $x^{4}y$ is
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Answer: A — Option A is correct: the term is $-{}^{5}C_{1}x^4(2y) = -5\cdot 2\,x^4y = -10x^4y$, exactly as printed. Option B is wrong: this keeps the factor $2$ but drops ${}^{5}C_{1}=5$, and the printed term carries the binomial coefficient too. Option C is wrong: this drops the minus sign, but an odd power of $y$ makes the term negative and the page prints $-10x^4y$. Option D is wrong: this raises the $2$ to the power $2$ instead of $1$, giving $5 \times 2^2 = 20$, whereas the index on $(2y)$ in this term is $1$.

NCERT Class 11 Mathematics · Ch 7 · §7.2.2 · p.130 ("= x5 -10x4y + 40x3y2 - 80x2y3 + 80xy4 - 32y5.")

Question 8 · 7.1 Introduction · Medium
Which binomials does Section 7.1 name when it recalls what was learnt in earlier classes?
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Answer: C — Option C is correct. The opening sentence reads: 'In earlier classes, we have learnt how to find the squares and cubes of binomials like $a+b$ and $a-b$.' Option D pairs $a+b$ with the product $ab$, which is not a binomial at all. Option A adds $a+b+c$, a trinomial, which the section never mentions. Option B names the binomials $1+x$ and $1-x$; those forms do appear later in the chapter as special cases, but they are not what this opening sentence names.

In earlier classes, we have learnt how to find the squares and cubes of binomials like a + b and a - b.

Question 9 · 7.2 Binomial Theorem for Positive Integral Indices · Easy
Taking $a=7$, $b=3$ and $n=4$ in Miscellaneous Exercise Q1, the value of $a^{\,n}-b^{\,n}$ is
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Answer: A — Option A is correct: $7^4-3^4=2401-81=2320$. Option C is wrong: it is what you get if you add the two fourth powers instead of subtracting. Option D is wrong: it is what you get if you compute $(a-b)^{\,n}$ instead of $a^{\,n}-b^{\,n}$. Option B is wrong: it is what you get if you use $n=3$ instead of $n=4$. And $2320=4\times580$, so $a-b=4$ is indeed a factor, as Miscellaneous Exercise Q1 predicts.

If a and b are distinct integers, prove that a - b is a factor of an - bn, whenever

Question 10 · 7.1 Introduction · Medium
Section 7.1 recalls that squares and cubes of binomials such as $a+b$ and $a-b$ were learnt in earlier classes, and uses them to evaluate numbers by a shift such as $(98)^2=(100-2)^2$. Evaluating $(999)^{2}$ the same way, as $(1000 - 1)^{2}$, gives
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Answer: B — Option B is correct. Writing $999 = 1000 - 1$ and applying the identity $(a-b)^2 = a^2 - 2ab + b^2$ gives $1000^2 - 2\times1000\times1 + 1^2 = 1000000 - 2000 + 1 = 998001$. Direct multiplication confirms $999\times999 = 998001$. Option C is $1002001$, which comes out when the cross term was taken as $+2\times1000\times1$, i.e. $(a+b)^2$ was expanded instead of $(a-b)^2$; $1002001 \neq 998001$, so it is wrong. Option A is $999001$, which comes out when the cross term was written as $-1000\times1$, without the factor $2$; $999001 \neq 998001$, so it is wrong. Option D is $999999$, which comes out when the square of a difference was confused with a difference of squares, $a^2-b^2$; $999999 \neq 998001$, so it is wrong.

we could evaluate the numerical values of numbers like (98)2 = (100 - 2)2, (999)3 = (1000 - 1)3, etc.

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